Step-by-Step Derivation of the Wilson Current Mirror Formula

I remember the first time I tried to build a precision current source for a bandgap reference. Everything worked beautifully on paper. Then I slapped it on a breadboard, and the output current drifted all over the place as the temperature changed. The problem? The classic two-transistor mirror just couldn't handle the real-world mismatch and finite beta. That's when I discovered the Wilson current mirror, and honestly, it changed how I thought about feedback in analog circuits.

Look, the Wilson current mirror formula isn't just another equation to memorize. It's the key to understanding why this circuit delivers dramatically better accuracy than its simpler cousins. If you're designing precision analog circuits—current DACs, op-amp biasing, or sensor front-ends—you need to grasp this derivation. Not because you'll solve it by hand every time, but because it reveals how the circuit works under the hood. Let's walk through it, step by grubby step, with the math and the intuition together.

Seriously, if you haven't looked at this circuit in detail, you're missing out. The Wilson current mirror formula elegantly captures a feedback loop that cancels base current errors. It's a beautiful piece of work. And by the time we're done, you'll not only know the formula—you'll feel why it works.

Why You Should Actually Care About the Wilson Current Mirror

Before diving into the algebra, let me tell you why this circuit matters. The basic two-transistor current mirror suffers from a fundamental flaw: it assumes all transistors have infinite beta. In reality, beta is finite and often poorly controlled. A beta of 100 means 1% of the collector current goes to the base—that's a 1% error. Worse, beta drops at low temperatures and high currents. Your "matched" mirror becomes an exercise in frustration.

The Wilson current mirror fixes this by adding a third transistor that acts as a feedback element. It senses the base current error and actively cancels it. The step-by-step derivation of the Wilson current mirror formula will show you exactly how this cancellation happens. The result? Output impedance that rivals a cascode, with a compliance voltage only one Vbe higher than the input. That's a big deal in low-voltage designs.

It's a big deal.

Here's the thing: you can't just blindly copy a circuit from a textbook and hope it works. You need to understand the assumptions buried in that formula. Is beta large enough? Are the transistors matched? What happens at low currents? The derivation forces you to answer these questions. It's the difference between being a circuit user and a circuit designer.

The Classic Two-Transistor Mirror (The Setup)

Let's start with the baseline. The standard mirror uses two NPN transistors with their bases connected and their emitters grounded. The input transistor is diode-connected (collector tied to base). You force a reference current into the input, and the output transistor mirrors that current. Simple, right?

But the current transfer ratio isn't unity. The input transistor draws a base current from the reference, and the output transistor draws its own base current. With the output current defined as beta times the base current, and the reference current shared between the collector and both base currents, you get:

I_out = I_ref * (beta / (beta + 2))

For beta of 100, that's about 0.98—a 2% error. And that's if the transistors are perfectly matched. Temperature gradients, process variation, and Early effect pile on more errors. The Wilson current mirror formula addresses the beta-dependent error directly.

I can't stress this enough: that base current error is the enemy of precision. The step-by-step derivation of the Wilson current mirror formula will show you how to reduce it by a factor of beta.

Injecting the Wilson Magic (The Feedback Loop)

Now, add a third transistor (Q3) between the input transistor's collector and the base node of the mirror. Q3's emitter drives the base of Q1 and Q2. Q3's collector connects to the supply. This might look like a cascode, but it's fundamentally different. Q3 provides negative feedback that regulates the base voltage.

Here's the intuition: if the output current tries to increase, the base current of Q2 increases. That increases the voltage drop across Q3's base-emitter junction, which reduces the drive to Q2's base. That's negative feedback in action. The Wilson current mirror formula quantifies this loop gain and shows the resulting current accuracy.

Q1 and Q2 are still the main mirror pair. Q3 is the feedback amplifier. The reference current flows through Q1's collector and diode. Q3's emitter current is the sum of the base currents of Q1 and Q2. The magic happens when you write the node equations and see the cancellation.

The Heavy Lifting: Algebraic Derivation of the Wilson Current Mirror Formula

Alright, let's get our hands dirty. I'm going to assume matched transistors with equal beta (all NPNs identical on the same die). This is a reasonable assumption for integrated circuits. We'll also ignore the Early effect initially—we'll come back to that.

