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Multiplexing signals discretely

I have two output signals that are to use the same line. One should override the other, it is a master in a sense but isn't active most of the time. It does not matter if the master interrupts the slave.

Slave---+----Master | Out

Here is a picture of the setup with my solution.

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The p-ch fet cuts power so the master can pull low(not sure of the output impedance of the master). There is a signal when the master is active which I use to drive the gate. The switch "A" simulate this effect. The diode prevents driving the bjt when it is saturated and the master is active.

I do not like the diode voltage drop but using fets seem to make it much more complex.

Anyone have a better solution = fewer & cheaper parts?

Ideally I would like to simply put a fet in replace of the diode and remove the p-ch. That would be 1 part count. Unfortunately it would require a floating driver

Reply to
Joo Blow
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The hardest part of a solution is defining the problem. The nature of the signals, source/load impedance, signal fidelity requirements, etc.

Your schematic suggests that you don't care anything about the nature of V1 except the edges. And it doesn't matter what happens at the transitions between the two modes.

Sum the two signals thru resistors. Split the V2 resistor in half and ground the center thru a NPN transistor. Turn on the transistor when V1 is present. V1 shouldn't degrade V2 at all because when it's not present, it's not present. If there is signal on V1 when it's "not present", you'll need two split-resistor/transistor combinations.

If this doesn't work for you, you need to be more precise with your requirements.

Reply to
mike

Look at using an SSR. Art

Reply to
Artemus

The 2nd simplest option is to use 2 resistors. The master signal gets

1/10th the impedance of the slave one, thus wipes out the slave more or less when master active.

The simplest is to use one resistor, and let the inherent impedance of the slave signal act as the 2nd R.

No active parts needed. Of course whether these give sufficient performance depends entirely on information you've not told us.

NT

Reply to
NT

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