# Zener Diode and comparator circuit question

#### cgha20@yahoo.com

Joined Oct 21, 2009
82
I am having a problem understanding the following circuit. The input across C1 should be between 22V and 36V. I am not sure how the zener is working in this circuit. It is a 15V zener but ive never seen a resistor at the anode this way, which is setting the comparator noninverting pin. I also am not sure what R9 and r11 is doing. basically the pmos needs to switch on after 22V. Any help would be appreciated

#### AnalogKid

Joined Aug 1, 2013
10,224
R9 assures rapid turn-off of Q1 by providing a discharge path for its gate charge.

NOTE: This analysis does *not* take into account the hysteresis created by R2.

D3 subtracts 15 V from the input, reducing it so it is within the common mode input voltage range of the comparator. Thus, the voltage at pin 2 is always 15 V less than the input. OTOH, the voltage at pin 3 is always 33% of the input. As the input voltage decreases, the voltage at pin 2 decreases faster than the voltage at pin 3. At an input of 22.5 V, the two inputs are equal. Below this point the output, goes high, turning off Q1.

With R2 in the circuit there are two transition levels (trip points), one when the input is decreasing, and one when it is increasing after the circuit has tripped. The math is a bit more messy, but it still is just series and parallel resistors plus Ohm's Law.

ak

#### DickCappels

Joined Aug 21, 2008
9,624
The Zener D3 needs several milliamps. The test current is 10 ma. With R4 being 100k you will not have enough current to get the Zener to be a Zener. I think you have the same problem with D4. Something more like 4.7k or even lower would give the circuit a much better chance of working.

#### cgha20@yahoo.com

Joined Oct 21, 2009
82
R9 assures rapid turn-off of Q1 by providing a discharge path for its gate charge.

NOTE: This analysis does *not* take into account the hysteresis created by R2.

D3 subtracts 15 V from the input, reducing it so it is within the common mode input voltage range of the comparator. Thus, the voltage at pin 2 is always 15 V less than the input. OTOH, the voltage at pin 3 is always 33% of the input. As the input voltage decreases, the voltage at pin 2 decreases faster than the voltage at pin 3. At an input of 22.5 V, the two inputs are equal. Below this point the output, goes high, turning off Q1.

With R2 in the circuit there are two transition levels (trip points), one when the input is decreasing, and one when it is increasing after the circuit has tripped. The math is a bit more messy, but it still is just series and parallel resistors plus Ohm's Law.

ak
Thank you for the very clear explanation. For some reason, the circuit is switching on at 29Volts instead of 22Volts. Do you think this is related to the hysteresis?

#### crutschow

Joined Mar 14, 2008
31,535
Do you think this is related to the hysteresis?
Yes.
R1, R2, and R3 generate hysteresis, raising the trip point about 7V when the op amp output is high.

If you want less, increase the value of R2.

If you want a lower trip point with only a small change in the hysteresis, reduce the value of R1.