Identify whether or not the mixing network is series or shunt. If the signal source
has one terminal on ac ground then:
a.) If the input active device has one of its input terminals on ac ground, then the
mixing network must be shunt.
b.) If the signal and feedback sources are applied to different input terminals of the
input active device, then the mixing network is series (this includes differential
amplifiers where two devices form the input active device).

when they say "if the signal has one terminal on AC ground".what is AC ground?

 

AC ground is the point used as the reference for all voltage measurements.

 

thanx

but is it including DC source (DC voltage and current source)? Or is there difference between DC ground and AC ground.

 

http://forum.allaboutcircuits.com/showthread.php?t=7640
a similar query.

 

Hmmm...

This is a very good question, and one that I do not think is covered in school. Unless one has a very good professor.

In cases like AC/DC ground or Analog/Digital ground I think the answer has to include an explanation of the current return path.

I have no expertise in making boards, but if you were to search USENET or other resources that discuss boad layout techniques you will find two lines of reasoning:

Digital and Analog ground planes must be separated. They do have to be connected somewhere in the layout, usually at a point at the power supply.

There is only one ground plane, but analog and digital circuitry are separated from one another. Although there is one plane, the devices are close together allowing a short return path to their analog or digital counterparts.

AC ground will supply a return path for AC current, but not DC. Think of a shunt capacitor to...well...ground. DC currents could not flow to ground, but AC currents could. Honestly, a schematic would help with a real world circuit.

 

Part of the difficulty may be the terminology. When books refer to AC ground and DC ground in an electronic circuit, they are actually refering to the paths to ground for AC and DC signals. Remember that any path has impedance regardless of how small that impedance is. If the path to ground is too narrow or too long, you could find a small AC signal at one end of the path due to the total impedance of the path to ground. The magnitude of impedance is dependent upon frequency. The high frequencies may appear on the AC path to ground because the reactance of that path is not zero at that particular frequency. It contains some reactance. If the AC path to ground comes too close to a DC circuit, that high frequency may appear in the DC circuit. Therefore, it is very important to keep the two ground paths separate and connect them at one point that is truly ground.

I once had a problem with a poorly functioning circuit until I realized that the signal ground pin on an IC component was not at zero volts. For some reason (probably desparation) I measured the potential difference between the signal ground pin on the component and the signal ground pin at the edge connector and read 0.2 volts! I'm having difficulty with the wording of this paragraph but basically, a faulty ground path was the cause of my problems.

John

 

chesart1,
I got your point. Nice observation.

I have a simple example to differentiate b/w AC and DC grounds. Take the case of a transistor as an amplifier - we have the signal component (AC), riding over the biasing voltage(DC). There is always a 'coupling' capacitor at some stage to give a 'return path' to the AC component to the ground while blocking the DC. The DC component is given the return path via the emitter to ground. (Remember, that the AC will prefer to take the path via the capacitor to the ground). Eventually, they meet at the same common ground.