Continue to Site

# Bridge Rectifier Problem when connected to Wall Power Outlet

Status
Not open for further replies.

#### richeek

##### New Member
Hi all
I know this is a long post but I tried to put everthing...

I am making a bridge rectifier to give DC voltage to another circuit. At the output of the rectifier I am connecting 0.47uF-250V, mylar capacitor and across the capacitor I am connecting resistive load.
The bridge should rectify 120 VAC from a WALL OUTLET to obtain a DC voltage. I am trying to acheive as close as possible to 170 VDC.
Now the problem is that I want to get this voltage with respect to circuit ground. What happens is if I try to connect other end of the capacitor to ground some diodes blow off. Even if I try to connect oscilloscope between two terminal of capacitor some diode blows off.
I searched it over the net and found out similar post: https://www.electro-tech-online.com/threads/full-wave-bridge-rectifier-problem.5568/
I think I am also facing the similar problem. Could anyone please tell me why this is happening? Why the other end of the capacitor is not always at potential 0V? What should I do now to get a 170V DC power supply?
I mesure the voltages at both ends of the capcaitor by just connecting positive terminal of a 10X probe to the circuit and floating the probe's ground. And I saw that capactor terminal that I think should be at 0V is actually oscillating between 0V and -170V.
At the same time other end of the capacitor is oscillating between 170V and 0V. Hence together they will give me 170VDC. But how to connect them to another circuit?
Kindly provide some inputs I am so confused.
Thank you.
Sincerely
Richeek Arya

Last edited by a moderator:
As neutral is connected to ground you can't connect the output of the bridge to ground, it will go BANG!!! as you've found out.

To do it with a bridge would require an isolating transformer, which sounds a good idea anyway.

Is it not neutral connected to ground? I have the notion that neutral is connected to ground in domestic power supply.

Look at the schematic for a bridge rectifier. The common of the bridge is not connected to the common of the source. If you connect the common of the bridge to the common of the source, then you will blow one or more of the rectifiers since there's a direct path from the hot side through the rectifier to common (ground).

Yes, neutral is connected to ground. That's why you blew the rectifier.

If you want a grounded connection from a line powered rectifier, then you can only use a half wave rectifier. But using any line powered circuit is dangerous since you are not isolated from the AC power and a lethal shock is possible if you touch any part of the circuit when also touching ground.

I am confused about one thing. If neutral wire is connected to ground then it means the potential of the Hot wire is changing between +120VAC to -120VAC during the cycle.
Potential of the neutral is constant 0V. So one end of the output is always at 0V. So how does it causing a short? I have attached a pic too to make myself more clear.**broken link removed**

If you look at the Full Wave Resistive Supply you will see the 4 diodes in the bridge.
If you connect the 0v output to say the lower AC line, you will short-out the left lower diode. This means the top left diode will work when the AC is in one direction but become a short when the AC reverses.
The only solution is to produce a half-wave supply as shown in the 4th circuit.
That's why it is important to draw a clear circuit diagram so that you can see exactly what is going on.

Can you explain me one thing and I am contrdicting myself. If the neutral is always grounded how does the rectifer circuit works in the negative cycle. Assume that it is driving a capacitive load?
My doubt is that during positive half cycle capacitor will get charged and when the negative cycle comes bottom right diode(in your figure) will be reverse biased as the anode is grounded and cathode is at poitive voltagevso there is not path for the current to flow.

Can you explain me one thing and I am contrdicting myself. If the neutral is always grounded how does the rectifer circuit works in the negative cycle. Assume that it is driving a capacitive load?
My doubt is that during positive half cycle capacitor will get charged and when the negative cycle comes bottom right diode(in your figure) will be reverse biased as the anode is grounded and cathode is at poitive voltagevso there is not path for the current to flow.

You're confusing yourself because you're trying to connect the output of the bridge to ground, you can't do that (as you've found out) - it has to be fully floating, and the fact that neutral and ground are connected makes no difference.

In your diagram the bottom right diode is shorted out, so the bottom left one will explode with a large BANG as it's directly across the mains.

Can you explain me one thing and I am contrdicting myself. If the neutral is always grounded how does the rectifer circuit works in the negative cycle. Assume that it is driving a capacitive load?
My doubt is that during positive half cycle capacitor will get charged and when the negative cycle comes bottom right diode(in your figure) will be reverse biased as the anode is grounded and cathode is at poitive voltagevso there is not path for the current to flow.
Your mistake is assuming that the bridge "gnd" voltage is as zero potential with respect to the 120VAC "neutral". It is not. If you look at the bridge gnd voltage with respect to the 120VAC neutral you will measure a pulsating voltage with a peak value of about -170V. That's why you can't connect the bridge gnd to the 120V neutral.

Thanks Carl. You are right. They are not at the same potential. The output potential of one terminal varies from 0 to 170V and that of the other from -170V to 0. However I still wonder how does after getting charged in the positive half cycle where one end is at +170V and other is at 0V how does the potential of the terminal changes to 0V and -170V respectively in the negative half cycle? **broken link removed**

Te problem is I can not decide which diodes will conduct?

Stare at this for a moment. R2 is there just to fool Spice into letting me rename Gnd to be "Neutral"

Note what V(Plus) and V(Minus) are doing. Also note what the differential voltage V(Plus,Minus) looks like.

Still think you can tie "Minus" to ground?

#### Attachments

• Draft26.gif
77.8 KB · Views: 683
Last edited:
Thanks Mike for sending the waveforms. Yeah they are not at the same potential I understood that. What I am still wondering is during the negative half cycle, V1 is going towards -170V. In this case how V(plus) is reduced to 0V. Just for learning purpose, I am curious by which path V(plus) is decreasing.
In the negative half cycle: D2 is reversed biased initially. (because V(plus) is higher than V(neutral). D1 is reverse biased too because of the same reason. Where is the path then??

Here are the currents through the four diodes

#### Attachments

• Draft26a.gif
58.7 KB · Views: 468
Ok. Currents are as expected theoretically.

... What I am still wondering is during the negative half cycle, V1 is going towards -170V. In this case how V(plus) is reduced to 0V. Just for learning purpose, I am curious by which path V(plus) is decreasing....

Message deleted!

Last edited:
Thanks guys for all the help. Problem is resolved now. Until I get a transformer I am using a half wave rectifer with all the necessary precautions.

Status
Not open for further replies.