domingo, 21 de marzo de 2010

Current mirror improves PWM regulator’s performance

  Power-supply designs requiring high-performance isolated feedback often use an error amplifier similar to the one, which relies on a second amplifier, IC1B, to provide the necessary inversion to keep the optocoupler, IC2, referenced to ground. To prevent bias-supply noise from entering the feedback path and causing oscillations, the amplifier relies on its ground reference and power-supply-rejection characteristics.

The power supply's output drives a voltage divider comprising R1 and R2 that maintains the amplifier's inverting input at the same voltage as the reference voltage that IC3 provides. C2, R3, and C3 comprise frequency-compensation components for the power supply's stable operation. This component-intensive error-amplifier configuration requires two operational amplifiers, one precision shunt-voltage reference, four capacitors and often a fifth in parallel with R6, and seven resistors.

Shows an alternative single-amplifier design in which IC3, an LM4040 precision-voltage reference, drives optocoupler IC2 with a "stiff" positive-voltage source over a wide current range. The voltage reference suppresses any noise present on the bias-supply rail. Variations in the reference and power-supply voltages appear in common mode at the amplifier's inputs and thus provide additional noise immunity. A resistive-voltage divider comprising R2 and R3 reduces the reference voltage to equal the power supply's regulated output voltage, which drives IC1's inverting input through R1. Given its single voltage divider, the error-amplifier circuit provides the same output voltage as the circuit and requires a single operational amplifier and precision shunt reference, four capacitors, and six resistors.

Miller-effect coupling of collector-emitter-voltage transitions into a typical phototransistor-based optocoupler's high-impedance, optically sensitive base region introduces a bandwidth-limiting pole, which dramatically slows the device's response time. Holding the phototransistor's collector-emitter voltage constant and allowing only its collector-emitter current to change provide an order-of-magnitude switching-speed improvement.

National Semiconductor's LM5026 active-clamp current-mode PWM controller, IC4, provides a convenient method of reducing an optocoupler's Miller-effect-induced slowdown. Shows the LM5026's internal current mirror driving what would normally serve as a frequency-compensation pin. Optocoupler IC2 connects directly between two constant-voltage sources comprising the current mirror and a voltage reference. The resultant decrease in response time relocates the bandwidth-limiting pole and improves the circuit's transient response.

The values of C2, C3, R3, and R1 apply only to this design and may require modification for other applications. Select R1 to provide equal impedances at both of the op amp's inputs. C2 forms a high-frequency noise filter. After you measure the converter's overall gain, calculate values for C3 and R3 that will provide proper gain and phase response. Several methods of calculation are available, most of which will provide adequate results.

Nombre: Rodriguez B. Joiver I.
Asignatura: EES

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Active Load-Pull Solution Tests RF Devices

The MB series is the first commercially available active load-pull solution, according to its manufacturer. The system uses a new measurement architecture to evaluate rf waveforms, power spectrum, s-parameters, and DCIV with unconditional stability and no wideband artifacts. The MB 20 tests devices and power amplifiers up to 20 W CW, while the MB 150 extends the power level up to 150 W CW.

The company's "Waveform Engineering," pioneered at Cardiff University, enables the replication of s-parameter concepts within the nonlinear domain. The solution simultaneously measures the actual current and voltage at the device, allowing designers to view and engineer their waveforms to match theory. The system uses a Tektronix sampling scope to enable simultaneous wideband measurements with a coherent alignment of all spectral components, including the fundamental and multiple higher-order harmonics.

Additionally, the system measures the dc and baseband response, essential in capturing the often seen memory effects in devices. The MB series includes a comprehensive user interface facilitating complete measurement automation and an intuitive introduction of waveform engineering. The system's modular design offers easy upgradability for higher power, additional harmonics, lower frequencies, and other changes in requirements

Nombre: Rodriguez B. Joiver I.
Asignatura: EES

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Use Current-Mirror Biasing To Avoid Squegging In RF Oscillators

If you've ever designed an RF oscillator, you've probably encountered squegging. Sometimes called "motor boating," squegging causes oscillators to start and stop at frequencies much lower than the frequency of interest. Viewed on an oscilloscope, squegging looks like bursts of oscillations. On a spectrum analyzer, it looks like a Christmas tree. In some designs, such as super-regenerative receivers or wildlife radio tags, this might be a desirable side effect. In most cases, though, it's a nuisance.

Squegging is inherently a nonlinear effect. As such, it's difficult to model mathematically, although it can be effectively simulated in your favorite flavor of Spice. The cause of squegging is usually a shift in the bias of a transistor or other active element responsible for providing positive feedback.

As oscillation starts, nonlinearities cause bias voltages to shift extensively, to the point where oscillation cuts off. The bias voltage then returns to its quiescent value and oscillation starts again. The cycle repeats at a frequency related to time constants within the bias network. Poor circuit layout or inadequate power-supply decoupling can also cause squegging. These causes are readily addressed, but the solution to the shifting bias problem is sometimes elusive. Changing bias points can help, but it's not always clear whether or not squegging is effectively squashed.

One way to avoid squegging is to use a current-mirror bias arrangement rather than the resistors found in traditional designs. As shown in the figure, the circuit is essentially an overtone Colpitts oscillator. Assuming transistors Q1 and Q2 are a matched pair, the current through Q2 is nearly identical to that of Q1, which is essentially the current through R1. As long as the supply voltage is constant, the bias current will not shift, since the base-emitter voltage is essentially constant.


