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I am having trouble with understanding how to design circuits involving op amps.

I need to design a circuit using as less op amps as possible which has the following characteristics:

  • high input impedance (greater than 100k)
  • low output impedance (smaller than 100)
  • can be adjusted with a single 47k potentiometer over the range 0 to -10
  • gain should vary linearly with potentiometer rotation

I really don't understand how to complete this...

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    \$\begingroup\$ We don't just do your homework for you here. Show that you've done some work, and explain exactly what you are stuck on. \$\endgroup\$ – Olin Lathrop Jan 14 '15 at 14:33
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If we can design a bit of a "silly" circuit, then the following would kinda work. (for one opamp) (Gain is not quite linear.) You've got to pick R4 to give a gain a bit more than 20... but I'm too lazy to work out the exact number.

schematic

simulate this circuit – Schematic created using CircuitLab

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Well, if you feed your signal into the pot you get a linear output level from 0 to 1 coming out - that sounds like a start. If you then buffer the signal before the pot you get >100k input impedance and, if you make that op-amp an inverting amplifier with Rin = 110k and Rf at 1.1Mohm you get inverting gain of 10. All that remains is to add a buffer op-amp (unity gain non-inverting) to the pot output and this should work. I'm struggling to see how you would achieve it without two op-amps unless the signal levels were small and you could use a 200 ohm potentiometer feeding the output directly.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ It's better to replace R2 by the pot R1 (47 k) connected as a rheostat; then R3 should be 4,7 k... but, unfortunately, the input resistance will decrease:( and a follower will be necessary. \$\endgroup\$ – Circuit fantasist Jan 14 '15 at 12:47
  • \$\begingroup\$ @Circuitfantasist - or use a 2Mohm pot \$\endgroup\$ – Andy aka Jan 14 '15 at 12:59
  • \$\begingroup\$ Another suggestion - connect the right end of R2 to the slider of the potentiometer R1, and use the op-amp output as a circuit output (i.e., implement the negative feedback network as a T-bridge). Thus R3 can be high enough while R2 and R1 - low enough. \$\endgroup\$ – Circuit fantasist Jan 14 '15 at 15:33

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