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Does anybody have any useful material on how to calculate unknown voltages in an amplifier circuit.

Thanks in advance.

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Fig.1.b.Positive feedback through 100k makes both op-amp inputs +5V potential.Input current is 10uA, input resistance 10k, so Ohms law says V1 = 10k * 10uA = 100mV = +5.0VThere is current flowing in both ideal op-amp input terminals. So current through 100k must also be 10uA.The input current of 10uA flows up on through 100k. Therefore V2 = +5.0V – (100k * 10uA) = +500mV. Therefore V2 = +5.5Vf

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Does anybody have any useful material on how to calculate unknown voltages in an amplifier circuit.

Thanks in advance.

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Yes. But there are many different types of circuits. It is best to attach a picture or post a link to the actual circuit.Does anybody have any useful material on how to calculate unknown voltages in an amplifier circuit.

We can then show you how to find the unknown voltage.

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Here are the circuit examples. ThanksYes. But there are many different types of circuits. It is best to attach a picture or post a link to the actual circuit.

We can then show you how to find the unknown voltage.

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There is an input current specified.

For the op amp to be operating in a linear mode what must the voltage be on the -ve input?

If the input impedance of the opamp is high, where does the input current go?

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Hi Thanks for the reply. I am just starting out with learning about these circuits so my original question was does anybody have any material that can help me understand how to calculate unknown voltages in these types of circuits. The image is an example of what I am working towards solving. The hints are much appreciated though.

There is an input current specified.

For the op amp to be operating in a linear mode what must the voltage be on the -ve input?

If the input impedance of the opamp is high, where does the input current go?

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An op-amp output voltage rises if input(+) is greater than input(–).

The op-amp output voltage falls if input(+) is lower than input(–).

That means that when input(+) is equal to input(–) the output will be stable.

Currents flowing through resistors drop voltages. Ohms law. V = I * R.

Chains of resistors form potential dividers.

Start here with op-amp circuits.

https://en.wikipedia.org/wiki/Operational_amplifier_applications#Amplifiers

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MIT has received permission to provide a older edition of "Op Amps for Everyone", which is Texas Instruments' stellar and easy-to-follow introduction to op amps. It covers almost everything you would need to know about op amps to become an expert, and will be really helpful in solving the kinds of problems you showed and can expect to encounter.

http://web.mit.edu/6.101/www/reference/op_amps_everyone.pdf

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Thanks for all of your help.

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Beginners should print a hard copy and put it in a 3 ring binder for frequent reference.

Likewise AN31 , opamp circuit collection.

http://www.ti.com/ww/en/bobpease/assets/AN-31.pdf

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http://www.ti.com/ww/en/bobpease/index.html?keyMatch=bob%20pease&tisearch=Search-EN-Everything

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Here is the first solution. Fig.1.a.

Negative feedback through 100k makes both op-amp inputs ground potential.

Input current is 10uA, input resistance 10k, so Ohms law says V1 = 10k * 10uA = 100mV = +0.1V

There is no current flowing in either ideal op-amp input terminal. So current through 100k must also be 10uA.

The input current of 10uA flows down on through 100k.

Therefore V2 = zero volts – (100k * 10uA) = –1000mV. Therefore V2 = –1.0V

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