Monday, August 3, 2020

LDO DC receiver cross feed vs shielding part 2 add DBM

Looking at the LO , RF and OUTPUT.

 I added an inductor in each input leg with no ill effect.

When I "LINK" the inductors the output is lost.
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.

LDO DC receiver cross feed vs shielding part 1 the oscillator


The tunnel diode oscillator would be ideal for our DC receiver but they are not readily available so I will be looking at the LDO.

Using 2 JFETs we have a LDO.


Friday, July 24, 2020

The cap divider tube amp performance

Last night I put a headphone on the amp and fed it with a antenna through a diode I received signals. Without the diode all is quite. For the test I used a tapped coil. I could move the tap and receive different stations. This would be a good amp for a hollow state version of the 40/80.

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.


Thursday, July 23, 2020

Ground to ground???

Joe had a dead battery and Tom offered to help. When they connected the jumper cable one had a bad end and it fell off. Joe says "Oh man!". Tom says "Not to worry". Tom connects one end to his battery and the other to Joe's battery. He pulls up to Joe's car until the bumpers touch. Boom Flash Spark. Tom backs up and look closer. He had connected his end of the cable to the hot terminal and the other to Joe's grounded terminal. When the bumpers touched BOOM! He moved the lead on Joe's battery and all was well when he touched the bumpers again.

This is the same as having two chassis on your work bench with the power lines grounded on one and hot on the other.

Always check a chassis to ground before betting your life on the circuit being wired correctly.

Use polarized plugs to prevent injury.

The schematic for the amp using the cap divider to power the tube heater

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. 
A tip of the hat to the DIT. A very useful circuit indeed.
One caution if you try it.
BE CAREFUL ABOUT THE NEUTRAL ON THE POWER LINE.
Use a polarized plug and be sure not to ground the hot line.

Wednesday, July 22, 2020

Cap divider part 3 proof of concept

DIT produced a winner with the Cap Divider. I put it on a breadboard and the results were impressive.

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.
I used a PI filter in the DC supply I will draw it and post next time if my internet does not crash. My server is not doing well this week I keep timing out.

Sunday, July 19, 2020

The cap divider revisted part 2

I took 6 .5ufd caps in parallel to replace the 5ufd cap and bingo! 12.56 volts feeding the heater. The tube has a nice warm cherry glow. It is a big advantage having a junk box with a bag of .5ufd high voltage caps in it.
This was the circuit from my last post. I used the parallel lamp to bypass some current around the tube. After replacing the 5ufd cap withe the 3.3ufd cluster I removed the lamp.
Question:
Note the B+ is 120vdc in sim. The real circuit reads 170vdc. Why?

 Anywho, I know have a dual triode with heater glowing and a 170vdc B+ so what to do with it?

The datasheet gives info on building an amp.


Saturday, July 18, 2020

The cap divider revisted

The DIT did good with the capacitive  divider. It had over 400 views and counting. I pieced together a little board with one powering a tube. I have some motor run caps in my shop. I had a 5ufd and a 7.5ufd on hand. I need a smaller cap but I made do for the test board.
This is the tube I'm using. It requires .15 amp. A 3ufd cap would be better but I have a 5ufd. I put a pilot light across the tube heater to share the current.
This ceiling fan cap would be ideal for testing the circuits. You can parallel the sections and have 1.5 ufd, 2.5ufd or 4ufd.
Here is my as built.With 120vac @60hz I read 12.7 vac across the heater. The lamp is a safety indicator. When it is glowing I am reminded to watch where I put my hands. I put a diode and 15ufd cap in series across the input terminal and have a 150 vdc supply. I will build the circuit in sim and post it.

Thursday, July 2, 2020

Earbud driver using a mosfet output stage

It is not pretty but it sings. I put a drop of glue where the part was located on the schematic and attached the parts. Then it was just a matter of connecting the dots with a piece of wire.

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.

Thursday, June 25, 2020

6-25-2020 Amp with MOSFET output stage.

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 printed the schematic and glued it to a board. I put a pin at each tie point on the schematic. I put a pin at the point between c4 and r12. You could just tie them together and save the pin but I like anchoring the components. All the components are sized to meet the TUP - TUN amp standard so locating parts will not be to difficult. After I connect the dots with parts we will see how it works compared to the simulation.

NOTE: I added a diode and coil to the last amp I built and can listen to my local AM station through the speaker. It draws about 8ma. This one delivers more output with less battery load according to the sim. Time to find out.

Wednesday, April 29, 2020

How do it know? What determines current flow?

The crew breaks for lunch. George pours himself a cup of coffee. He says, "That coffee is hot!". Lunch break the next day George pours himself a cup of tea. He says, "Wow that tea is so cold it gave me an ice cream headache.". George asked the crew,"How does it know?". Alvin says "What know what George?". George says, "When My wife sends coffee it keeps it hot. When she sends tea it keeps it cold. How do it know to keep it hot or cold?".
Of course it does not know but it is an interesting question. We have a container inside another container and an insulator between. In order to pass through the insulation the energy would have to overcome the resistance the insulator offers to energy movement. Thermal conductivity of the insulator will determine the rate the temperature equalizes.
What would happen if we drilled a hole through the insulator and inserted a metal rod? The thermal conductivity of metal is very high so the heat would travel the rod and the inter camber would loose heat much quicker. If we knew the temperature and volume of the liquid we could determine the number of calories needed to make the inside and outside reach equilibrium. If one 1/4" rod would bring equilibrium in an hour how long would it take with 2 1/4" rods? What if we used 2 rods that were 1/4" and 1/2"? The larger rod would convey the energy faster than the smaller. We could total the cross sectional area of the rods and determine the rate. A rod that is twice as large would have 4 times the cross sectional area.
Now consider a circuit with a battery supplying an open circuit. It would be like the first case. With no circuit the current would not flow. Add a large value resistor and it will allow a small flow. Add another resistor and you have a higher current flow and so it goes.
How do it know? The circuit must have a path and it will produce as much flow as it can. Adding more paths produces more flow.
It knows one thing push as hard as it can and find any path available. After that it self regulates.

Saturday, April 4, 2020

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.

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.

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 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