Get 27V AC from 220V AC

hello,
I have a PL504 vacuum tube (http://www.r-type.org/pdfs/pl504.pdf) and I need to drive its filament from 220V AC source without using a transformer, so a ballast resistor must be used.
The tube filament draws 27V 300mA

What value of resistor do I need? (Ohms, Watt)

There will be a fairly high voltage from the filament to the cathode that might cause hum or might cause breakdown.

The PL504 brings back memories...

The resistor value is (220 - 27) / 0.3 = 643Ω but it will have to be a big one, the power it will dissipate is 58W when the filament (heater) is at running temperature. You probably want to make this at least 75W to add a safety margin and to allow for the current rush of a cold filament when you switch on.

Alternatively, if your AC frequency is 50Hz you could use a capacitor of 5uF which would have an equivalent resistance of 636 Ohms and make up the difference with a smaller resistor. If you do this, the capacitor must be rated at 250V AC and you must be aware that a capacitor usually breaks down short-circuit so put a fuse in line with it.

As Audioguru states, be careful of the heater to cathode voltage which is rated at 250V peak. This is absolute peak, not RMS so assuming you are going to ground the cathode, put the ballast resistor is the 'live' side of the AC supply.

Brian.

I am thinking of completely isolate the filament from the cathode (they are electrically isolated inside the tube).
So making one circuit to feed the filaments (using AC) and then making another circuit to feed the anode and grid. (using a bridge rectifier, to avoid the unwanted situation of electrified when inserting the plug on the other way)

This way I could probably produce a quick and dirty transformerless transmitter.

By the way the resistor you mention confirms my calculations (I was a bit curious if the calculations are different if using AC or DC).
I will use 5x20W resistors to create a safe 100W one.

It is much easier to find and cheaper than the transformer, although I am aware of the AC hum. There are some ways to minimize this (like twist the cables and some coil-capacitor filters)

It is safe to treat RMS as DC for this kind of calculation. The load (the heater filament in the PL504) is resistive so there is no problem of phase shift, the voltage and current are in phase all the time.

The reason for using RMS is to give an 'equivalent' value that can be used in simple calculations when the real voltage is constantly changing and has an average of zero.

Still be careful that the heater to cathode voltage is kept as low as possible. I appreciate they are physically isolated but the filament is not supported centrally inside the cathode 'pipe'. The insulation if it touches the sides is very thin.

Brian.

If using a resistor in series with the filament, the filament voltage will me always 27V, so there shall not be a paroblem with the filament to cathode voltage, or will be?

The filament will always have 27 volts ACROSS it. The problem is that inside the PL504, the filament is inside a metal tube which is the cathode. The 250V limit is the maximum BETWEEN any part of the filament and the cathode. If you go higher than 250V you run the risk of the insulation layer on the inside of the cathode breaking down.

For example, if you have the cathode wired to ground and the filament wired to 220V AC, the peak voltage of the AC will be about 220 x √2 = 312 V. You can do it safely if you ground one side of the filament and put the resistors from the other side to 220V. If you wire the filament to 220V and put the resistors in the ground side the voltage will be too high.

Be VERY,VERY careful if you are running the filament directly from AC mains !!

Brian.

>>>
For example, if you have the cathode wired to ground and the filament wired to 220V AC, the peak voltage of the AC will be about 220 x √2 = 312 V. You can do it safely if you ground one side of the filament and put the resistors from the other side to 220V. If you wire the filament to 220V and put the resistors in the ground side the voltage will be too high.

Now I get it, If using DC (diode rectified) for the filaments, first connect the power resistor, and then the filament (to ground).


>>>Be VERY,VERY careful if you are running the filament directly from AC mains !!

I was thinking of not using any diodes, i.e. driving the filament directly from AC, through a power resistor. The filament will not be connected at any point on the ground. What do you think of it?

Instead of connecting it like this:
AC---RESISTOR---FILAMENT---AC

I should connect it like this?
AC---LOWER VALUE RESISTOR----FILAMENT---LOWER VALUE RESISTOR---AC

i.e connecting a smaller value resistor in each side of the filament. Will it be of any good?

Thanks a lot!

There is no problem running the filament directly from the AC mains as long as you can be sure that voltage between the filament and cathode can never go above 250V. The problem is not where the filament power comes from, it is the voltage difference between the filament and cathode that matters.

It would seem you are going to use a separate power source for the main supply line to the PL504. What is important is the voltage difference between the main supply and the AC to the filament. For example, suppose you use the resistors at each end of the filament like you suggest, the voltage will be something like this:

220V --- Resistor --- Filament --- Resistor --- 0V

so if you are using equal resistances at each side you have RMS:

220V --- 320Ω --- (123.5V) Filament (96.5V) --- 320Ω --- 0V

which is perfectly OK to run the filament.

But now suppose the main supply (HT to the PL504) is on the secondary of a transformer and is floating at around half the AC value. The voltage between the filament and cathode will reach around 280V and you risk the insulation breaking down. Consider what might happen if you short circuit the AC mains directly to circuitry on the secondary side of the HT transformer, it could be dangerous and even lethal if you have external connections that can be touched.

Can you tell me what the PL504 is being used for. It is designed to be a 'line scan output' device which handles short high voltage pulses.

Brian.

Yes, I am building a quick and dirty transmitter with the PL504 like this one **broken link removed**
but without using transformers (if possible) as they are hard to get and big. The purpose is to build it from commonly available shack materials.

