Filling the gaps.....

GroupDIY Audio Forum

Help Support GroupDIY Audio Forum:

This site may earn a commission from merchant affiliate links, including eBay, Amazon, and others.

radiance

Well-known member
Joined
Jun 4, 2004
Messages
3,228
Location
the Netherlands
Filling the gaps in my technical knowledge that is.....
Today I traced a signal (a square wave) through my GSSL clone to find a short and I came across something I did not understand.
Below you'll see a part of the gssl schematic. Between point A and B are normally the VCA's but I left them out (jumpered them) because I just wanted to test the In and Output part.

4ibrolj.gif


What I did not understand was when I probe between the 22u cap and the 27K resistor I see a square wave, but when I probe after the 27K resistor the square wave is gone. Then it reappears at pin 1 of the 5532, disappears after the 10K resistor (before pin 6) and reappears at pin 7 of the 5532.
What happened to the square wave?
 
The operational amplifiers will try to keep the negative inputs at or near "virtual ground", voltage-wise, since the positive input is referenced to ground. At that point, its all about current and your scope will not see the signal as a voltage.
 
Thanks for the answer. I understand it except for the current part. I can't really visualize how there can be current without voltage :?
 
The op amp attempts to keep the voltage differential between the two inputs very small. It also keeps the current flow into the inputs small, too, since they are (supposedly) extremely high impedances. Any amount of current at the input, across a very high resistance will produce a voltage, something the amplifier doesn't want to see, there.

The resistors in series with the input and in the feedback path serve as voltage to current converters. So the net effect of the feedback path and the input path is to sum the current at the input node to zero. Therefore, in the perfect op amp, at the inputs, there should be a net current in each input of zero and a net voltage between the inputs of zero. That is why the inputs are referred to as a "virtual ground". Of course, attaching a device that reads voltage levels at this point will see a net voltage of zero, even though the op amp is rapidly manipulating the balance of the input voltage and the output voltage to produce the net zero sum of current.
 
It all sound very logical and all....
The only thing I can't comprehend is where the square wave fit's in all this. How is the information of the square wave stored in this "current" ?

I'm stuck with this idea about waves beeing made up of voltage differences. 0 amplitude = 0v, and higher amplitude beeing (for example) 1v.
 
Resorting to the hydralic model of electricity, here, you can think of the square wave as a high pressure, low pressure signal. If you look at the pipe carrying the water (electrons) at any given point along the pipe, you will see the velocity of the water increase when the pressure is high and be less, when the pressure is low. So, the change in pressure (voltage) has been converted into a change in velocity (current). The variations are the information being carried.

The op amp throws a wrench into the works in that it introduces a servo mechanism that essentially measures the current by balancing it with an equal but opposite current flow which pulls the current (water) flowing into the node of the junction of the input and two resistors away from the node via a different path before it can flow into the input of the amplifier.
 
[quote author="burdij"]...... So, the change in pressure (voltage) has been converted into a change in velocity (current). The variations are the information being carried.
[/quote]

This did it.

Now I understand.
Thanks a lot for explaining.

:thumb:
 
[quote author="radiance"]Thanks for the answer. I understand it except for the current part. I can't really visualize how there can be current without voltage :?[/quote]

This is the fundamental basis of all negative feedback.

A basic opamp has very large open loop game, often on the order of 100,000 and more to 1. So 1 mV of difference between the + and - input pins will cause the output to swing volts in the direction of the input difference.

When we connect input and (negative) feedback resistors, these wild large voltage swings are tamed by a closed feedback loop. This has the practical effect of making the - input follow the + input very closely, give or take a very small error voltage equal to the output change divided by the open loop gain, often mV or less.

If the + input is grounded as in your example the practical effect of the negative feedback resistor trying to hold the - input close to the + input is that the - input becomes a “virtual earth” and any current flowing in the input resistor toward or away from the - input will be canceled by an equal but opposite current from the feedback resistor as the output swings volts in response to very tiny error voltages at the opamp's input pins.

If you were to look at that - input with your scope gain turned all the way up you might see a very tiny version of the input/output waveform.

This basic relationship in the virtual earth topology accommodates simple gain calculation from the ratio of input to feedback resistors. Since the opamp works to hold output current equal and opposite to input leg current scaling the resistance up or down scales the voltage to same amount.

This is a very powerful concept and the basis of much modern circuit design.

JR
 

Latest posts

Back
Top