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I need to design a non-inverting amplifier with gain from 1 (unity) to 15. From what I read, in theory this can be achieved by using a pot as the feedback resistor, but in practice this will only work depending on the open-loop gain of the op-amp. Below is the circuit which I intend to use, using the AD8544 as an op-amp on single supply and RV1 used to set the gain from unity to 15. The input signal will have Vcc/2 as the base reference, hence why R1 is connected to Vcc/2. Input frequency can be anywhere from 5 to 600Hz.

The datasheet of the AD8544 shows a graph of the open-loop gain VS frequency (figure 30). Does this graph show that the gain could be up to 40dB for frequencies 5 to 600Hz ?

Is there anything which can be improved in the circuit shown below?

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  • \$\begingroup\$ make sure vdd/2 is buffeded \$\endgroup\$ – vicatcu Oct 13 at 13:55
  • \$\begingroup\$ That's 40dB to several hundred kHz, mot Hz. \$\endgroup\$ – Brian Drummond Oct 13 at 13:55
  • \$\begingroup\$ @vicatcu what do you mean by "make sure Vdd/d is buffered"? \$\endgroup\$ – dritech Oct 13 at 13:58
  • \$\begingroup\$ like don't just use a voltage divider, add a unit gain buffer to it so that the vdd/2 output has low output impedance \$\endgroup\$ – vicatcu Oct 13 at 14:01
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Yes, it means the open loop gain is 40dB, i.e. ~100 V/V (20 log 100 = 40 dB), up to around 300kHz. It is a typical open loop gain of differential signal.

Since you want to amplify the input signal by a factor of 1 to 15 and the output signal is:

out = in * (1 + RV1/R1)

you have to make sure RV1 will vary from 0 to 14 * R1 = 14 * 3.48 = 48.72 Ohm, so 50 kOhm resistor will work.

Of course you cannot get more than Vcc on the output, the opamp will saturate. Therefore for your maximum amplification of 15, when RV1 = 14R1, your maximum input signal for linear amplification may be Vcc / 15, this will be then amplified to the value of Vcc. All the input voltages above Vcc / 15 will result in output of Vcc. In general, for any amplification factor of G the maximum input signal for linear amplification is Vcc / G, and opamp will saturate for voltages above that.

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  • \$\begingroup\$ So just to clarify, ill circuit work fine in practice? is there any components which I need to include to improve stability/performance of this circuit? \$\endgroup\$ – dritech Oct 13 at 14:50
  • \$\begingroup\$ Yes. Just remember you cannot get more than Vcc on the output, the opamp will saturate. Your maximum input signal in this case may therefore be Vcc / 15, this will be then amplified to the value of Vcc. Other than that I think your design is OK. \$\endgroup\$ – 4pie0 Oct 13 at 23:17
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It is a very slow opamp like a 741 so at only 10kHz its open loop voltage gain is 40dB which is 100 times. An OPA134 audio opamp has 10 times more open loop voltage gain at all frequencies but this circuit has no high audio frequencies.

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