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Old 9th January 2014, 05:18 PM   #1
wicked1 is offline wicked1  United States
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Default Rectifier question, voltage across diode.

I was reading an old thread about using thyratrons as rectifiers.

Basically, a person said the PIV is 300v, so can't use more than a 300-0-300tx. But then later went on to say he built a bridge rectifier w/ 2 thyratrons along w/ 2 SS diodes to get 500v from a 300 + 300 transformer.

But, doesn't each diode see the full 600v at a point through the cycle in a bridge?
I can see how in full wave they each only see 300v in reference to the center tap, where you'd end up w/ under 300vdc, but how does that work in a bridge to get 500v?


(This is the post if you care: thyratron rectifier)
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Old 9th January 2014, 05:44 PM   #2
DF96 is offline DF96  England
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The diode takes part of the PIV, thus protecting the thyratron.

I can't see the point myself, unless it is for cosmetic reasons. The best thing for rectification is a rectifier.
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Old 9th January 2014, 05:50 PM   #3
wicked1 is offline wicked1  United States
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I just set it up in a circuit simulator and see what I was expecting..
And looked up the definition of PIV, and for a bridge it says:
PIV=the full voltage of the transformer secondary, so in this example it would be 600v right?

I know it's a bad idea and I will probably not build a power supply w/ them.. I'm just trying to understand.
(I am thinking about an 'over the top' amp w/ many more glowing tubes than I need... It probably won't really happen, though)
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Old 9th January 2014, 05:51 PM   #4
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With a thyratron tube, it's possible to 'pulse' the grid and vary the output voltage of the power supply. RCA did this in their BTA 5G/10G and BTA 5H/10H AM transmitters. It was quite effective. With today's technology, I'm not sure why anyone would want to use thyratron tubes in a power supply.
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Old 9th January 2014, 05:53 PM   #5
fpitas is offline fpitas  United States
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Real men use Ignitrons, anyway ;-)
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Old 9th January 2014, 06:09 PM   #6
wicked1 is offline wicked1  United States
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Quote:
Originally Posted by Frank Berry View Post
With today's technology, I'm not sure why anyone would want to use thyratron tubes in a power supply.
Todays technology doesn't glow blue or give you the thrill of exposed 600v plate caps.
I know it's a bad idea, technically. But looking at pics of the big amps w/ all the blue MV rectifiers, Bright transmitting tube filaments, and plenty of glowing VR tubes, it gets tempting to build one. There are enough examples of such amps, it's clear I'm not the only one tempted by aesthetics.
I'm really just trying to understand the things I read here so I can learn.

well, Gluca is still active here and replied to me saying the same thing DF says.. The other diodes share the PIV w/ the thyratrons.
That is contradictory to the explanations and examples on simple scholastic "learn EE" type websites I just found.. But those may be very over simplified explanations, so I will go w/ what you all say here.
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Old 9th January 2014, 06:22 PM   #7
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The PIV is a property of the diode and stands for - surprise! - Peak Inverse Voltage.
This is the peak voltage the diode can withstand without breaking down.
So if the PIV is 600V that's the max the diode can withstand.

Now to the actual use: if you have a transformer winding with V volts RMS, you must realise that the cap on the output side of the diode gets charged to the transformer peak voltage which is 1.4 X V. Then when the output wave goes down, the diode starts to block and the other side of the diode goes down to - 1.4 * V. So the diode has now 2 * 1.4 * V volts across it.

Thus, if your diode PIV is 600V, the absolute max voltage from a transformer that can be used is 600/(2 * 1.4) = 214V RMS....

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Old 9th January 2014, 07:12 PM   #8
wicked1 is offline wicked1  United States
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Thanks, your explanation is much better than what I found, and I understand it!


OH, and I think the voltage drop thing just clicked....
it's because the bottom of the transformer isn't grounded. The other two diodes in the bridge make a virtual ground. So the drop across each diode is 1/2.. got it.
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