0
$\begingroup$

I was messing around with a Zener diode (having a reverse breakdown voltage of 5.6V) when i found out that on connecting it in reverse bias mode and providing an input voltage of 19V, the potential difference across the diode never comes out to be exactly 5.6V. My multimeter shows a reading of 5.4-5.5V.

I want to know why does this happen? Shouldn't it exactly be 5.6V?

If i increased it to a large enough value, would the potential difference across the Zener diode slightly be larger than 5.6V?

$\endgroup$

1 Answer 1

1
$\begingroup$

Zener diodes are not generally high precision devices. The voltage across them varies depending on the current through them and the operating temperature.

Here are two charts from the datasheet for the first part that came up in a google search for "5.6 V zener diode", which was the 1SMA5919BT3G from On Semiconductor. First the dependence on current:

enter image description here

There are multiple curves because the datasheet covers multiple part numbers for a family of devices with different zener voltage values. You can see that the variation due to current is much stronger for the lower-valued devices. The current dependence is also expressed as an equivalent output impedance in another table in the datasheet.

And here is the temperature dependence, expressed as a temperature coefficient as a function of temperature: enter image description here

As for the physical reason behind this variation, it's because the Zener effect comes from tunneling of carriers across the reverse-biased PN junction. The tunneling current is affected by the energy of the carriers relative to the states available on the far side of the junction and by the narrowness of the junction, both of which are enhanced as the applied voltage increases.

At 5.6 V, some of the reverse current likely also derives from the avalanche effect rather than the Zener effect. This effect requires spontaneously generated carrier pairs to be strongly accelerated as they cross the junction, so that they can "bash" additional carriers out of their trapped states, and again this effect is enhanced when the applied voltage (and thus the field strength in the junction) is increased.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.