An 8vpp signal swings 4v above zero and 4v below zero, so for the pot, if it is connected to the output with no other resistors and no load, the output should be adjustable from 4v peak down to 0v peak which is 8vpp down to 0vpp. If you dont get that then either something is not connected right, the pot is not working properly, there is a load resistance that is too small connected to the arm, or there is an output offset. If you have an output offset you should see it go down to zero when you adjust the pot down to a low value, but when you adjust the pot upward you will see a wave that is not symmetrical about zero so it may go higher than lower (for example +5v down to -3v instead of +4v down to -4v). You can check for this, and also maybe draw a little circuit showing how you have the pot connected exactly. Also, check the pot with a meter to make sure it works and it is about the right value. A pot that is too high in value or too low in value would cause problems too depending on how you intend to load the arm.

this part i understand and i fixed the problem.

First, the 180k was calculated based on a 20k feedback value to begin with, but with 22k to begin with we end up with 99k because 99k in parallel with 22k is 18k which is 2k less than 20k, and with 22k we have 2k more than 20k, so with an ideal diode we have a gain range that is roughly equally low as it is high meaning the circuit should be able to 'find' the right gain dynamically. We can also figure in the diode drop to make this estimate more accurate which will tend to change the value of the series resistance (for example 77k).

but this is still not clear for me. i understand that 20k is the ideal feedback resistance. 1/180+1/20 = 18k this is 2k difference with 20k
but we have a 22k resistor so it is 1/18-1/22=1/99 is 99k
This part i understand.

But what does the 99k say? and how the gain range is working. i understand that an ACG circuit increases the gain when the signal is too low and decreases when the signal is too high. but how is that working in the system.

 

this part i understand and i fixed the problem.



but this is still not clear for me. i understand that 20k is the ideal feedback resistance. 1/180+1/20 = 18k this is 2k difference with 20k
but we have a 22k resistor so it is 1/18-1/22=1/99 is 99k
This part i understand.

But what does the 99k say? and how the gain range is working. i understand that an ACG circuit increases the gain when the signal is too low and decreases when the signal is too high. but how is that working in the system.

Hi again,

20k is the ideal value for the feedback resistor, or at least near that value. We start out with 22k though because we want to make sure that the AGC action can force enough gain if needed. If we use 20k and not 22k, then the 20k considering tolerance might actually be 19.9k and then the circuit wont work. By using 22k, we ensure that we are past the tolerance issue because even even with a tolerance of -5 percent we still have a resistance of at least 20.9k, and so that would provide enough gain.

Now the ideal parallel resistance for 22k, if it really was 22k, would be the resistance that takes it down to 20k. This resistance happens to be 220k. But this goes in series with the diode which will also exhibit some dynamic resistance of it's own, so we have to lower that 220k. The question then becomes, what do we lower it to, not knowing the precise value of the dynamic resistance of the diode. We also want a value of 18k which is 2k below 20k, so that we can ensure that the AGC can cut back the gain when needed.

We dont know the precise value of the dynamic resistance of the diode, but we do know the min and max value of the diode when biased with some DC current. The min value is very low, on the order of Ohms, while the max value is very high, on the order of hundreds of megohms. For an estimate then, we can say that a value in the low single-digit Ohms as compared to 20k is zero Ohms, and that a value of 500 megohms as compared to 20k is infinite. This gives us the two end points: on the low end it is zero (0 Ohms) and on the high end it is infinite (an open circuit).

Now back to the 22k. We know that 22k is 2k above the ideal 20k, and we want to keep that value so that we are sure of getting a high enough gain if needed, and with a second resistor in series with a diode which is in parallel to that 22k (the AGC circuit) if the diode has infinite resistance we indeed get that value of 22k, so that condition is still satisfied, and the value of the resistor in series with the diode can be anything because with an open circuit replacing the diode we get infinite resistance, so we still have 22k. So the first mode of operation is satisfied no matter what the value of the series resistor is.

Next we have the 0 Ohm condition. With 0 ohms in place of the diode, we want to ensure that we can get down to 18k total resistance so that we can be sure that we can make up for the tolerance of the 22k resistor going toward the high tolerance end this time. So the question is, what value do we need in parallel with 22k to get 18k. The answer is 99k.

Now to get this more precise, we might consider the voltage drop of the diode which is around 0.65v. If swap out the diode with a 0.65 voltage source, calculate the voltage across the network, and then we can then calculate the required value of series resistance that would again take us down to a total of 18k across the network. This should end up being smaller because the diode resistance should be contributing to the total series resistance to some degree.

If we want to get better than that, we just replace the diode with a toned down spice model and calculate the network voltage and current, then the resistance at the operating point. That will lead to a more exact value however it will still be a little theoretical because we will have used a spice model anyway.
If you want to see how this is done we could probably do that here next. The solution to these kinds of problems usually end up being numerical only, but that isnt too much of a problem here i guess because we are using fixed values for the other resistors.

LATER:
Doing the calculation with the spice diode i get about 83.5k using a 1N4148 diode. The voltage of the diode at that point is about 0.42 volts with a current of about 27ua which should be checked with a test in real life.

LATER:
After reviewing the circuit again i think the simpler view is just that the diode causes clipping which limits the output voltage.

 

Wow i understand it but now i changed the 180k resistor with 99k but now it is much more unstable.
the combination of 180k and 22k in parallel gives 19.6k--> gives a gain of 2.96 so that is a bit lower than 3.
the combination of 22k and 99k gives 18k. --> gives a gain of 2.80

the strange thing is, is that the 180k resistor gives less THD
Fourier analysis gives:
THD 180k: 2.35%
THD 99k: 5.32%
So there is a big difference.

but maybe this is the simulation and in real, the 99k is better??

 

Wow i understand it but now i changed the 180k resistor with 99k but now it is much more unstable.
the combination of 180k and 22k in parallel gives 19.6k--> gives a gain of 2.96 so that is a bit lower than 3.
the combination of 22k and 99k gives 18k. --> gives a gain of 2.80

the strange thing is, is that the 180k resistor gives less THD
Fourier analysis gives:
THD 180k: 2.35%
THD 99k: 5.32%
So there is a big difference.

but maybe this is the simulation and in real, the 99k is better??

Hi,

Well actually we have not yet gotten into any theories that might be relevant to THD we were so far just talking about how to ensure oscillation more or less.

The biggest factor in THD would be the amount of 'pull' the diodes have to put in, that is, the amount of clipping. If they have to clip a lot then the THD would be bad, but with little clipping it should be good.
Less clipping would come about by having the linear elements as close to the right values as possible so that there needs to be little help from the diodes so the larger the value the better, as long as it is not too large because then it wont be able to adjust the gain enough.

In the real life situation however we have to ensure both of the above issues are considered, so we may not be able to get away with a larger value. It may work for some circuits and not others unless we start using 1 percnt resistors where we can ensure that we are already close to the ideal gain even without the diodes help.