A good reference to Figuring out Dropping resistors for Old radios

And then I will get into using capacitors to replace those resistors, The main reason is a lot less power Losses and heat.

You have to have some numbers values to work with, to solve for the resistor dropper value.

They are

Vin = Supply voltage

Vh = Heater voltage

Ih = Heater current

Rd = Dropper resistance

Vd = Dropper voltage

First find the filament voltages (accumulative) and heater current.

Look in any schematics you can find, or ask a fellow enthusiast here to see if they have the schematic.

Look also in tube charts for current demand and filament voltages.

We have to use the input AC voltage first, (Vin)

then subtract the sum of heater voltages 2nd. (Vh),

So Add all the filament voltages = 12v+35v+14v = **60.2 Volts (Vh)**

Now subtract the added filament voltages from the supply voltage.

Supply voltage (Vin) - Heater voltage (Vh) = voltage to drop(Vd)

120 (Vin) - 60.2 (Vh) = 59.8 (Rd)

Now Find the needed resistance required to drop that voltage.

To do that we take the voltage drop and divide by the heater current.

Rd = (Vd) / Ih

**59.8 (Vd)** / .15 (Ih)(ma. of tube current) = **399 ohms**(Rd)

Vin = Supply voltage (Volts) = 120

Vh = Heater voltage (Volts) = 60.2

Ih = Heater current (Amps) = .150

Rd = Dropper resistance (Ohms) = 399

Vd = Dropper voltage (Volts) = 59.8

We also need to consider the power that needs to be dissipated

and the formula for that is

The power dissipated by this resistor is

Ih^2*R. So, (.150Ih^2) * 399Rd = 8.98W. Pd

Ih^2 * Rd

Where: Pd = Dropper resistor power (Watts)

Vin = Supply voltage (Volts)

Vh = Heater voltage (Volts)

Ih = Heater current (Amps)

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To avoid wasted power try using a capacitor in its place

Capacitor dropper formula

First step is to find Xc

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

But we need to rearrange this formula to find C

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

Where: Xc = Impedance of the capacitor (Ohms)

F = Frequency (Hertz)

C = Capacitance (Farads)

Pi = Pi (3.14159)

Now use this formula because there is a 90 degree phase difference between the heater voltage and capacitor voltage using Pythagorean Theorem

Vin (squared) = Vh(squared) + Vc(squared)

120^2 - 60.2^2 = square root of 10775.96 = 103.8

now knowing the voltage across the capacitor and heater current we calculate the capacitor impedance.

Xc = Vc / Ih

103 / .150 = 686.6 Xc

Where: VC = Voltage across capacitor (Volts)

Vin = Supply voltage (Volts)

Vh = Heater voltage (Volts)

Now calculate the capacitor voltage

Vc = Square root of Vin(squared) - Vh(squared)

We can now solve for capacitor impedance (Xc), because we have the __capacitor voltage (Vc__) and __heater current (Ih).__

Xc = Vc /Ih

Where: Xc = Capacitor impedance (Ohms)

VC = Voltage across capacitor (Volts)

Ih = Heater Current (Amps)

Now that we have the value of Xc to find the capacitor value.

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

3.86 Mfd

1/(2Pi*60*686.6) = .000003863

now multiply by 1,000,000 to get microfarads

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Note:

It might be a good idea to add a surge resistor so lets say you want to add that by first using ohms law.

R1d = V1d / I1d

where R1d = surge resistor

V1d = additional voltage drop

I1d = filament current

So you might want to Add this to the heater voltage in your calculations

And add a discharge resistor to parallel the dropping capacitor in parallel of 100K ohms to 220 Kohms at 1 watt.

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I'm a Science Thinker, Radio Tinkerer, and all around good guy. Just ask Me!