so if i get a three transistor heat sink i should try and to mount it between the output pair in order to keep its temperature as close as possible to the output pair?

 

The third transistor is a small signal one, and doesn't require a heatsink of its own. It does need to be in good thermal contact with the output transistors, though, so it is usually glued (epoxy) to their heatsink.

It is called an amplified diode connection (I have never seen 'rubber diode' before).
I will post an article tomorrow.

 

so im leaning away from the rubber diode method as it seems more complicated, it would be a lot easier to get my head round if i could just connect it between q3 and R6 in place of the diodes in my current circuit, but i imagine i'd have issues with added gain in the negative channel doing this?
think i'm leaning towards a bootstrap capacitor and be done with it.

hoping this implementation is correct

where R6 + Rbs = collector resistor seen by C3, and Rbs >> R6. would there be any other reasoning to selecting values for these 2 resistors?
would this amp in its current form work given correct values?

 

Compare your circuit R11 with the Texas one which has R9 and R11 (bypassed by the C4 equivalent to your C1)

R11 is needed to ensure bias stability of Q2

 

was somewhat confused when adding this capacitor as it seems to vary in position somewhat between schematics, could i just bypass R11 or is the extra resistor (r9 in the texas schematic) a requirement?

 

You don't need the feedback loop that Texas connect the rubber transistor to.
You can just do a straightforward replacement of the diodes.
Note you would still need to stick the diodes to the heatsinks.
I said I would post details.

BTW I missed your last question as it was so close to your pic. It's very easy to do.
I try to leave some space
between an attachment and my text for this reason.








Thanks to Peter Williams.

 

Excellent, rubber diode it is.
Im going to start trying to calculate some values tonight/tomorrow so i can start some simulations. Little nervous as i have never done calcs for more than a single transistor amp, and they seem easier thanks to the volage devider biasing and collector emitter resistors. Probably going to have a few questions....

 

hi all, trying to start calculations.
haven't even started with values etc, was just trying to work out which resistances are which in terms of basic calculations ie which components make up Rc for q7, which make up Re etc.

i've already hit a brick wall with the first transistor (q7). here is the current circuit....

I moved the output resistors, initially had them on the emitters of Q7 and Q9, i noticed that this seemed wrong?
C2 has also moved, was done to make calculations easier but have no idea if its correct, and the calculations i was doing were totally useless anyway ^^

my problem so far is in considering the output at q 7. I was initally saying Rc= Rload + R9, Ve=Vcc/2? or just Vcc? Vceq=20V/2=10V
Ic = Vce/RL = 10/4=2.5A Ib= Ic/Hfe and Ie=Ib+Ic etc

however I realised that this is all considering the AC at the output. should i not be considering DC elements?
cant even work out what Rc would be for DC. so confused......

 

C2 now shorts the base of Q2 to earth at audio frequencies, removing any audio feedback.
That amp would sound horrible.

You need to stick with your old arrangement or follow the better Texas method with three resistors.

 

ah, ok, my bad. like i say it was somewhat off the cuff. ill change it.

am i heading down the right path considering the AC aspects? continuing down this route would mean for q6

Icq = Ib(q7) Ve = Vceq(q7) Vc = Icq x R7||Rin(q7)?

am i just way off the mark here?

 

that Vc is wrong.

 

feeling somewhat out of my depth, i have no idea if what im trying to do is even remotely correct and as a consequence i'm getting very stressed
can someone talk me through calculations for a simple 2 stage amplifier?
something along the lines of this

or even a purely class A, just anything when the output of the first stage becomes the input for the second.
i've been searching for an example for hours and cant find anything and so dont even know where to start.
trying to do it as i would a single stage, working backwards, but the loss of collector emitter resistors has me in a tiz

 

For post #92 only:

Your 2 left resistors set a DC level at the base of Q3. Q3 has no emitter resistor and so the current it allows is undetermined because that current is a function of base to emitter voltage and temperature. This undetermined current sets the voltage level across R6 at, "undetermined". The voltage at the base of Q4 is undetermined plus the voltage across 2 diodes at an unknown current. That voltage also depends on temperature. With the bases of Q4 and Q5 being held apart by an undetermined amount, and having no limiting resistors in their emitter circuits, their current is undetermined and probably causes smoke. If the voltage on the left of C2 is anywhere near the middle of the available voltage, you can get some signal to pass through to Rl. This will only last for a little bit of time because Q4 or Q5 will heat up while D1 and D2 do not heat up and thermal runaway will occur.

You have designed a short term smoke generator.

