why can't you measure the voltage on op amps inputs? I always get zero volts , why is that, what can I do to get the voltage on the input pins of Op amps?

 

Hard to diagnose without a circuit schematic.

 

It's like this on a lot of circuits , the inputs to op amps is zero volts even tho the output pin of the output of the op amps output has a AC or DC voltage

 

Could be you using the wrong range to measure the signal.
Otherwise, if the voltage on the input is too small, when you using the probe to touch then it may lost the signal, you can adding a op amp voltage follower between the signal and probe.

 

How are you measuring this? With a scope or a meter?

 

Either one, it's still zero voltage on the input pin but the output pin have a voltage or waveform

This is very common for Op amps

 

Where is the circuit?

 

If this is a classic inverting amplifier circuit, the inverting input is a 'virtual ground' the amp always tries to drive the pin back to zero, you'll never measure anything other than a tiny signal voltage there.

 

the inverting input is a 'virtual ground' the amp always tries to drive the pin back to zero, you'll never measure anything other than a tiny signal voltage there.

So what can I do to measure the voltage of a virtual ground?

Can I put a 1 ohm resistor on the virtual ground?

 

the inverting input is a 'virtual ground' the amp always tries to drive the pin back to zero,

Why is it driving the input pin back to zero voltage?

 

Why is it driving the input pin back to zero voltage?

Because that's what op-amps do, they always want the voltage between inverting and non-inverting inputs to be zero.

If you look at one of the technical descriptions of how the inverting amp works you'll see what's going on. Don't make the mistake of trying to consider the op-amp itself in isolation, the surrounding components are vital to operation and to your understanding of what's going on.

Imagine for a moment what happens if the -ve input goes slightly positive, the output immediately hammers to the -ve rail (open loop gain of several million) and the feedback resistor then pulls the input back to zero.

 

Because that's what op-amps do, they always want the voltage between inverting and non-inverting inputs to be zero.

Why is that?

How can you measure the input pin voltage?

If it measures zero volts , how can you know the input pin's voltage?

 

Why is that?

How can you measure the input pin voltage?

If it measures zero volts , how can you know the input pin's voltage?

We can't predict what you're doing. Post the circuit.

 

Indeed blah2222 we really need to see what Dexter is trying to achieve. If he's just wanting to measure the input voltage in a working op-amp circuit he will never achieve it because it's always zero.

Some light reading on op-amps http://en.wikipedia.org/wiki/Operational_amplifier

 

Missed the edit time, but I suspect this has something to do with this thread http://forum.allaboutcircuits.com/showthread.php?t=95738

 

Why is it driving the input pin back to zero voltage?

This is the classical effect of negative feedback.
(Remember the principle of control system loops: the feedback signal is used to sense and reduce the difference to the input signal by adjusting properly the output.)
But note that the differential voltage is NOT driven back to zero volts.
The voltage always is Vdiff=Vout/Aol.
The open-loop gain Aol is very large (range of 1E5) so that there are only µVolts between the input terminals. Hence, for inverting operation (pos. terminal grounded) the inv. terminal is not at ground potential but at "virtual" ground (which means: "near" ground).

 

If he's just wanting to measure the input voltage in a working op-amp circuit he will never achieve it because it's always zero.

Why is it always zero? because the op amps differential inputs want to default to zero volts? but how does it default back to zero volts?

How does a troubleshooting tech, get to measure the voltage that's going input the Op amp? without it measuring zero volts?

the feedback signal is used to sense and reduce the difference to the input signal by adjusting properly the output.)

The Feedback signal is a Differential voltage being feedback to the input pin of the op amp? this differential voltage is compared to the other input pin of the op amp which is Cancels out the voltages? which it is near ground in millivolts or microvolts ? ( virtual ground )

"virtual" ground (which means: "near" ground)

Virtual ground means the voltage is in millivolts or microvolts? or nanovolts?

 

Why is it always zero? because the op amps differential inputs want to default to zero volts? but how does it default back to zero volts?

No, there is no "default" value. As mentioned already, it is simply the feedback effect in case of a high-gain forward amplification.

The Feedback signal is a Differential voltage being feedback to the input pin of the op amp? this differential voltage is compared to the other input pin of the op amp which is Cancels out the voltages? which it is near ground in millivolts or microvolts ? ( virtual ground )

No, the feedback voltage is the voltage at the inverting terminal only.

Virtual ground means the voltage is in millivolts or microvolts? or nanovolts?

Assuming a dc output voltage of (1...5) volts and Aol=1E5 the input diff. dc voltage is
(1...5)/1E5 volts = (10...50)µvolts.

 

the feedback voltage is the voltage at the inverting terminal only.

The output pin of the output is the feedback voltage? if the output pin voltage is 5 volts, the feedback voltage is 5 volts? but the input pin on the op amp is zero volts or in millivolts, that doesn't make sense

Assuming a dc output voltage of (1...5) volts and Aol=1E5 the input diff. dc voltage is
(1...5)/1E5 volts = (10...50)µvolts.

So the GAIN and the output voltage = the input pin voltage?

 

The output pin of the output is the feedback voltage?

Who says that? There is only one "inverting" pin - and that is the inverting INPUT of the opamp.

So the GAIN and the output voltage = the input pin voltage?

The above sentence is neither an equation nor does it make sense.
I have given the formula in my former post.