# Make before break circuit

#### acheriti

Joined Apr 26, 2020
41
@AnalogKid @Ian0
Thanks a lot for your replies guys. Can you tell me if my understanding of the circuit is right ?
Case 1 : 50V only => D6 is off, thus D11 sees 50V which is higher than the breakdown votlage. D11, will conduct, and D7 will see the GND therefore will not conduct. At the mosfet Gate there is GND because of the conduction of D11 , therefore VGS=-50V and the mosfet will conduct the 50V
Case 2 : 24V bypasses the 50V => D6 conducts, therefore D11 sees (50-24)V which should be smaller smaller than the breakdwown voltage.The mosfet gate sees 50V , VGS=0V and thus no conduction of Q1 occurs. The 24V passes through D7

understanding of the simulation :
My diode D11 seems to behave in a weird way, conducting when it should not so this is the component I should check to change for making my simulation work.

Thanks for your time guys !

#### Ian0

Joined Aug 7, 2020
9,519
Given that D10 has a resistor across it and D11 does not, it doesn't make sense that of the two diodes in series, it will be D11 that somehow maintains 24 V across it.
It makes perfect sense. As the voltage across the two zeners and resistor increases, no current will flow (assuming a perfect zener) until the voltage reaches 24V, because D11 has no shunt resistor. Between 24V and 36V there is a current path through R5 and D11. When the voltage reaches 36V both zeners conduct.

As an alternative, Try this circuit using an optoisolator. (Needs CTR>50%)

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#### MisterBill2

Joined Jan 23, 2018
17,855
If therenismno current drawn by the load in the simulation then any leakage at all will deliver a lot of voltage. Try simulating the load with a resistor that will draw the same current.
And I am still asking what voltage does the useful part of the load require? That really does matter a bit as far as options go.