Line Cord Resistors and other Heater Droppers.

This page has come about mainly as a result of having open circuit line cord resistors in every single one of my Meck FM converters. Here, I look at the options for modern substitutes and other schemes for dropping the mains voltage for where a series heater power supply is used.
There is also much misinformation regarding diode and capacitor droppers on the internet, and sadly, some radios are having their valves damaged because of this. So, I hope to clear this up and show how to calculate the correct components.
Where a transformerless power supply is used for a radio or television set, the most common method of powering the valve heaters is to connect them all in series. I talk about the technique in some detail here.
In the early days, this series heater arrangement was used to permit operation from DC mains as transformer type power supplies are suitable for AC only. However, the scheme remained in use even when DC mains were no more. The main reason was that it eliminated the heavy, bulky, and expensive power transformer. This convenience comes at a disadvantage; the internal circuitry, and usually the chassis is connected to one side of the mains making it a shock hazard. Some apparatus was better than others in regard to the steps taken to reduce the shock hazard to the user.

Methods for voltage dropping.
Where the sum of valve heater voltages adds up to less than the mains voltage, the difference has to be dropped by some means. Methods to do this include resistors, barretters, light bulbs, or when the supply is AC only, capacitors, diodes, or transformers. Resistive devices are cheap and popular, as well as functioning on both AC or DC mains. They can be in the form of an ordinary wirewound resistor, a barretter, light bulb, ballast tube, or a line cord resistor. The disadvantage is all the voltage dropped is converted to heat; 30W is not atypical of the sort of dissipation. The overall power consumption is also higher than if a power transformer was/could be used. When the mains is AC only, other methods can be used. One well known one is to use the reactance of a capacitor. The advantage here is no heat dissipated, but there are drawbacks to be discussed later. Another method, very popular in UK made television sets, is to use a silicon diode, presenting a half wave current to the valve heaters. By using a diode this way, the dissipation in the heater dropper resistor is reduced because the diode only conducts on every half cycle of the mains sine wave. Again, there are drawbacks to this scheme. Of course a transformer can be used, and may be an auto transformer to reduce bulk and expense.


Resistive droppers.

Here, R is any of the resistive devices listed above. The value is found by ohms law where R=V/I. V is the sum of valve heater voltages. For example, lets assume the set uses a 14F8, 12AT6, and 35W4 and is to operate from 120V mains. The sum of heater voltages is 60.2. Now, the voltage to be dropped (V) is 120-60.2, or  59.8V. The current consumption (I) is 150mA. So, R= 59.8/.15 or 399 ohms. The power dissipated by this resistor is I^2*R. So, .15^2/399 =  8.98W.
Where an ordinary resistor is used and it fails, it's easy enough to use a modern replacement, using resistors in series or parallel to get the correct resistance, which is critical and should not automatically be substituted with the nearest preferred value.
Many circuits include a negative temperature coefficient (NTC) thermistor in series with the dropper resistor. The purpose of this is to remove the switch on surge current that occurs when the valve heaters are cold. As current flows, the thermistor gradually warms up because of the voltage drop across it. The resistance then starts to fall to a certain point at which the full heater current flows.
Where a thermistor is not used, it is worth noting that the less the value of dropper resistance, the greater the switch on surge. It goes to reason, for example, that if the dropper was 240 ohms, it would be impossible for more than 1A to flow with a 240V supply. If, however, the dropper was only 120 ohms, then the surge could be as high as 2A.

