How to reduce voltage from shift register to ESP32 MISO pin?

Question

In this circuit I have sensors that are rated for a minimum of 6V connected to a 74HC165 which passes its output to the MISO pin of an ESP32. I was hoping a simple divider to convert the 6V from the shift register to 3.3V for the ESP32 would work ... but no.

The ESP32 is unable to process the input consistently, delivering erratic output results. Interestingly, it all works perfectly if I remove the resistors for the voltage divider and power it all with 5V. The sensors can apparently handle the lower voltage fine ... but they need to be operated at 6V minimum.

Should I use a linear regulator or switching regulator instead of a voltage divider? Or is there something else I'm missing?

Much thanks,

Answer 1

Chances are high that in addition to feeding 6V into a pin that's not even 5V tolerant, you're trying to run a 74HC part with woefully inadequate pin voltages.

The easiest solution is the proper voltage divider, run your sensor at 5V, and use a 74HCT165 (note the 'T' in the part designation). The 74HCT line is made to run from 5V and accept TTL input voltages -- which are, happily, about the same as what comes out of CMOS being driven from 3.3V.

If you absolutely must run your sensor at 6V, then you'll need proper voltage translation between it (or them) and your 74HCT165.

For transitions that are slower than about 100Hz or so (and maybe up to the low kHz), this should work well -- and you can collect Q1, R1 and R2 into a pre-biased transistor, for more savings.

• Choose R3 to be around 1k\$\Omega\$ to 10k\$\Omega\$
• Choose R1 and R2 so that if they were an unloaded voltage divider, they'd have a voltage of around 1.4 - 2V when the sensor output is on (i.e., \$\frac{R1}{R1 + R2} \simeq \frac{1.4\mathrm{V}}{V_{out}}\$). That uses the transistor's Vbe to make a really soft threshold of around half the sensor output voltage.
• Choose R1 and R2 so that their parallel equivalent is equal to about five times R3 or less (i.e., \$\frac{R1\,R2}{R1 + R2} \le 5 R3\$). That makes sure that the transistor is well into saturation when its on.
• Don't sweat getting the numbers for R1 and R2 exact.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 2

You have to use a logic level shifter IC. You can use a dedicated IC which can translate the signal from one level to the other. you can also use the below circuit if there is only one signal to convert from one voltage level to the other.
The MOSFET can be logic N type MOSFET. Image and many other solutions from this link: https://next-hack.com/index.php/2017/09/15/how-to-interface-a-5v-output-to-a-3-3v-input/

As other answers have mentioned, there is also a need to check once the overall design and supply voltage level options to make sure you are following the recommended specification from the respective datasheets. The particular sample may work, but not always.

Answer 3

While the datasheet is not clear on what the absolute maximum ratings for the IO's are, I am certain that it's 3.6V. Usually "+0.3V" means that the inputs are protected with a diode that turns on at Vdd+0.3V. So this means don't tie Qh directly to the port of the ESP32, it could burn out the diode.

Source: https://www.espressif.com/sites/default/files/documentation/esp32-wroom-32_datasheet_en.pdf

The resistor divider isn't working because it's not dividing the output voltage from Qh, it needs to look like this:

The Voltage output specs for the SN74HC165 are Voh = 5.99V and Vol = 0.1V, after the resistor divider this would be 2.7V for Voh and 0.045V for Vol, using a resistor divider is compatible range of the ESP32's Vih of 2.475V and Vil of 0.825