I'd like to make a solid state replacement for some 6X5 rectifier tubes that I have in a couple of signal generators that I have. I have a couple of questions and was wondering if anyone had any input. First the Diodes, depending on which tube you're using the regular, G or GT the PIV could be between 1250 and 1600 but the diodes (I'm using IN5408's) have a PIV of 1000V. Do I need 2 of these in series for each leg to get an effective PIV of 2000V or is that just overengineering in this case?

For the resistance to bring B+ down to normal. Do i just take the voltage drop (difference between normal B+ and the B+ with just diodes installed) and divide by the DC current which I think is about 750mA, to get the resistance. Then just I^2 R to get the wattage.

Hi:

That is basically correct although I arrive at a resistance value somewhat empirically. I add a 10 ohm, 10 watt resistor in series with the diode, install the unit and check the value of B+. If it is high, I increase R. This may be trial and error but you may not exactly know the value of load current you are going to deal with. The resistor is sized from experience ( and I squared R plus some margin).

Lou

:I'd like to make a solid state replacement for some 6X5 rectifier tubes that I have in a couple of signal generators that I have. I have a couple of questions and was wondering if anyone had any input. First the Diodes, depending on which tube you're using the regular, G or GT the PIV could be between 1250 and 1600 but the diodes (I'm using IN5408's) have a PIV of 1000V. Do I need 2 of these in series for each leg to get an effective PIV of 2000V or is that just overengineering in this case?
:
:For the resistance to bring B+ down to normal. Do i just take the voltage drop (difference between normal B+ and the B+ with just diodes installed) and divide by the DC current which I think is about 750mA, to get the resistance. Then just I^2 R to get the wattage.
:
:

Lou, advises adding a margin to the I^2R power resistor rating. I would double it.

Not sure what you're up to - but NOS 6X5s go for about $5.

Yes - good advice. Double the rating to account for tolerances and ambient temp.

But Doug took the words right out of my mouth!! Why bother? The 6X5 rectifier is one of the CHEAPEST tubes you can find, and it is readily available too.

:Lou, advises adding a margin to the I^2R power resistor rating. I would double it.
:
:Not sure what you're up to - but NOS 6X5s go for about $5.
:

:Yes - good advice. Double the rating to account for tolerances and ambient temp.
:
:But Doug took the words right out of my mouth!! Why bother? The 6X5 rectifier is one of the CHEAPEST tubes you can find, and it is readily available too.
:
::Lou, advises adding a margin to the I^2R power resistor rating. I would double it.
::
::Not sure what you're up to - but NOS 6X5s go for about $5.
::
:
:
And here's a word of warning. If you wire diodes to a plug that plugs into a tube socket, don't forget that any exposed connections will be carrying lethal voltages on top of the chassis. Shock hazard.

Yes, maybe it's a bit of paranoia. The signal generator that I just repaired had a bad transformer so I immediately suspected the 6X5 tube. I don't want to blow out the new transformer and have to spend another $35. So it's really just an insurance policy for me. Although 6X5's are cheap, diodes are really cheap, 35cents each. So why not?

::Yes - good advice. Double the rating to account for tolerances and ambient temp.
::
::But Doug took the words right out of my mouth!! Why bother? The 6X5 rectifier is one of the CHEAPEST tubes you can find, and it is readily available too.
::
:::Lou, advises adding a margin to the I^2R power resistor rating. I would double it.
:::
:::Not sure what you're up to - but NOS 6X5s go for about $5.
:::
::
::
:And here's a word of warning. If you wire diodes to a plug that plugs into a tube socket, don't forget that any exposed connections will be carrying lethal voltages on top of the chassis. Shock hazard.
:

Thanks John, these are going to be for a signal generator and so its an enclosed unit. Thanks,

::Yes - good advice. Double the rating to account for tolerances and ambient temp.
::
::But Doug took the words right out of my mouth!! Why bother? The 6X5 rectifier is one of the CHEAPEST tubes you can find, and it is readily available too.
::
:::Lou, advises adding a margin to the I^2R power resistor rating. I would double it.
:::
:::Not sure what you're up to - but NOS 6X5s go for about $5.
:::
::
::
:And here's a word of warning. If you wire diodes to a plug that plugs into a tube socket, don't forget that any exposed connections will be carrying lethal voltages on top of the chassis. Shock hazard.
:

I made an octal plug with 2 diodes to replace a faulty 5Y3 tube. this way, I could sub a new tube in someday.
Several years later, I was doing something else in that radio and made contact with an exposed wire on the plug. I can now say my heart can withstand 350 volts at least once. But it was an enlightening experience I would not be cavalier about.

