Skip to content
Applied Electronics

Current mirrors in depth: Early effect, Wilson, cascode and Widlar proved

Error due to the Early effect and base currents, helper transistor, Wilson and cascode mirrors with output resistances proved, wide-swing cascode, Widlar mirror solved by iteration, MOS mirrors with lambda.

8 min read 0 comments

This page goes deeper into the lesson on current mirrors: it derives step by step the formulas that the lesson only states, namely the error of the mirror with a helper transistor and of the Wilson mirror, the output resistance of the Wilson mirror and of the cascode with the small-signal model, the Early effect on the copy ratio, and shows how to solve the Widlar mirror equation, which has no closed-form solution. The quantities are those of the lesson: \beta current gain, V_T = kT/q \approx 25.9 mV at 300 K, V_A Early voltage, r_o = V_A/I_C, g_m = I_C/V_T.

The simple mirror with the Early effect

In the lesson the error due to base currents was computed assuming equal collector currents. With the Early effect they are not: for the same v_{BE}, the collector current grows with v_{CE}:

    \[ I_C = I_S\,e^{v_{BE}/V_T}\left(1 + \frac{v_{CE}}{V_A}\right) \]

The reference transistor, diode-connected, has v_{CE1} = v_{BE}; the output one has the v_{CE2} imposed by the load. The bases are connected, so the exponential factor is the same and cancels in the ratio:

    \[ \frac{I_{C2}}{I_{C1}} = \frac{1 + v_{CE2}/V_A}{1 + v_{BE}/V_A} \]

Output characteristic of a BJT with the Early effect
The lines of the characteristic, extended, meet at minus V A. The reference transistor works at v CE equal to V BE, the output one at the voltage set by the load: with the same v BE the output current is larger.

Bringing in the base currents: at the reference node I_{REF} = I_{C1} + I_{C1}/\beta + I_{C2}/\beta. With I_{C2} = kI_{C1}, where k is the ratio just found,

    \[ I_{OUT} = I_{C2} = \frac{k\,I_{REF}}{1 + (1 + k)/\beta} \]

For k = 1 the lesson’s formula is recovered. Example. V_A = 100 V, v_{BE} = 0.7 V, v_{CE2} = 5 V, \beta = 100: k = 1.05/1.007 = 1.043, and I_{OUT} = 1.043/1.0204\cdot I_{REF} \approx 1.022\,I_{REF}. The two errors have opposite signs and here the Early effect prevails: the output current is 2.2 % higher than the reference.

The mirror with a helper transistor

A third transistor Q3, an emitter follower, has its base on the collector of Q1 and its emitter on the bases of Q1 and Q2: it supplies the two base currents.

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.

Theorem. With identical transistors, I_{OUT} = \dfrac{I_{REF}}{1 + \dfrac{2}{\beta(\beta + 1)}}, i.e. a relative error of about 2/\beta^2.

Proof. Q1 and Q2 have the same v_{BE}, so I_{C1} = I_{C2} = I_C and base currents I_C/\beta each. These two currents, 2I_C/\beta, come out of the emitter of Q3; its base current is \frac{2I_C}{\beta(\beta + 1)}, because the emitter current is \beta + 1 times the base current. At the collector node of Q1, the reference splits between the collector of Q1 and the base of Q3:

    \[ I_{REF} = I_C + \frac{2I_C}{\beta(\beta + 1)} \]

hence the formula. ∎

With \beta = 100 the error drops from 2 % to 0.02 %. The price is that the reference branch now has two v_{BE} in series, about 1.4 V, which reduce the headroom with low supplies.

The Wilson mirror

In the Wilson mirror the reference current enters the collector of Q1, which drives the base of Q3, the output transistor; the emitter of Q3 goes to the collector of Q2, diode-connected, whose base is joined to that of Q1. Q1 and Q2 thus form a simple mirror that “copies” the emitter current of Q3 into the reference branch.

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.

Theorem. With identical transistors, I_{OUT} = I_{REF}\left(1 - \dfrac{2}{\beta^2 + 2\beta + 2}\right), i.e. a relative error of about 2/\beta^2.

Proof. Let I_C be the collector current of Q1 and Q2, equal because they have the same v_{BE} (neglecting the Early effect). The collector node of Q2 receives the emitter current of Q3 and distributes it between the collector of Q2 and the two bases: I_{E3} = I_C + 2I_C/\beta = I_C(\beta + 2)/\beta. The output current is the collector current of Q3: I_{OUT} = I_{E3}\,\beta/(\beta + 1) = I_C(\beta + 2)/(\beta + 1). At the input node the reference splits between the collector of Q1 and the base of Q3: I_{REF} = I_C + I_{OUT}/\beta. Obtaining I_C = I_{OUT}(\beta + 1)/(\beta + 2) and substituting:

    \[ I_{REF} = I_{OUT}\left(\frac{\beta + 1}{\beta + 2} + \frac{1}{\beta}\right) = I_{OUT}\,\frac{\beta^2 + 2\beta + 2}{\beta(\beta + 2)} \]

So I_{OUT}/I_{REF} = \dfrac{\beta^2 + 2\beta}{\beta^2 + 2\beta + 2} = 1 - \dfrac{2}{\beta^2 + 2\beta + 2}. ∎

The output resistance. If the voltage on the collector of Q3 rises, the output current tends to grow because of the Early effect. But that current, through Q2 and the Q2-Q1 mirror, increases the collector current of Q1, which takes current away from the base of Q3 and pulls it back: it is negative current feedback. Small-signal analysis (model with r_\pi, g_m, r_o, lesson on transistor amplifiers) gives

    \[ R_{out} \approx \frac{\beta\,r_o}{2} \]

The factor 2 arises because half of the change in Q3’s collector current is returned as base current through the mirror, while the other half flows in Q3’s resistance r_o: the feedback multiplies the resistance by about \beta/2 rather than \beta.

