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The purpose of the circuit below is to get a V_cathode of 2.5V (Vref) when VDC is higher than 10.4V approximately. The problem is that when VDC>10.4V the voltage checked in V_cathode is 1.9V instead of 2.5V.

3V3_UC_PIN comes from a microcontroller pin that "enables" this circuit. It this case, it is always enabled (+3V3).

I tried with a lower R1 (33 ohms) as well but the result is the same.

enter image description here

What could be the problem here?

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    \$\begingroup\$ That circuit won't do what you expect it to. \$\endgroup\$
    – Andy aka
    Mar 17, 2022 at 9:35
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    \$\begingroup\$ You are running the device open-loop. Read the datasheet to find the correct closed-loop configuration. \$\endgroup\$
    – user131342
    Mar 17, 2022 at 9:54
  • \$\begingroup\$ "when VDC is higher than 10.4V approximately" Could you explain this in more detail? \$\endgroup\$
    – devnull
    Mar 17, 2022 at 10:46
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    \$\begingroup\$ When run open-loop like this, the on-state anode-cathode voltage will always be around 2V. If you just want a 2.5V reference you could connect the top of R2 to the cathode pin, and change the resistor values; to include your 10.4V check, you'd need more circuitry. \$\endgroup\$
    – jayben
    Mar 17, 2022 at 10:48
  • \$\begingroup\$ Where is V_cathode going to? If back to the microcontroller, couldn't you use an analogue input to measure the voltage across the potential divider? \$\endgroup\$
    – HandyHowie
    Mar 17, 2022 at 11:24

1 Answer 1

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It is hard to know from the description what is problem this circuit attempts to solve.

This TI application report is very comprehensive and presents the following solution to use the TL431 as a comparator with practically rail-to-rail output, without the voltage excursion limits of the TL431 output:

enter image description here

This short article presents a different solution with hysteresis (positive feedback):

enter image description here

The transition voltages are:

\$V_{T+} = 2.5 \times \left( 1 + \frac{R_1}{R_3} \right)\$

\$V_{T–} = 2.5 \times \left( 1 + R_1 \times \frac{R_2}{R_1 + R_2} \times \frac{1}{R_3} \right)\$

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