Choke L1 isolates Q1 from Q2 at RF; at dc, it's a short circuit. L2 serves two purposes: It keeps Q2's emitter at ground and, along with C2, determines the crystal overtone frequency. Q2's base presents a negative resistance to the crystal at frequencies above the resonance of C2-L2. Reactance values of 100 Ω or so work well. Larger reactances increase negative resistance and reduce bias current requirements. Choose the resonance of C2-L2 to be between the desired overtone and the next lower crystal resonance. Bias currents on the order of a milliamp or less are sufficient, given the reactance values above.
The only critical components are the matched transistors. Transistor arrays, such as the LM3046, are adequate for VHF. Transistor pairs like the Panasonic XN6537 or XN6543 are usable up to several gigahertz. The crystal can be replaced by an inductor, SAW resonator, strip line, or other resonator. Bias current depends only on the current through R1, so it's easy to control the transistor operating point.

Nombre: Rodriguez B. Joiver I.
Asignatura: EES

Current Mirror Speeds Up Balanced-Output Optoisolator

Circuit and system design requirements often mean compromises. For example, optoisolators are used almost everywhere, but they are highfrequency limited. This simple circuit allows for wider bandwidth at the cost of two transistors and two resistors. And it's all done with mirrors!
With conventional approaches, the resulting design isn't always optimized for high speed. In certain situations, the optoisolator load-resistor value might be higher than the value that provides the best high-frequency response. Changing the load configuration from a common emitter to a common collector doesn't affect performance.

The photons are coupled into the base region and the load connection doesn't affect them. Adding a speedup resistor from the base to emitter isn't always feasible, as some devices don't have an external base terminal. Also, this decreases the effective current-transfer ratio (CTR).

A current mirror offers low input impedance, which is good for high-frequency response, and high output impedance, which is good for voltage gain. In Figure 1, the driver for the diode part of the optoisolator is a differential amplifier made from Darlingtonconnected npn bipolar transistors, Q1 to Q4. Any other driver can be used. This is the one I used for the measurements.

The differential amplifier is biased by a 13-mA current source that's adjustable from 6 to 18 mA, a three-to-one range to accommodate the CTR spread. For simplicity, the schematic shows both circuits: the standard (Box A) and the one with the current mirror (Box B)—one each connected to the two differentialamplifier outputs.

Optoisolator U1 and R3 compose the standard circuit—the one with a simple load resistor. Figure 2, bottom curve, shows the frequency response for this circuit. The –3-dB frequency is 23 kHz. This value falls within the manufacturer's specifications for this circuit under actual operating conditions. Overall voltage gain is 37.7 dB at 1 kHz.

The improved circuit uses U2 and a current mirror made from two pnp transistors, Q5 and Q6, plus two resistors, R4 and R5. The resistors are used for degeneration. They equalize against VBE and other parameter variations between unmatched transistors. Optoisolator U2's output, the transistor collector, feeds the current-mirror input terminal, the base and collector of transistor Q5. The load resistor, R6, connects to the current mirror's output, the collector of Q6. With this connection, we have the same overall input-to-output signal polarity as for the conventional circuit.

Figure 2, top curve, shows the results for this circuit. The bandwidth is improved about 3.5 times to a value of 86 kHz. The voltage gain stays almost the same at 38.1 dB, a 0.4-dB difference. All of this is accomplished without increasing the driver current.

For a complementary output, just use the improved circuit with both optoisolators, U1 and U2. You get complementary signals, isolated from input to output and from one output to the other. Depending on your application, you may need to either match the optoisolators for equal CTR or add a balancing adjustment to the circuit.

But rarely do the system requirements mandate a fully isolated complementary output. If the complementary outputs don't require isolation between each other, use the complementary output circuit shown as Box C, the components around U3. It uses only one optoisolator, but two complementary current mirrors. The current source had to be readjusted at 16 mA to balance the dc between Outputs 3 and 4. Frequency response and gain, both measured differentially, are 45 kHz at –3 dB and 45.8 dB at 1 kHz.

These circuits are low-cost solutions and an alternative to using faster and more expensive optoisolators.

Nombre: Rodriguez B. Joiver I.
Asignatura: EES

Simple Current Sensor Features Galvanic Isolation

Applications that require a simple, low-cost current sensor with galvanic isolation can employ the circuit described here. If used as drawn, it's possible to sense currents up to 10 A with high precision (usually within %) and wide bandwidth (more than 500 kHz). The use of optocouplers provides 3000 V of isolation between the primary and secondary and 50 dB of CMRR. Transistors T1 and T2 form a coupled pair in current-mirror configuration that's unbalanced by the current (I) to be measured. Once this happens, transistor in OP1 balances the current-mirror biased by T3 that also biases OP2, making its transistor conduct the same current that balances the current-mirror. Because the balancing current is directly proportional to the current being measured, a direct measurement is achieved. It can be easily shown that the gain in current is given by the ratio of the sensing resistor (RS) and R1, producing a proportional voltage drop at the Output terminals.

For the circuit to work properly, it's important that the coupled transistors which form the current mirror share the same substrate and that the optocoupler be a dual version, in order to have matching characteristics. Variations on this idea may lead configurations that can be used in a wide range of applications, such as sensing current in a ground line (inverting the operating idea of the current-mirror using npn transistors) or offering different conditioning at the output (increasing voltage with increasing current). The power supplies needed can be implemented using Zener diodes since only a small amount of power is required—usually less than 50 mW.

Nombre: Rodriguez B. Joiver I.
Asignatura: EES

Designing a Current Source


  • Two NPN transistors -- models 2N2222 or 2N3403 recommended (Radio Shack catalog # 276-1617 is a package of fifteen NPN transistors ideal for this and other experiments).
  • Two 6-volt batteries.
  • One 10 kΩ potentiometer, single-turn, linear taper (Radio Shack catalog # 271-1715).
  • Two 10 kΩ resistors.
  • Four 1.5 kΩ resistors.

Small signal transistors are recommended so as to be able to experience "thermal runaway" in the latter portion of the experiment. Larger "power" transistors may not exhibit the same behavior at these low current levels. However, any pair of identical NPN transistors may be used to build a current mirror.

