Early effect in simple current mirror

Question

By "simple current mirror" I mean the following circuit.

Due to the Early effect the output current changes to some degree with output voltage. According to The Art of Electronics (3rd edition),

this is because of the slight variation of \$V_\text{BE}\$ with collector voltage at a given current in \$Q_2\$.

The way I understand the sentence above: Change in \$V_\text{CE}\$ at constant \$I_\text{C}\$ leads to change in \$V_\text{BE}\$, which, considering the Ebers-Moll equation, causes a change in \$I_\text{C}\$! By the way, the Ebers-Moll equation, according to the book, is \$I_\text{C}=I_\text{S}[\exp(V_\text{BE}/V_\text{T})-1]\$.

In the book, the Early effect is mathematically formulated for two cases: 1. Changes of \$V_\text{BE}\$ with \$V_\text{CE}\$ at constant \$I_\text{C}\$ 2. Changes of \$I_\text{C}\$ with \$V_\text{CE}\$ at constant \$V_\text{BE}\$. But in this circuit neither \$V_\text{BE}\$ nor \$I_\text{C}\$ is constant.

Another point: The two transistors have the same \$V_\text{BE}\$. So, applying Ebers-Moll equation, they must have the same \$I_\text{C}\$. The book says that output current can change as much as 25% due to change in output voltage. On the other hand, the current in the 14.4k resistor is almost constant (\$(15-0.6)/14.4\$). So about 25% of \$I_\text{C}\$ in Q1 must go to the bases via the wire connecting collector of Q1 to bases, which sounds strange.

Answer 1

As I have come to understand it, the Early effect is about:

If V_BE is held constant and V_CB is increased then I_C increases.

Alternatively, if I_C is held constant and V_CB is increased then V_BE decreases.

In either of the above cases, when V_CB is increased, beta (h_FE) increases.

If the circuit is configured in a way that allows both V_BE and I_C to adjust their value then, when V_CB increases, Early Effect is able to appear as a combination of both an increase in I_C and a reduction in V_BE & I_B with a corresponding increase in beta (h_FE).

That version of the Ebers–Moll which you have shown doesn't account for the Early effect. There is another factor added to the end of that simplified version of the Ebers–Moll equation to account for Early effect.

The multiplication factor is 1 + (V_CE/V_A) where V_A is the Early voltage of the particular transistor. The Early voltage is typically lower in PNP transistors than in NPN transistors leading to greater Early Effect (larger increase in I_C or reduction in V_BE) for a given increase in V_CB.

In your current mirror, there will be only a very small reduction in both V_BE and in I_B of Q2 as V_CB is increased and almost all of the Early effect will show itself as an increase in I_C. The relationship between V_BE and I_C of Q2 can be theoretically calculated using the extended version of the Ebers–Moll equation which includes the added factor to account for the Early effect.

In this situation, where V_BE is held almost constant, the Early effect can be modeled by an added resistance placed between the collector and the emitter of Q2 which represents the output resistance at the collector of Q2 known as 1/hoe where hoe is the output conductance. Intuitively it can be seen that, when V_BE is held constant, increasing V_CB will draw more current through this resistance (R_CE) resulting in an increase in I_C.

Now if we were to add low value degeneration resistors into the emitter circuits of both Q1 & Q2 then, when we increase V_CB, a good deal of the Early effect will happen as a reduction in V_BE of Q2 (with the V_BE of Q1 hardly changing). I_C will still increase a little but far less than without the emitter degeneration resistors. So, by adding those resistors we have improved the performance of the current mirror which will have less variation in I_C when the load varies and changes V_CB.

Answer 2

The transistors are forced to have the same BE voltage, but can have different BC voltages, depending on the load.

The Early effect causes the collector currents to differ, despite the same BE voltages. The base current difference of Q2 will flow down the 14k resistor. This difference is rather small, because it is a 25% variation of Q2's base current that adds to Q1's collector current. So the resistor current will stay constant to ~1% or better usually.

Answer 3

The "slight variation of V_BE with collector voltage at a given current in Q2" means that if collector voltage changes with fixed V_BE then the collector current will change.

It's a roundabout way of putting it, but it's not wrong.