You haven't satisfied all the requirements of your project with your circuit since both your compound transistor (Q1_Q4) and Q2 use voltage divider biasing and you have to have two types.

What happened to Q3?

Here are some suggestions based on what we have been saying. There are both single and split supply versions. Try these in MSim. Make sure you gets the capacitors the right way round, although I don't think MSim distinguishes.

If you like them we can move on to the limiter.

 

You haven't satisfied all the requirements of your project with your circuit since both your compound transistor (Q1_Q4) and Q2 use voltage divider biasing and you have to have two types.

What happened to Q3?

Here are some suggestions based on what we have been saying. There are both single and split supply versions. Try these in MSim. Make sure you gets the capacitors the right way round, although I don't think MSim distinguishes.

If you like them we can move on to the limiter.

I was just showing what has been figured out so far.

I don't think I've seen that particular set-up before. I can try to see if I can figure it out though.

 

The output stage is just the sort of thing you will see in amps from 0.1 watts t0 100 watts the world over, It is bog standard.

In the days before integrated circuits (op amps) the input pair configuration was pretty common in audio preamps, tone controls filters etc etc. R2 provides AC feedback and can be omitted. R4 provides the emitter feedback bias. This version has a gain of exactly 100 and an input impedance around 5k - nearly 10 times the output impedance of the signal generator.

 

The output stage is just the sort of thing you will see in amps from 0.1 watts t0 100 watts the world over, It is bog standard.

In the days before integrated circuits (op amps) the input pair configuration was pretty common in audio preamps, tone controls filters etc etc. R2 provides AC feedback and can be omitted. R4 provides the emitter feedback bias. This version has a gain of exactly 100 and an input impedance around 5k - nearly 10 times the output impedance of the signal generator.

Okay, I feel a bit lost again (sorry)

Is the voltage divider that I created this morning not going to be used now?

Also, the circuits you posted are way more complicated than anything we did it class.

The emitter feedback that we did in class looks different as well. I have attached the the layout of what we used as emitter feedback.

 

The output stage is just the sort of thing you will see in amps from 0.1 watts t0 100 watts the world over, It is bog standard.

In the days before integrated circuits (op amps) the input pair configuration was pretty common in audio preamps, tone controls filters etc etc. R2 provides AC feedback and can be omitted. R4 provides the emitter feedback bias. This version has a gain of exactly 100 and an input impedance around 5k - nearly 10 times the output impedance of the signal generator.

Okay, I've gotta go to a wedding for a bit. Thanks for your continued help.

 

You selected a voltage divider that biased the base at a voltage too high so the transistor is saturated (it is turned on hard and does not amplify).
I selected a voltage divider that biased the base properly so the collector voltage is near half the supply voltage so it can swing up and down.

 

The output stage is just the sort of thing you will see in amps from 0.1 watts to 100 watts the world over, It is bog standard.

No.
It has severe crossover distortion because the transistors are not conducting (class-B) until the input signal exceeds 1.2V p-p. The output transistors must be biased so they conduct a little at all times (class-AB).

If the emitter resistor values are 10 ohms then you are throwing away 2/3rds of the power. Try 0.22 ohms.

 

Thank you audioguru, just shows what happens when you are rushing.

I missed out two diodes. Sorry.

Here is the corrected sketch.

I don't think efficiency is an issue here. we have 30 volts available to develop 100 milliwatts in 8 ohms.

 

100mW is like earphones playing loudly.
With a 30V supply, a TDA2030A or LM1875 amplifier IC delivers 10W to an 8 ohm speaker at low distortion.

 

With a 30V supply, a TDA2030A or LM1875 amplifier IC delivers 10W to an 8 ohm speaker at low distortion.

But would fail this assignment.

Please read posts 22 and 25.

 

The arrangement in Audiogurus's post 46 has a voltage gain of about 10, for outputs from the collector (ie as a CE amp), but cannot supply the necessary 80 odd milliamps to the loudspeaker.

Do you feel my offering is too complicated? we can work further on yours if you like. I did say that before.

 

The arrangement in Audiogurus's post 46 has a voltage gain of about 10, for outputs from the collector (ie as a CE amp), but cannot supply the necessary 80 odd milliamps to the loudspeaker.

Do you feel my offering is too complicated? we can work further on yours if you like. I did say that before.

With yours I am afraid I may not be able to do the calculations. :-/ We are supposed to turn in calculations on each circuit. So far, the calculations on the circuit that audio got me started on should be easy. If you think you can walk me through the circuit you made that will be fine, otherwise I should probably stick with something more simple. We started on darlingtons on our last day of lecture if I recall, and thats about as far as we got. :-/

 

The arrangement in Audiogurus's post 46 has a voltage gain of about 10, for outputs from the collector (ie as a CE amp), but cannot supply the necessary 80 odd milliamps to the loudspeaker.

Do you feel my offering is too complicated? we can work further on yours if you like. I did say that before.

Well I'm sure you're in bed now. I will get some sleep to and be back on in the morning

 

My single transistor is just a preamp. It needs to have a complementary CC pair of transistors to drive a speaker but the output power in your assignment is almost nothing (0.0625W) so it will be easy.

 

My single transistor is just a preamp. It needs to have a complementary CC pair of transistors to drive a speaker but the output power in your assignment is almost nothing (0.0625W) so it will be easy.

