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Comments on this minimal amp?

Robyn

Apr 17, 2013
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Apr 17, 2013
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Hi crew!

So I breadboarded last night this Class B - Push Pull amplifier. I've tested it on a 3 way 8 ohms speaker cabinet and I'm very happy with the sound. Especially considering the general opinion about the sound quality of the TL08x ;)

The circuit comes from this page: http://www.allaboutcircuits.com/vol_6/chpt_6/10.html.

Like the first project I mentioned here I decided to build it because I love the way the author explains a lot about it's inner workings, and everything is justified and minimal. It's the Bauhaus of electronics!

I want to go to the next step by building the other channel (using the 2 op amps left in my TL084) and eventually build it into a nice old school wooden cabinet. But before I do it I would love to get your feedback (is this a pun?).

Talking about feedback, I would like to stay in closed loop mode in this discussion if that's OK :) Let me explain: I would love to avoid getting into design oscillation and add components until my brain saturates (going from pun to sustained metaphore). I one word I want to keep it simple. Not like this last paragraph.

So please let me know what you think could be done better. Are the components stressed in any way by a poor design? What is the power of this amp and how can I calculate it? How do I know if I can drive 4 or 8 ohms with this?

Attached is my version of the circuit. All I've done is add a 1K resistor (R2) in series with the pot (R3). I did this to avoid going in open loop on the TL084 2/2 (when to pot approaches full resistance) and clipping like crazy.

Edit: I am very bad at circuit analysis because I am a self taught beginner. It would be great to hear not only feedback but the reasons / calculations behind your feeback!
 

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BobK

Jan 5, 2010
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The main flaw I see with that circuit is that the TIP41/42 have their base current driven directly by the opamp. The opamp can typically supply no more than 40ma (min 10), and the typical beta of the transistor is 30 (min 15). So if we use the typical values, the collector current is limited to 0.04 * 30 = 1.2A peak or ~.85A RMS.

The power is I^2 R, so into 8 Ohms we get 2.89W or 4 Ohms we get 1.45W. At minimum numbers the limit is way below a watt. This is probably why he says you don't need a heat sink!

To get more power, you would need to use a darlington pair for the output transistors, and heat sink them.

Bob
 

Robyn

Apr 17, 2013
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Apr 17, 2013
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Hi Bob,

I just implemented that Darlington pair you're talking about. I had a BC547B and a BC558B handy so I used them.

- Rerouted the signals going the the bases of the TIP's to the bases of their respective BC's.
- Hooked the collectors of BC547 and TIP41 to V+ and the collectors of the other pair to V-
- The emitters of the BC's go to the bases of the TIP's

When I turn the amp on all I get is a buzz around 100Hz. I've tripled checked the connection, rebuilt the whole last stage, the buzz stays.

Now I calculated the gain of the Darlington pair using the simplified formula.
(by the way, my multimeter tells me that the hfe of my TIP42 is actually 220??? And 45 for the TIP41 which sounds more reasonable)

Let's assume that the hfe of the BC's is around 200 and 30 for the TIP's:
200 * 30 = 6000 (6230 to be picky)

Which for 0.04A would give 240A which sounds like something a industrial heater would consume. So I guess that's where my buzz comes from, is this right?

Should I use some sort of negative feedback to tame this gain?

Also, seeing as MOSFETS are voltage driven, wouldn't they be a better option as a follower to the op amp instead of the two darlington pairs?

Edit: I tried the Darlington with another pair of TIP's. It works this time but I don't hear or measure any increase in power. :(
 
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BobK

Jan 5, 2010
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The circuit you linked to had negative feedback. If you use darlingtons, you would want to use 3 diodes instead of 1 between the bases, because you now have 4 diode drops between them.

Did you try measuing the RMS output of the amp with a constant sin wave in? You could calculate the power out from that and the speaker impedance.

Theroteically you should be able to get about 10V p-p out of that amp, though it might be a volt or 2 less with darlingtons, so let's say 8V p-p practically. This would give you about 6 W at 4 Ohms or 3 W at 8. And don't forget heat sinking!


Bob
 

Robyn

Apr 17, 2013
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Apr 17, 2013
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Hey Bob,

I did the measurements with a sinewave, without the darlingtons I get 9.4V peak to peak and 1.33W @ 8ohms.

With the darlingtons (another pair of TIP41/42) I get... exactly the same thing! I have not added the extra diodes yet, because I am not sure if all 3 should all go from the output of the op amp all the the PNP or if I need to put one on the NPN side. Anyway, I should still get the power increase, just with crossover distortion right?

I feel like something is wrong with this circuit that is sort of bypassing the darlingtons. Is C3 in the right place? I also had a few misunderstandings in my previous post that I'd love to cover in order for me to get more confident with this circuit.

Sorry for being a pain, I just really want to understand everything so I can put one foot in front of the other without just blindly apply instructions. I don't want fish, I want to learn how to fish :)
 

BobK

Jan 5, 2010
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Did you turn up the input? The darlingtons will not increase the voltage gain, so with the same input I would expect the same output.

Also, how are you measuring p-p voltage? When I said 10V p-p I was actually using it wrongly to mean single sided, i.e. from ground to + or -. p-p actaully means from - peak to + peak. According to your power calculation, that is what you should be measuring.

And one other thing. You are measuring these voltages with the speaker connected, right? The voltage will easily go to the max with no speaker even without the darlingtons, it is a current limitation when you have the speakers connected that will limit the output.

And finally, you are using + and - 12V supplies, right?

Bob
 
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