Electronic – How does this overvoltage protection work


I was looking around for low cost and effective overvoltage protection, and I found this from this page:


With this part list:

  • RXE025 – 250 mA PTC resettable fuse
  • Zener diode – 5V6, 1 watt
  • Resistor – 1K ohm, 1 watt
  • Transistor – BD139 (NPN medium power 80V, 1.5A CC rating)

As reported by the author, this circuit seems to be effective and low cost, but I don't understand why the engineer placed a BD193 transistor in this circuit.

Can anyone explain the step-by-step operating method of this circuit?

Best Answer

It's pretty simple:

The Zener diode voltage is at (or slightly above) the normal Vcc voltage. For example, a 5.6V Zener for a 5V Vcc.

When Vcc is below the Zener voltage, no current flows through the diode, and the 1k resistor keeps the base of T1 low. No current flows through the collector to the emitter of the transistor.

When Vcc is higher than the Zener voltage, current flows through the Zener diode and through the base of T1. This allows current to flow through the collector and out the emitter of the transistor.

The current through the transistor is high enough to cause the fuse to open.

Short version:

Overvoltage causes the transistor to short and blow the fuse.

Having read the article and looked up the RXE025, I see I need to change my description somewhat.

This protection circuit doesn't shut off power to the protected circuit.

When the input voltage exceeds the Zener voltage the transistor conducts.

The RXE025 was chosen because even in a "tripped" condition it will pass enough current that the protected device (an Arduino in the original example) will continue to run.

The RXE025 turns in to a current limiting resistor to limit current through the Zener diode and the transistor.

The result is that Vcc is limited to a little more than the Zener voltage.

The "fuse" doesn't blow in the usual sense, it merely changes to a higher resistance.