EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: Falcon69 on November 12, 2013, 02:58:08 am
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Hello,
I am trying to create a very simple circuit.
I have a 12V source that goes into 2 diodes into series, then out to the load.
Then I have a 5V source that goes into a single Schottky Diode, then out to the same load.
I need to be able to switch between them with a single jumper. I don't have much space to do this on the circuit board, so the less components, the better.
I've heard this can be done with a PNP and an NPN transistor, but can't figure it out.
Can someone here help me out?
Thanks in Advance!
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You want both power supplies to be connected at the same time, and then choose which one is used?
The simplest solution is the diodes you already have. Turn on the 12 V supply and it will be used. Turn off the 12 V supply and the 5 V supply will be used.
If you want the 5 V supply to be used when both supplies are energized, connect the jumper as a switch in the 12 V supply path. Remove the jumper and the 5 V supply will be used.
Why does it need to be made complicated with transistors?
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no no no
One supply source, but Depending on which supply source is used from the other electronic device, will decide which diodes are used.
In a sense, I need to create a SPDT switch using a single jumper. I don't have the room for the switch, but I may for a few small SMD components and a jumper.
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Three jumper terminals...common in the middle terminal...12V and 5V on the ends.
In other words, create an SPDT switch with jumpers.
What are you doing, anyway?
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I don't have the room to add a 3 pin jumper. The jumper needs to come out the back of the switch, but the switch height is only 6.35mm. Not high enough to do a 3 pin jumper at a right angle.
A limit switch for a cnc machine. Trying to make it work for different breakout boards, some are TTL, some are Opto-Isolated, some are just plug and go.
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Well, if the Schottky and the other two diodes are wired parallel to each other, the drop across the schottky will almost certainly be much less than across the other two in series, so the other two will never conduct. Put the jumper in series with the Schottky. For 5V operation, leave the jumper in. For 12V, take it out.
I think that works, right?
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Well, if the Schottky and the other two diodes are wired parallel to each other, the drop across the schottky will almost certainly be much less than across the other two in series, so the other two will never conduct. Put the jumper in series with the Schottky. For 5V operation, leave the jumper in. For 12V, take it out.
I think that works, right?
Seconded. I was just about to write exactly the same thing when I saw John's comment.
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Simulator checks out. Don't know why I didn't even think of that. Number one rule of electricity, it takers the path of least resistance. DUH! :palm:
Thanks Guys, that should work just fine. Now to figure out how I'm going to put that jumper on the circuit.
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What are the diodes for? Just curious.