Barretters were popular in Australian and European AC/DC or DC only sets. These resemble a domestic light bulb but have an iron filament in a hydrogen atmosphere. Usually they have an E27 Edison screw or P base. Unlike a resistor or light bulb, they regulate the current over a wide mains voltage. So, a typical set could run from 200-250V with no need to adjust anything. Also, the switch on surge is reduced. It would be possible to use an ordinary light bulb and forego the regulation feature as a replacement, but more than likely, it will be necessary to provide a resistor as well to get the correct heater current.
Ballast tubes existed in some American sets from around the 1930's and 40's. One kind is a wirewound resistor assembled in what looks like a perforated metal valve, and has an ordinary valve base. It may be possible to fit modern resistors inside the enclosure. Even if not, they could be put elsewhere inside the set, above the chassis. The other kind of ballast tube is in a glass envelope and is simply a barretter under another name.
Light Bulbs can be used, but resistance varies depending on the current flow, so it isn't practical to calculate what wattage lamp is to be used and what the associated resistor (if required) will be. It will have to be done experimentally. As a starting point, it is easy to determine what the current consumption of a light bulb at its rated voltage is. P(power of lamp)=I(current)*V(voltage). For example, a 240V 60W bulb draws about 250mA. A 75W 240V bulb would probably be a good starting point for use in a set that had a 300mA heater string (You have hoarded a lifetime supply of incandescent bulbs haven't you? Because you won't be able to use a CFL in this application!). Keep in mind that the resistance of a light bulb is much lower cold than hot so the switch on surge could be a problem. A thermistor could be of use here.
Ordinary light bulbs have a tungsten filament so have the same temperature coefficient as a string of valve heaters. Interestingly, a carbon filament bulb performs much like a negative coefficient thermistor. This data was measured using two 240V 100W bulbs; one with a tungsten filament, and the other with a carbon filament:
 
 
Volts Carbon filament (mA) Tungsten filament (mA)
40 45 155
80 110 230
115 170 270
140 235 310
180 320 360
215 395 390
240 415 420
Cold resistance 786 ohms 40.6 ohms
Characteristics of 240V 100W carbon and tungsten filament bulbs.

From this, one could deduce that the carbon filament provides good inrush current protection, with such a high cold resistance. However it does not provide current regulation like a barretter, so is unsuitable for operating over a wide voltage range. As mains voltages have been standardised for some time, this is not as important as it once was. The tungsten lamp has less of a current variation over a certain voltage range, but offers no surge protection at all. The effect of using a tungsten lamp as a valve heater dropper is the same as if all the valve heaters added up to the mains voltage. For example, if the mains voltage rises 10%, then each of the valve heater voltages rises by 10%.
In comparison, an ordinary wirewound resistor does offer useful surge protection, but as the resistance does not vary with current, the valve heaters will be subjected to a greater variation in voltage than with a tungsten lamp dropper.
For these reasons it can be seen that the barretter is actually the ideal dropper.

Line cord resistors took over from ballast tubes and other dropper resistors mounted inside radio cabinets and lasted until the 1950's when they fell from use. By then, a series of valves had been developed for typical radio use which had 150mA heaters, and when used together added up to around 122V, thus dispensing with the need for any dropper. Line cord resistors were not used with Australian sets due to their safety hazards.
The reason for their popularity is that they allowed midget sets to be constructed, as all the heater dropper heat was dissipated outside the cabinet. Room was also not required for the large resistor or ballast tube.
A line cord resistor looks like an ordinary cloth covered appliance cord, but closer examination reveals three conductors. There are two ordinary conductors, one for (what is hopefully) the neutral, the other to feed the rectifier plate with the mains voltage. The third conductor is actually resistance wire wound around the length of the cord and provides the heater voltage. Needless to say, shortening this kind of cord will subject the heaters to excessive voltage. Provided the cord is left stretched out, it gets rid of the heat effectively. If left coiled up, it could be a fire hazard.
The insulation around the resistance wire is unreliable making them a shock hazard. Also, the resistance wire is not as flexible as the ordinary conductors. So, continual rolling up or moving the cord will eventually cause it to break. It's the usual cause of no heaters lighting in a set fitted with one.


Capacitive Droppers.

Where the mains supply is AC, an elegant idea is to use the reactance of a capacitor to drop the voltage. This resolves the heat dissipation issue as capacitors do not (and should not) get hot. These days the scheme is popular in some mains operated electronic devices such as time switches and PIR sensors. It allows a much more compact construction than  if a transformer was used. It seems to be also becoming popular with restorers of valve radios as a replacement for line cord resistors or other troublesome resistive droppers. As we are dealing with a reactive component, the frequency has to be taken into consideration, and of course the scheme won't work on DC as the capacitor presents an open circuit.
One would assume all that's necessary to use a capacitor as a dropper is to work out the value so that the reactance is the same as the resistance of a heater dropper that would otherwise be used.  However, there is a trap which many restorers are blissfully unaware of and which results in valve damage. Let's start with an important formula:

Formula for capacitive reactance , Xc= 1/(2PI*F*C) where F is in c/s and C is in Farads. By transposing the formula we can work out the capacitor value thus:

C= (1/Xc)/(2*PI*F)

For example, the capacitor value that would be equivalent to 400 ohms with a 50c/s supply would be:
C=(1/400)/(2*3.14*50)
C= 7.96uF

The trap is the capacitor presents that reactance only when there are no other components in the circuit. Once a resistive component is introduced (e.g. a string of valve heaters), then the current and voltage phase changes from 90 degrees and it all becomes more complex. The above formula is therefore not suitable. Without going into phasor diagrams and other complexities, here is the correct way to work out the required value. In this example, the heater string requires 84.8V at 600mA (it was actually a real situation with a TV set), and it is to run from 240V 50c/s.

