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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):

enter image description here

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

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  • \$\begingroup\$ Could you show what you mean by virtual ground? There is a circuit tool, draw the modified circuit. \$\endgroup\$ – laptop2d May 17 at 20:31
  • \$\begingroup\$ 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. \$\endgroup\$ – Kevin White May 17 at 22:22
  • \$\begingroup\$ Many errors on this diagram including schematic and formula including \$V_R=-\dfrac{R1}{R2}\cdot V_{ref}\$ \$\endgroup\$ – Sunnyskyguy EE75 May 17 at 23:07
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It might all become clear if you consider that the function of the differential amplifier is to measure the voltage across Rr and present the same voltage on the output of the differential amplifier with respect to ground. Using your virtual ground technique, you correctly (and easily) found the output voltage of the differential amplifier. Since this output voltage (with respect to ground) is the same as the voltage across Rr, the Error Amplifier output will be driven to whatever voltage is necessary to achieve this voltage across Rr. So you are sensing the voltage across sense resistor Rr and adjusting the output to give you a current source which is proportional to Vref.

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