Widlar current source:

Diagram from Widlar's original patentA Widlar current source is a modification of the basic two-transistor current mirror that incorporates an emitter degeneration

resistor for only the output transistor, enabling the current source to generate low currents using only moderate resistor values.

The Widlar circuit may be used with bipolar transistors, MOS transistors, and even vacuum tubes. An example application is the 741 operational amplifier, and Widlar

used the circuit as a part in many designs.

This circuit is named after its inventor, Bob Widlar, and was patented in 1967.

Diagram from Widlar's original patent

Analysis:

Figure 1: A version of the Widlar current source using bipolar transistors.Figure 1 is an example Widlar current source using bipolar transistors, where the emitter

resistor R2 is connected to the output transistor Q2, and has the effect of reducing the current in Q2 relative to Q1. The key to this circuit is that the voltage drop

across the resistor R2 subtracts from the base-emitter voltage of transistor Q2, thereby turning this transistor off compared to transistor Q1. This observation is

expressed by equating the base voltage expressions found on either side of the circuit in Figure 1 as:

$V_B = V_{BE1} = V_{BE2}+(\beta_2+1)I_{B2}R_2 \ ,$

where β2 is the beta-value of the output transistor, which is not the same as that of the input transistor, in part because the currents in the two transistors are very

different.[8] The variable IB2 is the base current of the output transistor, VBE refers to base-emitter voltage. This equation implies (using the Shockley diode law):

Eq. 1

$(\beta_2+1)I_{B2} =\left(1 + 1/\beta_2 \right) I_{C2} = \frac{V_{BE1} - V_{BE2}}{R_2} = \frac{V_T}{R_2} \ln \left(\frac {I_{C1}I_{S2}}{I_{C2}I_{S1}}\right)\ ,$

where VT is the thermal voltage.

This equation makes the approximation that the currents are both much larger than the scale currents IS1, IS2, an approximation valid except for current levels near cut

off. In the following the distinction between the two scale currents is dropped, although the difference can be important, for example, if the two transistors are chosen

with different areas.

Figure 1: A version of the Widlar current source using bipolar transistors.

Output impedance:

Figure 2: Small-signal circuit for finding output resistance of the Widlar source shown in Figure 1. A test current Ix is applied at the output, and the output resistance is

then RO = Vx / Ix.An important property of a current source is its small signal incremental output impedance, which should ideally be infinite. The Widlar circuit

introduces local current feedback for transistor Q2. Any increase in the current in Q2 increases the voltage drop across R2, reducing the VBE for Q2, thereby countering

the increase in current. This feedback means the output impedance of the circuit is increased, because the feedback involving R2 forces use of a larger voltage to drive a

given current.

Output resistance is found using a small-signal model for the circuit, shown in Figure 2. The transistor Q1 is replaced by its small-signal emitter resistance rE because it is

diode connected.[10] The transistor Q2 is replaced with its hybrid-pi model. A test current Ix is attached at the output.

Using the figure, the output resistance is determined using Kirchhoff's laws. Using Kirchhoff's voltage law from the ground on the left to the ground connection of R2:

$I_b \left( ( R_1 \parallel r_E ) + r_{\pi} \right) + (I_x + I_b ) R_2 = 0 \ .$

Rearranging:

$I_b = -I_x \frac{R_2}{( R_1 \parallel r_E ) + r_{\pi} +R_2} \ .$

Using Kirchhoff's voltage law from the ground connection of R₂ to the ground of the test current:

$V_x=I_x ( r_O +R_2) +I_b (R_2-\beta r_O)\ ,$

or, substituting for I_b:

Eq. 4

$R_O = \frac {V_x} {I_x} = r_O \left( 1+ \frac { \beta R_2} {( R_1 \parallel r_E ) + r_{\pi} +R_2} \right)$ $+ \ R_2 \left( \frac { ( R_1 \parallel r_E ) + r_{\pi}} {( R_1 \parallel r_E ) + r_{\pi} + R_2 } \right) \ .$

According to Eq. 4, the output resistance of the Widlar current source is increased over that of the output transistor itself (which is rO) so long as R2 is large enough

compared to the rπ of the output transistor. (Large resistances R2 make the factor multiplying rO approach the value (β +1).) The output transistor carries a low current,

making rπ large, and increase in R2 tends to reduce this current further, causing a correlated increase in rπ. Therefore, a goal of R2 >> rπ can be unrealistic, and further

discussion is provided below. The resistance R1//rE usually is small because the emitter resistance rE usually is only a few ohms.

Figure 2: Small-signal circuit for finding output resistance of the Widlar source shown in Figure 1. A test current I_x is applied at the output, and the output resistance is then R_O = V_x / I_x

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