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Virtual Model for Opamps in Negative Feedback Circuits

   From the above calculation, we have $v_+ - v_- = v_{out}/A \approx 0$ since the opamp gain A is very large, and so

 

v₊ = v_- . (10)

This says there is no voltage drop across the input terminals! This is quite a reasonable assumption since if $A \approx 10^6$, v_out =10V, then $v_+ - v_- = 10/10^6 = 10\mu V$, which is six orders of magnitude smaller than the output voltage.

Also, if $i_+, \ i_-$ denote the currents entering the two inputs, then since R_in is very large, say $10^6 \Omega$, then

$$i_+ = 0 \quad {\rm and} \quad i_- = 0$$ (11)

That is, the currents entering the input terminals is zero.

We now summarise these assumptions.

Virtual Short Assumption	v+ = v-
Virtual Open Assumption	$i_+ = 0 \quad {\rm and} \quad i_- = 0$

The virtual opamp model is illustrated in Figure 12.

  

Figure 12: Virtual opamp model.

This virtual model will always be used in the rest of the course, unless otherwise stated. It greatly simplifies the analysis of stable opamp circuits, and gives a very good first approximation to the actual behaviour.

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