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TOPIC: Who right

Re: Who right 7 years 7 months ago #10068

  • Jon G
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Remember, I have a limited knowledge of this... However, based upon what I've seen so far it still seems as though it's not the ideal solution for switching a load. Though I do realize after some reading that this is a necessity for creating some amplifiers.

What I mean is that, while it might save some parts, the disadvantages seem to outweigh the advantages. Am I wrong? I've done some Charlieplexing and I don't understand, which the exceptions of parts savings, why it would be better this way?

Re: Who right 7 years 7 months ago #10072

  • MrDEB
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Have seen this done both ways and on another forum several years ago it was suggested to go with the load on the emitter to act as the base resistor.
Doing a simulation in TINA the same load etc the circuit A draws 10ma less current if anyone is interested.
Looks like either circuit will work but if concerned about current draw, circuit A might be circuit of choice??
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Re: Who right 7 years 7 months ago #10073

  • jmessina
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Doing a simulation in TINA the same load etc the circuit A draws 10ma less current if anyone is interested.
Maybe the load is the same, but since everything else is different it's kind of hard to compare the two.

In general, if you're talking about switching, it's usually a lot easier and more flexible to follow Mike's advice and put the load in the collector and sink current thru an NPN low-side switch to GND, or source the current thru a PNP high-side switch.

Using an emitter follower, try switching 12VDC into a relay using the output of a 5V uC pin to drive the NPN and see how things work. Using the same parts, putting the load into the collector works a treat.

If it makes it easier, when you go to use a transistor, just think of it as two diodes... for a C-B-E NPN transistor it's an NP (C-B) and a PN (B-E), and the other way around for a PNP.

To turn on a transistor, treat the junctions just like you would a diode and :
- forward bias the base-emitter (B-E) junction
- reverse-bias the base-collector (B-C) junction
When you do that, you get current flow thru C-E
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