I receive quite a few emails asking about the calculations of component values for dropper resistors, dioderesistor dropper circuits, and capacitive droppers. I have therefore assembled all the information on this page, and have produced a spreadsheet to help with the calculations.
There is quite a lot of maths here. I have tried to keep it as simple as possible, and have given a worked example at the end of each section.
Single Resistor Dropper
This is the simplest type of dropper, and is the type fitted to the vast majority of AC/DC sets.
The resistor value is calculated using Ohms law. Subtract the total heater voltage from the supply voltage to obtain the voltage to be dropped by the dropper. Divide the result by the current (in Amps) to get the dropper resistor value (in Ohms):
Where:  Rd  = Dropper resistance (Ohms) 
Vin  = Supply voltage (Volts)  
Vh  = Heater voltage (Volts)  
Ih  = Heater current (Amps) 
To establish the total heater voltage, look up the heater voltages for the valves in a valve data book (or CDROM) and add them all together.
The power dissipated by the dropper resistor is:
Pd = (Vin  Vh) x Ih
Where:  Pd  = Dropper resistor power (Watts) 
Vin  = Supply voltage (Volts)  
Vh  = Heater voltage (Volts)  
Ih  = Heater current (Amps) 
Single Resistor Dropper Example
To calculate the required dropper resistor for a small AC/DC MW/LW set using UCH42, UF41, UBC41, UL41 and UY41 valves, operating from a mains supply of 240V AC.
From the valve data, the heater voltages are 14V (UCH42), 12.6V (UF41), 14V (UBC41), 45V (UL41) and 31V (UY41), giving a total of 116.6V. We will use a figure of 117V). Heater current is 0.1A.
To calculate the dropper resistor value:
Rd = 1230 Ohms
To calculate the dropper resistor dissipation:
Pd = (Vin  Vh) x Ih
Pd = (240  117) x 0.1
Pd = 12.3 Watts
Rectifier Diode and Resistor
By connecting a rectifier diode in series with the heater chain, only the positive half cycles get through to the heater and the negative half cycles are blocked by the diode. This reduces the heating of the heaters, so a lower series resistor is needed to achieve the correct heating effect of the heaters.
By removing half the waveform, we are effectively reducing the power applied to the heaters by half. The easiest way to calculate the resistor value is to assume that the power to be applied to the heaters is double the rated value; then adding the diode will halve it so we will be back to where we want to be.
Doubling the power applied to the heater is not as easy as doubling the voltage. If you were to apply double the voltage, the current would also double (ohms law). The power equals the current multiplied by the voltage, so if the current and voltage were both doubled then the power would be quadrupled.
From Ohms law:
Where:  Rh  = Heater resistance (Ohms) 
Vh  = Heater voltage (Volts)  
Ih  = Heater current (Amps) 
Rh is a constant, so Vh and Ih must increase proportionally.
We want to multiply the power by two, and we know that:
Ph = Vh x Ih
Where:  Ph  = Heater power (Watts) 
Vh  = Heater voltage (Volts)  
Ih  = Heater current (Amps) 
If we multiply Vh and Ih individually by the square root of two, the total power will be doubled, because of course:
We have therefore achieved the objective of increasing the heater voltage and current proportionally, and doubling the heater power.
Pulling all the bits together gives us:
Where:  Rd  = Dropper resistance (Ohms) 
Vin  = Supply voltage (Volts)  
Vh  = Heater voltage (Volts)  
Ih  = Heater current (Amps) 
Since:
We get something a bit easier to enter into a calculator:
The power dissipated by the dropper resistor is:
Where:  Pd  = Dropper resistor power dissipation (Watts) 
Vin  = Supply voltage (Volts)  
Vh  = Heater voltage (Volts)  
Ih  = Heater current (Amps) 
The reason for dividing this by two is because the diode halves the power being applied to the circuit (which of course is why we had to calculate the heater power for double the rated power in the first place).
This will only work with an AC supply, but since DC mains supplies no longer exist in the UK, this is no problem. To ensure reliability the diode should be amply rated for the job, because if the diode should fail shortcircuit the heaters would be overrun. I would use a 1N5408 diode which is rated at 3A, 1000V. Some people suggest the use of two diodes in series to protect against the risk of one failing, but I don�t feel this is necessary. If possible, do not mount the diode close to the dropper resistor or hot running valves, to make sure it remains cool.
Although the circuit would work with the diode connected either way around, the heatercathode stress on the rectifier valve will be less with the diode connected as shown in the diagram above.
If you measure the heater voltage and current with this circuit, the results will be confusing. This is because the waveform is neither a pure sinewave nor pure DC, so most meters will simply read an average value rather than the true RMS value. With an analogue meter switched to a DC range, multiplying the reading by 1.57 (1.11 x squareroot of 2) should give the correct value. It is fairly easy to multiply the reading by 1.5 in your head, which gives an error of less than 5%. With digital meters, it is more difficult to predict what they will make of this sort of waveform.
