I've build this circuit (that will be part of a more complex circuit): enter image description here

All works as expected and on the zener I have exactly 15 V rectified. Later, one day after, I've discovered that even with the AC power (230V) disconnected on the C1 capacitor there were about 5 residuals Volts. For this reason I've short-circuited the capacitor with the purpose of completly discharge it. This, probably, was my big mistake because after that the circuit stops working and I've discovered that the bridge rectifier was irrimediably broken and shot-circuited. Frankly I don't understand why this it happened and how to improve my circuit to prevent a similar event in the future. Any idea? Thanks in advance.

ADD: Reading all replays an comment I suspect that the circuit is not so good. Neverthless, I've forgetted to say that before that the bridge rectifier broke down, I've connected an oscilloscope probe across the zener and by this way I could see that there was a residual 5V voltage on the capacitor. I don't know if that can explain what is happen...

I've a significant update to my issue.

After I've added the bleed resistor and replaced the fried bridge rectifier, than I've replugged the AC power and all seemed to works fine too.

I've connected the multimeter and all measures are right.

At this point I've connected the oscilloscope probe and puff... the bridge rectifier was fried too.

The oscilloscope is a Tektronix TBS1000C Series, brand new, just purchased and works fine even after the issue described.

Nevertheless I've a suspect.

The circuit that mount the bridge rectifier is connected to the power wall outlet (230V) while the oscilloscope is connected to an APC Smart UPS.

All like the following layout:

wiring diagram

I don't know if this could be the reason of the issue and, especially, why. Any other idea?

  • \$\begingroup\$ If you want 15V in the end, better add a transformer stage before your rectifier. \$\endgroup\$
    – Eugene Sh.
    Commented Mar 2, 2021 at 17:43
  • 2
    \$\begingroup\$ This is a dangerous and inefficient circuit. Please consider using a step-down transformer. \$\endgroup\$ Commented Mar 2, 2021 at 17:50
  • \$\begingroup\$ What points did you connect together to discharge the capacitor? Shorting out the capacitor itself should not damage the bridge rectifier. Are you aware that both your DC+ and DC- are live? (It's a nasty circuit.) Also, your resistors need to be rated for 400 V peak. That's why most similar circuits use resistors in series, not in parallel. \$\endgroup\$
    – Transistor
    Commented Mar 2, 2021 at 18:29
  • \$\begingroup\$ Can you elaborate on how you discovered the rectifier was shorted? \$\endgroup\$ Commented Mar 3, 2021 at 10:28
  • \$\begingroup\$ Hi user253751! I've added a significant update to my original post \$\endgroup\$
    – M4Biz
    Commented Mar 4, 2021 at 17:03

4 Answers 4


This is an equivalent of your setup:

enter image description here

R1 represent your resistor, zener, diode and capacitor. The details of which is not important. The important thing is that you connect the gnd-clip on the scope to the negative terminal (zener anode). This clip is connected to PE which goes somewhere to the neutral as pointed out by @user287001. Now lets simulate what goes on before you connect the gnd-clip:

This is a simulation

enter image description here

As you can see the location where your did put the gnd-clip goes to minus 324V! Current is only limited by D3 and the resistance in PE and neutral. So no wonder the bridge fused. So the advices of isolation transformers in addition to utmost caution are better taken serious.

If an isolation transformer is applied the circuit would look similar to this:

Equivalent diagram when isolation transformer is inserted

Now the gnd clip can be connected as seen and the simulation now gives:

Simulation of circuit with isolation transfomer

Now the neg terminal is set to 0V (by definition of gnd). Notice that you still have up to 324V on the pos terminal. When ph1>ph2, D1 and D2 will conduct. When ph1<ph2, D3 and D4 will conduct. Finally notice that ph1 and ph2 are each clamped to gnd=neg. I.e. neither can go below a diode conduction drop (1V) below gnd, so there are no large negative potentials.

Here it is an ideal transformer. A real one will have some capacitive coupling from primary to secondary. This will inject a voltage/current into the secondary circuit which you will see if you remove the gnd clip and place the probe on neg terminal.

