 ### Network # Op amp configuration with 1 resistor ?

#### 24Volts

Mar 21, 2010
164
Hello,

I understand how to evaluate several op amp circuits now namely, non-inv, Inv, diff etc... , but today I have come across an untypical op amp configuration with only the rf resistor? Please view attachment.

At first I thought it looked simple but I was wrong. My question is, how is one supposed to calculate rf when all we know is that Vrf is 11 Vdc and that there is a gain of 10 by using the following gain formula:

G = vo/vi = 10/1

24v

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#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
There is something completely wrong with that.

It shows the inverting input at 0V, the non-inverting input at 1V.

In theory the output should be at negative infinity volts (or the supply rail -- whichever comes first).

There are only 2 ways I can reconcile this.

Firstly, the input is not actually 1V, it is 1V in series with a resistance of approx 9.1 ohms, or if the negative supply rail is around -10 volts.

#### 24Volts

Mar 21, 2010
164
It shows the inverting input at 0V, the non-inverting input at 1V.

Are we looking at the same diagram?
Confused!

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#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
Sorry, I failed to note that the supply rails are shown to be +/- 15V.

In other respects I think I have made no error. (apart from naming the inverting input the non-inverting input Doh!) Actually, I remember first thinking the output was +ve, so I must have read the diacram wrong the first time, changed it when I re-checked it but failed to swap over the references to the input.

One input is at 1V, the other at ground potential.

Even the calculations shown for the resistor suggest that.

If the op-amp had a gain of 1000 (and its gain will be much higher than this) the inputs would differ by 10 milivolts for an output swing of 10V.

I'm looking at this image: edit: it kinda works out if the op-amp has a gain of 10. But then the resistor is simply not required.

If the inputs to the op-amp are not equal in potential, then the op-amp is not acting like an op-amp.

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#### 24Volts

Mar 21, 2010
164
okay (*Steve*), I guess I am lost here... so I will have to build the
circuit and see whats happening.....

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
It won't work.

You need to understand op-amps.

When an op-amp is operating normally (i.e. the output is not bashed hard against an output rail) the voltage difference between the input terminals is typically immeasurably small (in fact we call it zero).

This is one reason why there are always 2 resistors. The flow of current from the output to either ground or the signal source causes the inputs to have the same voltage.

In your drawing the difference between the input terminals is 1V. This will force the output against a supply rail and the resistor you have shown does nothing.

Where did this circuit come from?

#### gorgon

Jun 6, 2011
603
The LM324 can't supply your 110mA load in the first place. You also need an input resistor to make the resistor divider for the calculated amplification. In short you'll need 2 resistors to make an amplification.

The Circuit you have drawn will only burn out the 324.

#### 24Volts

Mar 21, 2010
164
When an op-amp is operating normally (i.e. the output is not bashed hard against an output rail) the voltage difference between the input terminals is typically immeasurably small (in fact we call it zero).

In the circuit the output is not bashed against the output rail.... my negative
rail is -15Vdc and the circuit claims that the output is -10Vdc. I haven't tried this
circuit yet as I am away from my test bench until Tuesday

But, I know for sure if I remove rf, the the output will be bashed
against the negative rail as expected.

So rf is doing something, I just don't know why and what's causing
the output to presumably be specifically -10Vdc.

This circuit came from a tutorial from some remote forum. I saw
this circuit and was curious on how we would calculate rf. Like I say,
I didn't build this circuit yet so I am assuming the circuit is right. so if
this circuit is correct, then rf has to be doing something since it brings
the output exclusively to -10Vdc.

If you insist that Vout of this circuit can't be -10Vdc, then I will
build it myself and test it when i get back.

I sincerely thank you for your help
24v

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#### 24Volts

Mar 21, 2010
164
The LM324 can't supply your 110mA load in the first place. You also need an input resistor to make the resistor divider for the calculated amplification. In short you'll need 2 resistors to make an amplification.

The Circuit you have drawn will only burn out the 324.

