OP-AMP circuit starge problem!

[dr.power]

Hello guys,

I am wondering why the waveform in the first pic of below is like that???! I just wanted to design a non inverting amplifier using TL074 to amplify an electress mike.
I am wondering why the problem is solved by shorting the input capacitor (i.e C26)???! Changing the amplifier to inverting state soves the problem too without removing the C26 too!

Can you guys tell me what's the problem really?

Thanks a lot

 

[atferrari]

Applying power to the opamp?

 

[dr.power]

Applying power to the opamp?

Thanks for your input,

Yes, although you are not able to see it, but of course the power supply is connected to the op-amp.

 

[crutschow]

You've made a common mistake in using op amps. With the capacitor there's no path for the op amp input bias current. You must always have such a path on both inputs for an op amp to operate properly. You need a resistor across the capacitor.

 

[dr.power]

You've made a common mistake in using op amps. With the capacitor there's no path for the op amp input bias current. You must always have such a path on both inputs for an op amp to operate properly. You need a resistor across the capacitor.

Thanks Carl For all your helps,

Actually I do not know what do you mean by your above sentence, I had no such a problem when changed the op-amp to inverting, and of course I did not need to use an resistor across the input resistor in inverting op-amp, mode?!!

Why and when do we need to bias (DC bias do you mean?) the inputs of an op-amp???

I Know how to bias an op-amp to work as an amplifier in inverting/ non inverting mode:
https://www.elexp.com/t_gain.htm
But to remove any DC offset in the input (for instance the dc bias of an electret mike) I just simply add a capacitor in series with the mike and the input of an op-amp whether it is in inverting or in non inverting mode. So I am not able to understand what you mean by input bias current which seems to be something like DC biasing transistors??

 

[MrAl]

Hi dr.,

Why would you want to use such a large capacitor and no bleeder resistor anyway? If the op amp input impedance is 10 megohms that means that it will take about 150 seconds for any DC offset present in the input signal to die out. That seems too long.
Keep in mind that in the time domain there is also an exponential component that must be considered, so we can not simply reduce our analysis to frequency analysis only. The cap passes AC yes, but it also has a exponential charge and discharge time to consider.

Unless you are working with very low frequencies you should lower the time constant by at least connecting a 1 megohm resistor from pin 12 to ground. That will bring the total settling time down to about 15 seconds, but even that seems too long. Better would be a smaller cap and lower value resistor to ground.

How your circuit works in the simulator may not be the way it works in real life either when you do very rare circuit connections like this. Some circuit simulators will not have a problem with your circuit, but that still doesnt mean it's good. It really needs a lower cap and a resistor to ground.

 

[crutschow]

Actually I do not know what do you mean by your above sentence, I had no such a problem when changed the op-amp to inverting, and of course I did not need to use an resistor across the input resistor in inverting op-amp, mode?!!

Why and when do we need to bias (DC bias do you mean?) the inputs of an op-amp???

I Know how to bias an op-amp to work as an amplifier in inverting/ non inverting mode:
https://www.elexp.com/t_gain.htm
But to remove any DC offset in the input (for instance the dc bias of an electret mike) I just simply add a capacitor in series with the mike and the input of an op-amp whether it is in inverting or in non inverting mode. So I am not able to understand what you mean by input bias current which seems to be something like DC biasing transistors??

All op amps have an intrinsic DC input bias (operating) current as part of there normal operation, no matter how small (look at the data sheets). This requires a resistive path to carry this current (either to ground to to the output). When you changed the op amp to inverting you now either have a resistive path to ground or to the output for the bias current for each input, so it works properly.

So if you put a capacitor in series with an input there must also be a resistive path for the bias current.

Does that all make sense now?

 

[audioguru]

Like all opamps with Jfet inputs, the TL071, TL072 or TL074 opamp does not have any input bias current.
But like all opamps it needs a DC reference voltage which in this circuit (it has a dual-polarity supply) is 0V. A resistor from pin 12 to 0V will make it work the same as the second circuit that couples the 0V from the input source directly to pin 12.

I agree that the 3uF input coupling capacitor has a value that is much too high. If the input resistor to 0V is 100k ohms then the input coupling capacitor can be 100nF (0.1uF) to 330nf (0.33uF).

 

[KeepItSimpleStupid]

C...

I think you meant a resistor from pin #12 to ground.

For the OP:

You'll notice that there is a parameter called Ib, or Input bias current. That current MUST always have a place to go, otherwise odd things happen. In your circuit, the capacitor starts to charge. In a paractical circuit, your multimeter touching the terminal might make the circuit work properly because it adds 10 M ohms to ground.

 

[audioguru]

I think you meant a resistor from pin #12 to ground.

0V is ground. I said connect the resistor to 0V because some people don't know what voltage is ground.

 

[KeepItSimpleStupid]

Crutschow said in post #4:

You need a resistor across the capacitor.

Which doesn't even have a terminal that is connected to 0V or ground for that matter. It doesn't evan have a connection to virtual ground.

Pin #12 to 0V then NOT across the capacitor.

 

[MrAl]

Hi,

I tried to explain the need for a resistor from pin 12 to ground in post #6. Even if the IC input has no input bias current requirement it needs that resistor.
Some of you guys dont seem to read other posts

 

[unclejed613]

to minimize offset, put a resistor from the noninverting input to ground equal to the resistor from the inverting input to ground (in this case it would be 10k). this balances the input bias currents. the reason the offset goes away when shorting the cap, is that the signal source is a DC short to ground which provides a ground reference for the noninverting input. when you change this to an inverting amp, you ground the noninverting input, once again providing the ground reference to the noninverting input.

in the data sheet, the input bias current is 7nA. it's small, but it is there, and charges the input capacitor (which might not be the case in the real world, since an electrolytic cap has some leakage resistance in parallel (internally)).

in simulators, all "ideal" voltage sources have zero output resistance, and all "ideal" current sources have infinite output resistance.