I ended up using 1mv so i could read the gain directly from the voltage without disturbing the bias point too much.

In AC analysis you can use any voltage level you want. It won't affect the bias point (as you probably know).

 

In AC analysis you can use any voltage level you want. It won't affect the bias point (as you probably know).

Hi,

Oh i thought more about transient analysis but yes that's true.

 

Hello again,

Here is a way to measure the output impedance either by hand calculation or by sim

A formula for Rout when perturbing the output with a second load resistor of higher value than the original load resistor:

Rout=((Vout1-Vout2)*RL*RLT)/((Vout2-Vout1)*RLT+Vout2*RL)

where
Vout1 is the output voltage measurement with RL only.
Vout2 is the output voltage with both RL and RLT.
RL is the original load resistor.
RLT is the test resistor, typically 5 to 10 times higher than RL.
Rout is the output resistance.

The way to apply this is to take an output voltage measurement with RL only then another output voltage measurement with both RL and RLT in parallel, then compute Rout.

When i do this in the sim (with proper TIP models) i get the result:
Rout=3.28658 Ohms.

Note you can calculate Vout1 and Vout2 and apply this, or you can measure those two with a sim. I used a sim although for the input impedance i used both a sim and hand calculations in the normal way and then hand calculations using a perturbation method. They all come out within reason.

 

Hi,

I ended up using 1mv so i could read the gain directly from the voltage without disturbing the bias point too much. So if the gain was 123 i would read 0.123v output which is easy to convert.

I also found that the gain of the first stage is highly dependent on the dynamic resistance of the BE diode. 0 Ohms leads to a gain of 991 while 3 Ohms leads to just 161, which is quite different. This in turn tells me the first stage should maybe be redesigned.

Hi MrAl:

I have messed around with a re-design for stage one but I have had no luck. It seems to me that the driver stage
and the input stage have very low input impedances.

aac

 

Hi MrAl:

I have messed around with a re-design for stage one but I have had no luck. It seems to me that the driver stage
and the input stage have very low input impedances.

aac

Hi,

Yes you have to place a resistor in series with the 470uf cap.
The 470uf cap is the main reason the input impedance is so low. It also leads to a variation in gain over some different parameters which is also not usually desirable. A series resistor helps that also because it partly swamps the rBE.
Of course we lose some gain doing that too so it's a trade-off.

 

Hi,

Yes you have to place a resistor in series with the 470uf cap.
The 470uf cap is the main reason the input impedance is so low. It also leads to a variation in gain over some different parameters which is also not usually desirable. A series resistor helps that also because it partly swamps the rBE.
Of course we lose some gain doing that too so it's a trade-off.

Hi MrAl:

I have done some experimentation with a series resistor.

Cap Value vout Zin Series Res
470uf 5.66 252Ω n/a
47uf 3.45 404Ω n/a
470uf 1.22 969Ω 10Ω
470uf 0.32 2KΩ 47Ω

The input resistance is too low no matter what changes I make.
I believe that the input impedance to stage 2 (Q2) is only 840Ω.
Not sure where to go from here.

aac

 

The input impedance of Q1 and Q2 are low because the values of the base bias resistors are way too low!

 

The input impedance of Q1 and Q2 are low because the values of the base bias resistors are way too low!

Hi AG:

I have increased the value of both stage one and stage 2 bias resistors by 5 times. The
input impedance of stage one (by my calculations if they are right) is 3.8 KΩ but the overall
voltage gain really sucks at 66. I have attached the latest version of the sim in case you would
like to take a look at it. The output signal (with an input signal of 8 mV) has gone from
5.65 volts peak to 528 mV peak.

cheers

 

RE3 drops the gain of Q1 down to about 16, but adds some negative feedback that reduces distortion.

 

Hi MrAl:

I have done some experimentation with a series resistor.

Cap Value vout Zin Series Res
470uf 5.66 252Ω n/a
47uf 3.45 404Ω n/a
470uf 1.22 969Ω 10Ω
470uf 0.32 2KΩ 47Ω

The input resistance is too low no matter what changes I make.
I believe that the input impedance to stage 2 (Q2) is only 840Ω.
Not sure where to go from here.

aac

Hello again,

Oh i did not realize you were looking for exceptionally high input impedance.
For example when i add a 5 ohm resistor in series with the 470uf cap i get an increase from 260 to 639 ohms. I thought that would be enough as 600 ohms is a typical audio input impedance.
If you are looking for much higher input impedance then you probably have to add another stage such as an input buffer voltage follower, or add some positive feedback.
To add positive feedback, connect say a 10uf cap to the collector of Q2 with series resistor of say 300k and connect the other end of the 300k to the base of Q1. That might also raise the gain slightly. You should be able to raise the input impedance to maybe 5k ohms. However, you'll have to check for the effect this has on the distortion and that it does not oscillate. It is probably a good ideas to leave some small resistance in series with the 470uf cap too though, like 1 to 5 ohms.

