Hi RR
You also need to add emitter resistors in your 0.62W amp to stabilise the quiescent current. You get thermal runaway if the output transistors get hot and the bias stabiliser does not track the temperature. Since no thermal connection is ever 100% good emitter resistors are needed. If the current increases, the voltage across them increases, and this reduces the current, so helps to make sure that the current is stable with temperature.
For your 0.62W amp I would say you could use 1 ohm resistors.
Normally a transistor is used instead of two diodes because the voltage across it can be adjusted by using a variable resistor in series with a fixed value (to limit the range) between the base and emitter and a fixed resistor between the base and collector. The ratio is nominally 1:1 to give 2x Vbe but in practice will be less than this (Rbc < Rbe). This is known as the "Vbe multiplier".
John
You also need to add emitter resistors in your 0.62W amp to stabilise the quiescent current. You get thermal runaway if the output transistors get hot and the bias stabiliser does not track the temperature. Since no thermal connection is ever 100% good emitter resistors are needed. If the current increases, the voltage across them increases, and this reduces the current, so helps to make sure that the current is stable with temperature.
For your 0.62W amp I would say you could use 1 ohm resistors.
Normally a transistor is used instead of two diodes because the voltage across it can be adjusted by using a variable resistor in series with a fixed value (to limit the range) between the base and emitter and a fixed resistor between the base and collector. The ratio is nominally 1:1 to give 2x Vbe but in practice will be less than this (Rbc < Rbe). This is known as the "Vbe multiplier".
John
Yes, I tried 2N3055 with MJE2955 in the simulator and they work fine. I prefer plastic transistors though, because they are easier to mount on heatsinks (I think).
The first picture below shows the problem. Transistors only start to conduct when the base-emitter voltage reaches about 0.7V so in this circuit there is no output when the input is less than +- 0.7V. Even with a large input, there is bad crossover distortion.
In the second circuit we fix this by adding the diodes. Diodes also start to conduct at about 0.7V so in this circuit the diodes make sure there is enough voltage between the bases of the transistors to turn them on even when there is no input. Now there is much less distortion in the output.
But there is still a big problem. The transistors may be conducting too much current. Also, when the transistors get hot, their switch-on voltage gets smaller so they conduct even more current, which makes them even hotter. Current and temperature rise until the transistor burns out. This is called "thermal runaway".
To stop that, it is a good idea to put small resistors in series with the emitters, like in circuits I posted before.
Attachments
It is the base current for the transistors plus the current through the resistors.
If you bootstrap the resistors with capacitors, it is less.
Yes! There must be resistors at the emitters to avoid thermal runaway.
Attachments
Hi RR
You have forgotten about the 100 ohm loads on the bases of the output transistors. They place a 50 ohm load on the BC547 driver.
To drive an 8 ohm load from +/- 5V you need up to 625 mA peak current. 2N3904,2N3906 don't do this much anyway, so they will need more drive current and maybe go pop!
Start with transistors which have gain at 625 mA, although there are some nice ones like ZTX851 a parallel pair of BC337/BC327 with 1 ohm emitter resistors to share the current will do quite well. Throw away the 500 ohm load, and take the collector of the driver to the cathode of the lower diode (=pnp base), then disconnect and remove the lower 100 ohm resistor, wire it in series with the other 100 ohm resistor and connect a bootstrap capacitor between the junction of these two and the output rail, and you might just get an amp that works.
John
You have forgotten about the 100 ohm loads on the bases of the output transistors. They place a 50 ohm load on the BC547 driver.
To drive an 8 ohm load from +/- 5V you need up to 625 mA peak current. 2N3904,2N3906 don't do this much anyway, so they will need more drive current and maybe go pop!
Start with transistors which have gain at 625 mA, although there are some nice ones like ZTX851 a parallel pair of BC337/BC327 with 1 ohm emitter resistors to share the current will do quite well. Throw away the 500 ohm load, and take the collector of the driver to the cathode of the lower diode (=pnp base), then disconnect and remove the lower 100 ohm resistor, wire it in series with the other 100 ohm resistor and connect a bootstrap capacitor between the junction of these two and the output rail, and you might just get an amp that works.
John
Check the voltage at the base and emitter and collector of Q3. Then maybe you will begin to understand. Also check the current through R5 and R6 and R2 and R3.
You will learn nothing if you only look at the output.
If you want to understand, you must look at all the voltages and currents in the circuit and think about why they are what they are.
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