Charge capacitor from RF diode/coil?

[PaulQ]

I got a 1N34A diode crystal radio working (var. cap. not pictured) thanks to this site. As an experiment, I'd like to light an LED from a capacitor, which would be charged from the antenna/coil/diode. In the picture, the rectified voltage is .19 v (can reach .35 v). When I take out the capacitor, it measures (for a second) whatever voltage the circuit was measuring. I'd like the capacitor to charge enough to light the LED (1.8 v). I've experimented with different value capacitors and resistors, same outcome. Ground is wall socket screw--only ground available near window of my apt.
Thanks for suggestions.

 

[prcguy]

The capacitor is more for smoothing out the ripple in the resulting pulsed DC from the diode. You need enough RF to reach the required voltage to lite up the LED and enough current to sustain it. A capacitor will store energy but you need to replace it continuously. If you had a very large capacitor like several thousand or more uf, it could power the LED for a brief amount of time between charging the capacitor with RF.

What are you using for an RF source to run this thing?

 

[prcguy]

I just noticed the capacitor value is 1,000uf. You would need a very strong and continuous RF field to make that work and the coiled up wire is a very poor antenna. You should pick up a lot more signal with 100ft of wire outside. Even with that you would need an HF ham or CB operator very close by to provide enough RF to rectify to get above just a few 10ths of a volt.

 

[PaulQ]

Thanks for responding. Living in upper floor of apt. bldg I can't run wire outside. That goofy antenna receives better than 15' of wire across my room. Is it possible for the .20 v of RF to charge the capacitor? I've tried disconnecting the LED, also tried wiring capacitor negative back to diode--no difference in voltage.

 

[prcguy]

The components you have do not make electricity, they need a source of AC electricity and then it will rectify that AC and charge the capacitor to the peak level of the AC source and no more. It will also require the AC source to supply enough current to light the LED, which probably needs at least 5ma of current at over 1.5 volts and more like 20ma of current at 1.8 volts for full brightness.

If you have measured .2 volts somewhere in the circuit then that could charge the capacitor but it wont get anywhere near what it takes to light the LED. This is the kind of circuit you might use if you live across the street from a 10kW AM radio station where there is plenty of RF to charge a battery if you want. If you don't have any high power HF transmitters around there is nothing to power the circuit beyond what you are measuring.

Thanks for responding. Living in upper floor of apt. bldg I can't run wire outside. That goofy antenna receives better than 15' of wire across my room. Is it possible for the .20 v of RF to charge the capacitor? I've tried disconnecting the LED, also tried wiring capacitor negative back to diode--no difference in voltage.

 

[PaulQ]

Ok thanks. So I've learned that capacitors only charge to the level of voltage they're provided. I thought they might accumulate voltage until their capacity is reached, even if input is lower voltage than capacitor.

 

[RFI-EMI-GUY]

A voltage multiplier circuit will work better. It is a circuit that stacks several diodes in series along with AC coupling and DC storage capacitors so that the rectified voltage of each diode is added to the next. Usually used in high voltage circuits like CRT and night vision scopes , but the theory can be applied to RF rectification. But this assumed there is enough RF field to accumulate.

 

[RFI-EMI-GUY]

https://en.wikipedia.org/wiki/Voltage_multiplier

 

[PaulQ]

A voltage multiplier circuit will work better. It is a circuit that stacks several diodes in series along with AC coupling and DC storage capacitors so that the rectified voltage of each diode is added to the next. Usually used in high voltage circuits like CRT and night vision scopes , but the theory can be applied to RF rectification. But this assumed there is enough RF field to accumulate.

At night, the diode can produce up to .9 v; so a voltage doubler would produce 1.8 v which should power a red LED. So I modified circuit last night, using same 1N34A diodes, but voltage isn't doubling. Thanks for suggestions.

 

[Ubbe]

Even if you get enough energy to charge the capacitor the LED will start to conduct and draw too much current without emitting any light. You would need a trip circuit that disconnects the LED until the capacitor have charged fully and then connect the LED and it will make a short flash and will then have emptied the capacitors energy.

It's almost like Nicolai Tesla's system of air born energy collected from energy towers. Would be great if you could harvest all the RF floating around from cellular towers and other transmitters to power some household devices.

/Ubbe

 

[PaulQ]

Thanks Ubbe, I suspected the LED was continuously drawing current; adding resistors doesn't seem to change this.
Please suggest how (or schematic) to integrate trip circuit in the voltage doubler.

