Two-Stage Transistor Phase Splitter

Thread Starter

cebrax

Joined Jan 30, 2010
26
Hello!

I was studying transistor amplifiers at this page: http://www.tpub.com/content/neets/14180/css/14180_40.htm

I couldn't undestand how this circuit produces two outputs with same amplitude? One thing pops into my mind is that it is done with resistors.. But is it practical?

Here what this page says:

If the output signals must be larger in amplitude than the input signal, a circuit such as that shown in figure 1-28 will be used. Figure 1-28 shows a two-stage phase splitter. C1 couples the input signal to the base of Q1. R1 develops the input signal and provides bias for the base of Q1. R2 provides bias and temperature stability for Q1. C2 decouples signals from the emitter of Q1. R3 develops the output signal of Q1. Since Q1 is configured as a common-emitter amplifier, the output signal of Q1 is 180º out of phase with the input signal and larger in amplitude. C3 couples this output signal to the next stage through R4. R4 allows only a small portion of this output signal to be applied to the base of Q2. R5 develops the input signal and provides bias for the base of Q2. R6 is used for bias and temperature stability for Q2. C4 decouples signals from the emitter of Q2. R7 develops the output signal from Q2. Q2 is configured as a common-emitter amplifier, so the output signal is 180º out of phase with the input signal to Q2 (output signal from Q1). The input signal to Q2 is 180º out of phase with the original input signal, so the output from Q2 is in phase with the original input signal. C5 couples this output signal to the next stage. So the circuitry shown provides two output signals that are 180º out of phase with each other. The output signals are equal in amplitude with each other but larger than the input signal.
 

studiot

Joined Nov 9, 2007
4,998
Firstly the circuit in your link is a single transistor phase splitter.

This works because a transistor has two output points viz the collector and the emitter.
These outputs are 180 deg out of phase with each other. Another way to put this is to say that the output from the collector is inverted relative to the input, whereas the output from the emitter is in phase with the input. The actual magnitudes of each output are determined by the emitter and collector resistors, which can be set to make them equal.

Secondly your published image is incomplete and of a different arrangement as it does not show power supply configurations.

However it does contain sufficient information to say that there are two stages one fed from the output of the other. given the information I stated above can you now see how the two outputs differ?

This is the homework forum so you need to do some working yourself.
 

Thread Starter

cebrax

Joined Jan 30, 2010
26
Firstly the link I gave contains my question too. Click on "Next" on top and you'll see the other half that contains my circuit =)

So in the second circuit, you say that magnitudes are set to be equal with the help of resistors. But how is this practical? I mean if you don't use zero tolerance resistors it won't be the same I think. And as far as I know their resistance changes with the temperature, right?
 

studiot

Joined Nov 9, 2007
4,998
So in the second circuit, you say that magnitudes are set to be equal with the help of resistors
I really have no idea which circuit you are referring to.

can you now see how the two outputs differ?

This is the homework forum so you need to do some working yourself
I really have no idea if you understood my last query and comment.

How can you expect to hold a conversation or proceed if you don't respond to the other person?
 

retched

Joined Dec 5, 2009
5,208
I have to agree, I see no question on the link.. after clicking 'NEXT' or 'BACK'

Lets try this again, shall we?

You want to know if it is practical that the posted circuit segment can split a phase 180deg keeping the same amplitude, using resistors that are not 0% tolerance?

Yes. The signal is being amplified so the resulting wave can be split, and one inverted, resulting in said result.

Since Q1 is configured as a common-emitter amplifier, the output signal of Q1 is 180º out of phase with the input signal and larger in amplitude. C3 couples this output signal to the next stage through R4.
The coupled result is going to be the same amplitude.
 

Thread Starter

cebrax

Joined Jan 30, 2010
26
First of all, sorry to all of you..
I was not trying to be offensive, I am misunderstood

I understand the phases, still, I don't get why two outputs are at the same amplitude?
If Q1 amplifies the signal, why Q2 doesn't?
 
Last edited:

hobbyist

Joined Aug 10, 2008
892
If Q1 amplifies the signal, why Q2 doesn't?
Q2 doesn't, because
the base of Q2 is not connected directly to the input signal as Q1 is, so Q1 only gets the actual input waveform, inverts it and sends this amplified signal onto Q2, which then inverts the output of Q1, making it in sync to original input signal.

Your question arises because if Qi amplifies the original signal, then Q2 should be greater than the output of Q1, seeing that Q2, is getting Q1's output and amplifying it further.

Is this your actual question??


This could be so depending on the value of the bias resistors.

If Q2 stage is designed with a gain of around 1, then it could just send Q1's signal inverted to the output.

Again the results are figured with the value of the resistors used in both stages.
 

Thread Starter

cebrax

Joined Jan 30, 2010
26
Your question arises because if Qi amplifies the original signal, then Q2 should be greater than the output of Q1, seeing that Q2, is getting Q1's output and amplifying it further.

Is this your actual question??
Yep, this is my question. I thought to myself, if Q1 amplifes, Q2 should amplify too..
 

hobbyist

Joined Aug 10, 2008
892
Just a quick example.

The gain of a collector feedback CE. amp can be assumed to be around (RF / RS) where RF is the feedback resistor and RS is the series resistor, going into the base to increase input impedance.

In stage Q2 the ...RF=R5.....and RS=R4...

