operational amplifer (un expected behaviour)

[dark]

Hi board ,

I am using the following circuit. Now when the voltage at V2 starts to vary from 2.5V to 3.75V . I measure a shift at V1 as well from ( 2.5v to 2.65v) which is disrupting the output signal and is strange . Have I done something wrong ? I am using OPA2335 .

Thanks in Advance

-Dark

 

[audioguru]

The opamp has such a high amount of internal gain (three million) that both input voltages are the same in your circuit.
So when you increase V2 the non-inverting input voltage increases and the opamp increases the inverting input voltage to be the same.

When the inverting input voltage increases then V1's voltage increases too because it is not a low impedance voltage source.

It will not happen if you use another opamp as a buffer for V1.

 

[dark]

The opamp has such a high amount of internal gain (three million) that both input voltages are the same in your circuit.
So when you increase V2 the non-inverting input voltage increases and the opamp increases the inverting input voltage to be the same.

When the inverting input voltage increases then V1's voltage increases too because it is not a low impedance voltage source.

It will not happen if you use another opamp as a buffer for V1.

Hi , I have this on a PCB , though a prototype . If I reduce the setpoint voltage to 2.3V , then I get an output swing of 5V . but then the lower side gets offset with this approx 0.3v . I have a lot of margin in my sensor and measurement range . Is there any issue if I calibrate the proto this way .

Thanks for the help!

 

[audioguru]

Another problem with your high impedance for V1 is that the V1 resistor values (including the pot adjustment) affect the gain.

 

[unclejed613]

the - input node will always follow the + input node, because the op amp will do whatever it must to keep the two voltages equal. this is one of the basic rules of op amps. also be aware that many op amps will not tolerate very much forced difference between these inputs without damage. to keep V1 constant, you need a much lower source resistance so that the resistor drops the difference between V1 and the - input more effectively. either that or ground the + input, and bring V2 in through another resistor into the - input (assuming you can tolerate your output being inverted). in this way, all of V1 AND V2 will drop across their respective input resistors without interacting, because the - input becomes a "virtual" ground. so the op amp becomes an inverting mixer. i am guessing the reason for the variable resistor is to adjust for DC offset?

 

[dark]

the - input node will always follow the + input node, because the op amp will do whatever it must to keep the two voltages equal. this is one of the basic rules of op amps. also be aware that many op amps will not tolerate very much forced difference between these inputs without damage. to keep V1 constant, you need a much lower source resistance so that the resistor drops the difference between V1 and the - input more effectively. either that or ground the + input, and bring V2 in through another resistor into the - input (assuming you can tolerate your output being inverted). in this way, all of V1 AND V2 will drop across their respective input resistors without interacting, because the - input becomes a "virtual" ground. so the op amp becomes an inverting mixer. i am guessing the reason for the variable resistor is to adjust for DC offset?

Thanks for the help . Please advise solution ! . A voltage follower BJT from the 2.5V reference to opamap? . My simulator shows it to be ok anyways.

 

[AllVol]

Play-hookey.com (a very useful site) has the following to say about your circuit:

"The circuit (attached below) shows a non-inverting op amp circuit. In this circuit, the input signal is effectively used as the reference voltage at the "+" input to the differential amplifier, while the "-" input is indirectly referenced to ground. In order to keep the two input voltages to the amplifier the same, the amplifier must set Vout to whatever voltage is required to make the feedback voltage to the "-" input match the input voltage to the "+" input.

Since Rf and Rin form a voltage divider, the feedback voltage will be VoutRin/(Rf + Rin). The gain of this circuit, then, calculated as Vout/Vin, is (Rf + Rin)/Rin, or (Rf/Rin) + 1. Resistor Rz has no effect on the gain of the circuit. However, to balance out variations caused by the small input current to the amplifier, Rz should be made equal to the parallel combination of Rf and Rin."

Referencing your "-" input to ground may solve your problem.

 

[dark]

Play-hookey.com (a very useful site) has the following to say about your circuit:

"The circuit (attached below) shows a non-inverting op amp circuit. In this circuit, the input signal is effectively used as the reference voltage at the "+" input to the differential amplifier, while the "-" input is indirectly referenced to ground. In order to keep the two input voltages to the amplifier the same, the amplifier must set Vout to whatever voltage is required to make the feedback voltage to the "-" input match the input voltage to the "+" input.

