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I'm in desperate need of help here! No matter what I try, I can't seem to get this working!

I'm trying to drive a signal transistor with PWM on an AVR ATmega32. My PWM signal looks as it's supposed to:

PWM signal from ATmega32. Sweep Time/Div.: .5 ms | 2 Volts/Div.

I have it setup so that I can vary the duty cycle of the signal, and my initial thought was, that I would be able to control the amount of current being piped through the transistor. The idea is to use the small signal transistor in conjunction with two MJ15003 power transistors for regulating power in a high current power supply.

The crude schematic looks something like this:

Circuit of high current power supply.

All devices in the circuit share the same ground, but sometimes when I measure the base voltage on the signalling transistor (TIP50) it just sky-rockets above 15V! What is going on!?

The MJ15003's require 5V @ 5A base current, so I figure my signal transistor is too small.

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    \$\begingroup\$ Are you sure about the middle and right transistor's base connected to +5V? Start by adding a 220 ohm resistor between the left transistor's collector and +5V, then attach both other transistor's base to collector. At least the circuit should start working, disregarding meeting its output current spec for a moment. \$\endgroup\$ – jippie Apr 20 '14 at 11:08
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    \$\begingroup\$ As jippie say - look at the circuit you have posted - it makes no sense - both the power transistors will be permanently on because they are fed from 5V on their bases and you may well have damagbed them by now. Also, showing us a picture of a perfectly normal PWM signal just takes up space and wastes folk's time. If you are going to show a picture, take one of the sky-rocketing TIP50 signal. \$\endgroup\$ – Andy aka Apr 20 '14 at 11:45
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The circuit does not work because you have the bases of the MJ15003's strapped to +5V. Also, the TIP50 can not supply enough current.

Also, you confuse the device's MAXIMUM ratings with the dynamic characteristics. MAXIMUM rates tell you when the device will release the magic smoke.

You do not specify the maximum current you want to draw, I am assuming the maximum for a MJ15003 (about 20A). From the specs, figure 1, I see that the MJ15003 Hfe (current amplification) is about 20 at 20 Amps, meaning the base-emitter current should be about 1A. Figure 3 tells me that the B-E voltage will be about 1.4 Volts.

A TIP50 can supply at most 1A, so it is not powerful enough for this application. It's Hfe is about 30, so to feed 2A as required by the 2 power transistors, it will draw at least 66 mA from the AtMega. This is more than it can source (40mA), so you will need an extra stage.

Have a look at the Darlington configuration for how to make a power amplifier. You will probable need three transistors, not just two, and use one TIP50 for each MJ15003.

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1 - As shown, with no load each MJ15003 will get a continuous base drive of ~ 40 amps. That's assuming they don't blow out and you have a really hefty 5v power supply. Assume 1 volt Vbe at these current levels. Then the base current is (5 - 1) / .1 . Assuming a 3.5 ohm load, to give you 5 amps load current per transistor, your base current will drop to 35 amps - assuming you don't blow out transistor, which you will.

So I don't think you want to want to use this schematic.

2 - I've no idea what you mean by 5V @ 5A base current. It's clearly not true. Let's assume that you want to pull 5 A through each output transistor, with the transistors saturated, for a total load current of 10 A. Then you'll want a base current of 0.5 amps per transistor. As configured, your TIP50 won't do that. Even if you drive the bases of the output transistors from the emitter of the TIP50, you'll reach the maximum current output of the TIP50.

I'd suggest you replace the TIP50 with a PNP such as a TIP42. Emitter to +5, collector to R7, then drive each output transistor base from the collector of the TIP42 via a separate 6 ohm resistor.

Of course, now your TIP42 base needs 100 mA of drive, so you'll need a predriver. Almost anything will work.

This is all just off the top of my head.

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