So if R7 was made = to R3, then the voltage across R3 would feed into the input of Q2, and if Q2 was designed with gain of 1,
where, ....R4 was made = to R5...then that same voltage signal would get inverted in Q2 stage with NO amplification, and developed across R7, with a same amplitude only inverted.

Does that help clarify better.

The voltage gain with collector feedback is complicated more than emitter feedback, where emitter feedback, the gain is assumed more easily as.... RC/RE,.....


SIDE NOTE:
Q1 and Q2 stage has dual feedback, so gain for Q2 is assumed in line with ....RF/RS.... but only under certain conditions, while as you see Q1 stage has no series resistor, so gain is calculated differently, beyond the scope of this example....
 
Last edited:

KL7AJ

Joined Nov 4, 2008
2,229
The split load phase inverter as shown in the NEETS site is pretty much standard practice. The two transistor version is rather inefficient, because you have to ruduce the gain of the second stage to match that of the first stage...lots more parts and twiddling to do to.


If you need extra gain, it's better to just precede the split phase inverter with a normal voltage amp.

eric
 

JoeJester

Joined Apr 26, 2005
4,390
I've seen phase splitters (vacuum tubes) up to the end of the seventies. Of course the equipment was designed in the late 50s/early 60s.
 

Thread Starter

cebrax

Joined Jan 30, 2010
26
thanks to everyone and for the explanation about how the gain is set with resistors in positive feedback network.

Off topic, can anyone inform me about any documents on positive feedback in transistors? Google doesn't return much, in detail..
 

Audioguru

Joined Dec 20, 2007
11,248
The very simple and very old phase splitter and any audio amplifier does not use positive feedback.
Modern audio amplifiers have lots of negative feedback. The very old circuit you posted has no feedback, not negative and not positive.
 

hobbyist

Joined Aug 10, 2008
892
Positive feedback is used in transistor circuits, designed for oscillators.

While negative feedback is primarily used for designing transistor amplifying stages.

positive feedback feeds the signal at the output back into it's own input, in phase with the input so as to continue amplifying until it goes into oscillations.

Negative feedback is where the signal is reintroduced back into the input out of phase with the input signal, to cause a reduction in amplification, thereby keeping the transistor stage from breaking into oscilation.


A CE amp, with an emitter swamping resistor is a example of series current feedback, which means that when the input signal becomes to large with respect to the base emitter bias voltages, or if the transistor begins to heat up and begins to go into thermal runaway, then the voltage at the emitter begins to become more positive, thereby reducing the voltage between base and emitter, so then the transistor, begins to conduct less, as this lowers the collector current, this in turn lowers the emitter voltage to the point where the transistor conducts again, reaching an equilibrium.

This feedback is mainly for temperature stability, to keep the transistor from thermal runaway.

With your circuit in question, the same negative feedback is used but instead of it being in series with the signal as with emitter feedback, it is in shunt with the input voltage signal.

That's why it is called voltasge shunt feedback.
Used for the same reasons, to keep transistor parameters from affecting the amp stage.

This too is again negative feedback, because it;s primary purpose is to reduce gain of the amploifier to keep the stage from thermal runaway, and oscilating.

Collector shunt feedback, works as if the transistor begins to conduct more due to overheating, then the collector voltage begins to drop, and the voltage at the base in turn drops because it's voltage is determined by the voltage at the collector, so a drop in base voltage, causes the transistor to conduct less, as it conducts less then the collector voltage increases, and the transistoer reaches an equilibrium, to keep it conducting away from overheating.

Because of this reduction, the gain will always be lower due to negative feedback.

But the lower the gain, the more stable the stage is, so a stage is designed with negative feedback to keep the transistor parameters from interfering, and then compensated for with bypass capacitors, to increase the AC signal gain, of the stage.
 

Thread Starter

cebrax

Joined Jan 30, 2010
26
OK, I get it better now.

Can you explain me or ,direct me to somewhere, about the calculations of collector shunt negative feedback?
 

Audioguru

Joined Dec 20, 2007
11,248
The same current goes through the collector resistor and emitter resistor (but the emitter resistor has slightly more due to the base current also in it). So if the resistors have the same value then their signal levels are the same and are the same as the level of the input signal.
 

Thread Starter

cebrax

Joined Jan 30, 2010
26
Just a quick example.

The gain of a collector feedback CE. amp can be assumed to be around (RF / RS) where RF is the feedback resistor and RS is the series resistor, going into the base to increase input impedance.

In stage Q2 the ...RF=R5.....and RS=R4...

So if R7 was made = to R3, then the voltage across R7 would feed into the input of Q2, and if Q2 was designed with gain of 1,
where, ....R4 was made = to R5...then that same voltage signal would get inverted in Q2 stage with NO amplification, and developed across R3, with a same amplitude only inverted.
I have searched the internet and found some useful stuff, I had an idea and started trying to solve.
I have been trying to calculate and find, an equation that shows that the gain is related to rf/rs, but I can't. Can you help me out?
 

Audioguru

Joined Dec 20, 2007
11,248
A transistor has a voltage gain of about 100 to 200 if it does not have an unbypassed emitter resistor and its current is not too low.
If the value of Rf feedback is almost the same as the value of Ri input then its base has a low signal level due to the gain and the input level to the input resistor is almost the same as the output at the collector. Therefore total gain is Rf/Ri up to a gain of about 10. Higher gains have an error because the transistor does not have infinite open loop gain like an opamp.
 
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