I have to say though i dont really know what input impedance you are looking for, 1k, 5k, 10k, etc.

 

Hello again,

Oh i did not realize you were looking for exceptionally high input impedance.
For example when i add a 5 ohm resistor in series with the 470uf cap i get an increase from 260 to 639 ohms. I thought that would be enough as 600 ohms is a typical audio input impedance.
If you are looking for much higher input impedance then you probably have to add another stage such as an input buffer voltage follower, or add some positive feedback.
To add positive feedback, connect say a 10uf cap to the collector of Q2 with series resistor of say 300k and connect the other end of the 300k to the base of Q1. That might also raise the gain slightly. You should be able to raise the input impedance to maybe 5k ohms. However, you'll have to check for the effect this has on the distortion and that it does not oscillate. It is probably a good ideas to leave some small resistance in series with the 470uf cap too though, like 1 to 5 ohms.

I have to say though i dont really know what input impedance you are looking for, 1k, 5k, 10k, etc.

Hi MrAl:

I want to end up with a general purpose amplifier. I would like an input that would be suitable
for a microphone, tuner, cd player etc. I wasn't sure what input impedance would be appropriate
to handle those types of inputs. I also wasn't sure what output voltage those type of devices
would produce. So much to learn.

aac

 

Hi MrAl:

I want to end up with a general purpose amplifier. I would like an input that would be suitable
for a microphone, tuner, cd player etc. I wasn't sure what input impedance would be appropriate
to handle those types of inputs. I also wasn't sure what output voltage those type of devices
would produce. So much to learn.

aac

Hi,

I am guessing that 600 ohms would be enough for that stuff, but i guess something could come along that needs a higher input impedance. Maybe you would be better off adding an input buffer voltage follower, which is a common collector amplifier. I would say you should be able to get at least 10 times the input impedance that way.
Note that there are some constraints on how much gain you can get vs input impedance and remember the load impedance makes that more strict because the collector resistance can not be too high in order to drive the next stage properly. It might be interesting to look at the optimum values for gain and input Z, getting the max Zin and max gain.

Just to note, here are some Zin values and Vgain values for that single stage given the second stage load is still connected and it's the original circuit with the 5.6k lower bias resistor and 988.5 ohms lower resistor in the second stage.
Zin:
[343.2716167624145,423.0930291529124,499.1176708238551,571.5336074299129,
640.556546955865,0,0,0,0,941.8798034301926]
Gain:
[131.1215728830712,103.7488718342961,85.82443677081137,73.18801492530118,
63.80460770122632,0,0,0,0,38.95380682753324]

This is for that series resistor (in series with the 470uf cap) ranging in value from 1 to 10 ohms, but i skipped 6,7,8, and 9 ohms jumped from 5 to 10 ohms that is why some are zero.
So for Rs=5 ohms, we read from the above 640.6 ohms Zin and gain of 63.8 for example.
These were all calculated from nodal analysis with re=2.7 ohms as discussed before. So if you short the 470 ohm cap the resistance on the emitter when Rs=5 is 7.7 ohms in parallel with that lower 1k.

 

Hi,

I am guessing that 600 ohms would be enough for that stuff, but i guess something could come along that needs a higher input impedance. Maybe you would be better off adding an input buffer voltage follower, which is a common collector amplifier. I would say you should be able to get at least 10 times the input impedance that way.
Note that there are some constraints on how much gain you can get vs input impedance and remember the load impedance makes that more strict because the collector resistance can not be too high in order to drive the next stage properly. It might be interesting to look at the optimum values for gain and input Z, getting the max Zin and max gain.

Just to note, here are some Zin values and Vgain values for that single stage given the second stage load is still connected and it's the original circuit with the 5.6k lower bias resistor and 988.5 ohms lower resistor in the second stage.
Zin:
[343.2716167624145,423.0930291529124,499.1176708238551,571.5336074299129,
640.556546955865,0,0,0,0,941.8798034301926]
Gain:
[131.1215728830712,103.7488718342961,85.82443677081137,73.18801492530118,
63.80460770122632,0,0,0,0,38.95380682753324]

This is for that series resistor (in series with the 470uf cap) ranging in value from 1 to 10 ohms, but i skipped 6,7,8, and 9 ohms jumped from 5 to 10 ohms that is why some are zero.
So for Rs=5 ohms, we read from the above 640.6 ohms Zin and gain of 63.8 for example.
These were all calculated from nodal analysis with re=2.7 ohms as discussed before. So if you short the 470 ohm cap the resistance on the emitter when Rs=5 is 7.7 ohms in parallel with that lower 1k.

thanks,

aac

 

Well, it that case you must be using the different 2N3904JP model then I use, end of the story. But nevertheless, I already show you where did you make the error in your calculations (here re4 = RE31║RL ).

That doesn't change the value of re4.

 

That doesn't change the value of re4.

What??
The correct value is re4 = R31+ RL