Your circuit still does not have AC and DC negative feedback from its DC output to its DC input.
Negative feedback reduces its distortion and voltage gain.
OOPS! We missed the Monday deadline but I was away all weekend..
I fixed this amplifier:

Hi AG

Your modified amp in post #32 works nicely. Does it matter in which order C1 and R1 appear?
It doesn't appear to matter.

I am not clear on how to arrive at the value of the feedback resistor.

newbie

 

Hi AG

Your modified amp in post #32 works nicely. Does it matter in which order C1 and R1 appear?
It doesn't appear to matter.

My circuit does not have C1.

I am not clear on how to arrive at the value of the feedback resistor.

The datasheet of the transistor has written spec's for minimum and maximum hFE current gain, and shows a graph with the hFE at various currents. I simulated with an average hFE of 200 but the output clipped the bottom of the waveform because the hFE was actually higher than 200 in the simulation so I finally ended with a "typical" hFE of 244. Then Ohm's Law calculates the value of the base resistor and the ratio of the feedback resistor to the transistor's base impedance at the selected current and the input source impedance determines the voltage gain.

 

My circuit does not have C1.


The datasheet of the transistor has written spec's for minimum and maximum hFE current gain, and shows a graph with the hFE at various currents. I simulated with an average hFE of 200 but the output clipped the bottom of the waveform because the hFE was actually higher than 200 in the simulation so I finally ended with a "typical" hFE of 244. Then Ohm's Law calculates the value of the base resistor and the ratio of the feedback resistor to the transistor's base impedance at the selected current and the input source impedance determines the voltage gain.

Sorry, forgot that I re-named a couple of components.

 

Referring to post #32 which has the revised schematic. If the emitter resistors are omitted,
the current flowing into the base of Q4 is 246 μA and the current flowing out of the Q4
emitter is 24.87 mA. If you put 0.33 Ω emitter resistors in the circuit, then 193 μA flows
into the base and 19.46 mA flows out of the Q4 emitter.

How would you calculate the current reduction that an emitter resistor would have on
the circuit if you didn't have a sim program to do it?

thanks

 

Referring to post #32 which has the revised schematic. If the emitter resistors are omitted,
the current flowing into the base of Q4 is 246 μA and the current flowing out of the Q4
emitter is 24.87 mA. If you put 0.33 Ω emitter resistors in the circuit, then 193 μA flows
into the base and 19.46 mA flows out of the Q4 emitter.

How would you calculate the current reduction that an emitter resistor would have on
the circuit if you didn't have a sim program to do it?

thanks

Hi,

You could do a full analysis.
For example, for a one transistor stage with one base resistor one input E1 and emitter resistor changing from 0 to R3 we get:
dIc/dR3=(B*(B+1)*(E1-E3)*R3)/(R1*(B*R3+R3+R1))

where
E1 is input voltage assumed enough to bias the transistor,
E3 is the base emitter voltage maybe 0.7v,
R1 is the lone input resistor,
R3 the added emitter resistor,
B the transistor Beta.
dIc/dR3 the change in collector current when R3 is added.
There you can see the Beta is needed.

 

Hi,

You could do a full analysis.
For example, for a one transistor stage with one base resistor one input E1 and emitter resistor changing from 0 to R3 we get:
dIc/dR3=(B*(B+1)*(E1-E3)*R3)/(R1*(B*R3+R3+R1))

where
E1 is input voltage assumed enough to bias the transistor,
E3 is the base emitter voltage maybe 0.7v,
R1 is the lone input resistor,
R3 the added emitter resistor,
B the transistor Beta.
dIc/dR3 the change in collector current when R3 is added.
There you can see the Beta is needed.

I am lost. I was looking at Q4 in the schematic. I have found all the voltages
by inspection and I am now trying to determine what ICQ4 or IEQ4 should be
as a result of the emitter resistor.

 

I am lost. I was looking at Q4 in the schematic. I have found all the voltages
by inspection and I am now trying to determine what ICQ4 or IEQ4 should be
as a result of the emitter resistor.

Hi,

That was just an example of what you might come up with. For the actual circuit you have to come up with a current without the resistor R3 and with the resistor R3 then you can subtract and find the change in current with the change in resistance (from 0 to R3).
So you get a formula.
The reason for this is because the analysis yields a value that depends on the feedback caused by that resistor insertion. It's called emitter feedback. The voltage at the emitter increases as a result of that R3, and that reduces the voltage drop of the base emitter diode, and that causes less input current and thus less collector current.