Beware that not all transistors share the same terminal designations, or pinouts, even if they share the same physical appearance. This will dictate how you connect the transistors together and to other components, so be sure to check the manufacturer's specifications (component datasheet), easily obtained from the manufacturer's website. Beware that it is possible for the transistor's package and even the manufacturer's datasheet to show incorrect terminal identification diagrams! Double-checking pin identities with your multimeter's "diode check" function is highly recommended. For details on how to identify bipolar transistor terminals using a multimeter, consult chapter 4 of the Semiconductor volume (volume III) of this book series.


Lessons In Electric Circuits, Volume 3, chapter 4: "Bipolar Junction Transistors"


  • How to build a current mirror circuit
  • Current limitations of a current mirror circuit
  • Temperature dependence of BJTs
  • Experience a controlled "thermal runaway" situation




A current mirror may be thought of as an adjustable current regulator, the current limit being easily set by a single resistance. It is a rather crude current regulator circuit, but one that finds wide use due to its simplicity. In this experiment, you will get the opportunity to build one of these circuits, explore its current-regulating properties, and also experience some of its practical limitations firsthand.

Build the circuit as shown in the schematic and illustration. You will have one extra 1.5 kΩ fixed-value resistor from the parts specified in the parts list. You will be using it in the last part of this experiment.

The potentiometer sets the amount of current through transistor Q1. This transistor is connected to act as a simple diode: just a PN junction. Why use a transistor instead of a regular diode? Because it is important to match the junction characteristics of these two transistors when using them in a current mirror circuit. Voltage dropped across the base-emitter junction of Q1 is impressed across the base-emitter junction of the other transistor, Q2, causing it to turn "on" and likewise conduct current.

Since voltage across the two transistors' base-emitter junctions is the same -- the two junction pairs being connected in parallel with each other -- so should the current be through their base terminals, assuming identical junction characteristics and identical junction temperatures. Matched transistors should have the same β ratios, as well, so equal base currents means equal collector currents. The practical result of all this is Q2's collector current mimicking whatever current magnitude has been established through the collector of Q1 by the potentiometer. In other words, current through Q2 mirrors the current through Q1.

Changes in load resistance (resistance connecting the collector of Q2 to the positive side of the battery) have no effect on Q1's current, and consequently have no effect upon the base-emitter voltage or base current of Q2. With a constant base current and a nearly constant β ratio, Q2 will drop as much or as little collector-emitter voltage as necessary to hold its collector (load) current constant. Thus, the current mirror circuit acts to regulate current at a value set by the potentiometer, without regard to load resistance.

Well, that is how it is supposed to work, anyway. Reality isn't quite so simple, as you are about to see. In the circuit diagram shown, the load circuit of Q2 is completed to the positive side of the battery through an ammeter, for easy current measurement. Rather than solidly connect the ammeter's black probe to a definite point in the circuit, I've marked five test points, TP1 through TP5, for you to touch the black test probe to while measuring current. This allows you to quickly and effortlessly change load resistance: touching the probe to TP1 results in practically no load resistance, while touching it to TP5 results in approximately 14.5 kΩ of load resistance.

To begin the experiment, touch the test probe to TP4 and adjust the potentiometer through its range of travel. You should see a small, changing current indicated by your ammeter as you move the potentiometer mechanism: no more than a few milliamps. Leave the potentiometer set to a position giving a round number of milliamps and move the meter's black test probe to TP3. The current indication should be very nearly the same as before. Move the probe to TP2, then TP1. Again, you should see a nearly unchanged amount of current. Try adjusting the potentiometer to another position, giving a different current indication, and touch the meter's black probe to test points TP1 through TP4, noting the stability of the current indications as you change load resistance. This demonstrates the current regulating behavior of this circuit.

You should note that the current regulation isn't perfect. Despite regulating the current at nearly the value for load resistances between 0 and 4.5 kΩ, there is some variation over this range. The regulation may be much worse if load resistance is allowed to rise too high. Try adjusting the potentiometer so that maximum current is obtained, as indicated with the ammeter test probe connected to TP1. Leaving the potentiometer at that position, move the meter probe to TP2, then TP3, then TP4, and finally TP5, noting the meter's indication at each connection point. The current should be regulated at a nearly constant value until the meter probe is moved to the last test point, TP5. There, the current indication will be substantially lower than at the other test points. Why is this? Because too much load resistance has been inserted into Q2's circuit. Simply put, Q2 cannot "turn on" any more than it already has, to maintain the same amount of current with this great a load resistance as with lesser load resistances.

This phenomenon is common to all current-regulator circuits: there is a limited amount of resistance a current regulator can handle before it saturates. This stands to reason, as any current regulator circuit capable of supplying a constant amount of current through any load resistance imaginable would require an unlimited source of voltage to do it! Ohm's Law (E=IR) dictates the amount of voltage needed to push a given amount of current through a given amount of resistance, and with only 12 volts of power supply voltage at our disposal, a finite limit of load current and load resistance definitely exists for this circuit. For this reason, it may be helpful to think of current regulator circuits as being current limiter circuits, for all they can really do is limit current to some maximum value.

An important caveat for current mirror circuits in general is that of equal temperature between the two transistors. The current "mirroring" taking place between the two transistors' collector circuits depends on the base-emitter junctions of those two transistors having the exact same properties. As the "diode equation" describes, the voltage/current relationship for a PN junction strongly depends on junction temperature. The hotter a PN junction is, the more current it will pass for a given amount of voltage drop. If one transistor should become hotter than the other, it will pass more collector current than the other, and the circuit will no longer "mirror" current as expected. When building a real current mirror circuit using discrete transistors, the two transistors should be epoxy-glued together (back-to-back) so that they remain at approximately the same temperature.

To illustrate this dependence on equal temperature, try grasping one transistor between your fingers to heat it up. What happens to the current through the load resistors as the transistor's temperature increases? Now, let go of the transistor and blow on it to cool it down to ambient temperature. Grasp the other transistor between your fingers to heat it up. What does the load current do now?

