# Zener Diode Power question

#### alanj100

Joined Apr 26, 2019
19
SMD 5V6 Zener diode, The Zener diode I have chosen is a 5v6. The diode has an absolute max rating of 300mW.
If a dc level of 12V is put across it I get a power of 273mW across the diode.
So my question is will the Zener get too hot and burn out if I leave the 12v on it for some time

#### Papabravo

Joined Feb 24, 2006
18,400
SMD 5V6 Zener diode, The Zener diode I have chosen is a 5v6. The diode has an absolute max rating of 300mW.
If a dc level of 12V is put across it I get a power of 273mW across the diode.
So my question is will the Zener get too hot and burn out if I leave the 12v on it for some time
The devil is in the details. Show us the schematic diagram of what you are talking about. The information you have provided so far allows me to answer: "probably not, but maybe yes -- eventually". I'm guessing that sort of prevarication was not what you were looking for,

#### alanj100

Joined Apr 26, 2019
19
All we have is a 150R resistor and 5V6 Zener in series connected to a single gate xor chip. The max voltage of the chip is 6V6 volts. I just want to protect the chip incase someone accidently applies 12V rather than 5v

#### BobTPH

Joined Jun 5, 2013
4,896
Use a much bigger resistor, at least 1K, What type of chip is the input on (logic family)

Bob

#### alanj100

Joined Apr 26, 2019
19
All we have is a 150R resistor and 5V6 Zener in series connected to a single gate xor chip. The max voltage of the chip is 6V6 volts. I just want to protect the chip incase someone accidently applies 12V rather than 5v
The chip is M74VHC1GT86DFT1G. The problem with such a large value resistor is the current may well not be enough for the zener to conduct even if it goes above the max 6v6 volts

#### Alec_t

Joined Sep 17, 2013
12,803
How much current do you intend to source/sink from the IC output?

#### alanj100

Joined Apr 26, 2019
19
How much current do you intend to source/sink from the IC output?
uA as it is going to the gate of a MOSFET

#### Alec_t

Joined Sep 17, 2013
12,803
In that case the 1k resistor Bob suggested will be fine. The current through the Zener will be ~(12V-5.6V)/1kΩ = 6.4mA, which is good.

#### alanj100

Joined Apr 26, 2019
19
In that case the 1k resistor Bob suggested will be fine. The current through the Zener will be ~(12V-5.6V)/1kΩ = 6.4mA, which is good.
The problem I have is at 12v it would be fine but at 8 or 9 volts the current may well be too small for the diode to conduct

#### crutschow

Joined Mar 14, 2008
29,772
The problem I have is at 12v it would be fine but at 8 or 9 volts the current may well be too small for the diode to conduct
The diode will conduct down to zero current.
What makes you think it will stop conducting?

#### Papabravo

Joined Feb 24, 2006
18,400
All we have is a 150R resistor and 5V6 Zener in series connected to a single gate xor chip. The max voltage of the chip is 6V6 volts. I just want to protect the chip incase someone accidently applies 12V rather than 5v
I ask for a schematic and you give me another bit of "word salad" that is open to varying interpretations. Once more with feeling: "there is just no substitute for a for a complete schematic diagram".

#### crutschow

Joined Mar 14, 2008
29,772
A resistor in series with the gate, with a Schottky diode from the gate input (anode) to +5V (cathode), will also protect the input from overvoltage.

#### MrChips

Joined Oct 2, 2009
26,077
The chip is M74VHC1GT86DFT1G. The problem with such a large value resistor is the current may well not be enough for the zener to conduct even if it goes above the max 6v6 volts
You have not made it clear.
What is the 5V or 12V feeding?
Is it an input pin or Vcc of the 74xx gate?

#### sghioto

Joined Dec 31, 2017
3,271
The problem I have is at 12v it would be fine but at 8 or 9 volts the current may well be too small for the diode to conduct
Just use a 5 volt Low Dropout Regulator.

#### crutschow

Joined Mar 14, 2008
29,772
Just use a 5 volt Low Dropout Regulator.
View attachment 260650
That's not a good solution to protect the input to a logic gate, which is what the TS wants.

#### sghioto

Joined Dec 31, 2017
3,271
That's not a good solution to protect the input to a logic gate, which is what the TS wants.
I'm not so sure about that. The question was asked by MrChips in post #12 but have not received a reply as of yet.

#### MisterBill2

Joined Jan 23, 2018
11,876
A zener diode shunt regulator is terribly inefficient, except as a heat source. And, like all diodes it is very non-linear. The voltage regulation improves as the current increases, and at lower currents the voltage will be higher than the rated voltage. So for battery operated systems the zener is a poor choice for regulation . To protect a logic gate, where under normal conditions the zener will not be conducting, is reasonable. MY suggestion is to have the series current limiting resistor to be the lowest wattage available, so that in the event of an excess voltage being present for an extended time the resistor will fail open, and the gate will be protected.

Last edited:

#### alanj100

Joined Apr 26, 2019
19
A zener diode shunt regulator is terribly inefficient, except as a heat source. And, like all diodes it is very non-linear. The voltage regulation improves as the current increases, and at lower currents the voltage will be higher than the rated voltage. So for battery operated systems the zener is a poor choice for regulation . To protect a logic gate, where under normal conditions the zener wioll not be conducting, is reasonable. MY suggestion is to have the series current limiting resistor to be the lowest wattage available, so that in the event of an excess voltage being present for an extended time the resistor will fail open, and the gate will be protected.
Thanks , that is exactly what I thought

#### MisterBill2

Joined Jan 23, 2018
11,876
Another option that I have used is called "Diode Clamping", where there are two diodes in series invese across the 5 volt supply, annd the signal is tied to the junction of the diodes. So if the signal rises above the supply the top diode willconduct and prevent it from exceeding the supply by much. And if the input goes negative, the bottom diode will conduct and prevent the input from going far negative. This method does require a series resistor to limit the current, though. A one Kilohm resistor will usually be adequate.

#### crutschow

Joined Mar 14, 2008
29,772
Another option that I have used is called "Diode Clamping", where there are two diodes in series invese across the 5 volt supply,
For good protection the diodes should be Schottky types, so they start to conduct before the IC substrate diode does.