Voltage amplification?

Discussion in 'General Electronics Discussion' started by Number, Jun 12, 2013.

1. Number

65
0
Jun 9, 2013
I have a project where I am trying to boost the voltage from ~1.5V DC from a solar cell into something that's at least 5V DC. What I am trying to do is make a solar powered cell phone charger. My solar cell says it is 0.5 Watts & 6V DC but no matter how much light I give it I always get a 1.5V reading on the multimeter.

Awhile back I was reading up on transistors, and if i recall standard BJT's were for current amplification, and FETs were for voltage amplification. I could be wrong on this though. So how can I boost voltage from a solar cell into something more usable?

Op-Amps I believe can do this, but Op-Amps require another voltage source, which I do not think I can do with the solar cell I am using; so voltage dividing isn't really practical since 1.5V won't even power the Op-Amp, my memory says they're in the 4-5V for Vcc.

Any help is much appreciated.

2. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,501
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Jan 21, 2010
You need a boost DC-DC converter.

If you don't get more than 1.5V from the solar panel (that claims to be 6V) then something is wrong.

What is the short circuit current from the panel in full sun?

3. KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
As Steve says, you need a boost converter.

Your comments about transistors amplifying current and FETs amplifying voltage are correct, but amplification relates to signals, which contain just information, and not a significant amount of _energy_ or power.

For example, you can use a transistor, FET or op-amp to amplify a tiny signal from a microphone or similar device, to a higher level to power a loudspeaker, but you'll need a power source for the amplifier. The actual energy that drives the loudspeaker comes from this power source; the input signal only determines how this energy is used. This is why a transistor radio needs batteries!

What you're talking about here is boosting the voltage of a power source, i.e. a source of energy. That is fundamentally different from amplification of a signal, which is what's being referred to in your comments about current and voltage gain in transistors, FETs and op-amps.

You need to understand first that power output, for any circuit, is never more than power input. In fact, power output is equal to power input multiplied by efficiency, which for a typical voltage converter is no more than 0.8 (80%).

So, if your solar cell can provide 1.5V at, say, 0.1A, its power output is 0.15 watts. This figure can be calculated from the power law: P = V I, where P is power in watts, V is voltage in volts, and I is current in amps. If you feed your solar cell's output into a boost converter with 80% efficiency, the available output power will be no more than 0.15 x 0.80, which is 0.12 watts.

If the boost converter has an output voltage of 5V, the maximum available current will be only 0.024A (24 mA). I calculated this from the P = V I formula rearranged to I = P / V where P is 0.12 watts and V is 5V. So in the process of boosting the voltage (which is what a "boost converter" does), you reduce the current. This is referred to in physics as "conservation of energy", aka "there's no such thing as a free lunch"

If your charger can only provide 24 mA, charging a cellphone battery of, say, 1200 mAh would take about 50 hours! Actually, it wouldn't work at all, because when you connect your cellphone, it will try to draw hundreds of milliamps, which will make the voltage converter's output voltage sag and disappear.

If you would like to imagine that you can get more power out of a circuit than you put into it, there are many YouTube channels featuring "over-unity" devices that claim to do this, and will gladly accept donations from gullible science-impaired folks ;-)

So your first priority is to get as much _power_ out of your solar cell as possible. If it's specified as 6V but only ever generates 1.5V, even in direct sunlight, I would return it to the supplier. Did you buy it through eBay? Oh well, you know not to do _that_ again! Many eBay sellers hawk cheap low-quality products with specifications that only loosely correlate with reality.

Check the specifications on your cellphone charger. Most likely it's something like 5V, 1A. This is 5 watts of power; somewhat higher than what your solar panel can provide, even in direct sunlight. The answer is to buy a larger solar panel from a reputable supplier that can actually provide 5W of power in reasonable lighting conditions.

4. davennModerator

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Sep 5, 2009
Of course the solar panel rating of 6V is probably the open circuit ( no load) voltage
that is a very common thing.
And of course with a load you will never get 6V

Dave

5. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
Maximum power is typically achieved at around 80% of the open circuit voltage at which point you draw about 80% of the short circuit current.

Page 2 of this should clarify what I mean.

6. KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
Thanks for the rule of thumb and the link, Steve.

7. Number

65
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Jun 9, 2013
Kris,
That's what I was thinking. It didn't make sense that I could boost 1.5V to 5V. Although I'm not very familiar with different mechanisms, circuits, tips/tricks/trades etc. After all, a volt is just a potential difference in electrical energy. So I wasn't sure how it could be done to get more out then you put in, as you said. =]

I also bought a new solar panel, a 6V @ 0.5W. So it is essentially the exact same ratings as my other panel. I think what I an do is wire them up together to get more juice out of them. Much in the same way you connect multiple batteries together. The principle should be exactly the same.

If a supply source is say, 9V, and you put a load on it, then use a FET, what actually is boosting the voltage? And can it ever go higher than 9V? I reckon that it can be boosted much higher but doing so diminishes the amount of usable current, and thus less power.

So I stop asking you fine gentlemen questions, would you happen to have any links at hand; ones that discuss this material? So you know that I'm not lazy, I did view the content posted by this website, most of it I found quite useful, but more doesn't hurt. =]

You're information & replies are most appreciated, thank you.

Last edited: Jun 13, 2013
8. KrisBlueNZSadly passed away in 2015

8,393
1,271
Nov 28, 2011
Well, you can boost 1.5V to 5V, but you get less current. The important parameter is power, which is V * I, voltage multiplied by current. A DC-DC converter (boost, buck or whatever) converts one voltage to another, but it introduces losses. Assuming it's 80% efficient, the output POWER is 80% of the input POWER.
Again, the thing you can't create more of is power, not voltage or current. There are various ways to increase or decrease the voltage, but increasing the voltage decreases the available current, and vice versa, because the output power (V * I) can never be higher than the input power.
Yes, you can connect them in series to get a higher voltage. This won't necessarily solve your problem. I think you may need to buy a bigger, better-quality solar panel.
I'm not sure what you mean about using a FET. Are you talking about signal amplification again?
You're right that you can boost a voltage, and that doing so reduces the amount of available current. That's because power equals voltage multiplied by current; for a given amount of power, increasing the voltage decreases the current and vice versa.
I'm afraid I don't. But if you Google some of those important keywords (power, voltage, current, power law, watts, volts, amps, efficiency, voltage conversion, switching power supply etc) you should be able to find some good material.
You're welcome

9. Number

65
0
Jun 9, 2013
That's what I believe I'm going to have to do, unfortunately. The purpose of using a smaller solar panel was the compact size. This is what I was going to do, use the small compact size of the panel and mount it to an altoids can or small project box. Going bigger is not really helpful, but it will solve the input voltage so I can effectively charge a USB device.

10. BobK

7,682
1,688
Jan 5, 2010
A phone charger than can charge in about 2 hours is going to require about 5W of power. You are not going to get this with a solar panel the size of an altoids can. It would require a solar panel of about 50 cm (18 in) square.

Of course, if you want to charge in 20 hours you could use a 0.5W panel But that would be about 3 sunny days for a single charge. You would probably have to use an intermediate battery because the power coming directly from the solar panel would not activate the phone's charging circuit.

Bob

Last edited: Jun 13, 2013