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Applied Electronics

Solved current mirror exercises, from basic to exam level

Twelve exercises with solutions: simple BJT and MOS mirror, current ratios, Early effect, Wilson, cascode, Widlar, helper transistor, multi-output mirrors, active load and supply sensitivity.

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Exercises on current mirrors of the kind set in Applied Electronics and Analogue Electronics courses, in three levels: basic, for direct use of the formulas; intermediate, with design and errors; exam, with complete circuits as in written exams. The result is given straight away; the full solution opens with a click. The proofs of the formulas are in the lesson and the in-depth study linked at the top.

Unless stated: V_{BE} = 0.7 V, V_T = 25.9 mV, identical transistors.

Basic level

Exercise 1. The simple bipolar mirror

A simple mirror is supplied from V_{CC} = 10 V; the reference current is set by a resistor R = 9.3 kΩ between the supply and the collector of the diode-connected transistor. With \beta = 100, what are I_{REF} and I_{OUT}?

Result: I_{REF} = 1.00 mA; I_{OUT} = 0.980 mA.

Solution

The collector of the diode-connected transistor sits at V_{BE}: the drop across the resistor is 10 - 0.7 = 9.3 V, so I_{REF} = 9.3/9.3\,\text{k} = 1.00 mA.

Base-current error (lesson): I_{OUT} = I_{REF}/(1 + 2/\beta) = 1.00/1.02 = 0.980 mA.

Exercise 2. Mirror with an area ratio

An output of about 3 mA is wanted from a 1 mA reference using transistors with emitter areas in the ratio 3 to 1. With \beta = 100, what exactly is I_{OUT}?

Result: 2.885 mA.

Solution

With areas in the ratio n = 3, at the same v_{BE} the currents are I_{C1} and 3I_{C1}, and the base currents I_{C1}/\beta and 3I_{C1}/\beta.

Reference node: I_{REF} = I_{C1} + (1 + 3)I_{C1}/\beta = I_{C1}(1 + 4/\beta) = 1.04\,I_{C1}. So I_{C1} = 0.9615 mA and I_{OUT} = 3\cdot 0.9615 = 2.885 mA.

In general I_{OUT} = nI_{REF}/(1 + (1 + n)/\beta): the error grows with the copy ratio, because the large transistor needs more base current.

Exercise 3. The MOS mirror

A MOS mirror has (W/L)_1 = 10, (W/L)_2 = 40 and I_{REF} = 100 µA. Compute I_{OUT} neglecting channel-length modulation, and then with \lambda = 0.04 V^{-1}, V_{GS} = 1.0 V and V_{DS2} = 3.0 V.

Result: 400 µA; with channel-length modulation 431 µA.

Solution

Without modulation: I_{OUT} = 100\cdot 40/10 = 400 µA.

With modulation (in-depth study): I_{OUT} = 400\cdot\dfrac{1 + 0.04\cdot 3.0}{1 + 0.04\cdot 1.0} = 400\cdot\dfrac{1.12}{1.04} = 431 µA, an 8 % error, due solely to the difference in V_{DS} between the two transistors.

Exercise 4. Output resistance

A simple bipolar mirror supplies 0.5 mA, with V_A = 80 V. By how much does the current change if the voltage on the output collector goes from 1 V to 5 V?

Result: about 25 µA, 5 %.

Solution

r_o = V_A/I_C = 80/0.5\cdot 10^{-3} = 160 kΩ. A 4 V change produces \Delta I = 4/160\,\text{k} = 25 µA, i.e. 5 % of 0.5 mA. For an ideal current source it would be zero: the output resistance is the parameter measuring how close the mirror comes to ideal.

Intermediate level

Exercise 5. Design of a Widlar mirror

From a 1 mA reference an output of 20 µA is wanted with a Widlar mirror. What must the emitter resistor be? Compare with the alternative of a simple mirror with a 20 µA reference obtained with a resistor from V_{CC} = 10 V.

Result: R_E \approx 5.06 kΩ; the alternative would need about 465 kΩ.

Solution

From the lesson’s formula: R_E = \dfrac{V_T}{I_{OUT}}\ln\dfrac{I_{REF}}{I_{OUT}} = \dfrac{0.0259}{20\cdot 10^{-6}}\ln 50 = 1295\cdot 3.912 \approx 5.06 kΩ.

