In this last (long) weekend, not having handy any Darlington transistor I implemented this pair with discretes to PWM control a current up to 720 mA.

Having found the TIP41 hot to touch I recalled the necessity to facilitate the discharge of the internal capacity to reduce the rising time of the edge measured at the TIP's collector. Not sure of how to calculate that, I went by experimenting with progressively lower values.

From an initial rising time of 20 usec I came down to 3,1usec. Not bad.

Comments anyone?

 

Looks good to me. You can get some guidance by looking at the darlington transistors in the TIP line to see what resistors they use. I don't remember the collector current for a TIP41, but there probably is an equivalent darlington version.

Also, for possibly faster turn off, experiment with a small cap in parallel with R2. You might find that the cap lets you have a larger R2 for the same timing performance as your current resistor with no cap.

ak

 

If you really had 720mA and 10Ohms then the transistor is dissipating almost 3.5W. You could probably do just as well speed-wise and possibly better with a carefully chosen MOSFET and the power dissipation would be much lower as very low Rds values are available these days.

 

Have you checked the VceSat of the combined Darlington? It isn't very good (>2V), hence the high dissipation when turned on. A modern NFET does hugely better.

 

Looks good to me. You can get some guidance by looking at the darlington transistors in the TIP line to see what resistors they use. I don't remember the collector current for a TIP41, but there probably is an equivalent darlington version.

Also, for possibly faster turn off, experiment with a small cap in parallel with R2. You might find that the cap lets you have a larger R2 for the same timing performance as your current resistor with no cap. ak

Yes! I know I've read about that cap, maybe 20 years ago but in my mind I discarded it (confused it with the cap used in parallel with the feedback resistor in op amps).

Thanks. This afternoon I will give a try.

 

If you really had 720mA and 10Ohms then the transistor is dissipating almost 3.5W. You could probably do just as well speed-wise and possibly better with a carefully chosen MOSFET and the power dissipation would be much lower as very low Rds values are available these days.

What was burning my hand was the resistor. The TIP41 was just hot to touch. Later when I reduced the rise time it became a more or less warm transistor while the resistor kept burning my hand.

I am really aware about MOSFETs.

 

Have you checked the VceSat of the combined Darlington? It isn't very good (>2V), hence the high dissipation when turned on. A modern NFET does hugely better.

Sure Mike. Still interested to have this done given the time I spent in improving it.

 

You can fix the Vcesat problem by disconnecting the 3904 collector from the TIP41, and connecting it directly to the +12. Now the TIP41 can saturate. Should cut its dissipation by 75%.

ak

 

ak[/QUOTE]

You can fix the Vcesat problem by disconnecting the 3904 collector from the TIP41, and connecting it directly to the +12. Now the TIP41 can saturate. Should cut its dissipation by 75%.

ak

But if you do that remember to add a resistor in series with the collector of the 2N3904 to limit the drive current to the TIP41.

 

You can fix the Vcesat problem by disconnecting the 3904 collector from the TIP41, and connecting it directly to the +12. Now the TIP41 can saturate. Should cut its dissipation by 75%.

ak

That would then be an amplifier and not a darlington.

 

That would then be an amplifier and not a darlington.

I almost went into that level of detail, but I'm not in my normal environments these days. The result would be an emitter follower stage with current gain (but no voltage gain) driving a common emitter stage, in this case acting as a saturated switch. And because the output transistor is not operating in its linear region, the 3904 collector does not have to be connected to the load voltage, just something other than the output transistor collector that is high enough to supply the TIP41 base current.

ak

 

The circuit as below can be lower the Q2_Vce to ≤ 0.4V, I was calculated in the saturated status, Q2_Ib2 is 72mA and add I_R3 is 6mA, the I_R4 is 78mA, you can through your test to adjust the value, and the Q1 replace with 2N2222 or some other bjts has Ic over 5 times of 78 mA then it would be better.