Under the chassis, no problem. We expect the high voltages there.

:Thanks John, these are going to be for a signal generator and so its an enclosed unit. Thanks,
:
:::Yes - good advice. Double the rating to account for tolerances and ambient temp.
:::
:::But Doug took the words right out of my mouth!! Why bother? The 6X5 rectifier is one of the CHEAPEST tubes you can find, and it is readily available too.
:::
::::Lou, advises adding a margin to the I^2R power resistor rating. I would double it.
::::
::::Not sure what you're up to - but NOS 6X5s go for about $5.
::::
:::
:::
::And here's a word of warning. If you wire diodes to a plug that plugs into a tube socket, don't forget that any exposed connections will be carrying lethal voltages on top of the chassis. Shock hazard.
::
:
:

I'm guessing this has to do with "6X5 Paranoia", a widespread condition caused by taking internet lore as gospel.

:Lou, advises adding a margin to the I^2R power resistor rating. I would double it.
:
:Not sure what you're up to - but NOS 6X5s go for about $5.
:

Maybe I don't get it Andy.

What internet lore is this?

Meekly
Lou

:I'm guessing this has to do with "6X5 Paranoia", a widespread condition caused by taking internet lore as gospel.
:
::Lou, advises adding a margin to the I^2R power resistor rating. I would double it.
::
::Not sure what you're up to - but NOS 6X5s go for about $5.
::
:
:

That 6X5's are all bad, no matter how they are used.

:Maybe I don't get it Andy.
:
:What internet lore is this?
:
:Meekly
:Lou
:
::I'm guessing this has to do with "6X5 Paranoia", a widespread condition caused by taking internet lore as gospel.
::
:::Lou, advises adding a margin to the I^2R power resistor rating. I would double it.
:::
:::Not sure what you're up to - but NOS 6X5s go for about $5.
:::
::
::
:
:

Thanks Lou, I was thinking of putting an ammeter between the diode and the cathode pin on each leg so I can see what the load current is through each leg for the power rating. If I sum these 2 currents wouldn't that be the total load current at the cathode.
Steve

:Hi:
:
:That is basically correct although I arrive at a resistance value somewhat empirically. I add a 10 ohm, 10 watt resistor in series with the diode, install the unit and check the value of B+. If it is high, I increase R. This may be trial and error but you may not exactly know the value of load current you are going to deal with. The resistor is sized from experience ( and I squared R plus some margin).
:
:Lou
:
::I'd like to make a solid state replacement for some 6X5 rectifier tubes that I have in a couple of signal generators that I have. I have a couple of questions and was wondering if anyone had any input. First the Diodes, depending on which tube you're using the regular, G or GT the PIV could be between 1250 and 1600 but the diodes (I'm using IN5408's) have a PIV of 1000V. Do I need 2 of these in series for each leg to get an effective PIV of 2000V or is that just overengineering in this case?
::
::For the resistance to bring B+ down to normal. Do i just take the voltage drop (difference between normal B+ and the B+ with just diodes installed) and divide by the DC current which I think is about 750mA, to get the resistance. Then just I^2 R to get the wattage.
::
::
:
:

Steve

Sounds good but a regular meter won't give proper results in this circuit. Current through a rectifier is pulsing. The rectifier only conducts during peak of each wave. At that time high current.

Resistor required will be lower resistance and higher wattage rating than calculated.

Norm

:Thanks Lou, I was thinking of putting an ammeter between the diode and the cathode pin on each leg so I can see what the load current is through each leg for the power rating. If I sum these 2 currents wouldn't that be the total load current at the cathode.
:Steve
:
::Hi:
::
::That is basically correct although I arrive at a resistance value somewhat empirically. I add a 10 ohm, 10 watt resistor in series with the diode, install the unit and check the value of B+. If it is high, I increase R. This may be trial and error but you may not exactly know the value of load current you are going to deal with. The resistor is sized from experience ( and I squared R plus some margin).
::
::Lou
::
:::I'd like to make a solid state replacement for some 6X5 rectifier tubes that I have in a couple of signal generators that I have. I have a couple of questions and was wondering if anyone had any input. First the Diodes, depending on which tube you're using the regular, G or GT the PIV could be between 1250 and 1600 but the diodes (I'm using IN5408's) have a PIV of 1000V. Do I need 2 of these in series for each leg to get an effective PIV of 2000V or is that just overengineering in this case?
:::
:::For the resistance to bring B+ down to normal. Do i just take the voltage drop (difference between normal B+ and the B+ with just diodes installed) and divide by the DC current which I think is about 750mA, to get the resistance. Then just I^2 R to get the wattage.
:::
:::
::
::
:
:

Thanks Norm, that makes a lot of sense. So then if I instead measured the current from the junction of the 2 diodes to the cathode pin, I'd be measuring the peak of the fully rectified current. So if I just multiply that by .707, I should get close to the RMS value of the load current.

