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I want to amplify a voltage range (from 0.54 mV to 42 mV) in below circuit which the gain of the Op Amp is 10x, so the Vout will be (5.4 mV : 0.42 V) ..

and this Vout will be a Vin again in another non-inverting Op Amp to be amplified 10x so the final Vout will be (54 mV : 4.2 V)

Question What's the Vdd that must be supplied to the Op Amps in both amplifying stages to get these Vouts --the Supply has to be 2.5 volts only--?

Circuit

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  • \$\begingroup\$ That's very clear.. so in the 2nd stage where the maximum output will be 4.2v, the Op-amp can be supplied by 2.5 volts (let us assume 2.5v is above the minimum supply voltage), maybe this is a stupid question but I'm very new into op-amps \$\endgroup\$ – Mohamed Sayed Jan 5 '18 at 17:14
  • \$\begingroup\$ 0.54mV is below the input offset voltage of most opamps. Designing the circuit with the required accuracy is going to be challenging. Anyway, back on topic, why do you suggest the supply voltage should be 2.5V? Where does this figure come from? \$\endgroup\$ – dim Jan 5 '18 at 17:30
  • \$\begingroup\$ Also, why two stages? Why not have one stage with a gain of -100? \$\endgroup\$ – dim Jan 5 '18 at 17:32
  • \$\begingroup\$ First because the system going to run on a Li-ion battery which going to produce out 2.7v as a minimum discharging volts before it die & need to be charged .. Second no problem with using a gain of 100.. but I don't know much about op-amps and I think the output voltage is relative to the supply. \$\endgroup\$ – Mohamed Sayed Jan 5 '18 at 17:37
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    \$\begingroup\$ The supply voltage must be greater than the max output voltage you need plus some overhead specified in the opamp datasheet. So if you need 4.2V out, you can't supply only 2.5V. Does it answer the question? \$\endgroup\$ – dim Jan 5 '18 at 17:42
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To get an output voltage of 4.2V, your supply should be at least 5V. The output of the opamp cannot be higher than its rail voltage (and you won't even get that much out, even with a rail-to-rail type).

Further, with an input voltage of 0.54mV, you are going to have no luck with such a simple circuit. You will also need a negative supply with inputs that close to ground.

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