Digital Logic – Finding a Pass-Through Demultiplexer with Latch/Register

Your problem is your multiplex drive.

First, though, I have to point out that I believe your problem description is wrong. When you say, "the shift registers have no voltage across any of the output pins", I'm pretty sure that you mean that no LEDs are lighting, and that is not the same thing at all.

Second, your schematic is extremely misleading, since you show single LEDs in your displays. In fact, each display segment is composed of 4 pairs of LEDs connected in series, and this makes a big difference,

With that said, when you use NPN for your display selectors, you produce the following circuit for a selected segment:

^{simulate this circuit – Schematic created using CircuitLab}

Since Q1 is being used as an emitter follower, the LED Voltage cannot be more than about 4.3 volts, and so the LEDs are not lighting.

You need to replace your multiplexing transistors with an NPN-PNP pair, like so:

^{simulate this circuit}

Driving Q1 will cause base current to be pulled through Q2, turning it on. You do not need a 100 ohm in this case, 1k will be more than adequate, and the 2k base resistor will draw about 5 mA through the base, which will be more than adequate to turn on Q2 to the tune of 20 mA or less. You could actually use 5k and the circuit would work perfectly.

You may be looking for this:

CD4043/CD4044 3-state SR latch with common enable.

You might way to use the common enable in the CD4044 to implement the solution you're looking for. Connect all 4 R inputs to Vcc (HIGH), then use E as a common reset. For this to work you need a pull-down resistor on every output. Some MCUs inputs can be configured with internal pull-ups/downs, so you might be able to do this with no additional parts.

Most MCUs inputs can't be configured with internal pull-downs, only with pull-ups. So you may then want to consider another alternative. Use CD4043 instead of CD4044, then connect all 4 R inputs to GND (LOW) and use E as an active low common reset. You will then need pull-ups on every output instead of pull-downs, so just use the pull-ups of the MCU inputs by configuring it accordingly.

Why does this work?

If you look at the truth table of CD4044:

1. When E is HIGH it's equivalent to set all 4 R inputs to HIGH (= you get the same output from either E or R).
2. When E is LOW all the outputs will be in high-impedance (open circuit) and the pull-downs of the MCU will force a LOW on them regardless of the S inputs (= it effectively acts as a reset).

You can derive a similar deduction for CD4043.