Author Topic: Floating a bridge rectifier  (Read 7111 times)

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Offline AeternamTopic starter

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Floating a bridge rectifier
« on: May 12, 2016, 07:42:02 pm »
I feed a sine wave into a transformer which is followed by a full-wave bridge rectifier. I expect to see a rectified wave at the output with respect to local ground, but that's not the case. It's flatlining. Why?

My signal generator is not floating and I suppose that, by connecting the (also mains earth referenced) scope ground pin to the circuit, I create a short. But I was under the impression that the transformer would provide isolation from mains earth -- thus floating the part of the circuit on the right of the transformer and allowing me to probe around the circuit at my hearts content. Am I wrong?
 

Offline Delta

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Re: Floating a bridge rectifier
« Reply #1 on: May 12, 2016, 07:47:55 pm »
If your drawing accurately resembles your circuit, your diode bridge is wrong.....
 

Offline Mr.B

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Re: Floating a bridge rectifier
« Reply #2 on: May 12, 2016, 07:50:55 pm »
If your drawing accurately resembles your circuit, your diode bridge is wrong.....

How?
I approach the thinking of all of my posts using AI in the first instance. (Awkward Irregularity)
 

Offline retrolefty

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Re: Floating a bridge rectifier
« Reply #3 on: May 12, 2016, 08:05:00 pm »
If your drawing accurately resembles your circuit, your diode bridge is wrong.....

How?

 Ignore that. Have you taken a DMM reading in case you have a scope/probe/settings problem? Your expectations are correct but your measurement is somehow wrong. If not a measurement problem perhaps component problem(s) ?

 Your journey of knowledge is simply not complete yet, finish the journey.  :-+
 

Offline Delta

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Re: Floating a bridge rectifier
« Reply #4 on: May 12, 2016, 08:09:33 pm »
Sorry. It's not.  I must stop reading this forum after 8 pints of Guinness...
 

Offline rstofer

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Re: Floating a bridge rectifier
« Reply #5 on: May 12, 2016, 08:10:27 pm »
What do you get if you leave the probe ground disconnected?
Are you sure there is no conductivity between input and output?  There shouldn't be, but who knows?
It's worth remembering, and I'm sure you do, that probe ground is directly connected to earth ground.
You can also check it by setting your scope for A-B and using two channels.  Leave the probe grounds disconnected and put a probe at each location.
 

Offline IanB

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Re: Floating a bridge rectifier
« Reply #6 on: May 12, 2016, 08:10:42 pm »
The turns ratio of the transformer will be important. Do you know what voltage you expect on the secondary?

Also the 1 k resistor is important. A diode is a current operated switch, and with no load the rectifier will not function properly. You might consider trying a lower value resistor to increase the current flow through the rectifier.
 

Offline Mr.B

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Re: Floating a bridge rectifier
« Reply #7 on: May 12, 2016, 08:11:27 pm »
I must stop reading this forum after 8 pints of Guinness...

 :-DD
I approach the thinking of all of my posts using AI in the first instance. (Awkward Irregularity)
 

Offline AeternamTopic starter

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Re: Floating a bridge rectifier
« Reply #8 on: May 12, 2016, 08:16:39 pm »
The turns ratio of the transformer will be important. Do you know what voltage you expect on the secondary?

Also the 1 k resistor is important. A diode is a current operated switch, and with no load the rectifier will not function properly. You might consider trying a lower value resistor to increase the current flow through the rectifier.

Of course, OMG :) It's a step-down from 230V to 8V and I feed it 5 Vpp. The secondary voltage is completely dropped across the 2 diodes.

THANK YOU!
 

Offline alsetalokin4017

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Re: Floating a bridge rectifier
« Reply #9 on: May 12, 2016, 08:29:57 pm »
I feed a sine wave into a transformer which is followed by a full-wave bridge rectifier. I expect to see a rectified wave at the output with respect to local ground, but that's not the case. It's flatlining. Why?

My signal generator is not floating and I suppose that, by connecting the (also mains earth referenced) scope ground pin to the circuit, I create a short. But I was under the impression that the transformer would provide isolation from mains earth -- thus floating the part of the circuit on the right of the transformer and allowing me to probe around the circuit at my hearts content. Am I wrong?

