Look again at Mooly's post19.
Assume the output has moved to 1.0000Vdc
The resistor ladder forming the feedback is 3k0 and 1k0 to the gound symbol.
That 1.0000Vdc on the output divides down to 1.0000/(1k+3k) = 0.2500Vdc between the resistor and 0.0000Vdc at the ground symbol.
That 0.2500Vdc appears on the -IN pin of the opamp.
Using that "rule":
The output moves to a voltage to try to make the two input pins see the same voltage, we arrive at the same 0.2500Vdc apearing on the +IN pin.
You have three voltages on the signal pins.
Do they make sense?
The gain of a non inverting opamp is {Rupper/Rlower} + 1 = 3k/1k +1 = 4times
4times 0.2500Vdc = 1.0000Vdc
YES !!!!! it works.
Now look at the opamp datasheet.
It might say the voltage gain of the opamp is 25V/mV, i.e. 25000times (+88dB)
The voltage difference across the -IN & +IN pins will actually be 1.0000Vdc/25000 = 0.000004Vdc = 0.04mVdc = 40uVdc
That is the actual voltage diff at the input, if you had an instrument capable of reading that without affecting the input currents.
Assume the output has moved to 1.0000Vdc
The resistor ladder forming the feedback is 3k0 and 1k0 to the gound symbol.
That 1.0000Vdc on the output divides down to 1.0000/(1k+3k) = 0.2500Vdc between the resistor and 0.0000Vdc at the ground symbol.
That 0.2500Vdc appears on the -IN pin of the opamp.
Using that "rule":
The output moves to a voltage to try to make the two input pins see the same voltage, we arrive at the same 0.2500Vdc apearing on the +IN pin.
You have three voltages on the signal pins.
Do they make sense?
The gain of a non inverting opamp is {Rupper/Rlower} + 1 = 3k/1k +1 = 4times
4times 0.2500Vdc = 1.0000Vdc
YES !!!!! it works.
Now look at the opamp datasheet.
It might say the voltage gain of the opamp is 25V/mV, i.e. 25000times (+88dB)
The voltage difference across the -IN & +IN pins will actually be 1.0000Vdc/25000 = 0.000004Vdc = 0.04mVdc = 40uVdc
That is the actual voltage diff at the input, if you had an instrument capable of reading that without affecting the input currents.
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Look again at Mooly's post19.
Assume the output has moved to 1.0000Vdc
The resistor ladder forming the feedback is 3k0 and 1k0 to the gound symbol.
That 1.0000Vdc on the output divides down to 1.0000/(1k+3k) = 0.2500Vdc between the resistor and 0.0000Vdc at the ground symbol.
That 0.2500Vdc appears on the -IN pin of the opamp.
Using that "rule":
The output moves to a voltage to try to make the two input pins see the same voltage, we arrive at the same 0.2500Vdc apearing on the +IN pin.
You have three voltages on the signal pins.
Do they make sense?
The gain of a non inverting opamp is {Rupper/Rlower} + 1 = 3k/1k +1 = 4times
4times 0.2500Vdc = 1.0000Vdc
YES !!!!! it works.
Now look at the opamp datasheet.
It might say the voltage gain of the opamp is 25V/mV, i.e. 25000times (+88dB)
The voltage difference across the -IN & +IN pins will actually be 1.0000Vdc/25000 = 0.000004Vdc = 0.04mVdc = 40uVdc
That is the actual voltage diff at the input, if you had an instrument capable of reading that without affecting the input currents.
Thanks mate, now I have a better understanding of how they work,plus Ive been reading,the Handbook of op amp applications by Bruce Carter and Thomas R Brown from Texas instuments and Burr Brown. They have so many apps! It makes me wonder how the world wouldve got along without them?🙂
Op amps in the days of vacuum tubes were serious coffee boilers. Burr Brown goes back that far, also Philbrick. Of course only the names remain, not the old employees, owners, or worksites.
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Helo,guys ! How can I control DC offset of................................
New thread here:
DC offset problem with STK403 amplifier.