1
\$\begingroup\$

We are trying to use a zener diode shunt circuit to develop a voltage regulator.

  • Our input voltage is 5v.
  • Our desired output voltage is 3.3v.
  • The average load current is 80mA.
  • The peak load current is 150mA.

We assumed the load current to be 150mA.

We used a Zener diode of 3.3v and 0.5W. Based on this, we concluded that the current through the Zener diode is 150mA.

Thus, the total current is 300mA.

Based on this, we calculated the current limiting resistor value to be 5.6 ohms and 0.5W.

When we actually connected the circuit, we noticed that the output voltage across the Zener diode was coming out to be 4.3v instead of 3.3v and it starts to heat up gradually. Why is this happening?

We then replaced the 5.6 ohm resistor with a 220 ohm resistor and noted that the output voltage was now 3.3v but the current was insufficient (as expected). We then replaced the 220 ohm resistor with two 5.6 ohm resistors in series and found the output voltage to be around 4v. Similarly, with three 5.6 ohm resistors in series and found the output voltage to be around 3.7v.

Questions:

  • Are our calculations correct?
  • Our assumption is that the voltage across the Zener diode is independent of the current limiting resistor. Why is there an apparent dependence?
  • Are there any fundamental errors that we are making?
\$\endgroup\$
3
  • 3
    \$\begingroup\$ Welcome to EE.SE. There is a CircuitLab schematic button on the editor toolbar. Add in the schematic so that we're all clear about how it is wired. Please also add a link to the Zener datasheet. \$\endgroup\$
    – Transistor
    Sep 29, 2018 at 9:11
  • \$\begingroup\$ Data sheet link for the zener in question? \$\endgroup\$ Sep 29, 2018 at 9:11
  • \$\begingroup\$ As with any diode, because of the silicon region outside of the "junction" having non-zero resistance, at high currents the Ohms Law voltage will be significant. You found that region. \$\endgroup\$ Sep 29, 2018 at 17:31

2 Answers 2

1
\$\begingroup\$

Zener diodes are fairly imperfect devices; they are rated to produce their nominal output voltage at a certain current range and are about ten times better than just using a potential divider to create a regulated voltage but, as always, you must read the small print on the device and that is typically centred around using them for low power applications.

If you put too much current into them their terminal voltage will rise beyond their rated nominal output level so, you must use them as their data sheet defines or expect bad results. Here's a picture of what the output voltage can look like with varying current passing into the device: -

enter image description here

Picture source. As you can see the zener voltage does depend on current that flows into the device and you should not look at pictures like the one below and assume that it produces a constant output voltage: -

enter image description here

The picture above gives a false impression of perfect regulation once a certain current is exceeded.

I would say that your application requires a 3-terminal linear voltage regulator.

\$\endgroup\$
3
\$\begingroup\$

You have a fundamental misconception about how a shunt regulator works.

The key idea is that you size the series resistance for the peak load current by itself. Then, whenever the load does not actually draw that peak current, the Zener diode "shunts" the excess current to ground. In other words, when the load current goes up, the Zener current goes down, and vice-versa. The Zener dissipates the maximum power when the load is drawing its minimum current.

\$\endgroup\$

Your Answer

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

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