In this next phase of the experiment, we will intentionally allow one of the transistors to overheat and note the effects. To avoid damaging a transistor, this procedure should be conducted no longer than is necessary to observe load current begin to "run away." To begin, adjust the potentiometer for minimum current. Next, replace the 10 kΩ Rlimit resistor with a 1.5 kΩ resistor. This will allow a higher current to pass through Q1, and consequently through Q2 as well.
Place the ammeter's black probe on TP1 and observe the current indication. Move the potentiometer in the direction of increasing current until you read about 10 mA through the ammeter. At that point, stop moving the potentiometer and just observe the current. You will notice current begin to increase all on its own, without further potentiometer motion! Break the circuit by removing the meter probe from TP1 when the current exceeds 30 mA, to avoid damaging transistor Q2.

If you carefully touch both transistors with a finger, you should notice Q2 is warm, while Q1 is cool. Warning: if Q2's current has been allowed to "run away" too far or for too long a time, it may become very hot! You can receive a bad burn on your fingertip by touching an overheated semiconductor component, so be careful here!

What just happened to make Q2 overheat and lose current control? By connecting the ammeter to TP1, all load resistance was removed, so Q2 had to drop full battery voltage between collector and emitter as it regulated current. Transistor Q1 at least had the 1.5 kΩ resistance of Rlimit in place to drop most of the battery voltage, so its power dissipation was far less than that of Q2. This gross imbalance of power dissipation caused Q2 to heat more than Q1. As the temperature increased, Q2 began to pass more current for the same amount of base-emitter voltage drop. This caused it to heat up even faster, as it was passing more collector current while still dropping the full 12 volts between collector and emitter. The effect is known as thermal runaway, and it is possible in many bipolar junction transistor circuits, not just current mirrors.


Schematic with SPICE node numbers:

Netlist (make a text file containing the following text, verbatim):
Current mirror v1 1 0 vammeter 1 3 dc 0 rlimit 1 2 10k rload 3 4 3k q1 2 2 0 mod1 q2 4 2 0 mod1 .model mod1 npn bf=100 .dc v1 12 12 1 .print dc i(vammeter) .end 

Vammeter is nothing more than a zero-volt DC battery strategically placed to intercept load current. This is nothing more than a trick to measure current in a SPICE simulation, as no dedicated "ammeter" component exists in the SPICE language.

It is important to remember that SPICE only recognizes the first eight characters of a component's name. The name "vammeter" is okay, but if we were to incorporate more than one current-measuring voltage source in the circuit and name them "vammeter1" and "vammeter2", respectively, SPICE would see them as being two instances of the same component "vammeter" (seeing only the first eight characters) and halt with an error. Something to bear in mind when altering the netlist or programming your own SPICE simulation!

You will have to experiment with different resistance values of Rload in this simulation to appreciate the current-regulating nature of the circuit. With Rlimit set to 10 kΩ and a power supply voltage of 12 volts, the regulated current through Rload will be 1.1 mA. SPICE shows the regulation to be perfect (isn't the virtual world of computer simulation so nice?), the load current remaining at 1.1 mA for a wide range of load resistances. However, if the load resistance is increased beyond 10 kΩ, even this simulation shows the load current suffering a decrease as in real life.

Nombre: Rodriguez B. Joiver I.
Asignatura: EES

Revealing the Secret of Current Mirror

A current mirror consists of two consecutively connected inverse converters.


There are many professional publications about the famous current mirror circuit and its versions. Only, as a rule, they look formal, specialized and "sterile". These sources give a lot of details about current mirror versions but they do not show the basic ideas behind them.

The Simple but Complex Current Mirror

At first sight, the current mirror looks a simple bare circuit (Fig. 1); but actually, it is an odd, strange and exotic circuit that is never explained (there are only some guessworks). It is a great paradox that everyone knows what a current mirror is but nobody knows how it operates.

As a rule, the classical current mirror explanations begin with the assertion that the current-setting transistor Q1 acts as a diode. But it is very primitive and confusing to say "the input transistor Q1is a diode". Actually, it is not a diode; it is exactly a transistor operating in the active mode. It would be a diode, if its collector was disconnected. Then all the current IREF = VCC/R (R is omitted on the picture) would flow through the base-emitter junction acting really as a diode. But here the output collector-emitter part of the transistor is connected in parallel to the base-emitter "diode". In this way, it serves as a shunting regulating element that diverts the great amount (β/(1 + β)) of the current. But how and why the transistor does this magic?

There is something strange and confusing in this arrangement... As everybody knows, the base-emitter voltage VBE is the input quantity of the bipolar transistor and the collector current IREF is the output quantity. But here all is on the contrary - the collector current is the input and the base-emitter voltage is an output!?!? What an idiocy?

Questions to be Answered

In order to understand this odd circuit, we need to answer dozens of questions that are never answered. Here are some of them.

What does the transistor Q1 do in this circuit? What is its function there? What is the difference between it and a diode? Can we replace it by a bare diode or a base-emitter junction (leaving the collector disconnected) as it is shown in another current mirror story? Why and how the collector current IREF serves as an input quantity and the base-emitter voltage as an output one (we thought the base-emitter voltage VBE was the input quantity of the bipolar transistor and the collector current IREF was the output quantity)? What does the transistor Q2 do in this circuit? What is its function there? How do collector voltages vary when we change the input current IREF (the input voltage VREF , the "programming" resistor RREF and the load resistance RL )?
Since there are no satisfactory answers to these questions, let's forget temporarily all kinds "cut-and-dried" citations and to try disclosing the mystery of the famous circuit by ourselves. Just forget them and begin thinking by yourself!

