I'm currently designing a circuit that uses only one bit of a 74HC157 quad 2-to-1 multiplexer, like so:

However, I'd like to get rid of this chip. Since the B input can be tristated (it's coming from a latch), I was wondering if I could set it up like this:

I've seen this circuit used before, and it seems to do the right thing in Falstad's Java simulator, but I'm curious about a couple things. Wouldn't current actually flow back into A or B when B is enabled? Can they tolerate this? (A and B are both outputs from other 74HC-series ICs) Would I need diodes? How big should the resistor be?

 

What is the part number of the tristate circuit?

 

The tristate circuit is part of a 74HC573. (B is coming from one of the outputs, and S is the input to ~OE)

Can a few microamps of current flowing into an output pin damage the chip?

 

The tristate circuit is part of a 74HC573. (B is coming from one of the outputs, and S is the input to ~OE)

Can a few microamps of current flowing into an output pin damage the chip?

With a 5V supply, you can use a resistor as low as 1k (≈5mA) in your mux circuit and still get good logic levels at the output, and it will not damage anything. See the HC573 datasheet and look under Static Characteristics. You will find that, when vcc=4.5, you can drive up to ±6mA, so 1k is a conservative value. Higher resistor values will result in slower rise/fall times and more delay.
One disadvantage is that the resistor will draw current from Vcc even under static conditions if A and B are opposite logic levels. This won't happen with a CMOS mux.

 

Thanks, Ron. I'm more concerned with minimizing board real estate than static current draw, since I'm already using some old chips that draw plenty of current. Still, I'll do some experiments to see how large I can make the resistors.

 

Ah, I understand the situation better now. (It's been a while since my last electronics class...). An IC output pin is capable of sourcing and sinking current. I just have to pick a resistor that keeps source/sink current sufficiently low. However, it looks like this might make the signal delay too long, so I'll probably just use the CMOS mux.

 

Ah, I understand the situation better now. (It's been a while since my last electronics class...). An IC output pin is capable of sourcing and sinking current. I just have to pick a resistor that keeps source/sink current sufficiently low. However, it looks like this might make the signal delay too long, so I'll probably just use the CMOS mux.

Unless you are running at a high clock rate, the technique should work.
As an alternative, you can make a single 2-1 mux from a quad NAND package, but it will simply replace a 16 pin package with a 14 pin pkg.