Half the RMS Voltage times 1.414 will be the peak AC, but for the B+ supply, a 1N4006 or 7 will handle everything you can ever put in a receiver. For the A supply, any number of bridge rectifiers can be obtained that will laugh at any filament load you will ever put on it. What I'm trying to say is, just put a higher Voltage than you will have to rectify and forget about it. A 600 Volt diode will work perfectly well in a 300 Volt circuit.
Lewis
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Actually, it depends on the circuit. Is this a battery eliminator for 6 or 12 V DC car radio, or for something else?
Here's a good resource: http://www.hammondmfg.com/pdf/5c007.pdf
I don't know of any circuit where you would use half the AC rms in calculating the diode PRV. Make sure the diode can handle the current, also.
Rich
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Rich:
Wouldn't you, for a half wave rectifier, take one-half of the AC Peak to Peak, that being the negative half cycle you are not using, say for 100 VRMS, the PP would be 141.4 Volts, and half of that would be 70.7 Volts reverse Voltage on the diode? Anyways, a 1N4006 orter hold back anything coming out of a plate transformer used for a receiver.
Lewis
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http://3.bp.blogspot.com/-QY0ctFgbNQg/Tt_Z6NZ576I/AAAAAAAACxk/G4tD6FQ0psw/s1600/ccc110.jpeg
The diode needs to be rated for at least 2 times the peak value of the AC. I use 1N4007s for nearly all applications that need a couple of hundred mA max. You can always series a few 1N4007 for higher voltage margin.
Rich
A diode has to be able to withstand 2.828 times AC input voltage. Output at the cathode can be 1.414 times AC input voltage while anode end has to withstand peaks 1.414 in the negative direction.
For 100 volts AC cathode will have peaks to 141.4 volts while anode will have 141.4 negative peaks. That's 282.8 volts for every 100 volts rms.
Some diodes seem to have excessive PIV ratings but they are necessary for higher voltage. Even a tube like 5Y3 has a PIV rating of 1400.
Norm
:::::When selecting a bridge rectifier for a battery eliminator,what is the inverse voltage rating of the diode based on?
::::
::::Half the RMS Voltage times 1.414 will be the peak AC, but for the B+ supply, a 1N4006 or 7 will handle everything you can ever put in a receiver. For the A supply, any number of bridge rectifiers can be obtained that will laugh at any filament load you will ever put on it. What I'm trying to say is, just put a higher Voltage than you will have to rectify and forget about it. A 600 Volt diode will work perfectly well in a 300 Volt circuit.
::::Lewis
:::
:::Actually, it depends on the circuit. Is this a battery eliminator for 6 or 12 V DC car radio, or for something else?
:::
:::Here's a good resource: http://www.hammondmfg.com/pdf/5c007.pdf
:::
:::I don't know of any circuit where you would use half the AC rms in calculating the diode PRV. Make sure the diode can handle the current, also.
:::
:::Rich
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::Rich:
::Wouldn't you, for a half wave rectifier, take one-half of the AC Peak to Peak, that being the negative half cycle you are not using, say for 100 VRMS, the PP would be 141.4 Volts, and half of that would be 70.7 Volts reverse Voltage on the diode? Anyways, a 1N4006 orter hold back anything coming out of a plate transformer used for a receiver.
::Lewis
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:The diode still sees the peak voltage of the sine wave, which is 1.414 x rms. The first filter capacitor will charge to that value, assuming no load on the system.
:http://www.radio-electronics.com/info/circuits/diode-rectifier/half-wave-rectifiers-circuits.php
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:http://3.bp.blogspot.com/-QY0ctFgbNQg/Tt_Z6NZ576I/AAAAAAAACxk/G4tD6FQ0psw/s1600/ccc110.jpeg
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:The diode needs to be rated for at least 2 times the peak value of the AC. I use 1N4007s for nearly all applications that need a couple of hundred mA max. You can always series a few 1N4007 for higher voltage margin.
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:Rich
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You're right, Norm, I figgered it out last night while waiting for sleep. On the positive half cycle, the filter cap will charge to 141.4 Volts, and on the negative half cycle, another 141.4 Volta will be added to the charge, making the diode see 282.8 Volts as a PIV.
Amazing how fast you can forget stuff you knew so well fifty years ago, ain't it?
Lewis
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::::::When selecting a bridge rectifier for a battery eliminator,what is the inverse voltage rating of the diode based on?
:::::
:::::Half the RMS Voltage times 1.414 will be the peak AC, but for the B+ supply, a 1N4006 or 7 will handle everything you can ever put in a receiver. For the A supply, any number of bridge rectifiers can be obtained that will laugh at any filament load you will ever put on it. What I'm trying to say is, just put a higher Voltage than you will have to rectify and forget about it. A 600 Volt diode will work perfectly well in a 300 Volt circuit.
:::::Lewis
::::
::::Actually, it depends on the circuit. Is this a battery eliminator for 6 or 12 V DC car radio, or for something else?
::::
::::Here's a good resource: http://www.hammondmfg.com/pdf/5c007.pdf
::::
::::I don't know of any circuit where you would use half the AC rms in calculating the diode PRV. Make sure the diode can handle the current, also.
::::
::::Rich
:::
:::Rich:
:::Wouldn't you, for a half wave rectifier, take one-half of the AC Peak to Peak, that being the negative half cycle you are not using, say for 100 VRMS, the PP would be 141.4 Volts, and half of that would be 70.7 Volts reverse Voltage on the diode? Anyways, a 1N4006 orter hold back anything coming out of a plate transformer used for a receiver.
:::Lewis
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:::::
:::::
::::
::::
:::
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::The diode still sees the peak voltage of the sine wave, which is 1.414 x rms. The first filter capacitor will charge to that value, assuming no load on the system.
::http://www.radio-electronics.com/info/circuits/diode-rectifier/half-wave-rectifiers-circuits.php
::
::http://3.bp.blogspot.com/-QY0ctFgbNQg/Tt_Z6NZ576I/AAAAAAAACxk/G4tD6FQ0psw/s1600/ccc110.jpeg
::
::The diode needs to be rated for at least 2 times the peak value of the AC. I use 1N4007s for nearly all applications that need a couple of hundred mA max. You can always series a few 1N4007 for higher voltage margin.
::
::Rich
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