1. Work out the resistive value of the load: R=84.8/.6
                                                                                  =141.3 ohms

2. Work out Z, where Z=Vsupply/load current
                                       =240/.6
                                       = 400 ohms

3. Work out Xc by  (Z^2-R^2)^.5
                              =(160,000 - 19965.7)^.5
                              = 374 ohms

4. Now work out C with the original formula;
                           C=(1/374)/(2*3.14*50)
                           C= 8.5uF
This is the correct value of capacitance.

Now, what happens if one is ignorant and assumes the capacitor is selected merely by reactance equivalent to when a resistor is used? If the TV in the example used a resistive dropper, the value would be: R=(240-84.8)/.6 which is 258 ohms. That's a big difference, and if we select a capacitor with that value of reactance, the valves will be very much over run! Incidentally, the capacitor value representing 258 ohms is: C=(1/258)/(2*3.14*50), or 12.3uF.

Other problems:
One may think all troubles are now over having elegantly fitted the capacitor inside the midget set and done away with the heat problems. Alas, there's more. First is the switch on surge. A discharged capacitor presents a brief short circuit, and if the receiver is turned on at the peak of the mains voltage, then one would assume the heaters cop the lot.
Usually they do, and the closer the sum of valve voltages is to the supply, the less harm is likely. If possible, it's a good idea to have some of the voltage dropped by a resistor to reduce this. Secondly, if the capacitor fails short circuit, again the heaters cop the lot. I designed a circuit which elegantly overcomes this problem. It was used with an experimental three valve receiver:

Here we have a heater string requiring 18.9V at 600mA. As the valves used were not designed for series heater use, there was a slight variation in heater current between valves. The 100R resistors are used to swamp out this difference. The 20R 20W resistor goes a long way to reduce the switch on surge. The components of interest are the back to back 30V zener diodes and the Triac. Basically, the Triac is triggered should the voltage across it (and the heater chain) rise above about 32Vpeak, or 22Vrms. This automatically prevents the heaters being subjected to voltage surge. It also prevents damage should the capacitor go short circuit. In this situation, the 20R will effectively be across the mains and the fuse (not shown) will blow. Also, if one of the valve heaters goes open circuit, the Triac triggers preventing the full mains voltage appearing across the heater pins and burning out the 100R resistor. This circuit works exceptionally well, and is highly recommended. Even though it does use modern solid state parts, you can be assured the valves are protected.
However, we come to two more disadvantages. Next in line is that a capacitive dropper is frequency dependent. This is usually not a problem, but as many of my inverters provide a 100c/s output, the valves would be over run. Also if the circuit is designed for 50c/s and taken to somewhere where the supply is 60c/s or even 400c/s, then again the valves would be over run. The final problem is low power factor. With the public mains supply, it generally isn't a problem, although supply authorities don't like it. It is a problem with inverters. Presenting such a reactive load will result in much arcing across the vibrator contacts, or possible destruction of switching transistors or FET's in modern inverters.

Selecting the right type of capacitor.
If you've decided that the capacitive dropper is the way to go, then you need to select the right capacitor type. Do not use any kind of electrolytic! Even unpolarised ones are not designed for continuous AC across them. Neither are back to back polarised types. You must use one that is rated for continuous operation across the AC mains. Such examples are motor run, or phase correction capacitors as used with fluorescent lights. Some motor start capacitors are non polarised electrolytics. Do not use them as they are for intermittent use only. Do not use DC capacitors, no matter how high their voltage. A 400 or 630V DC capacitor is not suited for continuous AC mains operation! Because of the odd value likely, you'll probably have to parallel smaller capacitors to bring it up to value, or as I did above, introduce extra resistance to reduce the voltage from a slightly larger capacitor. It is essential that the heater voltages be checked once the circuit has been brought into operation. Capacitor tolerances and other things can result in non optimal voltages and this needs to be corrected. Finally, if you are concerned about shock hazard when the plug pins are touched after being unplugged, connect a resistor across the capacitor to discharge it. Typical values would be around 330K 1W. Too high and the capacitor takes too long to discharge, and too low, it dissipates more power.



