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How to use diodes

by Cspeed
Tags: circuit, diode, electricity
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Cspeed
#1
Dec29-08, 09:38 AM
P: 44
I have built an AC induction generator. The only diode I have I took off of an old computer, but I used my multimeter's diode test, so I'm pretty sure that it still works fine. I want to be able to connect a DC ammeter (with needle, not digital) to the generator (if I connect it just plainly to receive AC it doesn't work, as expected).

I think I should just be able to connect the diode in series with the generator and the ammeter, but the needle doesn't move still. The generator doesn't produce too much current, but it should definitely be enough to move the needle a little at least. The diode couldn't be backwards, right? I've also tried using several other diodes and LEDs (new, from school) and they don't provide any better results. They all seem to block all current from going through.

Also, it seems that when I measure the voltage across a lone diode (not connected to anything), it has a voltage like it were a capacitor or something (it isn't).

Any help is appreciated.
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Redbelly98
#2
Dec30-08, 07:27 PM
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What is the smallest current your meter is sensitive too? It sounds like the generated current is simply too small for your meter to register.

If you can measure an open-circuit DC voltage, that would mean your generator is working to some extent. I would try connecting different resistors in series with the diode, and then measure the voltage across the resistor. Starting with 1 MΩ, measure the voltage across the resistor. Then keep going down by 10x (eg. 100 kΩ, then 10 kΩ, 1kΩ, etc.).

Stop when the voltage is somewhere between 1% and 10% of the "open circuit" voltage. Use Ohm's law to calculate the current through the resistor, and you will be pretty close to the maximum current the generator can produce.

Also, it seems that when I measure the voltage across a lone diode (not connected to anything), it has a voltage like it were a capacitor or something (it isn't).
I have seen that too. Actually, the diode is acting like an antenna of sorts, probably picking up the 50 or 60 Hz voltages that are being radiated from the wiring and appliances around you. I can get readings of several hundred mV this way. The fact that a diode conducts preferentially in one direction means that a DC voltage can be generated and measured, even though these are AC signals. It's normal.
Cspeed
#3
Dec31-08, 09:28 AM
P: 44
When I measure the voltage with my digital multimeter, it is about .15 V AC (just simply connected in series with the generator). The DC ammeter should pick up .01 amps or maybe even lower. Is it possible that the diode doesn't work unless a minimum current is going through it? I really don't know much about diodes (I'm just an amateur), and my assumption was that it would simply block the negative voltages, creating pulses of DC voltage rather than AC.

I'll keep trying to figure this out and see if I can make any progress.

Redbelly98
#4
Dec31-08, 11:42 AM
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How to use diodes

Quote Quote by Cspeed View Post
When I measure the voltage with my digital multimeter, it is about .15 V AC (just simply connected in series with the generator).
Okay, that's useful to know and a good starting point.

The diode requires about 0.6 to 0.7 V in order to conduct a "reasonable amount" of current, of say 1 to 10 mA. So the 0.15V generator output is not enough to overcome the diode voltage, and the result will be an incredibly small current, perhaps of around 1 nA or less as a ballpark figure.

So the problem now is to figure out a way of generating significantly higher than 0.7 V without a diode, before adding the diode into the circuit. A few options are:
Turn the generator faster
Use stronger magnets
Use more turns of wire
Overall, you need perhaps 5 to 7 times the voltage you are getting now. That means 5-7 times as many wire turns, for example. Or some combination of the above that results in 5-7 times more.

Keep posting back with the questions, we are here to help. There's nothing like working at something and getting it to work finally.
Cspeed
#5
Dec31-08, 12:05 PM
P: 44
Thanks for your response; it answers a lot of questions.

The diode requires about 0.6 to 0.7 V in order to conduct a "reasonable amount" of current, of say 1 to 10 mA.
Why is this? Is this the same with all types of diodes? Assuming I can get the generator to go up to, say, 1.5 V AC (10x as much as now), can I expect to get a few deciamperes?
Pumblechook
#6
Dec31-08, 12:43 PM
P: 359
Silicon PN junctions need 0.7 V across tham before they will conduct. Barrier Potential.