Simulated with LTSpice

  • \$\begingroup\$ Hi Henning. Thanks for you reply. I've a questions too for you: in absence of failures the resistance R2 between Neutral and PE shouldn't be infinite? \$\endgroup\$
    – M4Biz
    Commented Mar 5, 2021 at 19:02

I am not condoning the construction of this dangerous and inefficient circuit, however if you must, it's better to have the resistors in series with the input. Your D3 is doing nothing of value so lose that.


simulate this circuit – Schematic created using CircuitLab

So the voltage across the bridge is reduced and any mains transients (which could destroy the bridge) will appear across the resistors. The resistor should be of a fusible flameproof type rated for appropriate mains surge voltage (kV not just hundreds of volts).

Better yet, as this circuit can cause lethal shock or cause a fire when inappropriately designed, constructed or used, just don't do it.


I guess your oscilloscope signal input GND is connected to the protective earth wire of scope's mains AC input. The PE wire is connected through the UPS to the PE of the 230VAC outlet. It is connected to the neutral wire in the power distribution box of your house. You shorted the circuit when you connected the probe GND.

If one makes test measurements in a circuit which is galvanically connected to the mains AC outlet he should feed the circuit under test through an isolation transformer. Get one if you do not already have it. Electronics repair workshops have it installed in their working desks.


  1. Do not even think to cut the protective earth wiring, not even temporarily for measurements. Or at least write your last will before starting to play with life. If the owner of those fine instruments happen to be someone else than you he also can affect negatively your wellfare after seeing that you have removed the PE. Isolation transformer between the mains AC outlet and the device under test is the acceptable route.

  2. Using an UPS as a substitute for the isolation transformer (mains AC input plug disconnected) cannot be accepted before researching how the UPS internally works. I guess at least the PE of the UPS input should be connected to mains AC PE to prevent unpredictable behaviour and voltage on the touchable metal parts of the UPS. That trashes the whole idea of getting isolation.

  • \$\begingroup\$ Hi . Thanks for you reply. As a matter of fact, I've connected the oscilloscope to the UPC only because I hadn't free AC outlets and the only available was one on the UPC line of my Lab. Do you thing that if I'll plug the oscilloscope in an "normal" AC wall outlet I'll don't have the issue? \$\endgroup\$
    – M4Biz
    Commented Mar 5, 2021 at 17:33
  • \$\begingroup\$ Unplug the scope. Take a multimeter and measure the resistance between the probe input GND and the PE pin of the mains AC plug of the scope. I have guessed it's zero ohms. If I was right plugging the scope directly to a wall 230VAC outlet doesn't help. Tell the result. If I was wrong the case becomes complex. BTW.Ungrounded mains AC operated complex electronic devices have often a high internal capacitance, say 200 nF between the signal GND and the mains AC parts of the device to limit rf emissions. I do not easily believe Tektronix uses such design. \$\endgroup\$
    – user136077
    Commented Mar 5, 2021 at 17:50

Assuming you had a low ESR cap and used the Gnd path that allowed current thru the diode bridge to discharge the cap a surge would occur. Yet the bridge is rated for 1A and 50A for one cycle so it seems unlikely to exceed spec. A better way is to add a 100k resistor across permanently. So the root cause of failure is unknown.

  • \$\begingroup\$ Hi Tony. Thank for you reply. Exactly where I'd must add the 100k resistor? \$\endgroup\$
    – M4Biz
    Commented Mar 2, 2021 at 17:52
  • \$\begingroup\$ loading the cap to decay in 100k*47u in 5 seconds \$\endgroup\$ Commented Mar 2, 2021 at 17:55
  • \$\begingroup\$ @M4Biz Straight across the terminals of the capacitor (i,e, in parallel with D5). \$\endgroup\$
    – Simon B
    Commented Mar 2, 2021 at 23:17
  • 1
    \$\begingroup\$ Yes that’s how you can bleed the charge on the cap without much power consumption. \$\endgroup\$ Commented Mar 2, 2021 at 23:32
  • \$\begingroup\$ @Tony Stewart Sunnyskyguy EE75, Hi Tony. I've just published an important update to this issue and because it was very detailed I've published an answer to my original post. If you want you can take a look at it \$\endgroup\$
    – M4Biz
    Commented Mar 4, 2021 at 17:30

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