Simple mirror with a 20 µA reference: R = (10 - 0.7)/20\,\mu\text{A} = 465 kΩ. On an integrated circuit a resistor of that value takes up an enormous area compared with a 5 kΩ one: this is the reason for the Widlar mirror.

Exercise 6. Analysis of a Widlar mirror

A Widlar mirror has I_{REF} = 1 mA and R_E = 2 kΩ. Find I_{OUT}.

Graphical solution of the Widlar mirror equation
The blue curve is the difference of the V BE values, the red line the drop across the emitter resistor: the output current is the intersection. The dashed orange line shows the iterations, which start from I 0 and approach the solution oscillating.

Result: about 41 µA.

Solution

The equation V_T\ln(I_{REF}/I_{OUT}) = I_{OUT}R_E is transcendental and is solved by iteration (in-depth study), written as I_{OUT} = \dfrac{V_T}{R_E}\ln\dfrac{I_{REF}}{I_{OUT}}, with V_T/R_E = 12.95 µA.

Starting from 50 µA: 12.95\cdot\ln 20 = 38.8 µA. Then 12.95\cdot\ln(1000/38.8) = 12.95\cdot 3.249 = 42.1 µA. Then 12.95\cdot\ln(1000/42.1) = 12.95\cdot 3.168 = 41.0 µA. Then 12.95\cdot\ln(1000/41.0) = 41.4 µA. Then 41.3 µA: the solution is about 41.3 µA.

Check: 0.0259\cdot\ln(1000/41.3) = 0.0826 V and 41.3\,\mu\text{A}\cdot 2\,\text{k} = 0.0826 V. The iterates oscillate around the solution, narrowing, as expected from a negative derivative less than 1 in magnitude.

Exercise 7. The helper transistor

With transistors of \beta = 50 and I_{REF} = 1 mA, compare the output current of the simple mirror and of the one with a helper transistor.

Current mirror with helper transistor
Q3, an emitter follower, supplies the base currents of Q1 and Q2 taking them from the supply: only the small base current of Q3 leaves the reference branch.

Result: simple 0.962 mA (3.8 % error); with helper 0.999 22 mA (0.078 % error).

Solution

Simple mirror: I_{OUT} = 1/(1 + 2/50) = 1/1.04 = 0.962 mA.

With a helper transistor (in-depth study): I_{OUT} = 1/\left(1 + \dfrac{2}{50\cdot 51}\right) = 1/1.000\,784 = 0.999\,22 mA.

The error is reduced by a factor of about \beta + 1 = 51: with low-\beta transistors, such as the lateral PNPs of integrated circuits, the helper transistor is almost indispensable.

Exercise 8. The Wilson mirror

A Wilson mirror with transistors of \beta = 100 and r_o = 100 kΩ. Compute the error in the copy ratio and the output resistance, and compare them with those of the simple mirror.

Wilson current mirror
The reference current enters the collector of Q1, which drives the base of Q3. The emitter current of Q3 flows through the diode-connected Q2, which copies it into Q1 through the common bases: the feedback loop makes the output current accurate and the output resistance high.

Result: error 0.0196 % (simple 2 %); R_{out} \approx 5 MΩ (simple 100 kΩ).

Solution

Error (in-depth study): 2/(\beta^2 + 2\beta + 2) = 2/10,202 = 1.96\cdot 10^{-4}, i.e. 0.0196 %, against 2/\beta, about 2 %, for the simple mirror.

Output resistance: \beta r_o/2 = 100\cdot 100\,\text{k}/2 = 5 MΩ, fifty times that of the simple mirror (r_o = 100 kΩ).

Exercise 9. The MOS cascode

A MOS cascode mirror uses transistors with g_m = 1 mA/V and r_o = 50 kΩ, V_t = 0.5 V and V_{OV} = 0.2 V. Compute the output resistance and the minimum output voltage in the simple and wide-swing versions.

MOS cascode and its small-signal circuit
On the left the circuit: M1 and M2 in series with the gates at fixed voltages. On the right the equivalent circuit with the test current i: r o1 brings the source of M2 to v s, and the gate-source voltage of M2 is minus v s.

Result: R_{out} \approx 2.6 MΩ; minimum voltage 0.9 V (simple) and 0.4 V (wide-swing).

