Widlar current source

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Widlar current source.

A Widlar current source is a current source attributed to the late Bob Widlar, in which an emitter degeneration resistor is only applied to Q₂ (designated as R2 in the schematic). This has the effect of turning off Q₂ relative to Q₁. The current source is commonly used to generate low currents.

1 Analysis
- 1.1 Approximate solution
- 1.2 Exact solution
2 Output impedance
3 References
4 See also

[edit] Analysis

Neglecting the base current of Q2, the output current $I C 2$ is also the current through R2. Hence,

$V B E 1 = V B E 2 + I C 2 R 2$ ... (1)

Since in general the base-emitter voltage is given by,

$V_{BE} = V_T\ ln\left(\frac{I_C}{I_S}\right)$ ... (2)

Rearranging (1),

$I C 2 R 2 = V B E 1 - V B E 2$ ... (3)

Substituting for $V B E$ from (2) into (3) we get,

$I_{C2}R2 = V_T\ ln\left(\frac{I_{C1}}{I_{S}}\right) - V_T\ ln\left(\frac{I_{C2}}{I_{S}}\right)$ ... (4)

$= V_T\left[ln\left(\frac{I_{C1}}{I_S}\right) - ln\left(\frac{I_{C2}}{I_S}\right)\right]$ ... (5)

So,

$I_{C2} = \frac{V_T}{R2}ln\left(\frac{I_{C1}}{I_{C2}}\right)$ ... (6)

[edit] Approximate solution

The current through R1 is the input or reference current given as,

$I R 1 = I C 1 + I B 1 + I B 2$ ... (7)

$= I_{C1} + \frac{I_{C1}}{\beta} + \frac{I_{C2}}{\beta}$ ... (8)

If $β$ is large, the last two terms of the above equation can be neglected. Hence,

$I_{R1} \approx I_{C1}$

Thus assuming the forward base-emitter voltage of Q1 to be 0.7 V,

$I_{C1} \approx I_{R1} = \frac{V_{CC} - 0.7}{R1}$ ... (9)

Substituting (9) in (6) we get,

$I_{C2} = \frac{V_T}{R2}ln\left(\frac{I_{R1}}{I_{C2}}\right) = \frac{V_T}{R2}ln\left(\frac{V_{CC} - 0.7}{I_{C2}R1}\right)$ ... (10)

The above equation is the approximate solution for the output current $I C 2$ and is non-linear. An iterative process has to be employed to arrive at a value for R1 which gives a value of $I C 2$ which is reasonably close to that desired.

Knowing the desired values of $I C 1$ and $I C 2$ , R2 can be easily calculated.

Note that with the circuit as shown, if V_CC changes, the output current will change. Hence the circuit should be driven by a constant current source instead of R1 to keep the output current constant.

[edit] Exact solution

The transcendental equation in (6) can be solved exactly in terms of the Lambert W function giving:

$I_{C2} = \frac{V_T}{R_2} W\left(\frac{I_{C1}R_2}{V_T}\right)$

[edit] Output impedance

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 around transistor $Q 2$ . The loop gain is $G = (1 + g m 2 R 2)$ where $g m 2$ is the transconductance of $Q 2$ . The feedback loop magnifies the nominal output resistance $r o$ of the transistor by the factor $(1 + G)$ . Hence the Widlar is a better current source than the traditional current mirror.