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I am working on an audio project in which a voltage controlled square wave oscillator drives a 5W 4Ohm Speaker. I am using a unipolar 12V supply. In order to limit the current going through the speaker I use a TIP31A (NPN power transistor) with a 10k resistor at it's base, which acts to limit the current across the speaker to ~0.4A (12V x 0.4A = 4.8W of power). However, I'm running into a problem where the transistor is getting very hot.

What I'm wondering is this: does the transistor get hot based on the fact that it is limiting current, or does it get hot simply based on the amount of current passing through it? If I reduce the size of the current limiting resistor at the transistor base and added another current limiting method, such as a power resistor in series with the speaker, would that reduce the amount of heat dissipated by the transistor?

Here is the part of the schematic relevant to power consumption, square wave flips 2n3904 on and off (this is to invert the signal, since the oscillator idles high) which controls power on the base of the TIP31A, which in turn drives the speaker:

Speaker Driver

Here is the full schematic. Keep in mind I am driving the speaker with a square wave (full on / off) so changing the amplifier topology to class B or AB will not help.

Full Schematic

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You have a 4 Ω speaker and 0.4 A. The power dissipated in the speaker is given by \$ P = I^2R = 0.4^2 \times 4 = 0.64 \ \text W \$ so your calculation is rather a bit off.

The voltage across the speaker is \$ V = IR = 0.4 * 4 = 1.6 \ \text V \$.

You have a 12 V supply so that means that the voltage across the transistor when on = 12 - 1.6 = 10.4 V. At 0.4 A that means the power dissipated in the transistor is \$ P = VI = 10.4 \times 0.4 = 4.16 \ \text W \$.

Note that both of these calculations are for when the transistor is on. When it is off (50% of the time for a squarewave) the current is zero so the power in the speaker and transistor will be zero also. This will result in an average power of half of the calculations above.

If I ... added another current limiting method, such as a power resistor in series with the speaker, would that reduce the amount of heat dissipated by the transistor?

Yes. You can afford to drop 6 to 8 V there at 0.4 A. You can work it out.


From the comments:

... but the voltage across the speaker is 12V, and current is 0.4A.

If that were true then the speaker resistance must be \$ R = \frac {V}{I} = \frac {12}{0.4} = 30 \ \Omega\$. Try measuring the voltage across the speaker terminals. It should be close to 1.6 V as calculated above.

So also by ohm's law we can arrive at the conclusion that: P = V x I = 0.4 x 12 = 4.8W.

That's the power for the whole circuit. 4.16 W of that is lost in the transistor.

I guess I'm just confused as to which applies in this case, when the transistor is acting as a resistor.

Your understanding is correct that the transistor is behaving like a variable resistor in that it limits the current.

Let's look at the current gain and do a little more maths.

enter image description here

Figure 1. DC current gain from the TIP31/32 datasheet.

  • Your 10k base resistor will give a base current of \$ \frac {12}{10k} = 1.2\ \text {mA}\$.
  • We're not using the 4 V of the datasheet (we're using 12 V) but let's assume that the hFE is between 25 and 50. That means that a base current of 1.2 mA should give us a collector current of 30 to 60 mA. Lo and behold, you measured 0.4 A.

Note that this is not a good way to set the speaker current if multiple such circuits were being built as the collector current would vary significantly from transistor to transistor.

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  • \$\begingroup\$ Hmmm... but the voltage across the speaker is 12V, and current is 0.4A. So also by ohm's law we can arrive at the conclusion that: P = V x I = 0.4 x 12 = 4.8W. I guess I'm just confused as to which applies in this case, when the transistor is acting as a resistor. If you could clarify how this works that'd be helpful. Thanks! \$\endgroup\$ – thegrinch Jun 14 '18 at 17:40
  • \$\begingroup\$ See if the update helps. \$\endgroup\$ – Transistor Jun 14 '18 at 18:37
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If the average current is about 400 mA then the load (including speaker) on the 12 volt rail will be 12/0.4 = 30 ohm. Given the speaker is 4 ohm we can make an estimate that the TIP31A is basically a 26 ohm resistor.

Power in this resistor (assuming the 400 mA is RMS) = 4.16 watts and, the transistor will get very warm.

If I reduce the size of the current limiting resistor at the transistor base and added another current limiting method, such as a power resistor in series with the speaker, would that reduce the amount of heat dissipated by the transistor?

Yes it would but all you are doing is shifting power dissipation from a transistor to a resistor.

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  • \$\begingroup\$ All voltage / current calculations are for peak values. As I said square wave is full on / off, so calculations are for on. \$\endgroup\$ – thegrinch Jun 14 '18 at 17:49
  • \$\begingroup\$ @thegrinch then please explain why you said this "12V x 0.4A = 4.8W of power" \$\endgroup\$ – Andy aka Jun 14 '18 at 17:57
  • \$\begingroup\$ Ohm's Law, P = V x I. As was pointed out below, I may have been mistaken to use this, since the power is divided between the individual components. \$\endgroup\$ – thegrinch Jun 14 '18 at 18:09
  • \$\begingroup\$ If you measured a DC average current of 0.4 amps and your voltage is 12 volts, then there is a power dissipation of 4.8 watts of which 4.16 watts is due to the TIP31A. If it were peak current i.e. average current is in fact 0.2 amps then power in the transistor is 2.24 watts and it's still going to get very hot without a heatsink. Without a heatsink, the thermal resistance is 62 deg per watt hence it will rise to close on 150 degrees and this assumes some air flow around the device and no localized waring of ambient. \$\endgroup\$ – Andy aka Jun 14 '18 at 18:52

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