Upon reading the data sheet this chip seems straight forward, but I'm having the hardest time hooking up the CD4052. The chip I'm using is actually a, "CD4052BE". Here is a link to the datasheet. I have hooked up the following circuit just to test the analog switching capability:


simulate this circuit – Schematic created using CircuitLab

Problem: There are two things happening that don't make sense to me when I test this same circuit:

  1. I started off with both A and B grounded. The DMM measurement reads the value 0.157V, where as when I measure the voltage at R1 (at the output for x0), it reads 0.1179V. That is actually where I would expect it to be. Why is there a voltage difference? It seems like it should just be similar to an open wire between the two.
  2. I soon realized that no matter what combination of A/B inputs I used, the voltage is ALWAYS as described in problem 1. It is as if the multiplexer isn't switching between x0, x1, x2 or x3. Why would this multiplexer appear to be stuck on a particular switch regardless of a/B inputs? While R1 reads 0.1179V, I would expect that the other combinations of A/B would yield: 1.09V at R2, 6V at R3 and 10.9V at R3. The R2, R3 and R4 all read 0v.

Please let me know if something about this seems incorrect or is I just flat out misinterpreted the use of this chip. I reviewed another post that references the CD4052 here, but it didn't discuss the specific issues I was having.


Your input logic to A,B should use the V1 voltage (12v) for a high, not 5v.

See page 7 of the data sheet for VIL, VIH.

For example, it shows that with Vdd at 10v you need at least 7v for a high. So at 12v Vdd you may need about a 9v minimum.

(Also see the switching limitations listed by jms.)

  • \$\begingroup\$ If you need to switch the circuit with only 5v logic you can use an NPN transistor for each input, with a resistor on each collector pulled up to the Vdd level. Then send 5v signals into the base of each transistor (with a series base resistor), and the emitter is connected to GND. (The control signals will be inverted from the base input to the collector output connection.) \$\endgroup\$
    – Nedd
    Mar 9 '16 at 4:36
  • \$\begingroup\$ It never occurred to me that logic high might be relative to Vdd. When I attached the A/B logic inputs to 12v instead of 5v, the switching occurred as originally expected. Thanks \$\endgroup\$ Mar 9 '16 at 11:14
  1. The analog MUX isn't equivalent to a short. It is most likely implemented with pairs of back to back MOSFETs as the switches, and they have a significant on resistance of 125 ohms or more. This resistance adds to the 1 k external resistance and causes the voltage drop you saw.

  2. I dont know why, maybe it fried?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.