:Steve
:
: Sounds good but a regular meter won't give proper results in this circuit. Current through a rectifier is pulsing. The rectifier only conducts during peak of each wave. At that time high current.
:
: Resistor required will be lower resistance and higher wattage rating than calculated.
:
:Norm
:
::Thanks Lou, I was thinking of putting an ammeter between the diode and the cathode pin on each leg so I can see what the load current is through each leg for the power rating. If I sum these 2 currents wouldn't that be the total load current at the cathode.
::Steve
::
:::Hi:
:::
:::That is basically correct although I arrive at a resistance value somewhat empirically. I add a 10 ohm, 10 watt resistor in series with the diode, install the unit and check the value of B+. If it is high, I increase R. This may be trial and error but you may not exactly know the value of load current you are going to deal with. The resistor is sized from experience ( and I squared R plus some margin).
:::
:::Lou
:::
::::I'd like to make a solid state replacement for some 6X5 rectifier tubes that I have in a couple of signal generators that I have. I have a couple of questions and was wondering if anyone had any input. First the Diodes, depending on which tube you're using the regular, G or GT the PIV could be between 1250 and 1600 but the diodes (I'm using IN5408's) have a PIV of 1000V. Do I need 2 of these in series for each leg to get an effective PIV of 2000V or is that just overengineering in this case?
::::
::::For the resistance to bring B+ down to normal. Do i just take the voltage drop (difference between normal B+ and the B+ with just diodes installed) and divide by the DC current which I think is about 750mA, to get the resistance. Then just I^2 R to get the wattage.
::::
::::
:::
:::
::
::
:
:

Not exactly. The diode current is feeding into a filter capacitor. During some or much of the cycle, the current is cut off because the diode ouput voltage is less than the capacitor voltage. So, the current waveform will not be a rectified sinewave - it will be sort of a truncated sinewave, so 0.707 isn't the correct factor.

Even most Fluke "true RMS" meters won't give the correct current, because the current waveform isn't symmetrical about the zero axis.

But the RMS current is only important for calculating the power rating of the resistor. But for that, estimate it or measure it anyway you want, and double for the resistor rating.

It's the B+ voltage that you are trying to match. For that, guess at the resistor dropping ohms, and go by trial and error - it shouldn't be all that critical.

:So then if I instead measured the current from the junction of the 2 diodes to the cathode pin, I'd be measuring the peak of the fully rectified current. So if I just multiply that by .707, I should get close to the RMS value of the load current.

It's the B+ voltage that you are trying to match. For that, guess at the resistor ohms, and go by trial and error.

:
:
::Steve
::

Thanks Doug, The engineer in me want to measure it but like you said, there's really no need to. I'll try the trial and error mothod.

:Not exactly. The diode current is feeding into a filter capacitor. During some or much of the cycle, the current is cut off because the diode ouput voltage is less than the capacitor voltage. So, the current waveform will not be a rectified sinewave - it will be sort of a truncated sinewave, so 0.707 isn't the correct factor.
:
:Even most Fluke "true RMS" meters won't give the correct current, because the current waveform isn't symmetrical about the zero axis.
:
:But the RMS current is only important for calculating the power rating of the resistor. But for that, estimate it or measure it anyway you want, and double for the resistor rating.
:
:It's the B+ voltage that you are trying to match. For that, guess at the resistor dropping ohms, and go by trial and error - it shouldn't be all that critical.
:
:
::So then if I instead measured the current from the junction of the 2 diodes to the cathode pin, I'd be measuring the peak of the fully rectified current. So if I just multiply that by .707, I should get close to the RMS value of the load current.
:
:It's the B+ voltage that you are trying to match. For that, guess at the resistor ohms, and go by trial and error.
:
:
::
::
:::Steve
:::
:
:

Why don't you leave the tube in the circuit and add two solid state diodes in series with each plate and the power transformer. If there is a plate to plate short in the tube, the diodes will protect the power transformer. You will also still have the voltage drop of the tube. The B+ would still have the correct voltage.

Thanks Tom, that's a good idea.

:Why don't you leave the tube in the circuit and add two solid state diodes in series with each plate and the power transformer. If there is a plate to plate short in the tube, the diodes will protect the power transformer. You will also still have the voltage drop of the tube. The B+ would still have the correct voltage.
: