# Anyone explaination how these two diodes contibute to the solution. (Reversed Engineered)

#### Calltronics

Joined Nov 20, 2020
19
What possably could be the purpose of this zenor and diode set up.
I am trying to reverse engeneer an obsolete power monitor board which is part of a larger control solution.

#### AlbertHall

Joined Jun 4, 2014
12,269
Impossible to say with so little information.
For starters, what is the rail at the top of the diagram connected to, and what is the coloured box surrounding D5?

#### Calltronics

Joined Nov 20, 2020
19
Impossible to say with so little information.
For starters, what is the rail at the top of the diagram connected to, and what is the coloured box surrounding D5?
The zener diode is depicted in its SOT23 packaging with the spare not connected terminal.
Say for example - the rail at the top is an ac supply rail with a voltage of 30v peak.

#### AlbertHall

Joined Jun 4, 2014
12,269
On the positive half-cycle of the supply, the diode wil be reverse biased so no current will flow and there will be no voltage across the zener.

On the negative half-cycle the diode will be forward biased and will conduct the supply voltage to the zener. The zener diode will be reverse biased and will conduct when the voltage across it reaches 7.5V. The two diodes will conduct and above about 8.5V will conduct as much current as the supply is capable of - there is no resistor to limit the current. So if the supply is capable of a large current then both diodes would be destroyed.

If there is nothing else connected to the junction of the two diodes then I can see no useful purpose for this arrangement. Did you invent this circuit or did you see it printed somewhere?

#### Calltronics

Joined Nov 20, 2020
19
On the positive half-cycle of the supply, the diode wil be reverse biased so no current will flow and there will be no voltage across the zener.

On the negative half-cycle the diode will be forward biased and will conduct the supply voltage to the zener. The zener diode will be reverse biased and will conduct when the voltage across it reaches 7.5V. The two diodes will conduct and above about 8.5V will conduct as much current as the supply is capable of - there is no resistor to limit the current. So if the supply is capable of a large current then both diodes would be destroyed.

If there is nothing else connected to the junction of the two diodes then I can see no useful purpose for this arrangement. Did you invent this circuit or did you see it printed somewhere?
Ahhhhh, damm old fool never thought that the rail would be negitavly biased. Oh boy am I getting old.
So all negative votages on the top rail will be clamped to around -8.5V, given that there is resistive load on the top rail!
Thank you AlbertHall for that heads up, appreciated.

#### AlbertHall

Joined Jun 4, 2014
12,269
Yep.

#### Calltronics

Joined Nov 20, 2020
19
Is there a way to PM you?

#### AlbertHall

Joined Jun 4, 2014
12,269
Is there a way to PM you?
There is a 'conversation' facility on AAC but it would be better for any discussions to be on the forum. I have only one little brain with knowledge of a few areas, the forum has many brains with knowledge over a much wider range of topics.