supplying variable 12v supply help.

[scsn]

Hi,
I have an project where I have to control a power supply using a processor. I was wondering if the following would work?

The power supply I have to control is a DC 12v (11.3v will work) , running at 600ma when 12v. I have to change the voltage from 12v to 3v at different intervals. I thought of using a 2A NPN transistor with emitter following. I will have a d/a convertor which will produce 0-5v, which goes into an opamp and converts to 0-12v . Would this work?

I have attached a drwing of what I though might work, and the pdf of the npn I was planning to use. Any help would be appreciated.

Many Thanks
Scott

 

[alec_t]

I don't see how you can get the full 0-12V output for a 0-5V input with R1/R2 connected as shown. But the attached tweak should work.

 

[scsn]

Thanks Alec for that, cheers.

I made a mistake on my drawing, it should have been GND on the end of R1. And R1 would be 10k, and R2 14k. And I think 3v to 11.7v is still okay of an output to drive the power supply.

Would my design work with the mistake in my drawing changed? and would I need the 10k resistor?

As the supply I am driving requires 600ma, will this manage to supply this amount of current (with the 10k)?

Thanks
Scott

 

[alec_t]

Would my design work with the mistake in my drawing changed?

Unmodified it is doubtful, using your chosen transistor, because the specified minimum gain is 120 at 100mA but may (I'm guessing) be only ~60 at 600mA. That means you could be asking the opamp to provide ~10mA base drive current. You would have to choose the opamp carefully to make sure it can source enough current, or else connect another NPN transistor (low power) in a Darlington configuration. The Darlington option would, however, reduce the output voltage range unless the opamp feedback is taken from the final transistor emitter (as in my circuit suggestion).

would I need the 10k resistor?

No. I added that as a current-limiting safety feature but on re-checking the simulated circuit I find it's unnecessary and should be removed.

 

[scsn]

Thinking about the power supply I have to drive. If the power supply the output will drive will be up to 12v @ 600ma, then isnt this the same as placing a 20R resistor between the emitter and gnd (as R=v/I) ?

So would my revised circuit work?

Thanks
Scott

 

[alec_t]

Our posts crossed in cyberspace. Correct; 12V/600mA = 20 Ohm. But see post #4.

 

[scsn]

Hi Alec,
Thanks for the reply, and making it clear.

If I have the opamp feedback as per your drawing, wouldnt that reduce the current going to the power supply?

Thanks
Scott

 

[scsn]

Hi Alec,
Just looked at the op amp I will use (I have attached it), the max 9944, which specifies it can supply 20ma . Do you think I would get away with using the 1 transistor?

Thanks
Scott

 

[alec_t]

If I have the opamp feedback as per your drawing, wouldnt that reduce the current going to the power supply?

No. R2 and R3 are in parallel with the load (the power supply), not in series with it. The advantage of taking the feedback from the transistor emitter is that transistor base-emitter voltage variation with temperatur does not affect your output voltage. R2 and R3 need to be in a ratio of 7:5. Your choice of 14k and 10k will be fine.

Do you think I would get away with using the 1 transistor?

Yes. From the datasheet it looks as though that opamp will do nicely to drive the 1 transistor directly. Omit R1 in my schematic.

 

[scsn]

Thanks for the reply.

Am I right in thinking that as the power supply will be having a resistance of 20R (as above) , and the 2 resistors will have a total of 24k, then the current will go through the power suppy more than through the resistors (therefore the power supply will get most current, as some has to go through the resistors)?

Thanks
Scott

 

[alec_t]

Well the power supply presumably will be powering a variable load, so you can't really say it's resistance is 20R, except when it's operating at 12V and supplying 600mA.

therefore the power supply will get most current, as some has to go through the resistors

Correct. The current through the resistors is negligible (only ~ 0.5mA).

 

[scsn]

Hi Alec,
Thats great, thanks for all your help on this.. cheers

Neil