Fig. 1 What does the transistor Q1 do in this odd circuit?
The General Idea Behind the Whole Circuit
It seems our story will be not based on "verifiable sources" just because there are not such sources; so, one can say it is an original research. Only, we will endeavor to explain the famous circuit so simply, clearly and evidently that there is not any need to verify these assertions. By the way, the very NOR stipulates such a possibility - to use sources, "the accuracy of which is easily verifiable by any reasonable adult without specialist knowledge".
The Problem: Current Direction Inverter
In circuitry, especially in the area of microelectronics (e.g., inside op-amps), sometimes we need to invert the direction of a current (to make the flowing in current a flowing out one and v.v. - the flowing out current a flowing in one) without changing its magnitude. In this way, the output current just follows, "copies" the input one but this copy is inverted, "mirror" one. In other words, the input current "programs" the output one and the whole circuit serves as a programmed current source (current-controlled or dependent current source).

Obviously, in order to do that, such a current direction inverter (having a more popular name - current mirror) has to be at least a 3-terminal device (Fig. 2). The reason of that is because the two entering currents have to "go out" from somewhere (Fig. 2a) and v.v., the two exiting currents have to "go in" somewhere (Fig. 2b). In this arrangement, the one lead serves as an input, the other - as an output and the third - as a common terminal (usually connected to the positive or negative rail and more rarely - to the ground).

Fig. 2. A current mirror is a 3-terminal device

How to Create a Current Mirror

Basic Structure

In electronics, we have two kinds of circuits producing or, more frequently, just controlling current - current sources and current sinks. The difference between them is that, regarding to some circuit point, sources "inject" current while sinks "absorb" current. Obviously, in order to implement the general arrangement above (Fig. 2) into a concrete current mirror, we need both the circuits. As a rule, these circuits are voltage-controlled; so, in order to control them by current, we need to connect current-to-voltage converters before them. Let's now draw the block diagram of the two possible current mirror arrangements.

If we have a current source, we convert the input current ("entering" the current mirror) into a voltage and then use this voltage to control a current sink; as a result, we obtain a current sink (Fig. 3a). Conversely, if we have a current sink, we convert the input current ("exiting" the current mirror) into a voltage and then use this voltage to control a current source (Fig. 3b); as a result, now we obtain a current source. We can already generalize this basic current mirror structure in a conclusion: A current mirror consists of two consecutively connected current-to-voltage and voltage-to-current converters.

Fig. 3a. A current mirror sinking the output current
Fig. 3b. A current mirror sourcing the output current

It is interesting that the two converters may be linear (then IOUT = V/R = IIN.R/R = IIN) but it is not obligatory. They might be non-linear devices having whatever transfer or IV characteristics that may even depend on other quantity (e.g., temperature); the only requirement is their characteristics to be reverse. In this way, if the one converter implements a function y = f(x) and the other represents the reverse function x = f -1(y) the whole function is y = f(x) = f(f -1(y)) . Note that the two converters might be connected in any order, including the direct after the inversed one (Fig. 4). At first sight, this result looks strange and nonsensical but in this way the main problem - reversing the current direction, is solved. So, we can formulate the next conclusion: A current mirror consists of two consecutively connected converters having reverse characteristics.

Fig. 4. The current mirror consists of two consecutively connected inverse converters
Simple Reversing

Usually, there is not a couple of direct and reverse converter and we have only one kind of converter. If this is a reversible converter, we can use it as a direct converter and then connect another such converter in an opposite direction to get a reverse converter.

Reversing by Negative Feedback

The problem is when the converter is not reversible and it is a bare one-way device. In this case, we cannot swap the circuit input and output ports; we cannot apply an input quantity to the circuit output and to get an output quantity from the circuit input (unfortunately, this is our case). Then, the only way to "reverse" the circuit is applying the ubiquitous negative feedback. Only systems with negative feedback have the unique property to reverse the cause and effect relations between the input and output quantities of the objects. They adjust their internal input quantity so that their internal output quantity to become equal to the "true" external input quantity. As a result, the internal input quantity follows, depends on the external input quantity; actually, the internal input quantity serves as an external output quantity. In this way, they can "reverse" the objects.

Implementing the General idea Into a BJT Electronic Circuit

Once revealed the general idea we can create as many as we want current mirror circuits (this is the power of this heuristic approach). In all these versions, only the electronic components (BJT, FET, op-amps, etc.) will be different; the general idea behind them will be the same. Well, let's begin with the most popular of them - the basic BJT current mirror.

An Output Part

We can drive a bipolar transistor by voltage or by current. If we change the base-emitter voltage as an input and use the collector current as an output (Fig. 5), we can think of a BJ transistor as a non-linear voltage-to-current converter having an exponential characteristic. So, we can use it directly as an output part of our simple BJT current mirror: The output part of the simple BJT current mirror is just a bipolar transistor acting as an exponential voltage-to-current converter.


How does the transistor T2 behave in this arrangement? In order to know, let's carry out an experiment - set a constant input voltage VREF = 0.5 ÷ 0.7V and then vary the load resistance RL (with the same success, you can vary the supply voltage V2 or even both the resistance RL and the voltage V2). The result is surprising - the transistor changes its present resistance RT2 between the collector and the emitter so that to keep up a constant total resistance Rtot = RL + RT2 = const (Fig. 6). As a result, the output current remains constant IOUT = V2/Rtot. In this way, the output collector-emitter part of the transistor T2 acts as a current-stable non-linear resistor. They usually say the transistor T2 acts as a simple current sink.

Fig. 5. Implementing the output part of a BJT current mirror
Fig. 6. Superimposed IV curves of the output part
An Input Part
Now, we need to make the BJ transistor serve as the needed opposite current-to-voltage converter. Only, we cannot "reverse" it directly since the transistor is a one-way device, whose base-emitter junction can control the collector current; the opposite is just impossible. What do we do then?
We have already known the remedy - its name is negative feedback. In our case that means to make the transistor adjust its base-emitter voltage "VOUT" so that the collector current to be IIN = V1/R. For this purpose, we have just to connect its collector to its base, in order to apply a "100% parallel negative feedback" (Fig. 7). As a result, although it seems strange, in this "reversed" transistor the collector current serves as an input quantity while the base-emitter voltage serves as an output quantity! The input part of the simple BJT current mirror is just a bipolar transistor with 100% parallel negative feedback.


