A while back, I posted a note about having found an old 1930's National SW3 receiver. The rig came without a power supply. I need to have 2.5 volts at about 3 amps to power the tube filaments. I have a 6.3 volt supply at 4 amps, but need to bring this down to 2.5 volts at 3 amps. So, I thought I would just use a voltage divider.
o+o +6.3V

R1 3.8 volts across R1

+o +2.5V

R2 Load = 3 amp @ 2.5 volt

o+o 0V
So, how to design a simple two resistor divider? Assuming that there is no load current:
Vout = Vin * R2 / (R1+R2)
The problem with this is that our load DOES draw current (about 3 amps) and therefore this will not work because the load can be considered another resistance in parallel with R2.
The 10 percent rule is a standard method for selecting R1 and R2 that takes into account the load and minimizes unnecessary power losses in the divider.
So, the first thing to do is select R2 so that I2 is 10 percent of the desired load current. This resistance and current is called the bleeder resistance and bleeder current. In this example, the bleeder current is:
Ibleed = I2 = 0.1 * 3 A = 0.3 A = 300 mA
Using Ohm's law to calculate the bleeder resistance:
Rbleed = R2 = 2.5V /0.3 A = 8.3333 ohm
Now, we need to select R1 so that the output is maintained at 2.5V. To do this, we simply calculate the total current through the resistor and use Ohm's law:
I1 = I2 + Iload = 0.3 A + 3 A = 3.3 A
R1 = (6.3 V  2.5 V) / 3.3 A = 1.1515 ohm
Now considering standard resistor tolerances and value, this would indicate the need for R1 = 8.1 ohm for a 5% resistor and R2 = 1.15 ohm.
In terms of power ratings:
P1 = V1^2 / R1 = 3.8^2 / 1.15 = 12.55 W
P2 = V2^2 / R2 = 2.5^2 / 8.1 = 0.77 W
Did I do this right?
I am not sure what happened, but the first time I did this exercise, I (somehow) came up with R1 = 7.5 ohm 3 watt and R2 = 5.6 ohm 5 watt...
More coffee...
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