So, is the complementary CC pair similar to what studiot was making? Would a darlington or a sziklai pair work? I've never actually solved a sziklai pair, so it'd probably be just as easy to do whichever you guys think will work best

 

I was reading this but I'm not exactly sure I understand what the VB batteries are.

 

OK

I found it hard to believe AG's post#46 that such a small change to the base bias chain would result in driving the transistor into saturation.

So I built the circuit and tested it.

As I expected, using a 3k3 resistor for R2 (since I only stock E12 resistors in the workshop) the circuit worked perfectly as an amplifier.

Using a 2N3906 and a 30 volts supply measured voltages were

Base -11.93
Emitter - -12.62
Collector - 8.5

beta @ .1mA base = 240; @1mA base =100 @10mA base = 20

Injecting a 12 mv 1khz signal through a 0.1 microfarad capacitor produced a collector waveform of 140mv, thus confirming the gain prediction of about x10

The upshot of all this is that two cascaded stages like this will get you the x100 gain you require. I would multiply all the resistances by 10 for the first stage so the current is 1.5 mA only.

These two stages satisfy part of your project aims as they use two biasing methods, potential divider and emitter biasing / emitter feedback. You already have the calculations for this configuration.

You will need to drive a CC output stage, such as AG mentioned or I described earlier. If you use a complementary pair as shown you will satisfy the requirement to use both NPN and PNP. The calculations are easy for a CC stage - just assume the voltage gain is 0.9 and compensate because you preceding stage is slightly greater than 100. You could probably persuade a single PNP device to work as a CC output but that would be too much for a 2N3906 so you would need to find a beefier transistor, capable of withstanding greater than say 40 volts Vce. A 2N3906/2N3904 pair would probably cope as they split the power load.

I was reading this but I'm not exactly sure I understand what the VB batteries are.

Vb batteries are the biasing for the transistors.

I have shown this in my (revised) sketch. As AG so eloquently put it I forgot the diodes first time but the bias Vb is provided by the two diodes and resistors I have called R7 and R8 in the top sketch. There is no magic in this.

This is not an elegant solution, but since the exercise was set before you have tackled the type of circuit I showed in the other sketch I wonder if it will be repeated when you have and your lecturer is trying to show how difficult it is to design with cascaded stages as opposed to multistage designs.

 

OK

I found it hard to believe AG's post#46 that such a small change to the base bias chain would result in driving the transistor into saturation.

So I built the circuit and tested it.

As I expected, using a 3k3 resistor for R2 (since I only stock E12 resistors in the workshop) the circuit worked perfectly as an amplifier.

Using a 2N3906 and a 30 volts supply measured voltages were

Base -11.93
Emitter - -12.62
Collector - 8.5

beta @ .1mA base = 240; @1mA base =100 @10mA base = 20

Injecting a 12 mv 1khz signal through a 0.1 microfarad capacitor produced a collector waveform of 140mv, thus confirming the gain prediction of about x10

The upshot of all this is that two cascaded stages like this will get you the x100 gain you require. I would multiply all the resistances by 10 for the first stage so the current is 1.5 mA only.

These two stages satisfy part of your project aims as they use two biasing methods, potential divider and emitter biasing / emitter feedback. You already have the calculations for this configuration.

You will need to drive a CC output stage, such as AG mentioned or I described earlier. If you use a complementary pair as shown you will satisfy the requirement to use both NPN and PNP. The calculations are easy for a CC stage - just assume the voltage gain is 0.9 and compensate because you preceding stage is slightly greater than 100. You could probably persuade a single PNP device to work as a CC output but that would be too much for a 2N3906 so you would need to find a beefier transistor, capable of withstanding greater than say 40 volts Vce. A 2N3906/2N3904 pair would probably cope as they split the power load.



Vb batteries are the biasing for the transistors.

I have shown this in my (revised) sketch. As AG so eloquently put it I forgot the diodes first time but the bias Vb is provided by the two diodes and resistors I have called R7 and R8 in the top sketch. There is no magic in this.

This is not an elegant solution, but since the exercise was set before you have tackled the type of circuit I showed in the other sketch I wonder if it will be repeated when you have and your lecturer is trying to show how difficult it is to design with cascaded stages as opposed to multistage designs.

I'll set it up and play around with it in multisim. Have we taken into account the source resistance? Wouldn't that mean we'd need a little bit more gain or am I misunderstanding?

 

So, is the complementary CC pair similar to what studiot was making?

Yes.

Would a darlington or a sziklai pair work? I've never actually solved a sziklai pair, so it'd probably be just as easy to do whichever you guys think will work best

A darlington or Sziklai pair are used for fairly high power output which your assignment does not have.

 

I found it hard to believe AG's post#46 that such a small change to the base bias chain would result in driving the transistor into saturation.

So I built the circuit and tested it.

As I expected, using a 3k3 resistor for R2 (since I only stock E12 resistors in the workshop) the circuit worked perfectly as an amplifier.

Using a 2N3906 and a 30 volts supply measured voltages were

Base -11.93
Emitter - -12.62
Collector - 8.5

Your circuit is alm,ost saturated. Every transistor is different so some will be saturated and some will not be saturated.
Why not design the circuit properly so it works with all 2N3904 transistors?

The circuit uses a 2N3904 NPN transistor but you tested it with a 2N3906 PNP transistor which is completely different.

The upshot of all this is that two cascaded stages like this will get you the x100 gain you require.

A single transistor can easily provide a voltage gain of only 100. Bypass the emitter resistor with a capacitor.