Really huh! Okay so I won't test it. thanks

#### KrisBlueNZ

##### Sadly passed away in 2015
Nov 28, 2011
8,393
That schematic is quite wrong in several ways. Obvious errors are (1) the inverting input should always be at the same voltage as the non-inverting input (this is normal for an op-amp circuit with negative feedback working within normal limits), not 1V higher, and (2) the current through the feedback resistor cannot be 110 mA because the LM324 cannot supply that much current from its output.

The design, with just one feedback resistor, is valid - it's an inverting current-to-voltage converter. Current from the input is countered by an equal and opposite current through the feedback resistor, so the inverting input has the same voltage as the non-inverting input.

The current fed into the input is converted into an opposite (negative) voltage at the output according to Ohm's Law as applied to the feedback resistor.

Where did this diagram come from? I would like to know, so I can p*ss in their milk bottles. Or at least send them a rude email.

#### 24Volts

Mar 21, 2010
164
KrisBlueNZ,

As misleading or awkward that circuit is I posted it in the hopes of learning something.
But it's nice to know that this is a non common topology in op amps and I thank you for your much appreciated assessment of the circuit.

Actually it's a print out I had from a while back which was
cut out and pasted onto paint for printing.
can't really recall ....

The design, with just one feedback resistor, is valid - it's an inverting current-to-voltage converter. Current from the input is countered by an equal and opposite current through the feedback resistor, so the inverting input has the same voltage as the non-inverting input.

So with an rf of 100 ohms is it fair to say that the current from V1 is cancelled
with the current from rf and therefore the current is equal to 0??

The current fed into the input is converted into an opposite (negative) voltage at the output according to Ohm's Law as applied to the feedback resistor.

So then mathematically (even though we know we would burn the op amp)
what is the current: isn't it:

I = Vdc/rf = 11/100 = 0.11 A??

thanks

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#### gorgon

Jun 6, 2011
603
The main problem with the drawing is the +1V input. With that you can't have -10V on the output, since the opamp is not stable (both inputs at the same level.) As drawn the output should be at -15V, or at minimum possible Vout.

Saying that this is a current amp is with a good portion goodwill, since the input is said to be +1V, not some mA value. Whatever is is, this opamp is working on voltage levels, not current. I still claim the the input resistor is missing.

Mar 21, 2010
164
okay I see
24v

#### KrisBlueNZ

##### Sadly passed away in 2015
Nov 28, 2011
8,393
So with an rf of 100 ohms is it fair to say that the current from V1 is cancelled with the current from rf and therefore the current is equal to 0??
No, the VOLTAGE at the inverting input is equal to zero, because the non-inverting input is tied to 0V. This is true for an op-amp connected with negative feedback, as long as the op-amp is operating within its specifications; that is, unless the op-amp's output is hard against its positive or negative voltage limit (which is determined by its power supply voltages) or too much current is being drawn from its output.

To see how the circuit works as a current-to-voltage converter, you need to apply current to the input. For example, you can apply a 1 mA positive current, using a +5V voltage source with a 5k resistor in series with it. Since the circuit's input (the op-amp's inverting input) is always 0V, there will be 5V dropped across the resistor and 1 mA will flow through it.

Any op-amp connected with negative feedback will use its output to try to keep its inverting input at the same voltage as its non-inverting input, i.e. 0V in this case. Its output will swing negative until the current through the feedback resistor is equal and opposite to the input current. If the feedback resistor is 100 ohms, this will happen when the output voltage is -0.1V because of Ohm's Law: V = I R, where I = 0.001A and R = 100 ohms.

Therefore an input current of +1 mA has been converted to an output voltage of -0.1V. It is an inverting current-to-voltage converter.

In this example I have added an input resistor, but it's not necessary to have one. The current can come from any component or subcircuit, such as a phototransistor or photodiode; no input resistor would be present. In these cases, the feedback resistor effectively becomes the current source's load resistor, across which the proportional voltage is developed.

