Tuesday, February 18, 2020

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.


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