There is the possibility that this circuit will give increased hum or buzz compared to the original arrangement. If this occurs, try connecting a 0.022uF Class X suppresser capacitor in parallel with the diode. If this doesn�t help, consider using a capacitive dropper instead.
To carry out the modification, it is probably easiest to connect a second resistor in parallel with the existing dropper resistor section, then connect the diode in series. The formula for resistors in parallel is:
Where:  Rd  = Dropper Resistance required (Ohms) 
R1  = Existing resistor (Ohms)  
R2  = Required parallel resistor (Ohms) 
Since we know R and R1, but don�t know R2, we need to rearrange this to:
The power dissipated in this additional resistor can be calculated, but it gets a bit complicated because we have to consider the proportions of current through the original and new resistors. The easiest way to work out power dissipated in this additional resistor, is to calculate the total dropper resistor dissipation as detailed above, then divide it between the two resistors in inverse proportion to their values:
Where:  Rd  = Overall dropper resistance (Ohms) 
R1  = Original resistor (Ohms)  
Pd  = Overall dropper dissipation (Watts)  
P2  = Parallel resistor dissipation (Watts) 
Rectifier Diode and Resistor Example
I will use the same example radio as above. Heater voltage 117V, heater current 0.1A, supply voltage 240V.
To calculate the dropper resistor value:
Rd = 527.3 Ohms
To calculate the dropper resistor dissipation:
Pd = 5.27 Watts
As you can see this is less than half the dissipation of a standard dropper resistor, yet the power applied to the valve heaters is still the same.
If we were modifying the original circuit with a 1230 Ohm dropper, we would want to work out the required resistor to connect in parallel to get 527.3 Ohms:
R2 = 923.4 Ohms
To calculate the power dissipated by this additional resistor:
P2 = 2.26 Watts
Capacitive Dropper
A capacitive dropper will run cold. When an AC signal passes through a capacitor, there is a 90degree phase shift between the current and the applied voltage, so therefore there is no heating effect.
The impedance (AC resistance) of a capacitor depends on the value of the capacitor and the frequency of the signal. The frequency is known and fixed (the mains frequency) and the capacitance is also fixed, so the impedance is constant.
The impedance of a capacitor is given by this formula:
Where:  Xc  = Impedance of the capacitor (Ohms) 
F  = Frequency (Hertz)  
C  = Capacitance (Farads)  
= Pi (3.14159) 
Since we are trying to work out the capacitance, the formula can be rearranged:
Many people think this is as far as we need to go, and simply put the calculated dropper resistance in place of Xc and work out the capacitance from that. WRONG! We need to take into account the 90degree phase shift.
If you connect a resistor and a capacitor in series and apply an AC voltage to the combination, then measure the voltage across the capacitor and resistor and add them together, the result will not be the applied voltage. It will be higher.
Do you remember all that stuff from school about rightangle triangles, and Pythagoras? See this page for a reminder. We now have a use for it!
Look at the rightangle triangle below. The hypotenuse (the long side that doesn�t connect to the 90 degree corner) represents the supply voltage (Vin). The other two sides represent the voltage across the capacitor (Vc) and the voltage across the heaters (Vh), which are at 90 degrees to each other.
Pythagorean Theorem states: "The square on the hypotenuse is equal to the sum of the squares on the other two sides". Taking our values, the formula is:
Where:  VC  = Voltage across capacitor (Volts) 
Vin  = Supply voltage (Volts)  
Vh  = Heater voltage (Volts) 
We know the supply voltage and the heater voltage, so by rearranging the formula we can calculate the voltage across the capacitor:
Knowing the voltage across the capacitor from this formula, and the heater current, we can calculate the capacitor impedance required using ohms law:
Where:  Xc  = Capacitor impedance (Ohms) 
VC  = Voltage across capacitor (Volts)  
Ih  = Heater Current (Amps) 
Then put the value of Xc into the formula for calculating the capacitor value (above) to get the capacitor value required. The result will be in Farads, so we need to multiply it by 1,000,000 to get microfarads.
As with the diode dropper arrangement, this will only work with AC supplies. In addition, the frequency is important, so for example a set modified with the calculations done for a 50Hz supply must not be used on a 60Hz supply, even if the voltage is the same.
The capacitor value will probably be somewhere in the range of 1uF to 2uF. Motor run capacitors are available in this range and have adequate voltage ratings. Alternatively, Class X suppression capacitors are available up to 1uF and possibly higher. Neither of these types are available in a vast range of values (for example motor run capacitors are available in 1uF, 1.5uF and 2uF). The tolerance is 10% or 20% so do not expect great precision.
Do NOT use electrolytic capacitors! The capacitor must be nonpolarised and must be rated for continuous mains voltage operation.
You will be unlikely to get a capacitor of exactly the correct value. You will probably need to use two in parallel:
For capacitors in parallel:
Ctotal = C1 + C2
For example, if you required 1.6uF you could use a 1.5uF motor run capacitor with a 0.1uF Class X suppresser capacitor in parallel.
