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I'm designing a circuit where there is an analog switch that turns a load on. I specifically need the load to turn on only when the voltage is above a certain level. The analog switch takes VCC and an enable pin that closes the switch when pulled high (0.7 x VCC) and opens the switch when low (0.3 x VCC.) Specifically, I want this switch to close when voltages are above 3.3V and open when below 3.3V.

A very simplified schematic is below, but effectively, I am using a Zener diode on the enable pin with a Zener voltage of 3.3V. My thinking is, when VCC > 3.3V, the Zener is in breakdown and the enable pin will see 3.3V (or around there accounting for minor voltage drop). When VCC < 3.3V, the pulldown will bring the enable pin low.

enter image description here

I'm definitely a bit iffy on this solution, though. I've not seen anywhere discussing the voltage at the anode of the Zener. I'm not sure if the Zener during breakdown will pass the full VCC voltage, or if it'll pass (VCC - 3.3V.)

I'm also aware that Zeners are not very accurate - but my application really does not need exactness, anywhere +/- 10-15% is fine. I just need some kind of "trigger" effectively.

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  • \$\begingroup\$ By "analog switch" do you just mean "mechanical switch"? Or do you mean something else? \$\endgroup\$ Dec 29, 2023 at 5:25
  • \$\begingroup\$ @MathKeepsMeBusy I mean an IC like the TI TS5A3159. ti.com/lit/ds/symlink/ts5a3159.pdf Seems kind of like a NC/NO relay that's not actually isolated and with very low enable current. \$\endgroup\$
    – Felix Jen
    Dec 29, 2023 at 5:32

1 Answer 1

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I'm not sure if the Zener during breakdown will pass the full VCC voltage, or if it'll pass (VCC - 3.3V)?

The voltage at the anode of the Zener diode will be \$V_{CC}\$-3.3 V, assuming \$V_{CC}\ge\$ 3.3 V.

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