What makes the Zener diode work in the breakdown region?

Thread Starter

AIBang

Joined Feb 3, 2024
26
As seen below, there is -12V coming from the pack and 48V supplied from above. What triggers the D5 zener diode to operate?
Because if the D5 diode does not enter the breakdown region, M2 does not conduct. Even if M2 does not conduct, M1 remains in the cut-off.
If pack- 0V, D5 zener diode does not work in the breakdown region.

I have an idea but I want to consult you.


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Papabravo

Joined Feb 24, 2006
21,225
IIRC a diode intended for use as a zener is doped differently. That and zener diodes have their breakdown region characterized and documented. Stabdard diodes may exhibit zener behavior, but you don't know where in the I-V curve that will occur. You also don't know if the diode will "flame out" if reverse biased.
 

crutschow

Joined Mar 14, 2008
34,454
Diodes that can tolerate reverse-bias breakdown without damage are "avalanche rated" but their actual breakdown voltage is not characterized, expect that it is higher than their rated operating reverse voltage.
 

Thread Starter

AIBang

Joined Feb 3, 2024
26
Diodes that can tolerate reverse-bias breakdown without damage are "avalanche rated" but their actual breakdown voltage is not characterized, expect that it is higher than their rated operating reverse voltage.
IIRC a diode intended for use as a zener is doped differently. That and zener diodes have their breakdown region characterized and documented. Stabdard diodes may exhibit zener behavior, but you don't know where in the I-V curve that will occur. You also don't know if the diode will "flame out" if reverse biased.
Your explanations are not very clear to me. Could you please explain your answer to my question more clearly?
 

Papabravo

Joined Feb 24, 2006
21,225
As I understand it, you are asking about the operation of the posted circuit, not how diodes work, right?

@Papabravo
@crutschow
I understood from the title that he was asking about what makes the zener diodes work in the breakdown region. The question seemed clear enough to me and I have little to no interest in a paicture of a circuit. If the TS wants me to examine his circuit it would be better to include the file he is using so I can run the simulation to see what he sees. The picture is as useless as a screen door in a submarine.
 

Papabravo

Joined Feb 24, 2006
21,225
Your explanations are not very clear to me. Could you please explain your answer to my question more clearly?
OK, so you did not understand my answer. Doping is a process where a pure slab of Silicon (Si, Atomic number 14, Atomic mass 28.0885) is alloyed with elements from Group III and Group V of the periodic table. This proces turns the silicon alloy into two useful types of material called n-type silicon and p-type silicon. One of them has an excess of negative charge carriers in the valence band of the atom in the crystal lattice. The other type has a deficit of electrons which are characterized as "holes" in the valance band of the atoms on that lattice. All types of diodes are made from these two types of silicon alloys and it is the amount of dopant (the group III or group V material) that accounts for the different properties of different diodes.

This is not necessarily stuff you need to know to design and use circuits. I just thought it was the essence of the question in your title. Maybe constuct a better title next time.
 

Danko

Joined Nov 22, 2017
1,835
What triggers the D5 zener diode to operate?
Case 1:
Voltage on gate M2 relates to source M2 is +9.64 V.
M2 is ON.
Case 2:
Voltage on gate M2 relates to source M2 is -0.01 V.
M2 is OFF.

But Zener D5 in both cases is not conducting and is useless.

EDIT: Changes M1 to M2.
 
Last edited:

Papabravo

Joined Feb 24, 2006
21,225
When a zener diode is used in a circuit and it has a sufficien reverse bias, the diode will operate in the breakdown mode. Reverse bias is the condition where the voltage on the cathode is greater than the voltage on the anode. Below a certain threshold the diode will conduct a very small leakage current -- think nanoamps. Once the reverse threshold is reached the voltage will be nearly constant at the threshold voltage and the current will be several 10's of milliamperes.

The only other diode I know anything about is a poem about death.

