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The Cascode Current Mirror

The Circuit

The cascode current mirror.

One way to add emitter resistance without using actual resistors or generating a large voltage drop is to stack one current mirror on top of another, as shown to the left. This arrangement is known as a cascode current mirror. The term "cascode" dates back to the days of vacuum tubes, before any sort of semiconductor devices were invented. In a vacuum tube, the equivalent of the emitter is called the "cathode," and the configuration of Q2 and Q4 to the left was described as "cascade to cathode," which got shortened to "cascode." The name remains unchanged regardless of the specific components connected in this way.

In this circuit, Q1 and Q2 form a current mirror in which both collectors are held to a voltage of VBE, so they show a close match in operating conditions. At the same time, they serve as emitter resistances for a second current mirror (Q3 and Q4).

Generally, all four transistors are matched, but that isn't strictly necessary. The transistors of each mirror pair should be matched to each other, of course, but Q1 and Q2 don't have to match Q3 and Q4. Each mirror operates independently of the other, although they necessarily interact with each other.



Base Currents

The lower current mirror, Q1 and Q2, operates normally, with IC1 = IC2, IE1 = IE2, and IB1 = IB2. This is not true of the current mirror composed of Q3 and Q4. In the upper current mirror, IE3 must include the base currents for Q1 and Q2. Thus, IE4 = IC2, but IE3 = IC1 + IB1 + IB2. For the same reason, IB3 > IB4.

When comparing IO to IREF, it is easiest to relate both currents to IE1, which is also IE2. Assuming that all four transistors are matched, and keeping in mind that IC/IE = β/(β + 1), we get:

IO = IC4  =  β  IE4  =  β²  IE2  =  β²  IE1



β + 1 (β + 1)² (β + 1)²
 
IREF = IC3 + IB3 + IB4 = IE3 + IB4 = IC1 + 2IB1 + IB4  =  β  IE1  +  2  IE1  +  β  IE1



β + 1 β + 1 (β + 1)²
 

With everything expressed in terms of IE1, we can calculate the ratio IO/IREF. This expression simplifies to:

IO  =  β²  =  1



IREF β² + 4β + 2 1 + 4/β + 2/β²


Output Resistance

Because Q2 is not connected as a diode, but is fully operational as a transistor, its output resistance rO serves as the emitter resistance for Q4. This means that ROUT for Q4 can be quite large. The mathematical expression for ROUT is:

ROUT = rO(2 + gmrO) ≈ gmrO²

Keep in mind that one rO is for Q2 while the other is for Q4. If the Early voltages of these two transistors aren't equal, the rO values must be handled separately. Either way, the effective output resistance is much higher than for the simple current mirror.


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