But how does the transistor T1 behave in this arrangement? In order to know, let's carry out another experiment - set a constant input (supply) voltage V1 and then vary the input current-setting resistance R (again, with the same success, you can vary the input voltage V1 or even both the resistance R and the voltage V1).

The result is still more surprising than before - now, the transistor changes its present resistance RT1 between its collector and emitter so that to keep up an almost constant resistance ratio K = RT1/(RT1 + R) = const. As a result, the output voltage remains almost constant VOUT = VCE1 = VBE = const (Fig. 8). In this way, the output collector-emitter part of the transistor T1 acts as a voltage-stable non-linear resistor. But this is the recipe of making various active diodes ("ordinary", "zener" or "rubber", adjustable...)! It is just a wonder! The parallel negative feedback has made a current-stable resistor (the output part of T1) behave as a voltage-stable one! This is the same transistor but in the first case it serves as a current-stable element while in the second case it serves as a voltage-stable element.

Fig. 7. Implementing the input part of a BJT current mirror

Fig. 8. Superimposed IV curves of the input part
A "Reversed" Transistor
At the same time, the input voltage source V1 and the current-setting resistor R form a composed current-source that "want" to produce a current IIN = V1/R through a voltage-stable element (the transistor T1). By the way, some mystic cascode circuits are based on the same arrangement (a current source supplies a voltage-stable element and v.v.). It is interesting that, in this situation, the voltage-stable element changes its present resistance, in order to "help" the current source to establish the desired current. For example, if we decrease the resistance R to increase the current, the transistor T1 will also decrease its present resistance thus helping us to increase the current and v.v. Doing that, the transistor T1 adjusts its base-emitter voltage so that the collector current to be always IIN = V1/R. As a result, although it seems strange, the collector current serves as an input quantity while the base-emitter voltage serves as an output quantity! The negative feedback has reversed the one-way transistor!
An Equilibrium Point
It is interesting to discover how the transistor T1 manages to reach the equilibrium point (a system with negative feedback always reaches the equilibrium). For this purpose, let's change firstly the magnitude of the input voltage V1 or the resistance R. The transistor responses to this "intervention" by changing its present resistance RT1... but till when? In order to understand, let's carry out an interesting experiment - break the feedback loop and drive the "true" base-emitter transistor input by a separate voltage source "VOUT" (Fig. 9). Then, let's begin increasing/decreasing the "true" transistor's input voltage "VOUT"; in return, the transistor will begin dropping/raising its collector voltage VC1 by decreasing/increasing its present resistance RT1. Figuratively speaking, the two voltages "move" against each other. In order to imitate the negative feedback behavior, we have to stop changing "VOUT" when the two voltages become equal (a zero indicator connected between the collector and the base can show this moment); this is the equilibrium point. Now, if we short the zero indicator (connect the collector to the base or, as they say, close the feedback loop), the system will remain at rest as it is at the point of the equilibrium.
Fig. 9. Discovering the equilibrium point of the current-setting transistor
Another Example: a Transdiode
By the way, there is another paradoxical circuit that does the same but it is an almost perfect - the logarithmic converter based on the so called transdiode (a BJT transistor connected in the op-amp feedback loop). As here, in this odd and also never explained circuit configuration, the collector current serves as an input quantity while the base-emitter voltage - as an output quantity?!? The only difference is that there an additional op-amp adjusts the transistor's base-emitter voltage so that its collector current to be exactly equal to the input one. The op-amp does this magic by observing the virtual ground point and keeping it (almost) equal to zero.
Assembling the Whole Circuit
Finally, we have only to connect the output of the input part (the base-emitter junction of T1) to the input of the output part (the base-emitter junction of T2) to build the famous BJT current mirror circuit!
Bjt current mirror in 1000.jpg + Bjt current mirror out 1000.jpg
= Bjt current mirror full 1000.jpg
Fig. 10. Assembling the whole BJT current mirror.
The only problem is that the transistor T2 "sucks" another base current IB from the input current. As a result, the output current is smaller than the input one.

IOUT = IIN - 2IB = β.IB - 2IB = (β - 2).IB.

Nombre: Rodriguez B. Joiver I.
Asignatura: EES

Current Mirrors

The next item is current mirrors, another little understood circuit. These are extremely useful in amplifier design, and in this section I will show where they can be used, and the benefits that can be obtained.
The basic current mirror is shown in the next Figure, and it can be seen that whatever current is injected into the left side is mirrored, and the right hand side is a constant current source (sink) reflecting the injected current. Should the input current at change, so will the output current, but it will remain constant, regardless of the actual voltage (provided it remains within the supply limits of course).

Figure 10
         A Basic Current Mirror

The problem with this circuit is that the current in the two halves is different - the mirrored current is too low, differing by 19 uA (it is actually 20 uA, but the simulation accuracy used was not great enough to show this). If we check, we will find that the emitter currents of both transistors are identical, so the 20 uA that "disappeared" is the base current that must be supplied to each transistor (10 uA each).

Adding emitter resistors does absolutely nothing to alleviate this, but is useful if the transistors are not matched. Even then the resistors do not really do a lot of good, unless the voltage developed across them is significant (at least 100 mV, and preferably more) but it helps a little bit.

Figure 11
     Buffered Current Mirror

A better solution is to use a buffer as shown in the previous figure. This removes the base current component of the error, and makes the current mirror matching a lot better. This simple addition has reduced the error dramatically, but it can be improved even more. While not generally needed for audio amplification, improved performance is essential for test and measurement, or other critical applications.