Once the circuit is assembled, check the heater voltage or current and adjust the capacitor values as required to get within 5% of the heater ratings. It would be best to start a bit low and work up to the correct value by trialanderror.
A small resistor should be connected in series with the capacitor to act as a surge limiter. I would choose one that will drop about 10 volts. Ideally, you should add the voltage drop of this resistor to Vh. Additionally a resistor of around 100K to 220K at 1W should be connected in parallel with the capacitor, to discharge it quickly when the set is switched off.
Capacitive Dropper Example
I will use the same example radio as before. Heater voltage 117V, heater current 0.1A, supply voltage 240V. The one additional thing we need to know is the supply frequency, which is 50Hz.
We want the surge limiter to drop 10V at 100mA. Using Ohms law:
R = 100 Ohms
Adding this onto the heater voltage (117V) gives a total of 127V.
To calculate the voltage dropped across the capacitor:
VC = 203.64 Volts
To calculate the impedance of the capacitor:
Xc = 2036.4 Ohms
To calculate the capacitor value:
C = 0.00000156 F
Multiply the farads value by 1,000,000 to get microfarads:
I would fit a 1.5uF capacitor, measure the heater voltage or current, and adjust with an additional small parallel Class X capacitor (probably either 0.047uF or 0.1uF) to get close to the correct reading.
We can also work out the power dissipation in the surge limiter resistor:
P = I x V
P = 0.1 x 10
P = 1 Watt
Since the capacitor is a �wattless dropper�, this 1 watt is the total power dissipated by the dropper circuit. Compare this to the 12.3 watts with the standard resistor dropper and you can see that the radio will run a lot cooler.
Summary
As there are three dropper methods, the following brief summary of the pros and cons of each arrangement (kindly provided by Jon Evans) may help you decide which is the most appropriate for your set.
Resistor: By far the highest waste heat generated. Most likely cause of failure is the resistor, which will almost certainly be wirewound. This would fail open circuit so is failsafe. Has the lowest initial turnon current surge when cold heaters are at much lower resistance (significant factor in heater lifetime for indirectly heated valves). Easy to get appropriate resistor values.
DiodeResistor: Much reduced waste heat. Same resistor failure mode but diode failure to shortcircuit will overrun valves and not necessarily be immediately obvious as a fault. Higher peak switchon surge current. Easy to get the appropriate resistor values. Exercise caution regarding accuracy of measuring resulting heater voltage.
Capacitor: Almost no waste heat, which may be crucial in a small midget set. Capacitor failure mode is typically short circuit, overrunning the valves. Highest switchon surge current. Limited range of capacitor values, but simple to measure results.
So the resistor is the safest option for the valves, but is also the hottest running. The other options have higher switchon surges and slight increased risk of damage if a component fails, but run cooler. Personally, I would regard the capacitive dropper as preferential to the diode dropper because of the reduced dissipation, the ease of taking measurements, and the reliability (overall I would expect a motor run capacitor to be more reliable than a semiconductor diode).
Spreadsheet
Although the maths is not that complicated if you follow the examples, the calculations (particularly for the capacitor dropper) are somewhat torturous, and leave plenty of opportunities for errors!
To make the job a lot easier and quicker, I have created an Excel spreadsheet containing calculations for all three options. The spreadsheet also takes into account and calculates the surge limiter resistor for capacitive dropper circuits.
It also contains separate worksheets for calculating Ohms Law and power, and for resistors and capacitors in series and parallel, which could be useful for some related calculations.
The spreadsheet file is saved in Microsoft Excel 5.0/95 format, and can be opened with any later version of Excel or with any version of OpenOffice. It will also load into Microsoft Works 4 and later, however you will only be able to view a single worksheet (you will be asked which one to use when opening the file).
It also works OK with Spread32 which is a freeware/shareware spreadsheet available here http://www.byedesign.freeserve.co.uk/. Scroll down and download spre32en which is the top left entry in the table. It's a tiny download  only 509k  so no problem on a dialup connection. The unregistered (free) version has nag messages but doesn't expire (registration costs $20)
To download click the link below.
Droper Calculations Spreadsheet (ZIP file  file size 6k)
To calculate dropper components, make sure the correct sheet is selected then simply enter the mains voltage and frequency, and the heater voltage and current in the relevant cells (note that the heater current is in Amps not milliamps). The results for the simple resistor dropper, the resistordiode dropper, and the capacitor dropper will be displayed in the results section.
This spreadsheet has been carefully checked and is believed to be accurate. However, it is offered asis and no liability can be accepted for any errors. The spreadsheet may be used and modified for your own personal use as required, but may not be redistributed in any manner.
I would like to thank Jon Evans for checking this page and spreadsheet, and for his many useful suggestions and contributions. I would also like to thank Douglas Dawson for highlighting an error in the section regarding the use of a test meter to measure voltages with a diode dropper.

