Can the Virtual ground cancel out the AC signal and just leave you with a DC voltage? or DC offset voltage?
zero volts on op amp input
- Thread starter DexterMccoy
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NO. A virtual ground has no AC signal component.
The way I finally understood why the pin voltage of an inverting Op Amp circuit (none inverting input grounded) was zero was to think of the Op Amp as a variable power supply feeding the top end of a 2 resistor voltage divider with the bottom end grounded. What voltage does it take at the top to give zero volts at the center node? The Op Amp will amplify and invert any input voltage and do whatever it can to make the top voltage go to the required value to force the input pin to zero.
I don`t think this is a correct statement.
As I have mentioed earlier, the tiny voltage at the inverting opamp input node (non-inv. input grounded) - as long as the whole circuit is in its linear mode - is exactly Vout/Aol (Aol: open-loop gain).
During synthesis and analysis of this inverter circuit we neglect in most cases this very small voltage (µV range).
That means: We treat this node as if it would have ground potential and this leads to the term "virtual ground".
However, in fact it is NOT ground potential but it contains all the components (AC and/or DC) that comprise the output voltage (divided by Aol). It`s a simple matter: Input=Output/gain.
Not considering the minute voltage errors described by LvW, the fact that you measure zero volts on the input pin does not mean you can't measure the voltage on the input pin. It means you are measuring zero volts on the input pin because it actually HAS zero volts on it.
How can it amplify zero volts on the input pin?
The input pin has millivolts? , microvolts? or nanovolts?
Zero volts = Zero current
That doesn't make sense to me
So, what you're saying is that every 120 volt AC appliance in the world uses no current because one side of their power cord is connected to ground. Zero volts = zero current, therefore, all machines with one wire grounded run for free.
so the input pin of an op amp = zero volts = a Reference? so it's a zero reference?
If you really want to see some voltage on the negative input of an inverting Op Amp circuit, tie the non-invetrting input to the wiper of a potentiometer. The ends of the potentiometer will then be tied to a positive voltage on one end and a negative voltage on the other end. The output voltage will change as the potentiometer is turned. As long as the output does not go into saturation, the voltage on the inverting input will match the voltage on the wiper of the potentiometer. The amplifier simply amplifies, with extremely high internal gain, any difference between the two inputs. The functional gain of the stage is regulated by the ratio of output/input resistance.
That's the definition of ground, a reference point, usually chosen by the designer. We could just as easily attach the positive terminal of a 9 volt battery to this alleged ground, see that the other end of the battery is 9 volts more negative than what we were calling ground, and name THAT point to be "ground". Now the op-amp will have +9 volts on both of its inputs, and it won't amplify the 9 volts, either. It will only amplify the current arriving through the input resistor.
How can I see some voltage on the negative input when using my DVM meter to measure the voltage?
Would I need a voltage follower circuit or transformer between the DVM meter probe and the Op amps input?
Connect the black lead to whatever we're calling ground today and touch the red lead to the pin you want to measure.
Sorry to say but this is DEFINITELY wrong.
Any amplifier - and this applies also to the high-gain device with Aol=1E5 - needs an input signal for delivering an output signal.
More than that, since the opamp is a voltage amplifier with a high-resistive input it needs a VOLTAGE to amplify and not a current, in contrast to other devices with a low-resistive input (Norton amplifier, CFA).
Remember:The well-known formula taking the finite opamp gain into consideration is derived based on the mentioned formula Vout=Vdiff*Aol (with Vdiff: diff. input voltage, other than zero).
EDIT: This is even true for the opamp used as a so called "current-to-voltage converter". The signal (input) current in conjunction with the opamps output voltage cause a small input voltage at the inverting terminal which fulfills the mentioned formula Vout=Vdiff*Aol.
>>>This is the only task the opamp can perform: To amplify a differential voltage between it´s input terminals. <<<
(The small input current required by a BJT input stage for proper operation is to be considered as a parasitic and unwanted effect - and has nothing to do with amplification properties).
It is a complete other question if and how this tiny input diff. voltage can be MEASURED. However, any simulation can show that between the input pins there is a voltage Vdiff=Vout/Aol.
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I have given a numerical example in my post#18 (at the end).
ramancini8
- Joined Jul 18, 2012
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If you assume (think?) that inverting voltage is ground, connect it to ground and observe the change in output. Use an instrumentation amp with a gain of at least 1000 to measure the virtual ground voltage. Another cute trick is to apply a low frequency sine wave voltage large enough to get a readable output to the input. Increase the sine wave frequency while keeping the input voltage constant, and you will notice that the virtual ground voltage becomes large as frequency increases. This is because the op amp gain decreases as frequency increases.
Right.
As another justification: Remember that the classical expression for the closed-loop inverting Gain=-Rf/Ro applies to large open-loop gains Aol only.
That means: It does not apply anymore for frequencies with - let´s say - Aol=10.
Why not? Because in this case the input diff. voltage cannot be neglected anymore during calculation of the closed-loop gain. That means: Usage of the virtual ground principle causes an unacceptable error.
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