A germanium detector diode needs less .. 0.3 V but will not survive too much current flow. OA90, OA95. You might find one of these in an old radio. OA47 may be the best.

http://members.tripod.com/baec/articles/crystal.htm


Alternative is a Schottky diode. As low as 0.15 V. But these can be as high as 0.45 V.

If the point is to drive a meter as a speed gauge or something and not actually use any of the power generated you could use an amplilfier.
Redbelly98
#7
Dec31-08, 03:41 PM
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A few years ago I measured current vs. voltage for a common silicon diode:



The blue dots are my measurements, the magenta line is a curve fit to the data.

The behavior follows an exponential growth curve quite well, from 1 nA up to 1A. The current grows by factor of 10 for every 0.10V increase in this range.

At -1V, the current is -1.5 nA, which can be considered to be zero in most applications.
berkeman
#8
Dec31-08, 08:22 PM
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Extra credit Quiz Question for OP Cspeed -- what is the "diode equation", and why does it support Redbelly's data?
Cspeed
#9
Jan1-09, 10:29 AM
P: 44
I(Diode)=I(saturation) * (e^qV/NkT - 1) seems to be the diode equation. Looking that up gave me some good insight. I would like to know about how diodes work, but at the same time I just want to get my generator working. I was hoping to build a bridge rectifier, which I had hoped would just convert x V AC to x V DC, but I'm seeing some info that makes me think that a diode will eat up .7 V (which is the forward voltage of silicon diodes).
berkeman
#10
Jan1-09, 12:43 PM
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Yes, you will always lose some forward voltage drop across a diode. By using Schottky diodes, you can minimize that drop, but they are a little more expensive than regular diodes.
Redbelly98
#11
Jan1-09, 01:49 PM
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Schottky diodes are 0.50 $US from Digikey:
http://search.digikey.com/scripts/Dk...=497-4547-1-ND

Voltage is about 0.25 V at 10 mA current. See Fig. 9-1 here:
http://www.st.com/stonline/products/...262/1n5817.pdf

You'd still need more than the 0.15 V presently generated. I'm wondering if either the generator could be spun around faster, or more coils of wire could be added to it. Either of those would increase the output voltage.

Mark
Pumblechook
#12
Jan1-09, 02:12 PM
P: 359
Thermistor-bolometer-thermocouple. OR just boost the voltage with a transformer.

OA47 (low barrier).
Redbelly98
#13
Jan1-09, 02:57 PM
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Quote Quote by Pumblechook View Post
... just boost the voltage with a transformer.
Yes! That's probably the easiest, quickest thing to do. Get a step-down transformer at Radio Shack, and use it in reverse to get a voltage boost.
Cspeed
#14
Jan1-09, 03:25 PM
P: 44
Just by turning it faster, I can get over .2 V, and I could probably get just over .25 with little extra effort, but this was really just meant to be a prototype. I could add more coils, magnets, etc. Thanks for all of your help, but I'm going to put the process of adding more components on hold for a while.
berkeman
#15
Jan1-09, 03:33 PM
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I agree with the transformer boost ideas, as long as your power budget works out.

Just FYI, another (more expensive and complicated) alternative would be to use synchronous MOSFET rectification. The name pretty much speaks for itself. Buck DC-DC regulators sometimes use synchronous rectification to increase efficiency, but I don't think I've seen a synchronous full-wave rectifier used in practical applications before. Still, if someone asked me to get all the power I could out of a 0.2Vpp waveform, I think that would be my main option.

Good experiments. Keep on experimenting and learning!
Pumblechook
#16
Jan1-09, 04:12 PM
P: 359
Two silicon diodes in series and a series 1.5 V cell. OR one LED.
Redbelly98
#17
Jan1-09, 09:57 PM
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Quote Quote by berkeman View Post
... if someone asked me to get all the power I could out of a 0.2Vpp waveform ...
It's 0.2 Vrms, so perhaps 0.25-0.3 V at peak maximum (see Post #3). So it's a tad somewhat slightly better than one might think.

Good experiments. Keep on experimenting and learning!
Well said.


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