Define: - I_ref = reference current (forced) - I_out = output current (what we want to find) - I_B1, I_B2, I_B3 = base currents of Q1, Q2, Q3 - I_C1, I_C2, I_C3 = collector currents - beta = common-emitter current gain

Start with Q1, which is diode-connected. Its collector current equals its emitter current minus its base current. But Q1's collector current comes from the reference current minus the base current of Q3. Wait, let's be precise.

Step 1: Write the node currents.

I_ref enters the circuit at the collector of Q1. That node also connects to the bases of Q1 and Q2 (through Q3's emitter) and the collector of Q1. The current through Q1's collector is I_C1. The current coming from Q3's emitter is I_E3.

At the input node (collector of Q1, base of Q1, collector of Q1): I_ref = I_C1 + I_B3.

Now, Q3's emitter current is the sum of the base currents of Q1 and Q2: I_E3 = I_B1 + I_B2.

Since all transistors are matched and V_BE is the same for Q1 and Q2 (same V_BE), their collector currents are equal: I_C1 = I_C2. Also, their base currents are equal: I_B1 = I_B2 = I_B.

Therefore, I_E3 = I_B1 + I_B2 = 2 * I_B.

The collector current of Q3 is I_C3 = beta I_B3. And the emitter current of Q3 relates: I_E3 = (beta + 1) I_B3.

So, (beta + 1) I_B3 = 2 I_B.

Now, I_B = I_C2 / beta. And I_C2 = I_out (the output current).

So, I_B = I_out / beta.

Step 2: Solve for I_B3 in terms of I_out.

From I_E3 = (beta + 1) I_B3 = 2 I_B = 2 * I_out / beta.

So, I_B3 = (2 I_out) / (beta (beta + 1)).

Step 3: Relate I_ref to I_out.

I_ref = I_C1 + I_B3.

But I_C1 = I_C2 = I_out.

Therefore, I_ref = I_out + (2 I_out) / (beta (beta + 1)).

Step 4: The Wilson current mirror formula.

Factor out I_out: I_ref = I_out [1 + 2 / (beta (beta + 1))].

The Wilson current mirror formula for I_out is:

I_out = I_ref / [1 + 2 / (beta * (beta + 1))]

For large beta (say, 100), beta * (beta + 1) is about 10,100. So the correction term is roughly 2 / 10,100 = 0.000198. That's a 0.02% error. Compare that to the 2% error of the basic mirror. That's two orders of magnitude improvement.

That's the power of feedback.

Step 5: The Simplified Approximation

If you don't need ultra-precision, the Wilson current mirror formula simplifies beautifully. For beta >> 1:

I_out ≈ I_ref * [1 - 2 / (beta^2)]

That's the takeaway: the current error scales as 1/beta^2 instead of 1/beta. That's why the Wilson mirror is so dominant in bipolar analog design.

Real-World Performance: What the Wilson Current Mirror Formula Tells Us

The formula is clean, but reality isn't. The step-by-step derivation of the Wilson current mirror formula assumed perfect matching and infinite Early voltage. Let's talk about the cracks.

First, the Early effect. Q2 and Q3 have different collector-base voltages. Q2's collector is at the output voltage (which can swing), while Q3's collector is fixed at the supply. This causes a mismatch in collector currents due to the Early voltage. The Wilson current mirror formula doesn't account for this directly. You need to add a correction term: the output current changes with output voltage. This limits the output impedance, though it's still much higher than a basic mirror.

Second, beta mismatch. Even on a monolithic die, beta can vary by 10-20% between transistors. The derivation assumed all beta values are equal. If they're not, the error term changes. The good news? The feedback still works, but the cancellation isn't perfect. You can analyze this by letting beta_1 = beta_2 = beta and beta_3 = beta* (1 + delta). You'll find the error scales as delta/beta instead of 1/beta^2. So matching matters.

