So the point is this.
The LO is producing millvolts of signal and the RF is microvolts so we must prevent the LO feeding the antenna connection.
What can we do?
Next post we will look at some "good practice" for the construction.
|Using the cap divider powered tube amp as the AF stage|
of this receiver would be a cool and interesting project.
It could use a lambda
diode oscillator or a tube oscillator.
|The as built schematic except one resistor. I put a 470 Ohm resistor in series with the diode to avoid inrush on power up.|
|I ran the sim to see how it would compare to the actual circuit.|
|The datasheet offered this circuit. I used 1.5k cathode resistor and 100k for the rest and .5ufd caps throughout.|
|I put a meter on the input and output. The signal generator on the input.|
|I adjusted the input to 10mv.|
|Apply power and have 10 volts output. Looks like a winner.|
|50uv input gives 800ua output. My eadbud is rated 1ma so that is enough. With 100uv input the signal distorts and will act as a "plate detector" receiving AM.|
|I changed some capacitor values for the build. This is the response for my build. I can add a 12" test lead to the input and receive my local AM station.|
|This could be my new favorite. It does need a receiver built to feed it.|
|With 10uv in I get 7mv out driving an earbud. The power consumption is less that 2.5ma. It should run for over 100 hours on a 9 volt battery. So it needs a build.|
|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.|
|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.|