# Level signal compatibility?

Discussion in 'General Electronics Discussion' started by 24Volts, Jul 2, 2014.

1. ### 24Volts

164
0
Mar 21, 2010
Hello,

I have an output from a quad 3 state buffer chip 74LS126 (Which operates at 5.0Vdc and its output will most likely be 5.0VDC when high) and this output needs to be fed into an input of a PIC32 MCU which its maximum input voltage tolerance is approx. 3.6VDC. Assuming the MCU takes in 3.3VDC, there is a voltage level difference of 1.7VDC.

Can I simply use a resistor to drop the 1.7 volts from the output of the 74126 to accommodate the 3.3 Vdc input of my MCU?

Thanks for all feedback!
8v

2. ### Gryd3

4,098
875
Jun 25, 2014
If all you need to do is read the 5V signal, you could use a voltage divider.
Using a resistor inline would be difficult to find the correct value, as you do not know the current flow to calculate resistance with, and I assume you do not know the output / input impedance of your devices.
You could also use a zener diode to take care of the excess voltage.

You have a couple simple options, this gets a little more complicated if you would like two way communication between the devices.

Edit: Link below does not show the simple 5V to 3V level shifting...
Here is an in-depth doc that will give you all sorts of info and examples:
http://www.nxp.com/documents/application_note/AN240.pdf

More Edits:
Resources from the web...
http://www.daycounter.com/Circuits/Level-Translators/Level-Translators.phtml

Last edited: Jul 3, 2014
3. ### KrisBlueNZSadly passed away in 2015

8,393
1,272
Nov 28, 2011
No, it most likely won't. The transistorised versions of the 74nn logic family (74nn, 74LSnn and a few others) use "totem pole" outputs and do not pull high strongly. This is why the low and high voltage levels for these families are skewed towards 0V - they are typically 2.4V and 0.8V - rather than being symmetrical, as they are for CMOS logic.

Probably your best approach would be a series resistor (say 1k) between the LS output and the MCU input, and a 3.3V 0.5W zener diode (BZX79C3V3 or 1N5226B) with its cathode to the MCU input and its anode to 0V.

Gryd3 likes this.
4. ### Gryd3

4,098
875
Jun 25, 2014
Exactly why I like the community already. The hands-on experience of others.
Take my voltage divider recommendation with a grain of salt. I was not aware that little quark, it would have thrown me off.

5. ### Harald KappModeratorModerator

12,330
2,930
Nov 17, 2011
Not exactly simple but effective is this circuit:

When the 7xLSxx output is low, the transistor is turned on and saturated, meaning the voltage drop from collector to emitter is ~0.1V. Therefore a good low signal is fed to the MCU.
When the 74LSxx output is high, the transistor is turned off. The voltage on the MCU's input is set by R2 to be 3.3V, a well defined high level.

Of course. more ressource intensive in terms of component count and PCB area...

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6. ### 24Volts

164
0
Mar 21, 2010
wow.... I have pretty much always used a logic
probe when checking logic outputs of my Ttl
chips and assuming that a high was 5vdc.

Thankyou all for your help !!!
24V

7. ### KrisBlueNZSadly passed away in 2015

8,393
1,272
Nov 28, 2011
Logic probes designed for TTL/LS used those same levels - 2.4V and 0.8V. Some logic probes had a switch to select between TTL/LS and CMOS thresholds.

When feeding a signal from TTL/LS to CMOS in a fully 5V system, a pullup resistor (typically a few kilohms) is used to pull the TTL/LS output properly high.

This is one of many good reasons to avoid TTL/LS.

8. ### 24Volts

164
0
Mar 21, 2010
cool KrisBlueNZ !!

thanks for your valuable insight!

24V