chapter  3
74 Pages

- Input Stages

Finally, in the last section we introduce current-mode amplifiers. In this chapter, we

focus on the circuit low-frequency behavior, while the effects of reactive components

are discussed in chapter 4.

Consider the circuit of Fig. 3.1 a), in which the impedance Z f is placed in parallel

to a voltage amplifier with gain A≫ 1 and infinite input impedance. Using the small signal equivalent circuit of Fig. 3.1 b), we can write the following equations:

 Iin+

= 0

Solving for Vout/Iin we find: Vout

Iin =


1+A Z f (3.2)

The circuit thus converts an input current signal to a voltage output and, if A is much

greater than one, the transimpedance gain is equal to Z f . It is interesting to calculate

the input and output impedance of the circuit. To do so, we can put a test voltage

source between the node of interest and ground and we measure the current absorbed

by the circuit. The ratio between the imposed voltage and themeasured current defines

the input impedance. Alternatively, we can force a current into the node and sense the

developed voltage. Fig. 3.2 a) shows the circuit that allows us to calculate the input

impedance with the first method. Since we are driving the input node of the amplifier

with an ideal test voltage signal of amplitude Vt , the output is −AVt and the current

flowing in Z f is given by:

It = Vt − (−AVt)

Z f → Vt

It = Zin =

1+A (3.3)

If A is sufficiently big, the input impedance of the transimpedance amplifier can be

made very small and thus the circuit is suitable to read current input signals. We

have assumed so far that the core amplifier output impedance is much smaller than

Z f . Fig. 3.2 b) shows a circuit that can be used to calculate the output impedance of

the transimpedance amplifier in case the core amplifier has a finite output impedance

given by Z0. The test voltage source is connected to the output, while the input is left

floating. Owing to the infinite input impedance of the core amplifier, no current can

flow in Z f , thereforeVin =Vt and all the current is sunk by the amplifier output port. In formula, we can write:

It = Vt − (−AVt)

Z0 → Vt

It =

1+A (3.4)

From (3.4) we see that the open-loop output impedance of the core amplifier is

dropped by a factor 1+Awhen the feedback element is connected. This usually leads to very low output impedance, which makes the circuit suitable to also drive small

loads, confirming that the output port of the circuit behaves as a good voltage source.