Yes, because the scope probe ground reference clip is connected back through the instruments and their line cords to the grounded side of the FG's output. So you are creating a groundloop short that bypasses the transformer. See the marked-up sketch below.

The way to do this measurement is to make a differential measurement using two scope probes with no ground clip connection, and use the scope's Subtract function (or, on many analog scopes, Invert and Add functions) to give the differential measurement on the output side. See the scopeshot below.

(And I just saw the transformer ratio. Yes... you'll need a bit more voltage on the output side of the transformer before you see anything sensible!)



The easiest person to fool is yourself. -- Richard Feynman
 

Offline JimRemington

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Re: Floating a bridge rectifier
« Reply #10 on: May 12, 2016, 08:36:15 pm »
Quote
I feed it 5 Vpp. The secondary voltage is completely dropped across the 2 diodes
Not quite true. Expect about 1.4 V drop across two diodes, so about 1.1V across the resistor.
« Last Edit: May 12, 2016, 08:41:34 pm by JimRemington »
 

Offline IanB

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Re: Floating a bridge rectifier
« Reply #11 on: May 12, 2016, 08:57:50 pm »
Quote
I feed it 5 Vpp. The secondary voltage is completely dropped across the 2 diodes
Not quite true. Expect about 1.4 V drop across two diodes, so about 1.1V across the resistor.

Ya gotta read the whole thread carefully before firing off a reply...  ;)
 

Offline wraper

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Re: Floating a bridge rectifier
« Reply #12 on: May 12, 2016, 09:03:23 pm »
I feed a sine wave into a transformer which is followed by a full-wave bridge rectifier. I expect to see a rectified wave at the output with respect to local ground, but that's not the case. It's flatlining. Why?

My signal generator is not floating and I suppose that, by connecting the (also mains earth referenced) scope ground pin to the circuit, I create a short. But I was under the impression that the transformer would provide isolation from mains earth -- thus floating the part of the circuit on the right of the transformer and allowing me to probe around the circuit at my hearts content. Am I wrong?

Yes, because the scope probe ground reference clip is connected back through the instruments and their line cords to the grounded side of the FG's output. So you are creating a groundloop short that bypasses the transformer. See the marked-up sketch below.

The way to do this measurement is to make a differential measurement using two scope probes with no ground clip connection, and use the scope's Subtract function (or, on many analog scopes, Invert and Add functions) to give the differential measurement on the output side. See the scopeshot below.

(And I just saw the transformer ratio. Yes... you'll need a bit more voltage on the output side of the transformer before you see anything sensible!)
I don't see any ground loop there. Yes, both sides of the transformer are connected to the GND, but transformer windings are isolated, therefore no ground loop possible. You cannot create ground loop with only one ground connection to the particular circuit which is otherwise isolated.

Quote
The way to do this measurement is to make a differential measurement using two scope probes with no ground clip connection
No way this is a correct thing to do  :palm:
« Last Edit: May 12, 2016, 09:15:31 pm by wraper »
 

Offline JimRemington

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Re: Floating a bridge rectifier
« Reply #13 on: May 12, 2016, 09:21:37 pm »
Quote
Ya gotta read the whole thread carefully before firing off a reply..
I suppose, depending on the interpretation of the word "it".
 

Offline Mr.B

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Re: Floating a bridge rectifier
« Reply #14 on: May 12, 2016, 09:26:57 pm »
Quote
Ya gotta read the whole thread carefully before firing off a reply..
I suppose, depending on the interpretation of the word "it".

The OP said the turns ratio was 230:8
That is 28.75:1
Feeding in 5Vp-p you would get approximately 0.174Vp-p out of the transformer.
Not nearly enough for two silicon diodes to start to conduct.
Also, as someone IanB said above, the 1K resistor is too high to get sufficient current flowing through the diodes. (Even if the p-p V was high enough.)

Edit: To acknowledge Ian.
« Last Edit: May 12, 2016, 09:29:14 pm by Mr.B »
I approach the thinking of all of my posts using AI in the first instance. (Awkward Irregularity)
 

Offline danadak

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Offline linux-works

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Re: Floating a bridge rectifier
« Reply #16 on: May 13, 2016, 02:59:36 pm »
that tek document was nice and to the point.  thanks for posting it.


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