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Reverse potential divider power with diodes only?

neazoi

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Hi I have a static potential divider, 1k at its top, fed to the DC. Then 10k at the bottom part, connected to the ground. Their middle point is the output of my divider.

Now I want to reverse the polarity of the devider. That is, connect the ground at the top pf the divider and the vcc at the bottom.

But I want this to be done using diodes. That is, when I apply say a voltage of 5v to a diode combination, the top of the divider is fed with 5v and the bottom connected to ground. And when I ground this diode combination, the reverse to happen.

Is that possible with any way?
 

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crutschow

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when I apply say a voltage of 5v to a diode combination, the top of the divider is fed with 5v and the bottom connected to ground. And when I ground this diode combination, the reverse to happen.

Is that possible with any way?
Not with diodes, but it's possible with a MOSFET H-Bridge.
 

    neazoi

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BradtheRad

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Your verbal description sounds similar to this trick for obtaining full-wave rectification using just one diode. (Source unknown.) (Useful as a frequency doubler, or absolute value detector.)

Not sure if it's what you're looking for. Perhaps if you replace the 100 ohm resistor with a second diode?

full-wave rectification using one diode with resistors.png
 

    neazoi

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neazoi

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Your verbal description sounds similar to this trick for obtaining full-wave rectification using just one diode. (Source unknown.) (Useful as a frequency doubler, or absolute value detector.)

Not sure if it's what you're looking for. Perhaps if you replace the 100 ohm resistor with a second diode?

View attachment 167776


This schematic is new to me. How does it work can you explain it's operation, I am interested!


I am trying to create an inverter gate in non-conventional ways.
My first experiment: http://qrp.gr/inductoramp/
My second experiment: http://qrp.gr/diodeamp/
Both of these experiments worked and tested on the scope

This time I am trying to think of the seemingly impossible task to make it using DC as a VCC. So, as posted in post 1, if one could reverse the polarity in a potential divider, he gets reversed levels at its output. Well that was one of the 10s of thoughts. So I am trying to figure out if your circuit will fit on this, can you please explain how it works?
 

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neazoi

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Dana thanks very much. The point however is to implement this crazy experiment without any transistors or opamps, just diodes, resistors, capacitors. I have done that in the links posted above, but they use an AC as the power source. Now I am trying to see if this can be done with plain DC.
 

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This schematic is new to me. How does it work can you explain it's operation, I am interested!

The instructions say to adjust the potentiometer so as to obtain symmetrical waveforms.

The pot sets a midway voltage at the rightmost end of the load. The left end always sees the more positive voltage, through each half of the AC cycle (made possible by the diode). Thus current always travels from left to right through the load.

Call it a novelty circuit which struck me as worth adding to my collection of circuits.
 

neazoi

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The instructions say to adjust the potentiometer so as to obtain symmetrical waveforms.

The pot sets a midway voltage at the rightmost end of the load. The left end always sees the more positive voltage, through each half of the AC cycle (made possible by the diode). Thus current always travels from left to right through the load.

Call it a novelty circuit which struck me as worth adding to my collection of circuits.

I tested it but with square wave at 3.5MHz and I could not see any rectification effect, no matter how I set the potentiometer. I do not know what causes it.
The way I am thinking it works is:, at the positive cycle the diode conducts and the voltage at the left of the load is more positive than that of the right side (due to the setting of the potentiometer. At the negative cycle, the diode does not conduct ant the left side of the load is closer to the ground than the right side of the load (pot set around the middle?). So it should provide some reversion of the voltage at the load and not a rectification. By connecting the probe to the left of the load resistor, I get a square wave on my experiment, but lower in amplitude
 

FvM

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I tested it but with square wave at 3.5MHz and I could not see any rectification effect, no matter how I set the potentiometer.
You apparently misunderstood the circuit in post #3. The full-wave rectified voltage appears across the 1 k load resistor., is has to be measured differentially.
--- Updated ---

More generally, it's not possible to get full-wave rectified output (absolute value) from a single endded source without involving a voltage inversion function (e.g. a transformer) or a difference operation.

In a restricted frequency range, LC or RC phase shift can do the trick.
 
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    neazoi

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neazoi

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voltage inversion function (e.g. a transformer) or a difference operation.

In a restricted frequency range, LC or RC phase shift can do the trick.

I am very interested in these techniques. How can they implemented? Do they all depend on AC or can they be made on AC as well?
 

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