So then mathematically (even though we know we would burn the op amp)
what is the current: isn't it: I = Vdc/rf = 11/100 = 0.11 A??
No, there is no 11 in the calculation. The 11 shown on the schematic comes from the (incorrect) idea that the inverting input is at +1V while the output is at -10V, giving a total of 11V across the feedback resistor.

#### 24Volts

Mar 21, 2010
164
wow okay, KrisBlueNZ,

when I hey back I will try these experiments !

#### gorgon

Jun 6, 2011
603
You should be aware that using the 100ohm feedback resistor and a 1mA input, will give you an output voltage of -0.1V.
If you increase the feedback resistor to 10kohm you'll get -10V from a +1mA input.

#### 24Volts

Mar 21, 2010
164
Hello guys,

This morning I ran some tests and all tests were inconclusive.

For example:

If you increase the feedback resistor to 10kohm you'll get -10V from a +1mA input.

Current measured at input = 1.5ma (I have 1.5 ma as an input current. How would I get 1ma input ?)
Non-Inv input: ground
Inv input = 1Vdc
Rf = 10K
Output = approx: -14.2Vdc and not -10Vdc ??? I find it weird that I can't use ohms law to make the right calculations for Vout.

Sorry for being slow at this.
thanks

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#### gorgon

Jun 6, 2011
603
You can't drive more than the -14.2 V from the output. To get it inside the drive range you need to reduce the feedback resistor, to, lets say 4k7 or 5k if you got that. A 1.5mA current will then give around -7.5ish V out.

How you get 1mA depends on how you generate the current. If you use a current generator you need to adjust this to a lower current. If you use a fixed voltage and a serial resistor, you can increase the resistance to reduce the current input.

You are still using the schematics that say +1V on the input, if you fix the input to a low impedance voltage source like that, the opamp will never work properly. The schematics is not complete since we can't see what you put into the circuit.

Ohm's Law is is present and working. You say the circuit put 1.5mA into the input, this is correct if you subtract +1 - -14.2 = 15.2V over 10k = 1.52mA.

You should realise now that the your circuit will not work if you either use a proper current generator at the input, OR add a Serial resistor to the input.

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#### 24Volts

Mar 21, 2010
164
How you get 1mA depends on how you generate the current. If you use a current generator you need to adjust this to a lower current. If you use a fixed voltage and a serial resistor, you can increase the resistance to reduce the current input.

I don't know what a current source means. The 1 volt is simply coming from a power supply adjusted to 1Vdc. I don't even know how to build a current source This is why I am having so much trouble getting any readings to make sense! If you use a fixed voltage and a serial resistor, you can increase the resistance to reduce the current input.

What do you mean by a serial resistor... is it a resistor from my power supply to the inverting input of my circuit?
How do I build a 1.5ma current source? To build a 1.5 ma current source, must I use an LM317?
Confused!

The schematics is not complete since we can't see what you put into the circuit.
There is nothing else to the schematic, I am just fiddling around with this circuit! Last edited:

#### 24Volts

Mar 21, 2010
164
To see how the circuit works as a current-to-voltage converter, you need to apply current to the input. For example, you can apply a 1 mA positive current, using a +5V voltage source with a 5k resistor in series with it. Since the circuit's input (the op-amp's inverting input) is always 0V, there will be 5V dropped across the resistor and 1 mA will flow through it.

Any op-amp connected with negative feedback will use its output to try to keep its inverting input at the same voltage as its non-inverting input, i.e. 0V in this case. Its output will swing negative until the current through the feedback resistor is equal and opposite to the input current. If the feedback resistor is 100 ohms, this will happen when the output voltage is -0.1V because of Ohm's Law: V = I R, where I = 0.001A and R = 100 ohms.

Humm! I think I understand now! We have to provide a 1 ma current!!! But now, we are back to the typical inverter op amp configuration !!! And so I guess its that or an LM317 so to get a constant current source of 1ma....

Thanks guys, sorry for being slow ....   24v

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