Since Rf and Rin form a voltage divider, the feedback voltage will be VoutRin/(Rf + Rin). The gain of this circuit, then, calculated as Vout/Vin, is (Rf + Rin)/Rin, or (Rf/Rin) + 1. Resistor Rz has no effect on the gain of the circuit. However, to balance out variations caused by the small input current to the amplifier, Rz should be made equal to the parallel combination of Rf and Rin."

Referencing your "-" input to ground may solve your problem.

Hi,

Nice suggestion , but the problem is my input signalV2 varies from 2.5V to 3.75V as stated earlier. Unless I offset the opamp by 2.5V I wong get a swing from 0-5V over the gain of x4 .

-Dark

 

[unclejed613]

well..... he does have it ground referenced, through the 1k pot and 3.9k resistor. the problem here is that a portion of V2 will always appear at the V1 terminal because the - input remains at whatever voltage the + terminal is at, and a portion of the output current through Rf is what causes it. what i was getting at (depending on whether or not the OP can tolerate having his output inverted) was using an inverting mixer to provide the isolation he wants between the sources. the - input becomes a "virtual" ground, and all of the input voltages are dropped across their respective input resistors

 

[dark]

well..... he does have it ground referenced, through the 1k pot and 3.9k resistor. the problem here is that a portion of V2 will always appear at the V1 terminal because the - input remains at whatever voltage the + terminal is at, and a portion of the output current through Rf is what causes it. what i was getting at (depending on whether or not the OP can tolerate having his output inverted) was using an inverting mixer to provide the isolation he wants between the sources. the - input becomes a "virtual" ground, and all of the input voltages are dropped across their respective input resistors

Hi ,
Cant invert the signal , as this goes to the ADC . How a bout using a voltage follower BJT buffer at the reference pin to ease the impedance .

 

[unclejed613]

or if he wants to isolate the inputs and keep the diff amp configuration he had at first, he can buffer the inputs so that they don't interact.

 

[audioguru]

In the differential amplifier circuit that has two added opamps, U1 is needed to provide a very low source impedance but U2 is not needed.

A BJT should not be used because it does not have a low enough impedance and its Vbe is affected by temperature.

 

[dark]

In the differential amplifier circuit that has two added opamps, U1 is needed to provide a very low source impedance but U2 is not needed.

A BJT should not be used because it does not have a low enough impedance and its Vbe is affected by temperature.

Thanks , any other kick start for prototype other than BJT / FET? .

I will add an OPAMP in the final circuit ,but cant do that at the moment .
will this opamp requires a simple unity gain arrangment or an additional resistance 100K something at its non inverting input.Also does this requires an output isolation resistance ?

 

[audioguru]

I will add an OPAMP in the final circuit ,but cant do that at the moment .
will this opamp requires a simple unity gain arrangment or an additional resistance 100K something at its non inverting input?

A unity gain non-inverting opamp circuit. If its input bias current is high then it can be matched to the other opamp with a series input resistor.

Also does this requires an output isolation resistance ?

Some opamps oscillate if they drive the capacitance of a long shielded cable. They need an isolation resistor at their output.
If you add a resistor then it will spoil the very low output impedance that is needed.

 

[unclejed613]

also, make sure the op amp used for the buffer is stable at unity gain. some (like the NE5532) are not and will oscillate. a TL072 is unity gain stable, and has a low enough input bias current that you won't need an input resistor to balance the bias current.

 

[Roff]

OPA2335 is a dual op amp. Are you using the other section?

 

[dark]

OPA2335 is a dual op amp. Are you using the other section?

Yes the other section is active LPF . Please have a look at the following Texas Instruments Design Note page 24 wont this suffer from the same problem ? . The op amps non-inverting input is being referenced with a pot or something is missing.

Thanks

 

[audioguru]

The non-inverting input of TI's opamp has an extremely high impedance. It is the reference voltage that has nothing to do with gain.

The reference input of your opamp circuit is the fairly low impedance of the inverting opamp circuit where the source impedance adds to the input resistance which affects its gain.

 

[Roff]

Message deleted.

 

[mneary]

All good but complex solutions.

What you're experiencing is gain error; you just failed to account for your 2.5V source's Thevenin impedance of 3.75k. If you don't use the 2.5V elsewhere, you just have to account for it.

Just reduce Rin on the (-) side of the op amp by 3.75k and the output will be as you expect. You'll have to calibrate it with the sensor output at 2.5V.

The other solution is to increase Rin and Rf. Is Rin currently about 4.0k? With sub-nanoampere bias and offset currents you should be able to increase Rin and Rf by a factor of twenty or more and reduce their influence on the 2.5V. You can still reduce the Rin on the (-) side by 3.75kΩ which should now be a small percentage.