 

was just a quick throw together, i forgot the emitter resistors. :S had got myself a little worked up as all attempts at calculations have left me feeling lost.

dont suppose you fancy talking me through the design decisions for something very basic, 2 stage that won't set itself on fire?
i've just never seen the thought process behind anything other than a single stage where all voltages are set by resistors.

 

You always start these backwards. Let's assume, 30 watts RMS into 8 ohms.
Convert that to peak voltage and current.
math math math 21.9 volts peak and 2.737 amps peak.
Each half of the output stage has to deliver 21.9 volts while delivering 2.737 amps.
So...lets assume a power supply of +/- 25 volts DC
The output transistors have to deliver 2.737 amps, and we need them using up very little voltage, so we'll drive them with 0.2737 amps. That's what's going on in your driver stage.
The driver stage starts with 50 volts that has to be split evenly at the + end and the - end with an adjustable spacer in the middle.
50 volts at 0.2737 amps is 13.7 watts. Too much waste for a 30 watt amplifier. (46%)
Let's throw in another output transistor on each side, at a gain of ten, and your driver stage current goes down to 0.02737 amps. Now we are wasting 1.37 watts in the driver stage. (4.6%)

Using the configuration in post #88, we need about (2) Vbe's and some waste current through R9 and R10 for the spacer. Let's waste a watt in each of R9 and R10 so we can use 2 watt resistors. How many ohms can we afford? The rms current through each half of the output stage is 2.737 amps over sqrt2 = 1.935 amps at full power. P = I^2R
1 = 1.935 squared times R
R = 0.267 ohms.
We can buy a .22 ohm resistor or a .27 ohm resistor considering we're calculating in a 100% over power factor for the resistor.

So, how much voltage degeneration do we need to make the transistors match at idle?
I just happen to have a datasheet for a matched pair rated for 6 amps at 80 volts.
Figure 10 says the Vbe can vary by 0.05 volts at a tenth of an amp, so lets try to swamp that out and avoid trying to use matched pairs. That isn't practical, so we'll compromise. How about 100 ma? 0.1 amp times .27 ohms = 0.027 volts for each resistor at idle and later we will provide a DC feedback path to compensate for any residual error.

We need about 0.63 volts for each Vbe and 0.027 volts for each emitter resistor. That's 1.314 volts for the adjustable zener.

By now, you know how to do an adjustable zener with 27 ma going through it. The collector of Q3 is going to hold up its half of the supply voltage, and R5 is going to have to hold up most of 25 volts. at 27 ma. Let's make it 1k ohms and adjust for 24 point something something ma idle current.

I'm getting tired now. I leave this to other people to pick at it and find ways to improve on my theories and mistakes.

 

ok, getting a little further. thank you.

calcs so far.

10 watts into 4 Ohms. 10W x 4 Ohms = 40 = V^2 sqrt 40 = 6.324Vrms 6.324 x sqrt 2 = 8.94Vpk 8.94/4ohms = 2.235A Ipk

Vcc required = 8.94 x 2 = 17.88V so lets say 20 giving 1.06V overhead for output stage.

aiming for minimal current waste at output 2.235/100 (paired output) = 0.02235A/22.35mA

20V x 0.02235 = 0.447W power consumption by output.

using 0.5 watt output resistors rms @ each output = 2.235 /sqrt 2 =1.58A
P = I^2 x R so p/I^2 = R = 0.5/1.58^2 = 0.20 Ohms

I checked farnell and i can get 0.2Ohm 0.5W power resistors, hurrah!

I settled upon a TIP 31A for the output transistor
http://www.farnell.com/datasheets/1702411.pdf

using this i think i can get away with 80 mA Ic

so 80mA x 0.20 Ohms = 0.016V@r9

The Vbe required is about 0.62V giving me 0.016V + 2 x 0.62 = 1.256V @ rubber zener.

this is for q6 and q7? with q3 supplying the voltage for q5 and q6? so i need to select a smaller transistor for q6 that will run of Vbe 0.62V, then calculate r7 to give 0.62V at the base of q7 using a Ve of my original 8.94V-0.016V =8.904V? or is this just miles off?

 

I'm too tired to mess with this today.
I got involved in so many threads yesterday that it's all I can do to clean up a few responses before I go back to my real life chores.
Maybe tomorrow, or around sunset.

 

no problem, thank you for getting me this far!

 

I settled upon a TIP 31A for the output transistor

Hopefully that's a TIP31 / TIP32 complementary pair?

Make sure when you order that you do not get confused with the TIPP31/TIPP32 complementary pair which are quite different and not power transistors.

Do you understand about mounting kits and isolation for these things?

 

not two much, i know i need to mount them to a heatsink, isolation wise not a clue. was going to mount them away from everything else if thats what your getting at ^^