Solution

Resistance (in-depth study): R_{out} = r_{o1} + r_{o2} + g_mr_{o2}r_{o1} = 50\,\text{k} + 50\,\text{k} + 10^{-3}\cdot 50\,\text{k}\cdot 50\,\text{k} = 0.1 + 2.5 = 2.6 MΩ, fifty times that of a single transistor.

Minimum voltage: simple V_t + 2V_{OV} = 0.5 + 0.4 = 0.9 V; wide-swing 2V_{OV} = 0.4 V.

Exam level

Exercise 10. Biasing a circuit with a multiple-output mirror

A multiple-output bipolar mirror is supplied at 5 V; the reference is set by a resistor. The outputs must have currents in the ratios 1, 2 and 4 relative to the reference transistor, which must conduct 0.5 mA. With \beta = 100:

  1. size the reference resistor neglecting base currents;
  2. compute the actual currents taking base currents into account;
  3. compute the error that would result from adding a helper transistor.
Multiple-output bipolar mirror
A diode-connected reference transistor and three outputs with emitter areas 1, 2 and 4 times its own: the collector currents are in the same ratios, and all the base currents load the reference branch.

Result: R = 8.6 kΩ; outputs 0.463, 0.926, 1.852 mA (7.4 % error); with the helper transistor an error of about 0.08 %.

Solution

1. R = (5 - 0.7)/0.5\,\text{mA} = 8.6 kΩ, giving I_{REF} = 0.5 mA.

2. The transistors have areas in the ratio 1 (reference), 1, 2, 4. The collector currents are I_C, I_C, 2I_C, 4I_C, and the base currents total (1 + 1 + 2 + 4)I_C/\beta = 8I_C/\beta. At the reference node I_{REF} = I_C(1 + 8/\beta) = 1.08\,I_C, so I_C = 0.463 mA. Outputs: 0.463, 0.926 and 1.852 mA, all 7.4 % below the intended value.

3. With the helper transistor the base currents 8I_C/\beta are supplied by its emitter, and only 8I_C/(\beta(\beta + 1)) is drawn from the reference: I_{REF} = I_C(1 + 8/10,100), error 7.9\cdot 10^{-4}, about 0.08 %. Mirrors with many outputs almost always need the helper transistor.

Exercise 11. The stage with an active load

A common-emitter amplifier is biased at I_C = 1 mA and has as its load a PNP mirror supplying the same current. Both transistors have V_A = 100 V. Compute the open-circuit voltage gain and compare it with that obtained with a 5 kΩ load resistor.

Result: about -1930 (66 dB) with the active load; -193 (46 dB) with the resistor.

Solution

g_m = I_C/V_T = 1/25.9 = 38.6 mA/V. The two output resistances are r_o = 100/1\,\text{mA} = 100 kΩ each.

With the active load the collector sees r_{o,npn} and r_{o,pnp} in parallel: 50 kΩ. Gain A_v = -g_m\cdot 50\,\text{k} \approx -1930, i.e. 66 dB.

With a 5 kΩ resistor (in parallel with r_o, almost negligible): A_v \approx -38.6\cdot 5 = -193, 46 dB. To get the same gain with a resistor 50 kΩ would be needed, with a 50 V drop at 1 mA: impossible with normal supplies. The mirror gives a high resistance for small signals and a small DC drop.

Exercise 12. Supply sensitivity

The reference of a mirror is obtained with a resistor from the supply V_{CC} = 5 V. Compute the sensitivity of the current to the supply, S = \dfrac{\Delta I/I}{\Delta V_{CC}/V_{CC}}, and the change in current if the supply rises by 10 %. What solution is used in integrated circuits?

Result: S \approx 1.16; the current rises by about 11.6 %.

Solution

I = (V_{CC} - V_{BE})/R, so dI/dV_{CC} = 1/R and

    \[ S = \frac{V_{CC}}{I}\cdot\frac{dI}{dV_{CC}} = \frac{V_{CC}}{V_{CC} - V_{BE}} = \frac{5}{4.3} = 1.16 \]

A 10 % increase (0.5 V) takes the current from 4.3/R to 4.8/R: +11.6 %, more than the supply itself.

Integrated circuits use self-biased references, in which the current is set by a V_{BE} difference across a resistor, V_T\ln N/R, independent of the supply (lesson, last section); combined with a V_{BE} they give the bandgap reference.

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