# Radio Theory and design

## Saturday, April 4, 2020

## Wednesday, February 19, 2020

### Cap divider feeding a low votage lamp.

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. |

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.

## 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

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.

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

## Sunday, February 9, 2020

### 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

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

## Tuesday, December 10, 2019

### Lambda Diode Oscillator

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.

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.

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