Third, compliance voltage. The Wilson mirror requires the output voltage to be at least two Vbe drops above ground (one for Q2, one for Q3). In low-voltage designs (1.8V or less), that's a problem. You might need a low-Vth process or a different topology.

Output Impedance and the Cascode Effect

The Wilson current mirror formula also tells us about output impedance. The feedback loop increases the output resistance by roughly beta/2 compared to a basic mirror. It's not as high as a true cascode (which gives beta * r_o), but it's a significant improvement. For most applications, it's enough.

Honestly, the output impedance is where the Wilson mirror surprises people. You get cascode-like performance without the extra voltage drop of a cascode. The step-by-step derivation of the Wilson current mirror formula doesn't directly give you r_out, but you can find it by small-signal analysis. The result is r_out ≈ (beta/2) * r_o2, where r_o2 is the output resistance of Q2. That's a big number.

A Quick Walk-Through of the Math (With Actual Numbers)

Let's plug in some real numbers so you can see the magic.

Assume I_ref = 100 uA, beta = 100.

Basic mirror: I_out = 100 * (100 / 102) = 98.04 uA. Error = 1.96 uA (1.96%).

Wilson mirror: I_out = 100 / [1 + 2/(100*101)] = 100 / [1 + 0.000198] = 100 / 1.000198 = 99.98 uA. Error = 0.02 uA (0.02%).

That's a 98x reduction in error. For a 12-bit system with 1 LSB = 0.024% of full scale, the Wilson mirror easily meets the accuracy requirement. The basic mirror fails.

Now, try beta = 50 (low temperature).

Basic mirror: I_out = 100 * (50/52) = 96.15 uA. Error = 3.85%.

Wilson mirror: I_out = 100 / [1 + 2/(50*51)] = 100 / 1.000784 = 99.92 uA. Error = 0.08%.

Even with low beta, the Wilson mirror holds up. That's why I use it in temperature-sensitive designs.

One more thing: the Wilson current mirror formula assumes the reference current is stable. If your I_ref comes from a resistor and a voltage reference that drifts, you've got bigger problems. The mirror can't fix a bad reference.

Common Questions About the step-by-step derivation of the Wilson current mirror formula

Why can't I use the basic mirror if my beta is very high?

Even with beta = 1000, the basic mirror gives a 0.2% error. That's fine for 8-bit systems, but for 10-bit or higher resolution, you need better. The Wilson mirror reduces the error to 2e-6 (0.0002%). Also, beta varies with current and temperature. The Wilson mirror makes your design robust against that variation. It's not just about the nominal error—it's about stability.

Does the Wilson current mirror work with PNP or PMOS transistors?

Yes, the step-by-step derivation of the Wilson current mirror formula is topology-independent. You flip the circuit vertically for PNP BJTs or PMOS FETs. For MOSFETs, you replace beta with the transconductance parameter, and the derivation changes because MOSFETs have no gate current. The Wilson mirror in CMOS is often called a "regulated cascode" and it uses an amplifier instead of a transistor for the feedback.

What happens if the three transistors aren't matched?

The derived Wilson current mirror formula assumes equality, but you can modify it. If Q1 and Q2 are matched but Q3 is different, the error term increases by the mismatch factor. Integrated circuits typically provide excellent matching, but discrete designs need careful selection. I always use a monolithic array (like the CA3046 or LM3046) for discrete prototype work.

How does the Wilson mirror compare to a cascode mirror?

A cascode mirror adds a transistor in series with the output, giving higher output impedance but requiring more voltage headroom (two Vce drops plus a Vbe). The Wilson mirror gives similar impedance with only two Vbe drops. For low-voltage designs, the Wilson mirror wins. For ultra-high output impedance (like in precision current sources), you might combine both in a "Wilson cascode".

Can I derive the output impedance from the Wilson current mirror formula?

Not directly. The step-by-step derivation of the Wilson current mirror formula gives only the DC current transfer ratio. The output impedance comes from a small-signal analysis that includes the Early effect. The formula for r_out is r_out ≈ (beta/2) * r_o2, assuming r_o2 is the output resistance of Q2. The exact expression involves r_o3 and the loop gain, but that approximation is good enough for most design work.