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 Post subject: Bridge Rectifier
PostPosted: Jan Mon 16, 2017 8:17 pm 
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I made a power supply (see link) and I am not able to decrease the 112v DC from the bridge rectifier. I have installed resistors of various sizes (2k and 20k) in the positive side as in the photo below but the voltage does not seem to be decreased. I hook them up with the load side + to the + on the BR and the negative load side to the opposite corner. The two remaining corners are marked with a sine wave and 120 volts is applied there. The output of the BR is correct. This is a test circuit with no load and no capacitors. It is a mystery to me. I have 112v DC but I need 90v DC.

viewtopic.php?f=12&t=311200
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Last edited by Ozo on Jan Mon 16, 2017 8:28 pm, edited 2 times in total.

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 Post subject: Re: Bridge Rectifier
PostPosted: Jan Mon 16, 2017 8:25 pm 
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The voltage would only decrease across the dropping resistors, if there's a load on the power supply.

No current through the resistor = no voltage drop across the resistor. The output side of the resistor is just floating to the same voltage as the input side.

You'll have to size your dropping resistor based on the current needed by the circuit you're powering.

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 Post subject: Re: Bridge Rectifier
PostPosted: Jan Mon 16, 2017 9:34 pm 
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What Rodney said!

Bridge rectifiers on AC lines scare me... You end up with 60VAC between neutral and your B+ connection and 60VAC between neutral and B-

Long ago I built an amplifier with this setup once and damaged an FM tuner i had... plus tripped a breaker. All repairable, but isolation transformers are your friend! :D

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 Post subject: Re: Bridge Rectifier
PostPosted: Jan Mon 16, 2017 10:00 pm 
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That's an interesting point, Steve, I hadn't really thought about it.

I'll definitely keep that in mind.


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 Post subject: Re: Bridge Rectifier
PostPosted: Jan Mon 16, 2017 10:09 pm 
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The early battery chargers had common connection to the line.

Chas

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Last edited by Chas on Jan Mon 16, 2017 11:47 pm, edited 1 time in total.

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 Post subject: Re: Bridge Rectifier
PostPosted: Jan Mon 16, 2017 10:25 pm 
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You can buy a 60 volt output transformer for less than $15 and use a full wave or half wave rectifier circuit to get 90 volts dc.
https://www.digikey.com/product-detail/en/triad-magnetics/F-59X/237-2005-ND/5032164

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 Post subject: Re: Bridge Rectifier
PostPosted: Jan Mon 16, 2017 11:22 pm 
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That's an extremely dangerous design.

It has no isolation between the ground-referenced AC line and the output.

Hope your insurance is paid up.

- Leigh

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 Post subject: Re: Bridge Rectifier
PostPosted: Jan Mon 16, 2017 11:40 pm 
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A typical radio running at 90v B+ might be drawing around 50ma.

So you should place a simulated radio load of about 1800 ohms 10 watts.
Place that dummy load across the two output terminals. Then you can adjust that series resistor.

But test your typical radio first to see how much current it needs.

For a 50ma load:

90v/.05A = 1800 ohms

P= Ix E
.05a x 90v = 4.5 watts .. double for safety = 10 watts
You could probably also use a 10 watt 120v light bulb or a 7watt night-light bulb.

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 Post subject: Re: Bridge Rectifier
PostPosted: Jan Mon 16, 2017 11:44 pm 
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The schematic is misleading. The power supply actually uses an isolated winding of a transformer for the AC input. If you had looked at the info in the link I submitted that would have been clear.

Thanks for clearing up the voltage drop question. :D


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 Post subject: Re: Bridge Rectifier
PostPosted: Jan Mon 16, 2017 11:54 pm 
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Pbpix wrote:
A typical radio running at 90v B+ might be drawing around 50ma.

So you should place a simulated radio load of about 1800 ohms 10 watts.
Place that dummy load across the two output terminals. Then you can adjust that series resistor.

But test your typical radio first to see how much current it needs.

For a 50ma load:

90v/.05A = 1800 ohms

P= Ix E
.05a x 90v = 4.5 watts .. double for safety = 10 watts
You could probably also use a 10 watt 120v light bulb or a 7watt night-light bulb.


Thanks, a dummy load sounds like the solution. As far as the series resistor. The original schematic I have been using advised a 4.7k resistor for a generic farm radio. I will let you know how it works out.

EDIT: I just happen to have a 1250 ohm 10 watt resistor. I will try that for now.


Last edited by Ozo on Jan Mon 16, 2017 11:58 pm, edited 1 time in total.

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 Post subject: Re: Bridge Rectifier
PostPosted: Jan Mon 16, 2017 11:58 pm 
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Ozo wrote:
If you had looked at the info in the link I submitted that would have been clear.
Why should I do that?

