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my radio is changing stations by itself!


mik3ca

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;D

What I wanted was a radio where I can change the stations, not have it change stations itself.

When I designed it on the circuit board, it worked perfectly.
As soon as I placed it on a circuit board and added the digital station changer (basically a homemade R-2R DAC attached to a homemade poor-mans varactor which is then connected to the tank circuit.) , I seem to get better quality audio out, BUT the major problem is that the stations consistently change. (this is after the digital station changer has been added).

In my R-2R DAC, I use 4.7K resistors and 10K resistors. I use the 4040 CMOS counter chip and I tied the lowest unusued bit to the RESET. The clock input is tied to +ve through a 62K resistor. I think the 62K resistor may be a problem.
I connected a button from clock to ground, so every time I press the button, the station should change because the voltage will be different which will then make the capacitance of the varactor different. I tried tying a capacitor across the button, but that made no difference. I also tried increasing the power supply capacitor from 1nF to 4.7uF, and I still have no luck.

Does anyone have any ideas to a solution?

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Rvar1 is fine as 62k. It could be any value higher than 33k.
1nF is fine for the varactor coupling cap if it is a ceramic disc with short wires. Anything more than about 200pF is fine.
Did you use low capacitance 2N3904 or higher capacitance 2N4401 transistors as varactors?
Did the circuit tune stations properly when just a DC voltage was fed to Rvar1?

I still think the high impedance tuned circuit is swamped by the low impedance of the super-regen transistor's emitter. The tuned circuit belongs at the transistor's high impedance collector like every other super-regen receiver circuit.

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Did you use low capacitance 2N3904 or higher capacitance 2N4401 transistors as varactors?

I used the 2N2222, because I wanted a common transistor that can give a higher range of capacitance when used as a varactor.


Did the circuit tune stations properly when just a DC voltage was fed to Rvar1?

Yes it does, provided that a minimum of 4K resistance is connected between the tap of the control and the varactor input. The control then is a 2M (2 megaohm) control.

I was wondering if my R-2R DAC is supplying too much current.
When I used 62K and 120K as my R-2R DAC resistors, it worked better, but I didn't like those numbers because they make timings larger, and I only want to wait milliseconds between each button press.
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I was wondering if my R-2R DAC is supplying too much current.

Supplying current to what? The varactors are turned-off transistors that don't conduct current.

When I used 62K and 120K as my R-2R DAC resistors, it worked better, but I didn't like those numbers because they make timings larger, and I only want to wait milliseconds between each button press.

I don't see any large value filter capacitors that would take a long time to charge. You have 62k plus the DAC so a total of about 100k ohms charging the 1nF coupling capacitor from the varactors. So the varactor's voltage will be at 98% in 1nF x 100k x 5= 0.5ms. One two-thousandth of a second, pretty quick.

Is the supply voltage regulated?
What is the voltage range coming out of the DAC? 
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Is the supply voltage regulated?

I think so. I'm mainly using a DC adapter. Seldomly, I use batteries.


What is the voltage range coming out of the DAC?

I haven't measured it. but shouldn't the range of any homemade R-2R DAC go from 0 to full voltage?

and what does voltage range got to do with continuous tuning?

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I'm mainly using a DC adapter.

The voltage from unregulated DC adapters varies all over the place. Then your varactor's tuning varies all over the place. You need a voltage regulator for your project.

What is the voltage range coming out of the DAC?

I haven't measured it. but shouldn't the range of any homemade R-2R DAC go from 0 to full voltage?

and what does voltage range got to do with continuous tuning?

The CD4040 has a very low amount of output current. If the resistor values in your DAC are too low then the output voltage will be too low and then the varactors won't tune in wide enough frequency steps.
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The CD4040 has a very low amount of output current. If the resistor values in your DAC are too low then the output voltage will be too low and then the varactors won't tune in wide enough frequency steps.


oh boy, now I have to go and buy new value resistors.
So this must mean that these resistors are hogging the current in such a way, that the outputs from the cmos chip are meaningless.  :(

According to http://www.electronicproducts.com/ShowPage.asp?SECTION=3700&PRIMID=&FileName=sepana1.sep2003, they state there are two types of R-2R ladders. Current mode, and voltage mode. Which is better for my case and why?
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Why did you subtract 0.4V from 5V?
....
According to the datasheet, I found a line where VoH is 4.6V and the current is -0.51mA minimum.

The output of a Cmos logic IC goes to the supply voltage when high, only if it is not supplying current to a load. The outputs are tiny little Mosfets that have an internal resistance when they are turned on. So if the load draws 0.88mA then the output voltage typically drops 0.4V. A weak one (but still within spec) with a load current of only 0.51mA will have its output voltage drop 0.4V.

Therefore the voltage feeding your DAC isn't 5.0V but is whatever voltage that the 4040 output can produce with the DAC's current from it.

Look again at the datasheet. If you draw an output current of typically 4.2mA then the output voltage drops to only 2.5V. With a weak one then the current is 2.4mA.
So if the DAC resistance on an output is 2.5V/2.4mA= approx 1k then the voltage from the DAC could be only half what you want, a 50% error.
If the DAC resistance to an output is 4.6V/0.51mA= approx 9k then the voltage from the DAC could be 4.6V or more which is an error of only 8%.

post-1706-14279143121162_thumb.png

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