ETA: As @Danko has pointed out, a diode which does not conduct, because it is neither forward biased nor reverse biased is indeed useless.
 

Thread Starter

AIBang

Joined Feb 3, 2024
26
When a zener diode is used in a circuit and it has a sufficien reverse bias, the diode will operate in the breakdown mode. Reverse bias is the condition where the voltage on the cathode is greater than the voltage on the anode. Below a certain threshold the diode will conduct a very small leakage current -- think nanoamps. Once the reverse threshold is reached the voltage will be nearly constant at the threshold voltage and the current will be several 10's of milliamperes.

The only other diode I know anything about is a poem about death.

ETA: As @Danko has pointed out, a diode which does not conduct, because it is neither forward biased nor reverse biased is indeed useless.
Case 1:
Voltage on gate M1 relates to source M1 is +9.64 V.
M1 is ON.
Case 2:
Voltage on gate M1 relates to source M1 is -0.01 V.
M1 is OFF.

But Zener D5 in both cases is not conducting and is useless.
So if that zener diode is not conductive, what is its function there? When you said that it works neither in the forward nor in the reverse region, I looked in the simulation and there is almost no current flowing through it. Even though the diode is not working, I can continue to receive a signal from my CHG_Wake label. When I increase the voltage of the V2 supply(that is a charger), the current can find mA through D5. Could you please clarify this?
 

Thread Starter

AIBang

Joined Feb 3, 2024
26
When a zener diode is used in a circuit and it has a sufficien reverse bias, the diode will operate in the breakdown mode. Reverse bias is the condition where the voltage on the cathode is greater than the voltage on the anode. Below a certain threshold the diode will conduct a very small leakage current -- think nanoamps. Once the reverse threshold is reached the voltage will be nearly constant at the threshold voltage and the current will be several 10's of milliamperes.

The only other diode I know anything about is a poem about death.

ETA: As @Danko has pointed out, a diode which does not conduct, because it is neither forward biased nor reverse biased is indeed useless.
Case 1:
Voltage on gate M1 relates to source M1 is +9.64 V.
M1 is ON.
Case 2:
Voltage on gate M1 relates to source M1 is -0.01 V.
M1 is OFF.

But Zener D5 in both cases is not conducting and is useless.
I added the ltspice file I was working on. I'm waiting for your help.
 

Attachments

BobTPH

Joined Jun 5, 2013
8,989
I understood from the title that he was asking about what makes the zener diodes work in the breakdown region. The question seemed clear enough to me and I have little to no interest in a paicture of a circuit. If the TS wants me to examine his circuit it would be better to include the file he is using so I can run the simulation to see what he sees. The picture is as useless as a screen door in a submarine.

Seriously, you think this is a a question about the semiconductor physics of Zener diodes:

As seen below, there is -12V coming from the pack and 48V supplied from above. What triggers the D5 zener diode to operate?
Because if the D5 diode does not enter the breakdown region, M2 does not conduct. Even if M2 does not conduct, M1 remains in the cut-off.
If pack- 0V, D5 zener diode does not work in the breakdown region.
 

Thread Starter

AIBang

Joined Feb 3, 2024
26
Yes it does. M2 gate is at ~0V and its source is at -12V. So Vgs=12V.
D5 ensures that Vgs does not exceed 18V.
Thank you. How can I mathematically explain the voltage drop across resistor R1? When Pack- = -12V, there is a voltage drop of over 1Meg, but I cannot explain this mathematically and electronically, can you help?
 

Alec_t

Joined Sep 17, 2013
14,318
How can I mathematically explain the voltage drop across resistor R1? When Pack- = -12V
R4 and R5 are in parallel, equivalent to one 5 meg resistor (Rpar).
R12 is negligible compared to Rcom+R1, so ignore it.
When M2 conducts its Vd will be very close to its Vs, i.e at -12V.
You now have a voltage divider R1, Rpar, with 48V at the top and -12V at the bottom.
I leave it to you to find from that the voltage across R1.
 
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