Wednesday, February 19, 2020

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.


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.

Tuesday, February 18, 2020

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
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.

Sunday, February 9, 2020

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

40/80 final build

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 oscillator

TUN amplifier

DC receiver

Tuesday, December 10, 2019

Lambda Diode Oscillator

This simple oscillator will drive a converter to feed DBM or Gilbert Cell. The first step in building a new radio will be the LDO. It can be made with plug in coils and be multi band. It can hardly get any simpler. A couple of JFETs and a resistor feeding a tank circuit will work. A series resistor to limit the current is a good idea.

The resistor is being changed to determine what size pot will work.
The plan is to put a 470 Ohm resistor in series with a 1K Ohm pot.
A low value resistor allows the current to soar and a high value will kill the oscillator.  The 470 Ohm to 1470 Ohm should be a good trade off.

Thursday, December 5, 2019

relating a sine wave amplitude to the phase angle

Rotating the phasor creates a sine wave. The amplitude of the phasor is the sine of the angle (Assuming a max amplitude of 1). If the max was 100 the phasor would be 100* sine(theta) etc.

The angular velocity can be measured in degrees and using a sine table the amplitude can be determined. As an example let's look at 45 degrees. Sine is the first column. sine 45 degrees is .7071. Assuming a 100 volt peak the amplitude at 45 degrees would be 100*.7071 = 70.71.

Using the sine chart it would be possible to determine the value of the signal at any angle or conversely to determine the angle at and signal level. Now you can assign values to the waveform and determine the angle using the chart.

Wednesday, December 4, 2019

A mind experiment into the world of Electro - Magnetics. Part 6

In this series of screen shots I show a transmission line with different feed Z and load Z.

What effect does matching the load Z to the feed line Z have?

What effect does lowering the generator Z have?

Consider R1 as an antenna, when the signal is radiated (absorbed by the load) we see no reflected signal and the generator can handle more power. The rule of thumb often quoted is for maximum power transfer we need to match the load to the generator. This is true BUT the generator drops half the power.
 Food for thought.