Stagg AB20 as a preamp?

[polkat]

I have a small Stagg BA20 amp, and I wish to tap the circuit board to create a preamp output (use this amp as a preamp). The output on these amps is driven by a CM1875T audio power amp IC. I know that pin one of this IC is the positive input, so, if I tap into that input pin and wire out to a jack on the panel (so that the board input to that pin now goes to the panel), would I then have a preamp output?

If incorrect, what should I do?  Thanks!

[J M Fahey]

Yes, something like that, but to make it better/safer follow the track connected to pin 1 backwards: you should find audio comes through a capacitor, connect your preamp output wire to "the other side" of it.
Solder a 2200 to 4700 ohm resistor to the preamp out jack hot pin, and connect the signal wire you just added to the resistor's free end.
Both measures are to isolate the LM1875 input from the outside world; the capacitor isolates DC and the resistor avoids that a shorted cable mutes your amp.

[polkat]

J M Fahey, thanks for the response! Hard to follow this thing as the traces are tiny and there doesn't appear to be any schematic available, but I'll see what I can. Of the choice between the 2.2K~4.7K resistor, which might be a better choice?

[J M Fahey]

it's the same, toss a coin.
the idea is that if the output cable gets shorted or whatever, the series resistor avoids it sucking the signal

[polkat]

Thanks! Actually, my intentions (apparently not stated very clearly in my OP), is to make this little amp a preamp only and to kill any signal to it's own power stage. I'd do this by cutting loose and then shorting pin one of the CM1875T to ground, after tapping the preamp output as you describe above. That make any difference in the way you'd do it?
Thanks.

[J M Fahey]

May work, but in that case I would leave the LM1875 in the signal path, to add some of its own power amp clipping if overdriven.
In this case, instead of getting the preamp signal straight out, leave everything as-is, simply disconnecting the speaker, and adding from speaker out to ground a voltage divider, which will pad the strong speaker out signal into something manageable.
You can try a 47K resistor feeding a 10K pot, which now becomes your "Power/Master" control.
You drive your new power amp with that signal.
You can use the 1875 clean or overdrive it, and then reamplify it at will.
You will not hurt it by running it without speaker , although for peace of mind you might add a 100 ohm, 1/2 or 1W resistor replacing it.
You will be surprised by the good sound.

[polkat]

My past involved some electronics. I've built a few tube amps and done some tube repairs, but the transistor/IC stuff I'm a bit leary about. So lets see if I understand this.

Your offering TWO ways of doing this: First, leave the power amp chip connected, then run the speaker out + line to a voltage divider, and use that as a signal for the power amp. OR, run a 47K resistor (I assume from the tap-in point we discussed before) to a 10K pot and feed the power amp from that (to give a wider variety of sound (distortion).

On the first idea, I am guessing that the voltage divider idea will use the whole amp as a preamp for the new power amp....right? I like this idea, but if the speaker out goes through the voltage divider, then to ground, where do I get the signal from? I assume it goes speaker out->voltage divider->output of divider to power amp->and common grounds. Sorry, I don't know too much about voltage dividers. Everything turned up, this amp supposedly puts out 20watts rms. Not sure how much voltage that converts to, but most power amps look for 1volt peak-to-peak, which I assume is what the amp inputs to it's LM1875.

Am I getting way off track here?

[joecool85]

I run my Dean Markley K-20X with a line out from the speaker level like Juan suggested.  I used this as a schematic for it: http://singlecoil.com/docs/out.pdf

It says to leave the speaker plugged in, but I assume this is for tube amps.  I've run mine for hours without the speaker plugged in and it was fine.

[J M Fahey]

Both suggestions I made are resistive "voltage dividers" or attenuators.
Since an image is worth 1000 words, here you have 2000 words.

[polkat]

Thanks guys! I've got it now and will give it a try. Some final questions: Using the variable level idea, what wattage would be best for the 47K resistor? And will there be any impedance problems between the attenuator and the line-in of most power amps? If you can advise me on that, I can take it from here.

Thanks again for the help!!

[J M Fahey]

The 47K dissipates almost nothing, 1/4 or 1/2W will be ample.
No impedance problems.