Bypassing and Switching with the CD4053 CMOS Analog MUX

Copyright 2000 R.G. Keen. All rights reserved. No permission for local copies or serving from web pages other than

I get a lot of questions on how to do an electronic bypass that approaches the performance of a true bypass mechanical switch. This has been hard in the past due to limitations of the switching elements available and a lack of understanding of how to use them. Here's an electronic switch that approaches the performance of mechanical true bypass switches. It's cheap and easy. Try it!

The basis for the switch is the CD4053 analog multiplexer IC. This is a CMOS logic chip that contains three SPDT CMOS analog switches and the attending control logic to make them work. People who have tried these before have often complained that they pop or distort, and have then moved on and forgotten them.

There are some tricks to using this chip. First - the analog inputs and outputs need to be held somewhere near the middle of their power supply. The 4053 does have a "Vee" supply pin that lets it control signals near 0V, but that is often a fair amount of trouble to use. the simplest thing to do is to AC couple the in/out pins and bias them to the middle of the 4053's power supply. In a +9V system, this only needs a two resistor divider and a cap, then a resistor and capacitor per in/out pin.

Yes, this is a pain, eats up board space, and is complicated. It works, though. When biased this way, the CD4053 has little or no distortion and no perceptible switching pops - exactly what we want.

The three SPDT switches are independent. As noted in the illustration, when "A" (pin 11) is low, "X" (pin 14) is connected to "X0" (pin 12). When "A" is high, "X" is connected to "X1" (pin 13).  "B" and "C" do the same switching control for "Y" and "Z". "Low" means below 1/3 of the Vdd voltage, or 3V in a 9V battery system. "High" is over 6V. Like all CMOS, it's important to NEVER leave an input pin unterminated, so all control pins that are not being used must be tied to ground or some definite logic level.

When a switch is connected, it looks like a moderately nonlinear resistor between pins. The resistance varies with the signal voltage from pin to pin, which really means that for low distortion, you have to keep the current through the switch low. The switch looks like a 120 to 500 ohm resistor. So if it's switching things that are themselves 10's of K ohms or higher, the distortion will be low.

So - let's make a bypass switch.

In the illustration at right, I've shown the basic hookup for a bypass switch. We don't need one of the three SPDT's, so I've tied "C" to ground and left the switching pins open. The other two are hooked up in the old standard "true" bypass fashion. Notice that it would actually be better to use one of the alternate true bypass setups to avoid the signal passing through two of the switches when bypassed, but this hookup does work OK in most cases.

The in/out pins are all biased to half the 9V supply; notice that the "bypass" path from pin 12 to pin 2 is only biased once, and needs no capacitor, so we save some parts there over the general case.

The input/output capacitors of the effect can also be presumed to be adequate to keep the bias DC value out of the effect, so we save two more caps there.

All we need now is a control signal. I've hooked the A and B controls together. When both are driven high, the effect is heard. When both are low, the effect is bypassed. Notice that if we had other uses for the third switch in the CD4053, we could use that as well, for either AC or DC switching.


A simple and easy way to control a CMOS logic chip is with another one. The CD4013 is a very cheap, highly available dual D flipflop. We can use one section of it to control the 4053. It's guaranteed by design to drive the two logic inputs properly.

We use a momentary switch to ground the base of an ordinary transistor. This turns the transistor off, and a capacitor on the base of the transistor prevents any switch bounce from tripping the flipflop more than once. The transistor's collector goes high cleanly, clocking the flipflop. The D input to the flipflop is connected to the "-Q" output, which is the opposite of the logic state the flop was clocked to the last time. Hooking it up this way means that every clock pulse it will switch to the opposite state from where it is sitting now.

From there, we just connect the "Q" output to the A and B control lines, and the bypassing works. We'd like to have an indicator of the status, so we add a transistor to buffer the Q output so when it's high, an LED is lit.

But there are also many ways to do this.

Here are two more. On the left, I've used a CD4049 hex inverter IC to make a toggling flipflop and an LED driver. The two inverters on the left form a latch. When the power is turned on, one of them will come up a little bit before the other one, and so its output will driver the input of the other inverter high, which drives its output low, which reinforces the higher inverter's output at being high. The thing is guaranteed to lock up with either pin 6 or pin 4 fully logic high, and the other one low.

If we guess that pin 6 comes up high, pin 4 will be low. The 0.1uF capacitor charges to +9V through the 100K capacitor. It's perfectly happy to stay this way forever. However, if we press the momentary switch that connects the capacitor to pin 7, the +9V on the capacitor pulls pin 7 high. The 100K resistor from pin 4 to pin 7 prevents the logic low output at pin 4 from fighting off the capacitor, so pin 7 goes high. That forces pins 5 and 6 low, which makes pin 4 high, and now pin 4 is holding pin 7 high also. All this happens in the first 100nSec or so when the switch is pressed. Switch bounce can only reinforce the action. Both inverters change state, and hold each other in the new condition. The switch opens before the capacitor can change voltage much, and the capacitor begins charging to 0V through the 100K resistor. The flip has flopped.

We have four more inverters, might as well use them. There is no reason to choose the high or low state of pin 4 for effect on or off, so we can just attach two of the four remaining inverter inputs to pin 4; these are pin 3 and pin 14. Pin 2 then drives the A and B control signal, and pin 15 drives a higher current LED.

And manual non-momentary switches also work. The circuit on the right shows how to provide the control signal and drive an LED with a single switch.