Calculating the voltage in an OP-amp circuit with a current source

In summary, the current flows like the image below, but calculating the output voltage for the opamp is a bit confusing because there is no current flowing through the resistor between the - input and R1.
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DenDanne
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Summary:: Find the voltage in an OP-amp circuit with current source

I(in) = 1 uA.
1649702563724.png


What I'm confused about is if there is any voltage flowing through R1. Because if there was and let's say it went downwards, then where would it go, I mean it cannot just disappear. And if it goes upwards, then it has to go to the right through R3 (right?). But then we have I(in)+I(out) = I(in) and that does not make sense that more current can just appear out of nowhere.
So my guess is that the current passes like the picture below. But then how would I calculate U(ut). I can of course calculate the voltage between R2 and R3, but the problem then is that I suppose there isn't flowing any current through R1, so I cannot really use Ohms law?
https://www.physicsforums.com/attachments/299760
 
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  • #2
DenDanne said:
What I'm confused about is if there is any voltage flowing through R1.
There's no such thing as voltage "flowing through" something. Currents flow through paths, and there are voltages between nodes in a circuit (or "across" a component).

DenDanne said:
So my guess is that the current passes like the picture below. But then how would I calculate U(ut). I can of course calculate the voltage between R2 and R3, but the problem then is that I suppose there isn't flowing any current through R1, so I cannot really use Ohms law?
In a circuit with an ideal opamp, it's important to remember the fundamentals. An ideal opamp IC has very high input impedance at its +/- inputs, and the negative feedback causes the voltage difference between those +/- inputs to be zero.

So in this circuit, the - input terminal is at the same ground potential as the + input, right? And since the input impedance of the - input is high, all of that source current is going to flow through which resistor first? And doing that causes the voltage between R1, R2 and R3 to be what? And how do you keep going then to finish calculating the output voltage for the opamp?

See if you can use those hints to get farther in this problem... :smile:
 
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Related to Calculating the voltage in an OP-amp circuit with a current source

1. What is an OP-amp?

An operational amplifier, or OP-amp, is an electronic device that amplifies the difference between two input signals. It is commonly used in electronic circuits for signal processing, filtering, and amplification.

2. How do I find out the voltage of an OP-amp?

To find out the voltage of an OP-amp, you will need to measure the voltage at the input and output terminals using a voltmeter. The voltage gain of the OP-amp can then be calculated by dividing the output voltage by the input voltage.

3. What is the ideal voltage gain for an OP-amp?

The ideal voltage gain for an OP-amp is infinite, meaning that the output voltage is always perfectly amplified from the input voltage. However, in practical applications, OP-amps have a finite voltage gain that is determined by the specific design and components used.

4. What factors can affect the voltage gain of an OP-amp?

The voltage gain of an OP-amp can be affected by various factors, such as the input and output resistances, the power supply voltage, the frequency of the input signal, and the temperature. These factors can cause the voltage gain to deviate from the ideal value and may need to be considered in circuit design.

5. How can I increase the voltage gain of an OP-amp?

The voltage gain of an OP-amp can be increased by using external components, such as resistors and capacitors, to create a feedback loop. This can be done through non-inverting or inverting configurations, which can increase the overall gain of the OP-amp. However, it is important to note that increasing the voltage gain may also introduce other trade-offs, such as decreased stability or increased noise.

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