Advice needed for amp design

[ifrythings]

Well, this is my first amplifier that I am making that is descrite, so I'm not even close to the smarts of some of our local gru's.

So I guess my first question is will this circuit work?
If it will work than we can go onto the values I may need.

If this thing is a horrible site to the eye's please tell me how to make it better.

This is ment to be a low power learning amplifier for me so thats while I have an op-amp as the input and pre-drive stage (As I don't understand differential amplifiers yet).

Also I would like it simple and not a lot of parts. I don't care at the moment if it has 10% THD, like I said earlier, it's a learning experience.

[ifrythings]

edit: found a misstake in schematic, added resistor to feedback network to make resistor divider.

[AndrewT]

Hi,
the opamp can only drive about reg voltage-1v. About 14Vpk ~=10Vrms for a +-15V opamp.
This darlingtons are set up as followers and so the maximum output voltage is still the same as the opamp less another pair of Vbe drops about 12Vpk to 13Vpk. The +-24Vrails do not get used to full potential.

You need to add a bit of gain in the output stage.

The amp has DCgain = AC gain. Any offset error in the opamp will appear at the output times the circuit gain.

How about adding an RF input filter?
You have already put in the series resistor for that filter.

You will probably need to add a high frequency compensation capacitor around the opamp and/or in parallel with the feedback resistor. Leave traces and pad locations to add these at time of debugging.

Well spotted:- the omitted NFB resistor.

[ifrythings]

How about adding an RF input filter?
You have already put in the series resistor for that filter.

I don't understand what you mean about that statement, you got and example?

The darlingtons I have, have a minimum gain of 1000-10 000, I thought that was enough?

Is there anyway to improve this so the outputs can reach close to the main B+/- rails?

BTW I've been looking at other schematics of amps and they all seem to use a P driver for N output device and N driver for P output device why do they do that?

[darkfenriz]

You will not have voltage gain provided by transistors, so if op-amp can swing from -14V to +14V output also will. Just will be able to provide high current to drive a speaker.
Of course circuit can work and be reliable.
RF filter can be easily added to input, but isn't nessesary (see schematic).
Depending on parts you use this may or may not oscillate. If there are any troubles add a small cap in feedback circuit (schematic again)

[CBS240]

Good sugestions I see here... This circuit is basically a power op-amp that can handle lower impeadences like a speaker. I personally like the emitter follower output stage for this reason, but as stated previously, it has voltage gain less than 1, but significant current gain. The circuit should work well, but may produce less unwanted heat by using +/- 18V instead of +/- 24V since the output swing cannot exceed the op-amp output as is. However, it does not hurt to have improvements as this is just another lesson for learning, eh?
In the Vbe multiplier, a voltage divider with the pot between base and emitter is probably better than using the wiper to bias the base as pots wear out and have bad connection. You want when this happens to have that transistor saturate and not cut-off. If it cuts off, your outputs may saturate and the magic smoke might get out. Also, the bias 'turn on' resistors could be replaced with current sources, this way the op-amp isn't driving these resistors(AC model) and are just driving the output transistors.(search for current source)
Also a DC blocking cap wouldn't hurt either. (between ground and divider resisitor of feedback resistors.) This way, DC output is fedback with no voltage division and the op-amp(differential amp) can correct offset more accuratly.

[CBS240]

Originally posted by ifrythings


Is there anyway to improve this so the outputs can reach close to the main B+/- rails?

BTW I've been looking at other schematics of amps and they all seem to use a P driver for N output device and N driver for P output device why do they do that?

This is usually because those drivers produce voltage gain as the output is taken from the collector.

Emitter follower(common collector; input to base, output is emitter) follows the signal minus one PN juntion drop. Therefore there is no voltage amplification, just current amp.

Common emitter(input on base, output is collector) amplifies both voltage and current but voltage is 180 degrees from input. Typically doesn't have such low output impeadence and usually followed by a follower stage. Also has higher input impeadence.

Common base(input on emitter, output is collector) amplifies both voltage and current, but output is in phase with input. Input impeadence is lower than common emitter.

search on these topologies...
As you can see the output arrangement you have is emitter follower. Darlington arrangement gain is the product of both transistor gains, but is still a follower(with 2 PN junction drops). Just means it requires much less current from the op-amp to drive. There are drawbacks to darlington packages though.

[ifrythings]

There are drawbacks to darlington packages though.

What mite those be?

Thanks for the advice on the bias, never thought about the wiper going open, though there would still be a resistor in series if the wiper went open but I see that the ratio would be way too off and would cause PSU short/transistor saturation.

So it would be better to put a cap on the op-amps output before it goes into the "power" stage?

I don't understand why you would want one in the feedback line.

So this would be considered a current amplifier right? or is it voltage and current?

would this be better?

[AndrewT]

Thanks Darkfenriz,
for showing where those two caps go.

As Cortez keeps telling me a pic is worth at least a few hundred words.

[CBS240]

Originally posted by ifrythings



I don't understand why you would want one in the feedback line.

Like this....

The cap presents an infinate impeadence to 0Hz or DC. This means that any DC offset present at the output willl be compared to the input to the op-amp and the circuit will compensate as the two inputs will strive to be equal. Without it, the DC will have gain, set by the ratio, and may leave some offset at the output. To AC signals the cap will conduct giving them gain, hince DC blocking cap.