Figure 11B
    Four Transistor Current Mirror

As you can see from the above, previous figure is almost perfect the current balance is extremely good. While this arrangement is used in analogue opamps and other circuits requiring high precision, there would be no advantage using it in a power amplifier. There simply is no need for such precision. It will not generally improve distortion, bandwidth or dynamics, but may give a marginal improvement in DC offset (which can be up to 100mV without causing any problems whatsoever in most power amps used for audio). All we need to do now is find a use for these circuits.

Differential Pair Amplifier

Consider the differential pair (aka long tailed pair or LTP). Most of the time, we are losing half the gain of the circuit, since the output is taken from only one collector as shown in the next figure. This configuration also suffers from linearity problems, unless the output is current only - as is the case when driving the base of a transistor (this is shown below).

Figure 12
The Long Tailed Pair As A Voltage Amplifier

The circuit as shown (without the essential biasing components, which were omitted for clarity) has a voltage gain of 285 (again using transistors with a hFE of 100), and is quite linear at low output voltages. The linearity will suffer badly as the level increases, and even with the +/-20 volt supplies used is not satisfactory (10% THD) as a voltage amp for outputs greater than 1.35 Volts RMS (this is at an input voltage of 5 mV).

Using a current source / sink in the "tail" is very common in amplifier circuits, and this variant is shown below. It is commonly (but entirely mistakenly) assumed that this increases the gain (in the circuit shown, gain is reduced to 168), but the real purpose is to improve the common mode capability of the circuit. Common mode signals are those that are applied to both inputs in the same polarity, and are generally required to be rejected. Using a simple resistive tail severely limits the common mode voltage that can be accommodated before severe distortion occurs, and indeed the common mode rejection of the circuit is almost useless.
In the example above, the common mode rejection is well under 1dB but with a current sink tail the rejection is almost 65dB. In most amplifier circuits common mode signals (of the undesirable kind) are not an issue if the input stage is properly designed. Although a high common mode ability is usually considered necessary, this is not always the case. For a typical power amplifier, the common mode voltage cannot exceed the input voltage for full power. There are other good reasons to use a current source/sink though, one of which is to ensure that circuit stabilises at a low voltage, eliminating (or at least minimising) switch-on / off thump.

If a current mirror is used as the load, gain is increased by a very useful amount, and the next figure shows the arrangement used. The stage gain is now 850 and the use of a current sink as the tail has no effect (provided that the current is maintained at 4 mA). This circuit is shown in the next figure, and although useful in certain applications, it is not suitable to drive the output stage of a power amplifier.

Figure 13
Using A Current Mirror The LTP Load

If we really wanted to get silly (and I have seen it done), we can put all the bits together in one place, and finish up with an input stage + Class-A driver, with a total open loop gain (i.e. without feedback) of 33,800 or 90dB (but still loaded with 100k). This will not increase dramatically when the output is buffered - the output impedance is actually reasonably low, at about 4k. The buffered current mirror does not help the gain, but reduces output offset. The complete circuit is shown in the next Figure, and is a useful example of the techniques discussed in this article. Open loop distortion is about 5% at 6V RMS output, but will fall dramatically when feedback is applied.

Figure 14
                Combination LTPs and Current Mirror

This circuit uses an input LTP that drives a secondary LTP as the Class-A amplification stage. The load for this second LTP is a current mirror, and this arrangement has excellent linearity. I suspect that it could be a cow to stabilise in a real amplifier circuit, and quite frankly, I do not see any reason to go this far. Many amplifiers have been designed using this arrangement, and it's very common with MOSFET output stages. Initial measurements on a MOSFET amp using this drive stage show that stability is not as good as I would like to see (there are traces of oscillation at some output levels), but overall stability seems to be acceptable. I expect that the wide bandwidth of the MOSFET output devices might make this arrangement a little more tractable than would be the case with bipolar transistors. I have not found it to be necessary though, and the P101 MOSFET amp does not use this input/driver stage combination.


Nombre: Rodriguez B. Joiver I.
Asignatura: EES

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Luces de emergencia para el hogar Romero Loren. C.I: 18.762.881

Luces de emergencia para el hogar

En cada página de electrónica donde exista un grupo de circuitos útiles para construir, siempre encontraremos uno llamado "Inversor 12VCC – 110/220VAC". A muchos de nosotros nos ocurre que no nacimos sabiéndolo todo y que hay cosas de las que no tenemos un real conocimiento de su funcionamiento o su aplicación específica. Además, otra de las situaciones que ocurren en estos casos es que encontramos sólo un circuito sin siquiera dos renglones explicativos de las condiciones más elementales del desarrollo que allí vemos dibujado. Sucede que quien deja allí el circuito, intuye que sabemos lo que quiso hacer y cómo funciona eso que sólo a él le funcionó. Por lo general, luego de darle decenas de vueltas terminamos abandonando todo porque la frustración se apodera de nosotros al no poder lograr lo que el circuito promete.
En este artículo vamos a ver como construir un sistema de iluminación de emergencia con todos los detalles explicados paso por paso para que su realización sea exitosa y por sobre todo, útil para tu vida diaria. El esquema en bloques de la imagen superior intenta ser muy claro. El corazón del desarrollo se basa en un oscilador que se encargará de generar una señal de 50 o 60 ciclos por segundo para activar los transistores MOSFET quienes se encargan de conmutar, a través de cada uno de los bobinados del transformador T1, la energía que le suministra la batería Bat1. T1 es un transformador convencional de 110 o 220VAC (de acuerdo al país donde vivas) a 9Volts + 9Volts. El bobinado correspondiente a los 9Volts se conectarán a los MOSFET y la parte de 110/220VAC será la salida hacia una lámpara de bajo consumo.

