The following is an exam question I encountered.

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In the above circuit the current across E is 180 mA. Since the current across 300 Ohm resistor is 13.33 mA, there will approximately be a current of 166 mA through the zener diode Z1.

Then the resulting power dissipated would exceed the power rating of the diode Z1.

My question is; what exactly happens in a situation like this? How will a circuit behave when the power rating of a diode is exceeded (both theoretically and practically)

Thank You

  • 1
    \$\begingroup\$ Zener1 is dissipating 10v * 0.2 amps, or 2 watts. One of the leads will be WELDED to the metal case, the other lead enters thru a glass seal. If the two leads are VERY SHORT and soldered to large PCB traces with air blowing across the traces to remove heat, the Zener may survive and even last a long time, depending on how many temperature cycles (on/off) occur. \$\endgroup\$ – analogsystemsrf Jul 21 '19 at 10:13
  • \$\begingroup\$ As peufeu notes - the question appears to have a fatal error if implemented in reality but is slightly worse than marginal for example purposes. . If Vz1 is 10V and Rload = 0 (tales max current) I_Rs2 is 33 mA. BUT Vrs1 = 90V so Irs1 = V/R = (100-10)/500 = 180 mA. Z1 conducts 180 - 33 mA =~ 150 mA so PdZ1 = V x I = 10 x 150 = 1.5W. \$\endgroup\$ – Russell McMahon Jul 21 '19 at 13:31

Assuming that enough current flows through both diodes so they regulate, and assuming they're perfect, then the voltage across them is their Vz.

Thus current in the 500R resistor is (100-10)/500 = 180mA

Current in the 300R resistor is (10-6)/300 = 13.3mA

Current in Z1 is (180-13.3) = 166.6mA and power is 1.66W

So, your calculation is correct. Z1 dissipates more than what it is rated for. Note that as analogsystemsrf says, this does not necessarily mean it will blow. This depends how the power rating is calculated. What determines lifetime is maximum chip temperature and thermal cycling / dilatation of materials... and that depends on how the heat is removed from your part. So the same part (for example a TO220 package) can have wildly different power ratings depending how it is mounted (free air, board, heat sink...) or if a fan blows on it or not.

If we assume the power rating is correct, then the diode will overheat. It can melt its solder joints and fall off the board, or simply blow. Leaded zeners mainly fail shorted, at least the modern 'nailhead' types that don't have wire bonds. A SOT-23 type zener does have bondwires, so can blow open after it shorts if enough current is available.

Now, to answer the second question, even with RL=0 ohms, we have 33mA through the 300R resistor, 147mA through Z2, and it still dissipates too much. There's probably a typo in the value of the 500R resistor.... the resistor dissipates 16W which is no good. It should be 5000 ohms.


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