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Behaviour of diode 1n4007 at 220V ac

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Junior Member level 1
Oct 25, 2012
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My bridge rectifier(1N4007) is giving some voltage at ground terminal of bridge while measuring with the ground of oscilloscope at 220V 50 Hz power supply.....

I found a reason for this the breakdown voltage of diode is 120V.
dts y all diode conducting beyond d 220V and gives reverse voltage level at ground....

My Question is that
why all of the SMPS circuit use Bridge rectifier(1N5399) directly 220V ac with a single fuse? Does reverse voltage at ground not harm to the Capacitors?
What is the reason behind this?

The break down voltage for 1N4007 is 1000V and it is the same for 1N5399.
**broken link removed**

Do you have any transformer between the mains and the rectifier?
no i dnt hv any transformer between mains and bridge.....

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@Nitish: Diode PRV is 1000V but reverse breakdown voltage is 120 V at room temperature.

U can prefer in reverse characteristic of diode.....

I think you are referring to Fig 4 in

The is in % and not in Volts.

The ground voltage of the rectifier is not at supply ground. If you connect the rectifier ground to supply ground (through oscilloscope) this creates a short through one of the diodes during the negative half cycle.


oh u want to say that there is ground voltage of rectifier and supply totally different.....

then how could i measure the signal which i dropped using the voltage divider at rectifier output terminal.

I am assuming you have a rectifier + filter connected to the mains supply and you have a resistor divider to measure the voltage.

You will have to use an isolation transformer for either the oscilloscope or the rectifier. Then you can check.
Otherwise you should be seeing some smoke coming from one of your diodes( just draw the rectifier circuit and short the supply ground and the rectified output ground and check for one of the half cycles).
Isolation is a must unless you know what you are doing exactly and where you are measuring. Otherwise you might damage the oscilloscope since the short circuit current flows through it.
it can be cleared if you can post the circuit you made and mark between which points you mesured the voltage
thanks for ur reply nitish.....

I just want to know that why diode breakdown at 220V reverse voltage while the breakdown voltage is given V(RRM)=1000V. and V(rev RMS)=700V.

and there is no transformer isolation provided in circuit.

one more question if i connected a capacitor across it then it should be damaged because it get reverse voltage at ground.


You should remember that the negative point of the resistor (bottom node) is not the same as the -ve node of the 220V sine source. The former is the supply neutral while the other is circuit ground.
If you short both the terminals (like when you use an oscilloscope probe across the resistor to check the voltage) you can see that the top left diode in your schematic will be forward biased in the negative half cycle and will act as a dead short between the +ve and neutral of the supply. This will burn up that diode since the current will be much much more than its limits.

The diode breakdown in your case is not the reverse breakdown but due to very high current in forward bias.

In normal operation, you do not connect the two terminals. In this case, the voltage across the resistor is always unidirectional since it gets the rectified voltage. When you put a capacitor across it, the voltage across the capacitor is also always unidirectional. There is no reverse voltage across the capacitor either. As long as the the voltage across the capacitor does not go beyond its limits you are OK.

You can try this out in simulation first and use two terminal probes which are available in most software check the waveforms. Also try to observe what happens when you short the rectified output ground with the supply neutral pin.
thanks for ur reply.....
my actual arrangement like that.....


in this figure i want to ask you that what is the reason for negative voltage at rectifier ground while in datasheet the breakdown voltage is given V(RRM)=1000V. and V(rev RMS)=700V.


i think current is again responsible for this reverse voltage because I(RM)=1mA.
is it the correct reason for that

1. What is the wire voltage 103V measured with respect to?
2. Which is the circuit ground that you would have set for the simulations?
3. What is the voltage at the top terminal of the 220k resistor?

hi nitish,

1. I'm using oscilloscope ground. which are giving me this difference.
2. Oscilloscope ground.(0Volt)
3. 203 V at top terminal of 220k resistor.

203V at the top of the resistor means that there is a positive voltage (of about 100V) across the resistor w.r.t the bottom node of the resistor. Hence there is no reverse/negative voltage at the rectifier output at all.
The rectifier ground will show negative voltage in the negative half cycle of the 220V input, if it is measured w.r.t. the 220V supply ground.

Which simulation software are you using?

The issue here is that you are not able to see the waveforms at the output of the rectifier since the diodes go bad. This would not be possible without an isolated oscilloscope since otherwise the oscilloscope negative terminal is connected to the supply ground through the oscilloscope.
nitishn5 is right This mesurement is not possible without an isolated oscilloscope

omshankar you try to power the oscilloscope from an UPS with the input plug compleately removed (important no connections to the ups other than the scope input or output)this makes the scope isolated from the mains power supply

you can connect your circuit to the mains and test the points find the results and post. Hope this helped you if so dont forget to click the helped me button below
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