Even if you get enough energy to charge the capacitor the LED will start to conduct and draw too much current without emitting any light. You would need a trip circuit that disconnects the LED until the capacitor have charged fully and then connect the LED and it will make a short flash and will then have emptied the capacitors energy.

It's almost like Nicolai Tesla's system of air born energy collected from energy towers. Would be great if you could harvest all the RF floating around from cellular towers and other transmitters to power some household devices.

/Ubbe

 

[prcguy]

You have to consider the LED takes anywhere between 5ma and 20ma to light up and the RF energy has to be at a high enough level to support a continuous 5 to 20ma current and its probably not. The circuit you are making is something an instructor might use in a high school electronics class to demonstrate certain principals and he may be using a portable transmitter suitable for the task.

This is not a useful circuit for the average home since there is probably nothing to charge the circuit up unless your patient and wait for that high power HF ham operator to transmit while he drives past your house. Then the circuit is dead until the next time he passes by transmitting.

At night, the diode can produce up to .9 v; so a voltage doubler would produce 1.8 v which should power a red LED. So I modified circuit last night, using same 1N34A diodes, but voltage isn't doubling. Thanks for suggestions.

 

[PaulQ]

Hi prcguy I'm 56 years old, this experiment is interesting to me, albeit high-school level.

I'll measure the current, after rewinding the coil...the middle tap broke last night (too much tweaking).

 

[RFI-EMI-GUY]

At night, the diode can produce up to .9 v; so a voltage doubler would produce 1.8 v which should power a red LED. So I modified circuit last night, using same 1N34A diodes, but voltage isn't doubling. Thanks for suggestions.

The circuit you have drawn does not resemble in any way a voltage multiplier.

 

[prcguy]

What you might do is try and figure out what frequency some strong signals in your area might on be then make a specific antenna for that. You can easily gain or loose 100X the energy by using the wrong antenna and the one you have is probably not great at anything. If you have a nearby FM transmitter you could make a dipole for the FM broadcast band with the diode and bits right at the feedpoint. For a nearby AM station I might consider a large ferrite rod antenna tuned to the specific frequency and so on.

Hi prcguy I'm 56 years old, this experiment is interesting to me, albeit high-school level.

I'll measure the current, after rewinding the coil...the middle tap broke last night (too much tweaking).

 

[RFI-EMI-GUY]

Also as PRC says, the RF field might not have appreciable current. So test it unloaded . You can always make a LED flasher that blinks slowly as the voltage and current builds up.

 

[PaulQ]

Thanks for helpful replies. I'll return to this after improving antenna for max voltage (given space limitation) and working out the voltage divider circuit (on left, "a"), which I adapted from this article, by Ray Marston. My adapted circuit is attached to this post.

RFI Guy, please indicate how R. Marston's circuit or my adaptation of it does not in any way resemble a voltage multiplier. This is a great learning experience.

 

[RFI-EMI-GUY]

Thanks for helpful replies. I'll return to this after improving antenna for max voltage (given space limitation) and working out the voltage divider circuit (on left, "a"), which I adapted from this article, by Ray Marston. My adapted circuit is attached to this post.

RFI Guy, please indicate how R. Marston's circuit or my adaptation of it does not in any way resemble a voltage multiplier. This is a great learning experience.

It would be easier to explain how it should work:

In the drawing below C1, C3 and C5 (edited) are used to charge each doubler circuit. Pairs of D1+D2, D3+D4 and D5 + D6 are full wave rectifier/doublers that independently charge C2, C4 and C6 in series.

In your circuit, D1 blocks the full wave rectification, so only a half of the waveform is rectified. See figurea 3 a and 3 b of his article as to how a single doubler should be wired.


Simplified circuit Figure 3 a and 3 b

 

[RFI-EMI-GUY]

Back to prcguy's point, your challenge will be to glean enough power to overcome circuit losses. For example, the capacitors should be low leakage so that they are not dissipating more power than absorbing as a charge. Could this be workable to charge some very small battery? Maybe, but it may take a big antenna and a lot of time. There is actually a lot of energy in the HF band at certain times of the day. But it is not at all coherent, meaning the voltages of the waves will not be greatly additive.

 

[RFI-EMI-GUY]

I have a field strength meter from the 1960's a Monarch FSI-1 that has a super sensitive mechanical meter movement. It works great with one simple half wave diode rectifier. I have wondered how much more sensitive a meter I could make using voltage multiplier circuit. I did modify that meter to light an LED when the sensitivity control is set to off. It will light with a 5 watt UHF portable nearby.