The Diode Dropper.

This was a very popular scheme with British TV sets during the 60's and 70's. The idea seems very simple at first glance. Ignoring D2 for the moment, it can be seen that because of D1, the heater string is presented with a half wave rectified sine wave. What this means is that D1 provides the circuit with half the power that would otherwise be applied. As a result, the heater dropper, R, dissipates much less power than in the conventional circuit. In this circuit, only the positive half of the mains cycle is used for heating, but it is immaterial what polarity is used in terms of the heater operation.
If the mains supply is 240V 50c/s, then at point Vh will be a 10mS long half sine wave with a peak of about 340V. The other 10mS of the waveform will be flat at 0V. A common misconception is that the rms voltage at Vh will simply be half of the mains supply. Nothing could be further from the truth, and circuits designed so will result in valve damage. It's the power that's halved, NOT the voltage. Why is this so you might ask? Think of a resistive load connected to the 50c/s sine wave mains supply, like a light bulb. The entire power to light the bulb comes from the area under the curve of both the positive and negative half cycles, each of which take 10mS to go from 0 to peak. Thus one complete cycle takes 20mS.  It's the power that makes the filament glow, not just the voltage or current. Now, if we chop off one of these half sine waves, then obviously the area under the curve over the 20mS period is halved, and thus the lamp is fed half power. A diode is the ideal device with which to do this, and in fact "light bulb savers" were sold on this principle. They simply had an adaptor containing a diode that could be plugged into a light socket. Hairdryers and brush motor power tools also use the scheme, with a diode connected in series when half power operation is desired.

Having established the load is fed half power by introduction of D1, we need to work out what Vh will be before calculating the value of the heater dropper resistor. As it happens, Vh only has to be calculated once; it is determined solely by the mains supply and not anything in the rest of the circuit.
1. As an example, lets say we have a 240R resistor across the 240V mains (R) . Current will be 1A (I) , and power 240W (P). Simple electrical theory there.
2. Now, insert a diode, and power will now be 120W, as we have halved the power by removing one half of the mains cycle.
3. Using the formula for power, P=I^2 * R, we can then determine what the rms current will be:
                                                  120=I^2 * 240
                                                  I^2 =120/240
                                                   I^2=.5
                                                      I=.707A

4. Now to calculate Vh; using V=I*R,
                                                  V=.707A * 240 ohms
                                                  V= 170V
You can use any load and any voltage and you'll get the same result.

We can clearly see that Vh will be 170Vrms, NOT 120Vrms. Note that ordinary voltmeters will give an erroneous reading due to the unusual waveshape and presence of DC. Either use a CRO or a true rms meter if you want to actually measure it. For those that use 120V mains, Vh will be 120 * .707 = 85Vrms.  I've seen two examples on the internet of American restorers replacing the line cord resistor with only a diode, thinking the heater chain will be fed from 60V. Sadly, the valves in those sets are destined for early failure. Not convinced? You can easily demonstrate by connecting a mains light bulb first through a diode, and then to a variac set at half mains voltage. The lamp fed via the diode is somewhat brighter isn't it?

Now that we know Vh is .707 of the mains voltage, the heater dropper is calculated in the usual way.
The advantage of this circuit can be illustrated with a simple example. A television has a heater string requiring 150V at 300mA. For the conventional circuit with no diode, the resistor for 240V operation is 300R and dissipates 27W. Introduce the diode, and the supply becomes 170Vrms. Now the resistor is 67R and dissipates only 6W. If you're lucky and the heater voltages add up to 170V (240V supply) or 85V (120V supply) you won't need any resistor.

