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I was constructing the UC3843 on PSIM from scratch. The circuit diagram and the application note state that if the input voltage is greater than 34 V, the Zener diode does not allow the circuit to run. This is how it seems in the block diagram of the UC3843:

enter image description here

However, in my PSIM, the circuit still works normally when I design it like this:

enter image description here

This means that the output at that point is still VCC even if the VCC is >34 V. In that case, what's wrong here and how to modify this part so that the VCC >34 V will turn off the circuit just like the application note says?

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    \$\begingroup\$ Please quote what the website says verbatim. From memory the zener clamps and, if pushed hard enough, the zener burns then the internals of the chip burn. \$\endgroup\$
    – Andy aka
    Mar 24 at 18:40
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    \$\begingroup\$ It is a Functional Block Diagrams, not an electrical circuit. It is representative of how the chip works, not the exact details of how it does it's magic. \$\endgroup\$
    – StainlessSteelRat
    Mar 24 at 18:43

2 Answers 2

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The Zener diode is intended to clamp the supply voltage. You're supposed to place a resistor in series with VCC to limit the current flowing through the Zener diode.

In other words, if you supply the chip with more than 34V, it will pull the supply voltage down to 34V and turn the excess voltage into heat. It will still operate with more than 34V, but if you omit the series resistor, the chip will blow up.

The resistor required for 100V operation should be on the order of 47k Ohms (resulting in 2mA startup current).

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That circuit will blow up the Zener diode. You need something (typically a resistor) to limit the current from the 100 V source to the Zener diode, down to below the Zener diode's maximum current rating.

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