I have an SN74HC595N shift register that I want to have output 5V on the Q outputs. If I power it with 5V on VCC, can I tie the SER, SRCLK, etc. pins directly to the 3.3V digital outputs on my microcontroller? I wasn't sure if I needed a transistor between each microcontroller pin, or if the lower voltage would be acceptable directly to the IC.


3 Answers 3


It is not guaranteed to work. You could use a level converter chip xx145, or you could use a 74HCT595, which will accept valid TTL levels at the inputs (and a 3.3V CMOS chip will give you that).

Edit: The margins are fine for 3.3V input, as alexan_e mentioned. See below.

enter image description here

Note that you should not connect any outputs form the HC595 or HCT595 directly to the 3.3V circuit.

A simple single-transistor level shifter was discussed here:

If you use a MOSFET you don't need the base resistor, so it can be even simpler.

  • \$\begingroup\$ You mention the 0.5v and 3.15v levels in a way that seems to suggest that they refer to the 74HCT595 ("the chip I linked to"). This was probably not the intention so I think it would be better to clarify this by mentioning that these refer to 74HC595 and to add the 74HCT595 transition levels ( <0.8v for LOW and >2v for HIGH ) that will work fine with 3v3 signals. \$\endgroup\$
    – alexan_e
    Mar 4, 2014 at 18:56
  • \$\begingroup\$ Thanks for the answer - the HCT version seems like it would be the easiest way to accomplish this. On the other method, instead of a level converter, could I use a transistor with the collector tied to 5V, the emitter tied to the '595 pin, and the base connected to the 3.3V out from the MC? Thanks again. \$\endgroup\$
    – user38201
    Mar 4, 2014 at 19:46
  • \$\begingroup\$ @alexan_e Thanks very much for catching that, I'll edit my answer appropriately. \$\endgroup\$ Mar 4, 2014 at 19:48
  • \$\begingroup\$ @user38201 The HCT is the best and easiest way. I've edited the answer to include a reference to a single transistor level shifter. \$\endgroup\$ Mar 4, 2014 at 19:55
  • \$\begingroup\$ Thanks - that link is helpful as well. I'll probably go with the HCT solution. \$\endgroup\$
    – user38201
    Mar 4, 2014 at 20:04

I tested this last night to see what would happen. Here was the set up I used.

5 VDC from external power supply to bread board positive rail. Ground from same power supply to negative rail.

NodeMCU Dev board 2.0 powered via USB cable (HiLetgo brand).

  • GPIO14 (Pin 5) (3.3 VDC) to SN75HC595 SER (PIN 14)
  • GPIO0 (Pin 18) (3.3 VDC) to SN75HC595 ST_CP (PIN 12)
  • GPIO2 (Pin 17) (3.3 VDC) to SN75HC595 SH_CP (PIN 11)


  • Connected to NodeMCU (Listed above)
  • VCC (PIN 16) to 5 VDC power rail
  • MR (PIN 10) to 5 VDC power rail
  • OE (PIN 13) to Ground rail
  • GRD (PIN 8) to Ground rail
  • Q0 though Q7 to positive legs of LEDs (Resistors to ground.)

When pushing a simple loop (for int i = 0; i < 255; i++) the led output acted erratic. LEDs would light up out of order and with different results than what the micro-controller was instructing it to do.

I then started pushing the same value with a 100 ms delay. At first I was pushing the value 255 (b11111111). All the LEDs stayed on. I then switched to 0 (b00000000) with a 100 ms delay. Sometimes the lights would stay off, sometimes they would all come on, sometimes only a few would come on.

I'm not sure why the IC was reacting this way. When I removed the 5 VDC power supply from the breadboard and used 3.3 VDC the Shift register, LED outputs acted as expected. I got the proper results each time I latched the IC.

So the conclusion I came too was to use the same voltages for all inputs of the SN74HC595 shift register. If you need to up the voltage of Q0 - Q7 do so after via level shifting.

I hope this helped.


You might be able to make this arrangement work just by adding pullup resistors from the '595 inputs to +5V. However, you can only add these pullups if your particular microcontroller has 5-volt-tolerant I/O pins. The \$V_{IH}\$ level for the '595 is too high for a 3.3V processor output, but adding pullup resistors may allow this to work for you. I wouldn't recommend this for a production design, however.

Edit: I was assuming that you really had the HC version of the '595. If you have the HCT version instead then please see the comment by @alexan_e instead.

  • \$\begingroup\$ Thanks for the answer. My microcontroller is actually 5V tolerant, so I could do it this way. I guess it wouldn't be the best way though. \$\endgroup\$
    – user38201
    Mar 4, 2014 at 19:42

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.