I will feed the HV of the tube with a circuit similar to this **broken link removed** but I will probably use a bridge rectifier to avoid getting electrified if the phase of the 220v is at the chassis side.
The mains voltage measured is actually 240V

Many amateurs use this technique for the HV side of the tubes but I have only seen feeding the filaments directly from mains, in old television sets. Something like this is what I firstly have tought **broken link removed**

I would greatly appreciate a solution proposition on my problem (if it exists)
Thank you all

I would guess the amplifier with the mains powered filaments is from the USA where the mains is usually around 115V and therefore less likely to arc inside the filament.

The first schematic has a serious error - there is an electrolytic capacitor wired across the filament supply which will almost certainly explode !

I can see what you are trying to do and why you need the 27V. The design with the '504 uses an EL504 not a PL504 which has a 6.3V filament. I can not see any safe alternative to a transformer unfortunately, for the HT and the filament. You could try using two 220/12 transformers with the primary sides in parallel and secondaries in series. That would give you 24V which should be near enough and the 0.3A load is quite small so they would not have to be very large.

Brian.

The purpose of the capacitor (and the coil) in the first schematic in the filaments is to filter out the hum. It should NOT be an electrolytic though as there is AC there.

Using transformers IS the only safe way indeed. I was just tried to reduce cost and weight at a great risk though.
I have many spares of PL504, do you think I should try the resistor configuration or not?

betwix,

There is no problem running the filament directly from the AC mains as long as you can be sure that voltage between the filament and cathode can never go above 250V. The problem is not where the filament power comes from, it is the voltage difference between the filament and cathode that matters.

But now suppose the main supply (HT to the PL504) is on the secondary of a transformer and is floating at around half the AC value. The voltage between the filament and cathode will reach around 280V and you risk the insulation breaking down. Consider what might happen if you short circuit the AC mains directly to circuitry on the secondary side of the HT transformer, it could be dangerous and even lethal if you have external connections that can be touched.

Click to expand...

If I run the tube filament using AC and in a configuration like this:
220V --- 320Ω --- (123.5V) Filament (96.5V) --- 320Ω --- 0V
(i.e. 220v---resistor---filament---resistor---220v)

How can the cathode voltage (cathode connected to ground) ever go more than 250v? I am asking because the filament will always "see" 27v because of the resistors in both ends and the cathode will be connected to the ground. (filament and cathode are isolated internally)

just a hint, I do not use a transformer for the high voltage neither, but a bridge rectifier and then a capacitor-resistor-capacitor (PI) filter. And also 3 fuses, one for the 220v mains one for the high voltage and one for the filament circuit.


I am sorry I flood you with so many points

If you are connecting the cathode to the negative side of the supply (not ground !!!) directly from your bridge rectifier, I would suggest you do this:

1. Connect cathode directly to negative rail.
2. Connect one side of heater filament directly to the negative rail.
3. Connect the other side of the filament to the 640Ω resistors.
4. Connect the 640Ω resistors to the positive side of two new diodes wired like the positive side of the bridge rectifier.

I means the two new diodes will carry the extra 0.3A to power the heater filament but 1N4007 or similar will handle enough current easily. Doing it this way will ensure the heater filament to cathode voltage will stay well within safe limits.

You would be duplicating the positive output of the bridge rectifier but without the capacitor filter - just using raw rectified DC. The negative side of the bridge would also be carrying the filament current but most rectifiers will handle an extra 0.3A without difficulty.

Be very careful if you wire a bridge rectifier across the mains, remember that the negative side is NOT ground and it must never be wired directly to ground !!

Brian.

Ok,
You mean something like the picture?
(the first bit is the high voltage I am thinking of and the second bit is the filament psu based on your instructions)

Please correct any errors you believe.

That is correct, but you do not need two bridge rectifiers. The diodes in the top bridge already connect to the negative line so you do not need to duplicate them.

Just use one diode from each side of the AC to provide the positive voltage for the filament. Let the negative side return through the bridge providing the HT supply.

Be careful what you call ground. The chassis is the negative side of the supply but it will not be at ground potential. If you connect it directly to ground you will short out the mains AC supply. The chassis will be 'live' which ever way you connect the AC supply wires. For RF purposes, if you want to connect the chassis to earth you must do it through high voltage capacitors.

Brian.

Just to confirm,
You mean something like the attached picture?

Also I am worried about the value of the 640Ohm resistor. Because it is connected to rectified DC the input voltage will be 220x1.41 now and not 220v. So I need to do the calculations for this new input voltage?


Because the EL504 transmitter antenna is inductively coupled to the transmitter (**broken link removed**), should I connect the earth side of the antenna to the real ground, and not to the negative side of the supply? (to avoid get electrified when I touch the earth connectors of the cables

The problem with doing it that way is there is an extra 300mA load on the HT supply so the capacitors must be made much larger.

I have tried to attach a picture of what I mean but forgive me if it is not clear. Due to family problems at the moment, I am connecting through a mobile phone so I do not have my usual tools available.

Yes, as long as no part of the unit is connected to 'Earth' ground you should be OK. Remember to couple the RF drive inductively as well as do not have any part of the circuit, including the chassis itself where it can be touched.

Brian. (GW6BWX)

Thenk you very much! the image is very clear OK.

What about the value of the 640Ohm resistor. Because it is connected to rectified DC the input voltage will be 220x1.41 now and not 220v. So I need to do again the calculations for this new higher input voltage?

No, it's rectified without a reservoir capacitor, so the RMS voltage is still about 220. By the way, what's your motivation to use a bulky power resistor instead of a small transformer?