Podemos utilizar cualquier lámpara para la tensión de línea domiciliaria. El único secreto en este desarrollo será que debe ser de bajo consumo o CFL aunque también podemos utilizar las actuales lámparas a LED. Con un transformador T1 de la tensión antes especificada y de 2 Amperes de capacidad de suministro de corriente podemos colocar cargas de hasta 30W sin problemas. En este punto debemos hacer una aclaración que casi nadie hace. Si utilizas transformadores chinos, de los que llaman la atención por lo pequeño de su tamaño, procura sobredimensionarlo hasta 4 o 5 Amperes (la tensión continuará siendo la misma) para evitar excesos de temperatura luego de una o dos horas de funcionamiento. Por último y a la izquierda del diagrama se puede ver el bloque correspondiente al cargador de la batería Bat1. Aquí utilizaremos un sistema que cargará la batería hasta el valor correcto y luego cortará la carga de manera automática hasta que la tensión de la batería descienda y vuelva a reiniciarse el ciclo de carga. Al circuito del cargador lo veremos en la segunda parte de este artículo, junto a otras posibilidades de implementación de iluminación de emergencia. Por ahora consideramos una batería cargada a pleno y nos dedicaremos al funcionamiento del "Inversor".

En la etapa de potencia, a la salida del circuito, encontramos los ya conocidos transistores MOSFET IRFZ44N (los mismos que utilizamos para el puente H), el transformador T1 y un capacitor de 100nF X 630V para suavizar los picos de conmutación inducidos en el transformador. A este capacitor se le puede dar cualquier valor hasta 1uF (siempre utilizando capacitores de poliéster) o se pueden colocar varios en paralelo tratando de no superar ese valor máximo recomendado. Un fusible en la alimentación, al punto medio del transformador, siempre será una buena medida de seguridad y precaución. Remarcamos entonces lo importante a saber y entender: lo que habitualmente es el bobinado primario de un transformador, aquí es la conexión de salida hacia una lámpara de bajo consumo. Lo que siempre es el bobinado secundario de un transformador, aquí se conecta a los transistores de conmutación y a la batería según como indica el circuito.

Los transistores Q1 y Q2 serán activados por un circuito integrado CMOS CD4047B que estará conectado para funcionar como un oscilador libre y que nos entregará una onda cuadrada en cada una de sus salidas (Q y /Q/)(/Q/ = Q negada), invertidas 180º entre sí. Esto resultará en que una de las salidas estará en estado lógico alto mientras la otra se encuentre en estado lógico bajo y viceversa. Este modo de funcionamiento alternará la conducción de Q1 y Q2 induciendo una tensión alterna útil en el secundario de T1, muy cercana al valor nominal de tensión del transformador. Recuerda siempre que la tensión de salida dependerá de la carga que le conectes a la salida del transformador T1. Si dejas la salida libre, puede que obtengas hasta 300V de tensión, pero al conectar una carga notarás que ese valor disminuirá y se establecerá entre los 175 y los 250 Volts, insistimos: de acuerdo a la potencia consumida por la carga. Si colocas cargas mayores a 30 o 40W la tensión caerá a valores inoperables. Si deseas obtener mayor potencia de salida, debes incrementar la capacidad en Amperes de T1 e intentar disminuir la tensión de 9 + 9VAC a 6 + 6VAC del mismo transformador para obtener mayor potencia de salida. También aumenta el valor nominal del fusible de protección.
El circuito formado por P1, C1 y R3 se encargan de fijar la frecuencia de trabajo del CD4047B y responden a la fórmula T = 4,40 * R * C, donde R se expresa en Ohms, C en Faradios y T en segundos, siendo 4,40 un valor constante. De este modo, podemos "jugar" con distintos valores de capacitores y resistores hasta logar una frecuencia que nos entregue el mejor rendimiento de T1, el que se observará logrando la máxima tensión de salida. Esto significará que el núcleo será capaz de transmitir la mayor cantidad de energía desde un bobinado hacia el otro y que la frecuencia lograda será la óptima para este propósito.

El circuito de funcionamiento automático
Para comprender de manera sencilla el funcionamiento automático de la luz de emergencia, tomaremos a T1 (BC640) como un interruptor que para conducir (para comportarse como una llave cerrada) debe tener baja tensión en su base. ¿Cómo logramos una baja tensión en la base de T1? Haciendo conducir, como si fuese una llave cerrada, a T2 (A1015). De este modo, la tensión de la batería alimenta el circuito integrado (a través de R4), éste comienza a oscilar, actúa sobre los MOSFET y éstos hacen trabajar al transformador que se encargará de generar la tensión necesaria para encender la lámpara conectada a su salida. T2 conduce porque R14 pone su base a un potencial bajo. Cuando vuelve la energía eléctrica y el circuito se dispone a comenzar el ciclo de carga de la batería, la base de T2 recibe tensión positiva a través del divisor resistivo formado por R15 y R16, interrumpiendo la conducción de T2. Al pasar a cortarse la conducción de T2, también se interrumpirá la corriente que circula por T1, deteniendo la oscilación y apagando la luz de emergencia.

Es decir, el circuito inicial del cargador de baterías con el transformador y el rectificador de entrada nos servirá de monitor para determinar cuándo la luz de emergencia se debe encender y cuándo se debe apagar. Por lógica, al interrumpirse el suministro eléctrico, la luz encenderá y al retornar la energía por la red, la luz se apagará de manera automática. Comienza a experimentar con diferentes transformadores y juega con los valores de P1 – C1 – R3 hasta lograr los resultados más eficientes. Nosotros te dejamos los valores para 50/60Hz pero recuerda que existen núcleos que trabajan mejor a frecuencias más elevadas (200, 300Hz.). En la próxima entrega te mostraremos el cargador con corte automático para la batería y su implementación en este circuito. Es decir, el cargador funciona hasta que la batería está completa y luego se interrumpe la carga para no dañar la batería. 

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