# Amplifying DC mV to Volts (Help needed)

Discussion in 'The Projects Forum' started by Electro Monkey, Jul 2, 2012.

1. ### Electro Monkey Thread Starter New Member

Jul 2, 2012
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Hi,
I am working on a project where my power source only produces 50-100mV and I would like to amplify this voltage so that I can power a white LED light.
Currently I have built something called a 'Joule Thief' that lets me run a ~3.6V white LED off of a 1.5V AA battery. Im not sure if I am on the right track?

Any advice would be greatly appreciated!!!

2. ### wayneh Expert

Sep 9, 2010
12,396
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You've already tried the "standard" approach, but as you see it cannot take you that far. 2X is one thing, but you're asking for almost 100X.

What you need is referred to "energy harvesting" and there are EH modules you can buy for this purpose. They're too expensive (\$20?) for casual hobbyists and DIYers.

Perhaps if you tell us what you are really trying to do, clever solutions might follow. If you have another power source, all you need is an op-amp.

3. ### mcgyvr AAC Fanatic!

Oct 15, 2009
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How much current can this power source provide?
How much current does your LED require?

afternath likes this.
4. ### afternath New Member

May 7, 2012
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Good question mcgyvr.

You cannot use only your 100mV source + opamp to powering your led since you will have to supply your opamp to make it working, moreover your opamp will only amplify voltage and not current and this circuit will become useless!!
Your LED need power to work properly, and power is voltage x current.

I watched for Joule thief on wiki and found sth interesting:

Example of a joule thief circuit driving an LED. The coil consists of a standard ferrite toroid core with two windings of 20 turns each using 0.15 mm (0.006 inch) diameter wire (38 swg) (34-35 AWG). The circuit can utilize an input voltage down to about 0.35 V and can run for weeks using a 1.5 V LR6/AA. The battery voltage is usually 1.5 V. The resistor is ~1 kΩ, 1/4 W. The transistor could be a BC547B, 2SC2500, BC337, PN2222, 2N4401 or other NPN. Vceo = 30 V, P = 0.625 W. A white light-emitting diode with Vf = 3.2 V might be used.[1]

Its saying :
"The circuit can utilize an input voltage down to about 0.35 V"
350mV is still high for you so you can try to adapt this circuit to make it working with 100mV input voltage

5. ### takao21203 Distinguished Member

Apr 28, 2012
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463
Not much point trying to make it work from 100mV.
It won't even start up from 350mV. This is the sustain voltage, means the circuit can sustain operation downto this voltage.

Using germanium components, you can make it work from 300mV, but they are uncommon these days.

6. ### afternath New Member

May 7, 2012
17
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Yes, definitively this will be a hard task to find a circuit which power a 3.6V LED with 100mVDC. Find another power source ...

7. ### Electro Monkey Thread Starter New Member

Jul 2, 2012
8
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Thanks for all the replies.

I was thinking of using a Op amp but i think there must be a easier way.
I saw this video on YouTube:

In it they show some sort of circuit which they call a 'boost converter', around 1:20 seconds into the video. It appears to be able to use the small amount of electricity produced from a thermoelectric module to light up a LED.

Does anyone know what a 'boost converter' circuit is or how i can build or buy one?

8. ### absf Senior Member

Dec 29, 2010
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You haven't answered mcgyvr's question. Where is this 100mV of yours coming from? Can it maintain the voltage if you're drawing 300mA from it. Without this vital information, discussing about boost converter ans lighting your white LED is meaningless....

Allen

Oct 15, 2009
4,791
976
10. ### wayneh Expert

Sep 9, 2010
12,396
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The good news is, prices for this sort of thing have come down. At less than \$5 for the IC mcgyvr referenced, I stand corrected on my earlier statement.

I think a lot more people will start to play with such things. It may become the new solar cell, with people coming here demanding a circuit to power their refrigerators with a 2" pelltier.

11. ### wayneh Expert

Sep 9, 2010
12,396
3,246
An op-amp would be a great solution if you just wanted to amplify the voltage 100X, but it needs its own power supply and wouldn't help you at all to capture the small signal as a power source. That's a completely different story.

12. ### Electro Monkey Thread Starter New Member

Jul 2, 2012
8
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to answer your questions I am unsure how much current my power source can provide. I assume a white LED would require around 20mA.

13. ### absf Senior Member

Dec 29, 2010
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OK, A white LED needs at least 3.2V and since your current needed is 20mA,
The power needed would be 3.2 x .02=64mW (P=V x I).

Since your power supply is 100mV, I = P / V, so
I = .064 / 0.1 = 640mA, assuming the efficiency of the converter is 100%.

Are you capable of supplying 640mA from the 100mV source?

Allen

14. ### absf Senior Member

Dec 29, 2010
1,493
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Wow, this is new stuff to me. Would download the datasheet and take a closer look. The low power PIC from microchip might be able to run with a lemon battery using this chip.

Allen

Last edited: Jul 2, 2012
15. ### absf Senior Member

Dec 29, 2010
1,493
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At first I also though that he wanted to use the 50-100mV as an indication to switch on something. That the voltage level is too low for some logic family to indicate as logic High.

Allen

16. ### wayneh Expert

Sep 9, 2010
12,396
3,246
The OP (and absf) might be interested in this. There's a kit at the bottom of the page.

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17. ### absf Senior Member

Dec 29, 2010
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That's awesome! But why it has to be so expensive? What is its secret?
Is the SMD chip by Burr Brown?

Allen

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18. ### #12 Expert

Nov 30, 2010
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The implications are scary! Not long until personal position transmitters can be implanted at birth.

19. ### Electro Monkey Thread Starter New Member

Jul 2, 2012
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I think this is exactly what I am after. Does anyone know how this circuit works? From what I have read it seems it uses a cap to store up the energy.