Problems:
Diodes do fail, and when they do it's almost always in short circuit mode. This means the valve heaters would be over run, and probably without the user realising it. This is where D2 comes in. Should D1 fail short circuit, D2 will conduct when Vh goes negative and blow the fuse (not shown). Some British TV sets used other schemes to alert the user, but they didn't actually protect the heaters. It would be sensible to include a small capacitor, say .01uF, across the diode to minimise high voltage spikes damaging it. Like the diodes, it needs to be rated at a suitably high voltage and be suited for continuous mains operation. Many restorers don't bother with D2 to their possible detriment.
One serious problem with this scheme is that it loads the mains supply asymmetrically, introducing a DC component which can saturate transformers and cause electrolysis in the mains distribution system.
Inverters might not like it either, but at least this time the power factor isn't reduced and the supply frequency is irrelevant.
In Australia, the supply authorities disliked TV sets with half wave rectifiers because of this, and is why there were very few here, apart from the safety issue with live chassis construction. One wonders what it was like in the UK during peak viewing time with all those live chassis TV's taking not only a 300mA bite every half cycle for the B+, but when diode droppers were introduced, this was doubled. It surprises me so many sets were made like this when in actual fact it is possible to largely cancel out the asymmetrical loading.
If we reverse D1 and D2, the heaters draw power only on the negative cycle, and the B+ only on the positive cycle. And, as British series heater TV valves draw 300mA, then the loading is much more even. The last series heater live chassis TV set sold in Australia, the Thorn R2M did just this. I discuss the TV set power supply problem with the Ekco TX287 article.
Of course the diode dropper works on AC only. Fed from DC mains, and depending on the polarity, the heaters would either not work at all, or be over run. However, the chance of plugging into DC mains is remote enough not to worry about this possibility these days.



Transformers.
No introduction needed here.  Efficiency is high and there's no asymmetrical loading of the mains or reducing the power factor. Not frequency critical within limits either. As we're talking about live chassis sets here, an auto transformer can also be used. Obviously, a transformer is the best way to power the heaters.



And so to the line cord resistor replacement:
Having covered the possibilities, it really depends on the individual set as to what to use. Line cord resistors have not been made for many years, and with today's regulations, probably never will be again.
However, it occured to me that it should be possible to make a replacement using wire from an electric blanket heating element. This has the required insulation (in fact a good deal better than what was originally used), and flexibility requirements (again, vastly improved on the original). By shrouding it in hollow shoe lace along with two ordinary conductors, it should therefore be possible to make an authentic replacement. Indeed, it turned out that electric blanket element wire also had the right electrical characteristics, and the idea has been successfully implemented on two receivers so far.
Thus there is no need to butcher the original radio with non original droppers.
The first reproduction line cord resistor was used on one of my Meck FM converters. See the full description here
The second was made for one of my Emerson CF255's. This set requires a tap for the dial lamp, but this was easily provided.



Using droppers for other loads.
Finally, I'll talk about the options for running things besides valve heaters. A typical example is someone who wants to run something designed for 120V on 240V.
For purely resistive loads, like light bulbs, brush motors or heating elements, any of the droppers described will work. However, with the exception of using a transformer, all the droppers will give higher voltage on no load, so are only suitable for constant loads.
For inductive loads like small shaded pole induction motors, anything but the diode dropper will work. The capacitive dropper will work, but the calculations are not the same. Suffice to say, it's easiest to use trial and error. I had to replace the fan motor in one of my fan heaters. As it happened I had U.S made motor that fitted perfectly. However, I had to drop 120V to use it here.
I used a 2.8uF 440VAC capacitor in series which worked perfectly. Note that because of resonance effects, the voltage across the capacitor may be much higher than the mains supply. This is why motor run capacitors are 440V and not 250VAC.
With the exception of a double wound transformer, none of the droppers isolate the mains supply. This is why it is dangerous to use a 120V live chassis radio on 240V mains by means of a line cord resistor, as seems to be done in the U.K. For one thing, with the radio switched off, there will be 240V present at the end of the line cord resistor and this could stress the insulation of aerial isolation capacitors and switch contacts. Secondly, the chassis could be at 240V and as I've shown elsewhere on this site, radios made for 120V operation don't have much in the way of insulation, if any, isolating the user from shock. It's tempting also to convert U.S made series heater radios or TV sets to run off 240V, thus dispensing with the outboard stepdown transformer. I have actually tried it with my 17" Philco, but with the safety hazards as well as the non originality of the conversion, soon converted it back to 120V operation. Again, not recommended. A double wound stepdown transformer is the best way of powering these appliances, and will actually make them safer than when their original American owners used them.
 

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