# Odd op-amp circuit with a potentiometer

Im supposed to calculate the gain and impedance (in & out) of the inverting amplifier below. I also need to simulate the circuit with Ltspice using different $V_s$ values.

Im aware that the gain is calculated as $\cfrac{V_o}{V_s}=-\cfrac{R_2}{R_1}$. The problem with calculations and simulating is the circuit inside the red ellipse. I suppose the voltages +15, and -15 are for the amp (usage voltages).
But how to take that part into account calculating voltage $V_s$. Im not quite sure how the potentiometer in this case (or in overall) works. Also if possible, is there some equivalent circuit for understanding and simulating on Ltspice?

Any help will be appreciated.

• If you are required to find only the ratio Vo/Vs you do not need the actual value of Vs. Hence, the variable voltage divider plays no role (for the gain). Something similar applies to the input impedance at the node labelled Vs. In this case, the problem reduces to the question how to simulate a variable resistor, correct? – LvW Jun 21 '18 at 12:26
• Yes, thats likely the main issue. – M.P Jun 21 '18 at 12:31
• The variable voltage divider does provide another ~27K of input impedance, lowering the overall gain as it's in series with the 10K. Or you can consider the 10K as a load on the voltage divider, even though the ground is virtual. Draw the voltage divider as a Thevenin equivalent. – Cristobol Polychronopolis Jun 21 '18 at 13:01
• No - I do not think, that the resistive path left from the node Vs has any influence on the gain Vo/Vs. Considering the 10k resistor as a load to the divider circuit is relevant only for computing the actual value for Vs . – LvW Jun 21 '18 at 14:51

M.P..... Now - after some answers and some related comments - I think, it is clear to you that the resistive chain (at left) does NOT influence the ratio (gain) Vo/Vs.

However, if you want - in addition - to find the value of Vs as a function of the potentiometer position, you simply have the problem of a loaded voltage divider, which is relatively simple to solve. The load is simply the 10k resistor because in this caculation, you can assume the inverting opamp input to be grounded.

Simulation: I am not familiar with LTSpice, however in the PSpice program you can do the following (perhaps something similar works in LTSpice):

• Replace the pot by two resistors Rx and Ry.

• Do not allocate any values, but write instead {Rx} and for Ry write {5k-Rx}

(the form of the brackets is important in PSpice).

• In addition, declare Rx as a variable parameter (in PSpice using the .param command).

• Now, you can do a DC sweep, however, not for a DC source but for the quantity you have declared as a parameter (in your case: Rx). Sweep Rx from zero to 5k.

+15V and -15V are for the signal you are generating. Those three resistors form a variable voltage divider. The output of that goes to the opamp, and is multiplied by the gain you have calculated.

Figure out the maximum and minimum voltage you can get out of the voltage divider. From that you can see if the +-15V supply would be adequate to power the opamp. (Calculate output voltage of the opamp using max and min values from the divider and the gain.)

Hints only since this seems to be homework.

• Just read about the voltage dividers. So we can suppose R on the mid can vary between 0..5$K\Omega$. So voltage at $V_s$ is between $15V\cfrac{51K}{51K+51K+5K}$ and $15V\cfrac{51K+5K}{51K+51K+5K}$ ? – M.P Jun 21 '18 at 12:29
• Again, the 10K input resistor on the amp stage is a 10K load to [virtual] ground from the wiper. – Cristobol Polychronopolis Jun 21 '18 at 13:03
• @M.P remember the divider has a total of 30V difference. For your calculations replace the 15V with 30V. Then subtract 15V to shift your answer to what would be seen at Vs. – Clipboard_Waving_Enginerd Jun 21 '18 at 16:51