If other information is relevant to your question, you should present that information in your post.

I'm not about to waste my time hopping all over the net looking for stuff.

- Leigh

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 Post subject: Re: Bridge Rectifier
PostPosted: Jan Mon 16, 2017 11:58 pm 
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I know, many don't like math, but it really does help to learn and understand Ohms Law.

E = I X R
or
Voltage = Current X Resistance

https://en.wikipedia.org/wiki/Ohm's_law

:D

-Steve

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 Post subject: Re: Bridge Rectifier
PostPosted: Jan Tue 17, 2017 12:04 am 
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azenithnut wrote:
I know, many don't like math, but it really does help to learn and understand Ohms Law.

E = I X R
or
Voltage = Current X Resistance

https://en.wikipedia.org/wiki/Ohm's_law

:D

-Steve
Thanks, that is constructive. I know what ohms law is . It is the "understanding" part that escapes me. And I do not mind hopping on over to Wiki to read your links info. I can not guarantee a full understanding of it however.


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 Post subject: Re: Bridge Rectifier
PostPosted: Jan Tue 17, 2017 12:22 am 
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Ozo wrote:
Pbpix wrote:
A typical radio running at 90v B+ might be drawing around 50ma.

So you should place a simulated radio load of about 1800 ohms 10 watts.
Place that dummy load across the two output terminals. Then you can adjust that series resistor.

But test your typical radio first to see how much current it needs.

For a 50ma load:

90v/.05A = 1800 ohms

P= Ix E
.05a x 90v = 4.5 watts .. double for safety = 10 watts
You could probably also use a 10 watt 120v light bulb or a 7watt night-light bulb.


Thanks, a dummy load sounds like the solution. As far as the series resistor. The original schematic I have been using advised a 4.7k resistor for a generic farm radio. I will let you know how it works out.

EDIT: I just happen to have a 1250 ohm 10 watt resistor. I will try that for now.

I= E/R

If you have 90v with a 1250 ohm resistor it will draw 72ma

90/ 1250 = .072A

It will get you in the ball park.
But if the actual radio draws 50ma you'll have to adjust the series resistor again later.

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 Post subject: Re: Bridge Rectifier
PostPosted: Jan Tue 17, 2017 12:35 am 
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Another option would be to put a 90 volt Zener across the output to regulate the voltage.

If none can be found, you can use six 15 volt Zeners strung in series.

http://www.digikey.com/product-detail/e ... ND/1114294

The resistor will need to be brought down to about 2000 ohms 5-10 watt (I know, 10 watt is overkill but it will run cooler)

-Steve

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 Post subject: Re: Bridge Rectifier
PostPosted: Jan Tue 17, 2017 12:36 am 
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If this P/S is to be used on a battery tube radio, B+ current will very little... Most are 12ma and less...


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 Post subject: Re: Bridge Rectifier
PostPosted: Jan Tue 17, 2017 12:41 am 
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True, in that case, the 4700 could be left alone.

-Steve

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 Post subject: Re: Bridge Rectifier
PostPosted: Jan Tue 17, 2017 12:46 am 
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35Z5 wrote:
If this P/S is to be used on a battery tube radio, B+ current will very little... Most are 12ma and less...

Thanks.
If that's the case you can use a dummy load of 7.5k ohms 2 watts
E/I =R
90v /.012A = 7,500

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 Post subject: Re: Bridge Rectifier
PostPosted: Jan Tue 17, 2017 12:52 am 
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Pbpix wrote:
35Z5 wrote:
If this P/S is to be used on a battery tube radio, B+ current will very little... Most are 12ma and less...

Thanks.
If that's the case you can use a dummy load of 7.5k ohms 2 watts
E/I =R
90v /.012A = 7,500


The schematic details for my Philco 39-75 farm radio say "Battery drain for B+ is 8.5 M. A."

http://www.nostalgiaair.org/pagesbymode ... 013306.pdf


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 Post subject: Re: Bridge Rectifier
PostPosted: Jan Tue 17, 2017 1:34 am 
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Ozo wrote:
Pbpix wrote:
35Z5 wrote:
If this P/S is to be used on a battery tube radio, B+ current will very little... Most are 12ma and less...

Thanks.
If that's the case you can use a dummy load of 7.5k ohms 2 watts
E/I =R
90v /.012A = 7,500


The schematic details for my Philco 39-75 farm radio say "Battery drain for B+ is 8.5 M. A."

http://www.nostalgiaair.org/pagesbymode ... 013306.pdf

So... simple:
To calculate your dummy load:
E / I = R
90v / .0085A = 10k ohms 2 watts

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