The output resistance of the cascode

The MOS cascode consists of two transistors in series: M1 at the bottom, with its gate at a fixed voltage, behaves as a current source with resistance r_{o1}; M2 above, with its gate at a fixed voltage, carries the current to the output.

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.

Theorem. The output resistance of the MOS cascode is

    \[ R_{out} = r_{o1} + r_{o2} + g_{m2}r_{o2}r_{o1} \approx g_{m2}\,r_{o2}\,r_{o1} \]

Proof. Apply to the output (the drain of M2) a small test current i, with all constant sources turned off: the gates are at fixed voltage, i.e. grounded for small signals. In M1 gate and source are grounded, so its g_mv_{gs} source carries nothing, and M1 reduces to the resistance r_{o1}: the voltage at the source of M2 is v_s = i\,r_{o1}. In M2 the gate is grounded, so v_{gs2} = -v_s = -i\,r_{o1}, and its source carries the current g_{m2}v_{gs2} = -g_{m2}r_{o1}i from drain to source. By Kirchhoff’s current law at the drain node, the current in r_{o2} is i - g_{m2}v_{gs2} = i(1 + g_{m2}r_{o1}), and the output voltage is

    \[ v_{out} = v_s + r_{o2}\,i\,(1 + g_{m2}r_{o1}) = i\,\big[r_{o1} + r_{o2} + g_{m2}r_{o2}r_{o1}\big] \]

Dividing by i gives the formula. ∎

The physical meaning: M2 holds its own source almost still, because any attempt to raise it reduces its v_{gs} and hence the current. The resistance of M1 is “multiplied” by the intrinsic gain g_{m2}r_{o2} of M2, typically 20-100. The same calculation for the bipolar cascode gives R_{out} \approx \beta r_o: here the multiplication is limited by the base current of Q2, which escapes towards the fixed base.

The minimum voltage of the cascode

For all transistors to stay in saturation (in MOS, v_{DS} \ge V_{OV}) the cascode needs a minimum voltage across it. In the “simple” cascode mirror the gates of the two diode-connected reference transistors, one above the other, fix the gate of M2 at 2V_{GS} = 2(V_t + V_{OV}). The source of M2 is therefore at V_t + V_{OV}, and the output drain must be at least V_{OV} above it: v_{out,min} = V_t + 2V_{OV}. M1 works with v_{DS} = V_t + V_{OV}, more than necessary: one V_t is wasted. The wide-swing cascode mirror biases the gate of M2 at V_t + 2V_{OV} with a dedicated branch, so that M1 works at the edge of saturation with v_{DS} = V_{OV}, and the minimum voltage drops to 2V_{OV}. Example. With V_t = 0.5 V and V_{OV} = 0.2 V: 0.9 V against 0.4 V, a decisive difference with a 1.2 V supply.

The Widlar mirror equation

For design, i.e. finding R_E given the current, the lesson’s formula applies directly. For analysis, i.e. finding I_{OUT} given R_E, the equation

    \[ V_T\ln\frac{I_{REF}}{I_{OUT}} = I_{OUT}R_E \]

is transcendental: the unknown appears both inside the logarithm and outside, and cannot be isolated with algebra. It is solved by iteration. Rewritten as I_{OUT} = \frac{V_T}{R_E}\ln\frac{I_{REF}}{I_{OUT}}, one starts from a trial value, puts it on the right and computes a new value, until it settles. Convergence is guaranteed because the right-hand side varies slowly: its derivative with respect to I_{OUT} is -V_T/(R_EI_{OUT}), which at the solution equals -1/\ln(I_{REF}/I_{OUT}), less than 1 in magnitude when the current ratio exceeds e \approx 2.7, i.e. always in the Widlar’s use cases (for a function with derivative less than 1 in magnitude the iteration is a contraction and converges, Numerical Calculus lesson on roots of functions). A complete example is in the exercises.

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.

MOS mirrors and channel-length modulation

For the MOS mirror with channel-length modulation \lambda, the same reasoning as for the Early effect gives

    \[ \frac{I_{OUT}}{I_{REF}} = \frac{(W/L)_2}{(W/L)_1}\cdot\frac{1 + \lambda v_{DS2}}{1 + \lambda v_{DS1}} \]

with v_{DS1} = v_{GS} for the diode-connected transistor. Since \lambda is inversely proportional to the channel length, precision mirrors use long transistors (L several times the process minimum), or make v_{DS2} equal to v_{DS1} with a cascode or with an amplifier regulating the output voltage.

Exercises on the topic, with exam-style design and analysis problems, are on the exercise page linked at the top.

Related articles

Subscribe
Notify of
guest
0 Comments
Oldest
Newest Most Voted
0
Would love your thoughts, please comment.x
()
x