6
\$\begingroup\$

I have a small sinusoidal signal, with an amplitude of 10 mV (the signal goes from -5mV to 5mV) that I'd like to amplify.

I'd like to use an op amp, but my problem is that I only have +3V and GND voltages to supply the op amp. Which means the classical amplifying circuit would not work:

enter image description here

Any idea how I could amplify my input signal?

\$\endgroup\$
  • \$\begingroup\$ Normally you'd AC-couple it into a supply divider, but the signal is so low that it could easily get lost in the PSU noise. How clean is the supply? \$\endgroup\$ – Ignacio Vazquez-Abrams Oct 11 '16 at 8:27
  • \$\begingroup\$ What do you mean by "clean"? Signal comes from an excited piezo transducer. \$\endgroup\$ – Vincz777 Oct 11 '16 at 8:29
  • \$\begingroup\$ How clean is your power supply voltage. Is the noise on the power supply < 1mV? The simple solution is to put a small series capacitor and then use two resistors to set the DC voltage on Vin to about 1.5V. But if you have a noisy power supply then all you will amplify is the power supply noise. \$\endgroup\$ – Andrew Oct 11 '16 at 8:41
  • \$\begingroup\$ Power supply is a 3V battery. I thought about shifting of 1.5V the Vin, but the 1.5V offset is going to be amplified too, am I right? \$\endgroup\$ – Vincz777 Oct 11 '16 at 9:30
  • \$\begingroup\$ @Vincz777 Not if you set your op-amp V- to 0v and V+ to +3v. That's called floating ground -- it makes the opamp operate "as if" ground was at +1.5V. \$\endgroup\$ – Florian Castellane Oct 11 '16 at 11:59
13
\$\begingroup\$

The basic circuit just needs a few extra components.

enter image description here

Adding C2 gives it a DC gain of 1 but keeps the AC gain set by Rf,Rg (G= 1+ Rf/Rg)

Adding R1,R2 creates the mid point DC offset at the non inverting input.

Adding R3 and C4 creates a smoothed supply (low pass filter) for this offset.

Adding C1 decouples the input.

Adding C3 decouples the output

Additional edit

R3 and C4 ensure that the voltage to the potential divider (R1,R2) is as noise free as possible. Any noise on the potential divider would appear at the (non-inverting) input and be amplified.

With regard to values:

R3 was chosen to be less than 1% of R1,R2 so that it would not drop too much voltage before the potential divider. C4 value was chosen because it is a commonly used value and gives a low break frequency .

These values are not particularly critical and others could be substituted if needed

\$\endgroup\$
  • 1
    \$\begingroup\$ P.S. Be careful about capacitor types. You may want to use a film capacitor for C1 to avoid microphonic effects (though if you are using the circuit as a microphone amplifier anyway that may not matter so much). \$\endgroup\$ – Peter Green Oct 11 '16 at 17:05
  • \$\begingroup\$ Is R3 supposed to say 10k instead of 1k0? \$\endgroup\$ – Brian Rogers Oct 11 '16 at 18:01
  • \$\begingroup\$ @Brian the prefix is used like a decimal point, so 1k0 means 1.0k. \$\endgroup\$ – 2012rcampion Oct 11 '16 at 18:30
  • \$\begingroup\$ @2012rcampion Thank you, I didn't know whether that was a typo or just a notation I wasn't familiar with. I haven't seen 1k written as 1k0 before. \$\endgroup\$ – Brian Rogers Oct 11 '16 at 18:34
  • \$\begingroup\$ Just to avoid any confusion I confirm that 1k0 is 1000 Ohms. \$\endgroup\$ – JIm Dearden Oct 12 '16 at 11:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.