# Using transistors as SPDT switch?

• posted

Hello,

I would like to build the switch in the following circuit using small signal transistors:

5V +-------+ +---- GND | | | | | | o /o / / o | | .-. | | | | LOAD '-' | | | 12V ---------+

Thanks,

Jon

• posted

It's easier to visualize stuff like this if it's drawn with the current flowing down, namely with the +12 at the top.

John

• posted

signal

Sorry 'bout that. Thanks for the help; that is a very nice and simple solution.

Thanks,

Jon

• posted

In this circuit, it looks like current passes through the load in both the ON and OFF positions. Assuming an N-channel enhancement mode FET:

When switch is closed with application of Vg > Von to gate, FET does not have opportunity to conduct do to exponential characteristic of diode limiting drop to ground to approximately 0.7v, so current through load is [12-Vdiode)]/Rload.

When switch is open with application of Vg < Von to gate, FET is off, and current through load is [12-Vdiode]/Rload.

Assuming NPN BJT insted of NFET, you would also have to limit current through the base-emiitter diode.

Depending on your load, which probably has relatively high resistance, I don't see what a FET with a low Rds(On) wouldn't work.

-Le Chaud Lapin-

• posted

Just pretend you're a Physicist ;-)

...Jim Thompson

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|  James E.Thompson, P.E.                           |    mens     |
• posted

When the fet is on, the diode is back-biased and out of the circuit. The fet saturates as well as it can, so current is (12-Vsat)/Rload or maybe 12/(Rload+Rdson).

No, the current is (7-Vdiode) / Rload

The 5 volt supply does need to be able to sink Iload.

John

• posted

I thought the same thing until I noticed that the OP has a GND separating the 5-volt branch from the rest of the circuit, which if meant to be present, would make that part of circuit irrelevant, hence chance for misinterpretation.

I guess it would help to know whether the GND was meant to be there or not.

-Le Chaud Lapin-

• posted

Or an arabic engineer, with signals flowing from right to left!

I like to draw with positive supplies high on the page, and current flowing down. That places PNP emitters up and NPN emitters down. Signals flow to the right, feedback to the left.

John

• posted

I think it's a font problem. I think he meant...

5V +-------+ +---- GND | | | | | | o /o / / o | | .-. | | | | LOAD '-' | | | 12V ---------+

John

• posted

Yes, that is what I had meant to paste in originally, thanks.

Jon

• posted

[snip]

Same here, actually. I was just yanking your chain ;-)

I'm really picky about schematic format... I like to draw so that it's quite obvious how signals flow and how circuits work.

So I spend quite a lot of time cleaning up drawings to avoid clutter and confusion.

...Jim Thompson

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|  James E.Thompson, P.E.                           |    mens     |