# Current mode operation of a Source Measurement Unit (SMU)

I'm studying semiconductor parameter analyzers. In particular, I have difficulties in fully understanding the circuit in the following slide, which represents an SMU operating in I-mode (i.e., it injects a current in the device under test and measures the voltage on it):

I start assuming ideal "error amplifier" block, thus the concept of virtual ground applies; as a consequence all the current which flows in R1 must flow in R2 as well. In this way, I easily find the output voltage of the "differential amplifier" block of the picture.

Question 1: why is it written that this voltage is also the voltage on Rr? The voltage on Rr should be de-amplified by the gain of the "differential amplifier" block

Question 2: It can be a stupid question, but this circuit has created in me a doubt: what about if I apply virtual-ground concept for the 2 inputs of the "differential amplifier" block? Rr would be short-circuited and the circuit would have no meaning, then the question: why am I allowed to use virtual-ground concept for the "error amplifier" block and not for the "differential amplifier" block?

Thank you

• Could you show what you mean by virtual ground? There is a circuit tool, draw the modified circuit. – laptop2d May 17 at 20:31
• The differential amplifier block is not an ideal opamp with infinite gain - it has internal feedback to give a defined gain and present an output that is relative to ground equal to some multiple of the voltage across the sense resistor Rr. The error amplifier however can be represented by an ideal I-amp with infinite gain for which your virtual ground concept applies. – Kevin White May 17 at 22:22
• Many errors on this diagram including schematic and formula including $V_R=-\dfrac{R1}{R2}\cdot V_{ref}$ – Sunnyskyguy EE75 May 17 at 23:07