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This is my first question ever so please don't kill me :)

I have built a stereo hi-fi (end) amplifier 2x10W based on a IC TDA2009. I have bought it as a construction (learning) kit ready to solder.
Schematic:
Amplifier circuit schematic Bill of parts

(the pictures are from the manual that came with the kit, please don't mind the Czech headers)
Parameters for the circuit (from the manual) are:

  • Us power supply voltage: 8-28V (I use 9V power supply)
  • Peak output current Io (freq. >= 20Hz): 3.5A
  • ...and more but IMHO (now) not important values

Now, everything worked great (although I had to replace the C8 capacitor, no idea why it was broken), so I decided I want to replace the R1 and R3 resistors with a potentiometer for variable gain (gain control). I bought stereo potentiometer, linear, 1kOhm, wired it up to R1s and R3s place (first pin ground, second and third pin to the PCB). When I turned the circuit on, the pot started smoking and burned.
If I get it right, I understand that the potentiometer couldn't handle the load when set to little resistance. I didn't really had a choice when selecting the pot as the one I used was the only stereo 1kOhm pot they had in the shop.
So I need to know:

  • Is my approach correct here? Can I replace the two resistors with a stereo pot?
  • Am I right thinking that the pot couldn't handle the current/load, or was it just a bad piece?
  • How do I calculate the power that can be on the resistance R1/R3?

Thank you for help!

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  • \$\begingroup\$ You need to connect one end to pin 8/10 respectively, the other end to ground, and the wiper to C6/7 respectively. \$\endgroup\$
    – user207421
    Jul 27, 2015 at 11:18

2 Answers 2

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The usual approach with power amplifiers is that you have the volume control on the input, and a constant gain of the actual power circuit. This would mean putting the potentiometer on the inputs of the circuit.

Also, the chip is not guaranteed to be stable at all gains, so if you adjust the gain in the feedback loop of the chip-amp, it might oscillate and start misbehaving.

If you connect the pot up front then you don't have to worry about wattage of the potentiometer, high wattage potentiometers are expensive.

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  • \$\begingroup\$ Thank you! In the light of my attempt this makes much more sense. Could you suggest me which pot should i use (resistance, linear/log, power rating)? And should the pot be before or after the input capacitor? \$\endgroup\$
    – Petr Krýže
    Jul 27, 2015 at 12:07
  • \$\begingroup\$ @PetrKrýže I would put the potentiometer before the input capacitor, as I've read that some pots can become "scratchy" if exposed to much DC current. The pot law should be logarithmic, as our hearing is logarithmic. Use a 10k or 100k as they are easy values to get, and are high enough so they don't load down your music source. If you have to use a linear pot, read sound.westhost.com/project01.htm for info on how to approximate a logarithmic response. \$\endgroup\$
    – user55924
    Jul 27, 2015 at 12:27
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Not a good idea.

If you refer to the the specification sheet http://www.micropik.com/PDF/TDA2009.pdf you'll see on page 9 a table which shows components, typical values and what they do. (Note R2/R4 - 18k should read 18R)

By replacing R1 and R3 with a 1k variable you actually lowered the gain.

Also the resistor chain R1/R2 is directly connected to about half the DC supply level at the output of the amp so when R1 (or R3) is about zero you only have 18 ohms in the circuit.

What you could have done is left the fixed resistors in circuit and added the 1k0 variable in series with it giving you a range of 1k2 to 2k2. This would boost the amplification by a factor of about 2.

Volume control with this type of amplifier is normally done by adjusting the size of the input signal through a variable resistor from zero signal to maximum.

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  • \$\begingroup\$ Thank you for info, but I actually wanted to lower the gain (in other words, I wanted the 1k to be the max). The idea with the 1k in series is nice and I wouldve done it if it wouldnt double the gain :) \$\endgroup\$
    – Petr Krýže
    Jul 27, 2015 at 12:03

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