Op amp configuration with 1 resistor ?

[KrisBlueNZ]

Gorgon has explained it pretty well. You can't apply a voltage to the input. That will simply force the op-amp to slam its output against one of the supply rails and the circuit will do nothing useful. You have to apply a current to the input, which can be opposed by the current through the feedback resistor so that the op-amp can bring its inverting input to the same voltage as its non-inverting input, i.e. 0V.

If you force a voltage onto the inverting input, you will prevent the op-amp from bringing its inverting input to the same voltage as its non-inverting input, and the op-amp will simply compare the two voltages and swing its output as hard as it can towards one of the supply rails, depending on the polarity of the voltage difference between its inputs.

The current for the input of this circuit can come from any circuit or a device that generates a current, positive or negative. I suggested a photodiode or phototransistor, because that circuit would make a good amplifier for a photodiode or phototransistor, but there are various other circuits and components that operate as current generators that you could use.

The simplest way to generate a current is using a voltage source with a series resistor, as I described. In this circuit it's convenient that the inverting input is forced to 0V (the non-inverting input voltage) by the action of the op-amp acting through the feedback resistor, because it means that the current through the series resistor will always be directly proportional to the voltage applied to the left hand end of it.

This is exactly how an op-amp-based inverting voltage amplifier stage works. The inverting voltage amplifier is the same circuit but with a series resistor added between the circuit's input point and the op-amp's inverting input. The input voltage generates a proportional voltage across the series resistor, and the op-amp's output sends an equal but opposite current through the feedback resistor to oppose that input current and bring the inverting input to the same voltage as the non-inverting input, i.e. 0V. In the process, the voltage at the output of the op-amp has to be proportional to the inverse of the input voltage. Therefore it is an inverting voltage amplifier.

The voltage at the inverting input is always held equal to the voltage at the non-inverting input, so the current through the input series resistor is always proportional to the input voltage (at the left end of the input resistor), according to Ohm's Law as applied to the input resistor. The inverting input is a summing node, where current from the input and an equal opposing current from the output (through the feedback resistor) cancel each other out, resulting in zero voltage (since the inverting input is connected to 0V).

The ratio of the input resistor to the feedback resistor determines the gain of the circuit. (Actually it's a negative gain because the circuit always inverts.)

The inverting amplifier circuit can be extended to a summing amplifier by adding more inputs, each with its own series resistor connecting to the summing node at op-amp's inverting input. Because this summing node always stays at 0V, because of the op-amp's action through the feedback resistor, there is no unwanted interaction between the inputs. Each input simply generates a proportional current into the summing node, and the op-amp's output is forced to the appropriate voltage to cancel out the sum of those currents. This gives a very simple and effective circuit for combining several input signals.

Gorgon is also right that the feedback resistor value, 100 ohms, is unusually low and will constrain the output voltage range because a general purpose op-amp like the LM324 cannot source or sink much current from its output. The load on an LM324 should be limited to around +/- 10~15 mA, so a feedback resistor of 1k or higher is more sensible. Something in the range 3k3~100k would be typical.

 

[24Volts]

Hello KrisBlueNZ,

thank you very much for the explanation.

So if i omit the series resistor, I could also use an LM317 to do some tests for this right?

thanks

 

[KrisBlueNZ]

Yes, you can use that circuit without an input resistor, and feed a current into it from an LM317 wired as a current source, and see what happens.

Remember that all the current you feed into the circuit has to flow through the feedback resistor and into the output of the op-amp, and the op-amp isn't designed to supply more than about 10~20 mA, so don't feed more than that amount of current into the input.

 

[24Volts]

Yes, you can use that circuit without an input resistor, and feed a current into it from an LM317 wired as a current source, and see what happens.

Remember that all the current you feed into the circuit has to flow through the feedback resistor and into the output of the op-amp, and the op-amp isn't designed to supply more than about 10~20 mA, so don't feed more than that amount of current into the input.

understood ! when I have a chance I will try it out.

I thank you very much for your assistance .... I really appreciate it.

This will be fun to try

Thanks
KrisBlueNZ

 

[gorgon]

If you use a +1V power supply, you can generate 1mA just by putting a 1kohm resistor between the opamp and the power supply.

 

[24Volts]

okay thanks gorgon!

 

[24Volts]

Hi guys,

One last question....

I have a power supply that can generate Dc voltage and Dc current. I read that if I put my voltage supply at max and adjust my current knob, the power supply can actually act as a constant current source. See pic of my power supply in attachment.

Now, to actually measure the current with my VOM, I connected the + lead of my VOM to the plus prong on my power supply and the negative lead of my VOM to the negative prong of my power supply.

I can then adjust the "Fine" current knob on the power supply to read 1.5 ma. And this is okay as I am able to measure the current being sourced out from the power supply. I then disconnect my VOM and connect the power supply's prongs to my op amp circuit... which is the positive of power supply to V1 and minus of power supply to ground. (See op amp step by step schematic below).

I then measure Vout from the op amp and I read the correct voltage. But then, I try to re-connect my VOM to the power supply and re-measure the current out of my power supply with my VOM. At this point the fuse in the VOM blows?? This actually happened twice???

Help... why is this happening??
thanks everyone

 

[KrisBlueNZ]

There's probably a large capacitor (something like 1000 uF) across the output terminals of the power supply. This is commonly done to improve the effective series resistance and transient response of the power supply and it's helpful when the power supply is in voltage regulation mode, but it can cause this problem when the power supply is used as a current regulator.

What happens is: when you disconnect the load from the power supply, the 1.5 mA coming from the current regulation section of the power supply no longer has an external path to flow through, and it charges the output capacitor up to the voltage that the power supply is set for. Then when you connect your meter across the power supply outputs, the capacitor discharges into the multimeter in a brief surge of unlimited current, which blows the fuse in the multimeter.

So the output capacitor of the power supply is actually a liability when it's used as a current source.

To avoid this problem, you need to make sure that the path for the output current from the power supply does not get interrupted when you make changes to the circuit. You can do this with careful use of alligator clip leads.

Before you turn the power supply on, set up your external wiring so the power supply's positive output goes to the positive lead of the multimeter (on milliamps range); the negative multimeter lead goes to the input of the circuit; and the 0V rail of the circuit returns to the negative of the power supply. That's a complete path for the power supply's output current.

Then whenever you want to disconnect any part of that circuit (i.e. either the multimeter or the circuit under test), bypass it with a short circuit first, using an alligator clip lead. Once the circuit element is bypassed, you can disconnect it without interrupting the current path and causing the power supply's output capacitor to charge up and store a damaging amount of energy.

You can set up a switch to bypass the multimeter if you don't want to use alligator clip leads.

Every time the voltage across the power supply's output changes, there will be a brief surge in current because of the output capacitor, but the surge will be a lot smaller than the surge that occurs when the output capacitor is charged to 20V, for example, and a low-resistance load is suddenly connected across it.

The surge can be reduced by adding a resistor in series with the positive output from the power supply. If you're sourcing 1.5 mA then a 1k series resistor will drop 1.5V which sounds reasonable.

 

[24Volts]

ah!!!! I see now!

Thanks KrisBlueNZ, I think u saved me lots of fuses

24v