What is Zener diode max current?

Pv is the wattage - 1.3W in this case.

It's just basic electronics formula - you know the wattage, you know the voltage, so it's trivial to calculate the current.

Watts = volts x amps

Reconfigure: amps = watts divided by amps.

They can't give a specific figure, as it varies depending on the actual zener voltage.

When the Zener diode's specification is given as P/V (power per voltage), it refers to the Zener diode’s power-to-voltage relationship under specific conditions, typically in terms of how much power is dissipated for each volt of Zener voltage. This is often used to express maximum power dissipation per volt of voltage regulation.

Here's why it might be given this way:

1. Power Dissipation and Voltage Rating Relationship​

• P/V is essentially the power dissipation per unit of Zener voltage. By specifying this, manufacturers help engineers understand how the diode behaves across different voltages.
  For example, if you know the Zener voltage (Vz) and the maximum power dissipation (P_max), you can figure out how much current the diode will be handling and whether that falls within safe operating conditions.

2. Efficiency of Power Handling​

• P/V gives you a way to estimate how much current the Zener diode will need to handle for a specific voltage to achieve a certain level of power dissipation.
• It’s a simplified specification that allows engineers to calculate and scale the Zener diode’s capacity for larger or smaller voltages. It essentially tells you how much power the Zener will dissipate relative to the voltage it's regulating.

3. Scaling Power for Different Voltages​

• When a Zener diode is used in different voltage-regulation applications, you can use the P/V value to scale the power dissipation for a range of operating voltages.
  For example, if the specification says "P/V = 20mW/V," this means for every volt of Zener voltage, the diode will dissipate 20 milliwatts of power. So:
  • If the Zener voltage is 5V, the diode will dissipate 5V × 20mW/V = 100mW.
  • If the Zener voltage is 12V, it will dissipate 12V × 20mW/V = 240mW.

4. Simplifying Design Decisions​

• By giving the P/V value, manufacturers allow engineers to easily estimate how the diode will behave across different operating conditions without needing to calculate the current manually for every situation. It's a shorthand method for understanding the diode's power dissipation relative to the voltage it regulates.

Example:​

If the datasheet of a Zener diode specifies P/V = 10mW/V, this means for each volt of Zener voltage, 10mW of power will be dissipated.

• So, for a 5V Zener voltage, the diode would dissipate:5V×10mW/V=50mW5V \times 10mW/V = 50mW5V×10mW/V=50mW
• Similarly, for a 12V Zener voltage, it would dissipate:12V×10mW/V=120mW12V \times 10mW/V = 120mW12V×10mW/V=120mW

It’s a quick way to understand the power-handling capability of a Zener diode in relation to its Zener voltage.

Conclusion:​

The P/V specification is a shorthand way of communicating how much power the diode dissipates for each volt of Zener voltage. It allows you to scale the power dissipation for various voltages, making it easier to design and assess how the diode will behave under different operating conditions.

Simple math for the 1N4728A would give 1.3/3.3 = 393mA as absolute maximum. Data sheet shows a test current of 76mA, and Maximum regulator current of 276mA.
So, the "safe" area (rule of thumb) of zener regulation would be between those two numbers, provided the zener is at ambient temperature. The closer it is to the test current, the less heat it will dissipate. It may regulate below the test current as well, but you would have to experiment with that.
Unfortunately that datasheet does not give an actual P/V value (specification)

Be aware though, that the data sheet’s rated power is valid for an ambient temperature of 25C.

If the diode is enclosed and/or close to a heat generating device, like a regulator, the ambient that the diode sees may be higher.
You have to derate accordingly.

Also, most 1N4728 are rated 1W, not 1.3W. Some datasheets show the 1.3W rating only if the leads are very short (1/8") to a "heatsink" (copper trace/pad large enough to sink heat from the zener)

As a designer, I would never run any real current through a zener, but rather I would setup a current mirror using a zener through a transistor that could deliver the current I needed for the application.

Beau Schwabe said:

As a designer, I would never run any real current through a zener, but rather I would setup a current mirror using a zener through a transistor that could deliver the current I needed for the application.

Click to expand...

Yes.
For best voltage accuracy/stability you want to operate the Zener somewhere near the current used in the data sheet to measure it's rated voltage.
For example all the Zeners below are rated at 5mA current:

Vishay datasheet ( Motorola design) defines several max currents with assumptions for each.


Motorola defined voltage rise from IZT1 =1 mA with a high knee resistance to the low knee resistance at max power of 250 mW at IZT2.
If you cannot apply a heatsink to the leads and keep the leads at 25'C then you use a low current like 5 mA which became an industry standard by other suppliers to dissipate about 10% of this. The reason is that the tempco. of voltage changes from NTC at low Vz to PTC above 5V and keeps rising. The regulator current is the maximum at 900 mW but only if you regulate the temperature of the leads @ 25'C.

The IR is the max pulse zener current for 10 ms is the must not exceed max value.

You do not want to use 250 mW unless you predict the temp rise from 110 deg C/W above ambient and know the tempco for that Vz.


Thus 5 mA became the test method used for manufacturers other than Motorola. With dynamic resistance specs you can predict the error voltage with a dynamic load and interpolate voltage between the Izt1 and Izt2. The rest of the world decided later to standardize on 5mA so that you use a constant current source when it matters or is needed.

I understood this 25 yrs ago when I started comparing the Vf of 5 mm LEDs. The wide variation of Vz is all due to knee resistance and not the threshold for each colour. Mfg quality controls the tolerance on these nominal knee resistance values which also varies with Zener Vz and LED colours.

Using the std. 20 mA LED test the best R/Y LEDs have a Zzt of <10 Ohms and the best 20 mA white LEDs are < 15 Ohms . There is also a direct quality relationship with knee resistance times Pmax @ 85'C that is predictable.

Each Iz test method has its pros and cons.

FWIW Vishay has grown by acquisitions such as General Semiconductor in 2002 and I speculate has purchased the rights to produce many OEM components that have expired on patent rights such as Sharp and Panasonic Opto's and light sensors.

I found that my select suppliers of LED's have much lower knee resistance than all the rest. This allowed me to get LED's better than Zeners in two price ranges +/-50 mV or +/- 100 mV @ 20mV. rather than 2.9 to 3.8 for white.