Cap divider feeding a low votage lamp.

For the test I'll use a 15 Volt 80 ma lamp. With 120 VAC at 60 Hertz.

I made the little test board with 2 lamps. the caps are .5ufd so I used 3. Looking good.

One lamp has 12.7 volts.

25.14 Volts across both lamps.

I put a jumper across 1 lamp and have 13.2 volts across a single lamp.

A tip of the hat to thr DIT. It is a very useful circuit for someone who needs to reduce the voltage but has no transformer. The only draw back I see is being able to find the capacitor you need for a specific voltage / current requirement and you need to make sure the cap is rated for line voltage.
There is a ready supply of 50 volt caps but they will not work in this application.

The radio in the background has been in the dark for a long time. The pilot lamps were blown and I could not find replacements. That is not a problem any longer. I can put the 2 I'm testing in it.

Addendum:

The as built matches the circuit. One more thought. In the original we had a limiting resister. I am using the pilot lamp for the limiter. Using it with tubes the surge could be a problem without the limiter. The higher the current the more surge. A rule of thumb is inrush will be 8 to 10 times the run current. The inrush can last 15 - 20 seconds in a motor circuit. Just a little extra info.

Expanding the capacitive divider

Remember the sine wave is generated by the rotating vector. The amplitude of the peak can be determined as the sine of the angle at any given instant.

The voltage across the components will not add to the value of the input at any given instant because they are not in phase with each other. When we add the signals the using the vector they will equal the input. When we plot the signals the result is a right angle triangle. We can use HYP = SQRT( A^2 + B^2) to solve for the values.

Now back to this one. The load and Xc are equal and producing equal voltage drops. Because of the phase angle they will not add to the supply.
Ec = 170
Ert = 170
Et = SQRT(170^2 + 170^2)
Et = 240
I = E / R
I = 170 / 800
I = 212.5 ma
Back to our original circuit. We need to know the number of tubes, their voltage rating, and their current rating. As long as we can make a string with the same current we can produce a working circuit. We could use a 300ma tube in series with 2 150ma tubes in parallel. We would transpose the formula to calculate the cap value.

EXc = SQRT( Et^2 - Ert^2)
Xc = Exc / It
etc
The original circuit would work as long as the load voltage was low. As the load increases the phase angle increase and Zt increases reducing the current.  With one or two tubes it would be close enough. In the last example it is 212ma. The tubes may still function but at reduced capacity.

Tip of the hat kiddo. I have an old radio in need of a panel light. The 15 volt lamp in my junk box may find a use.

Capacitive divider

The designer in training has a new project to consider. This one is really cool. It does have a couple of underlying factors that can easily be overlooked. First I give the simulations to show the complexity of the design then we'll look at things to consider. The problem being addressed is the lack of available power transformers and how to over come that. We want to build a tube (valve) radio and need 300ma heater current. If we use dropping resistors we waste a lot of power. The unit could double as a heater in cold weather.  R=E/I R=240/.3 = 800R so we find a capacitor with 800R reactance at 50Hz is 4ufd. A 12 tube will have 40 Ohms resistance. You can see in the sim that works fairly well. I have 12 volts at 300 ma. Xc is much greater than R and we can use simply design. What happens if I want to power several tubes?

I use 400R for ten tubes. You would expect E=IR E=400*.3 = 120Volts. But look the current is now .25A? Well we added series resistance so we would expect less current. Let's see 800R + 400R = 1200R and I=E/R =240/1200 = .2A. Wait a minute the sim says .25A? How can this be? Remember the reactive stores energy and injects it back into the circuit. It also produces a phase shift. We will need to use vector (sometimes call phasors) to solve riddle.

With 800R we should see half the supply on the resistor. Let's see 157.5 * 2 = 315. What? The source is 240V. Remember that phase shift. We must account for it to solve the riddle.
** note the 157.5 is not the peak. 

If we only wanted to power two 12V tubes the original circuit would be close enough. My simple rule of thumb is to use a 10 to 1 ratio of resistances and go with the simple design. With 800 to 80 it is close enough. Next post I will explore the angles involved and how to resolve them.

The new 40/80 using TUN amp and LAMDA diode oscillator

40/80 continues


My friend and I designed the 40/80 in 2018. We used NE612 for the mixers and a logic chip for the oscillator. It was a unique design and worked well. Today I start on a new version using all discrete components. A TUN RF amp, Gilbert cell for the mixers, LAMBDA diodes oscillators and a TUN AF amp.

 lambda diode

TUN amplifier

DC Receiver