Current flow V Signal flow

[Mick B]

I might have bit off more than I can chew. I put this amp together and  the output/Power transistors on one side get to hot to touch. I was hopping someone would walk me through one channel of this I am having a hard time following the flow I get lost in the differential part of this.
 thanks VERY VERY noob   

[basinstreetdesign]

I hope you have large heatsinks on all of the power transistors, Q11 through Q14 inclusive and at least SOME heatsinking on Q7 through Q10 inclusive.  That may not guarantee to save your bacon but may at least give you enough time to kill the power before the Q's roast.

I will try to give a terse theory of operation of the circuit.

OK, so here goes:
Looking at the right channel, the input signal comes in the connector and is AC coupled and biased to ground with R3.  It goes to one side of a differential pair which will have a large amount of signal gain.  If the circuit is working normally and you put a scope on the collector of Q1 to compare the signal level with the input, you may not see a lot of gain but that's because the circuit uses a lot of feedback (more on this later).  The differential pair, AKA a "long-tailed pair" uses R7 to deliver a fairly constant current (about (Vee-.7)/R33 Amps) which is split more-or-less evenly between Q1 and Q3.  The current through Q3 is entirely wasted as Q3 has no collector resistor.  The amplified signal of this pair is developed across R5 and sent to the next stage, Q5.

Transistor Q5 is a straight-forward CE stage with no emitter resistor so it also has a fair amount of gain.  The collector current of Q5 develops Q5's output voltage across R13.  Because this current also goes through two diodes, two output voltages are developed to be passed on the the output stages.  These voltages are in-phase and the same amplitude but separated by about 1.4 V DC.  This separation is necessary because the two output stages require their driving signals to be so separated so that the output stages are not simultaneously turned off with near-ground signal level.

On positive excursions of the audio signal, Q7 and Q11 turn on and act as a conventional Darlington pair, or high-current-gain emitter follower and drive current into the load.  During this time Q9 and Q13 are turned off.

On negative excursions of the audio signal Q7 and Q11 are turned off and Q9 and Q13 turn on.  They form a Sziklai pair.  When Q9 conducts it develops signal voltage across R19 and pulls the base of Q13 higher (farther away from the negative rail).  This turns Q13 on and delivers negative current to the load.  If R17 and R19 are not chosen with care these transistors can burst into oscillation (usually 1 - 10 MHz) and waste a lot of power.

R21 and C9 form an impedance -correcting network which can aid in making the power stage stable against spurious oscillations.  This network is called a Zobel Network and are not well understood by lots of designers (including me).  Their values also depend on what the load looks like impedance-wise.

Now, lastly let's look at R11, R9 and C3.  This is the feedback from the output to the other side of the differential pair.  The differential pair subtracts the feedback voltage on the base of Q3 from the input signal on the base of Q1.  This reduces the gain of the whole circuit by a lot and also reduces its signal distortion by the same amount.  This whole circuit has the classical architecture of an op-amp.  And, just like an op-amp it requires gain-setting feedback network.  The gain set here is calculated just as with an op-amp as (Rf/Rg + 1).  In this case it is (R9/R11 + 1) = 71.  So with +/- .7V Pk signal in it should deliver about +/- 50V Pk out to speakers.  C3 AC couples the feedback to the high-gain stage so that residual DC offsets in the circuit will not be amplified as much.

You said that the Power transistors get hot.  So this means that they are passing too much current during no signal times.  Possibly the bases of Q7 and Q9 are more than 1.4 V apart (one diode in the wrong way?).  Or one of those bases pulled too close to the rail?  Or you have a wrong/bad connection on the bases of Q11 or Q13.  Perhaps wrong resistor value in R19?

Hope this helps.

[basinstreetdesign]

AMA

[ArthurDent]

A common way to gracefully discharge big caps is to put a 'bleeder' resistor across them. I would not advise discharging them quickly because the current from a shorted caps can be quite high. If you have two 2200uf caps in parallel, just one resistor across the pair will do. It looks like you have around 20 volts on each rail so trying a 500 ohm 2 watt resistor might be in the ballpark.  Try this calculator to see if that would work for you.

https://www.engineersedge.com/instrumentation/capacitance_discharge/capacitance_discharge_calculator_12909.htm

[rstofer]

Unless you have some kind of switching, those bleeder resistors will be presenting a load to the PS at all times.  They will be dissipating heat at all times.  There is a tradeoff between discharge time and  PS load or heat.

One way to do this would be to use a DPDT power switch that switched the resistors across the caps when the device is switches off.

[Audioguru]

The LED and its resistor should be connected to the amplifier circuit at all times. Then to turn off power to the amplifier the LED also gets turned off.

[Mick B]

Well I tried the link you gave me, I couldn't seem to get it to do much of anything. I don't have that resistor and I don't have a switch yet, and I have no idea how to over-complicate the amp with a mosfet or something like it. I think i'm going to go with a DPDT that seems the easiest seeing how I don't have a SW yet. Thanks

[Mick B]

I had thought of that, I was looking to see someone would come up with a mosfet or something like it solution. The only sane way is to use DPDT and be done with, Thanks for your input.