# BEAM Solar Engine Circuit Help

Discussion in 'General Electronics Discussion' started by henrib736, May 25, 2013.

1. ### henrib736

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May 25, 2013
Hi Everyone ,

I am new to electronics, and I'd like some help understanding this simple circuit if you don't mind.

Here's the circuit:

My understanding of the basic function of this circuit is to charge the capacitor using the solar cell, and when the FLED reaches a certain voltage, it goes on which causes the PNP transistor to conduct which causes the NPN transistor to conduct. Then current from the capacitor flows through the battery until the voltage is low enough for the transistors to go "inactive". That's my basic understanding.

I'm confused with the orientation of the solar panel and the capacitor. If the electrons in the negative side of the solar panel are flowing to the negative side of the capacitor, how would those electrons flow through the FLED in the right direction?

Also, how would the capacitor release its charge if the solar panel is still connected?

Thanks for reading. Any help would be appreciated!

-Henri

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May 25, 2013
3. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
Call the transistor on the left Q1 and the one on the right Q2.

The BE junction and the LED are forward biased.

When the voltage on the capacitor gets high enough to allow current to flow through them, this turns on Q1 a little.

With Q1 turned on a little, Q2 begins to turn on. As it does so, it pulls the base of Q1 more negative, turning it on harder, and hence Q2 is turned on harder.

Because Q2 is turned hard on, current flows through the motor.

When the capacitor is discharged, the transistors turn off and the solar panel starts charging the capacitor up again...

The capacitor releases its energy because the transistors turn on. The solar panel cannot supply power fast enough to keep the capacitor charged. (If it could, the motor would stay turned on.

I think this circuit would be improved by a resistor between the collector of Q1 and the base of Q2. This will limit the base current of Q2 to a safe value (100R should be sufficient). I'd also put a reverse biased diode (the arrow would point UP in this circuit) across the motor to save the transistors from voltage spikes.

4. ### poor mystic

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Apr 8, 2011
Hi
You can think of the solar cell together with the capacitor as the "power supply". There is no problem with the charge stored on the capacitor. It's like taking a battery out of a flashlight - the electricity doesn't mind at all.
As for the "flashing led" integrated circuit, it seems OK to me - the right way up, et cetera.
Does the circuit work?
Mark

5. ### henrib736

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May 25, 2013
Thanks for the responses.

Poor Mystic, doesn't the current have to flow backwards to release the charge from the capacitor? If the electrons are moving from the negative to the positive end of the solar cell, then how will the capacitor release the charge?

When the capacitor does release its charge, where would the electrons from the negative side of the capacitor go since the transistors and LED are forward biased?

6. ### davennModerator

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Sep 5, 2009
dont confuse yourself switching between electron flow ( negative to positive) and convention current flow ( positive to negative)

Steve described the circuit operation using conventional current flow which is all you need to use and understand for this an any other circuit

Dave

7. ### henrib736

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May 25, 2013
That makes a lot more sense!

I have another question.

Can the capacitor have more voltage than the source?
My source is around 0.5-1 volts, but the FLED needs around 2.4 volts to flash. Is it possible to charge the capacitor to a higher voltage than the source in the circuit?
If not, do I have to get a better solar cell?

Thank you.

8. ### henrib736

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May 25, 2013
Nevermind, I found out that capacitors can't have more volts than the source.

9. ### henrib736

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May 25, 2013
Poormystic, the circuit works.

10. ### poor mystic

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Apr 8, 2011
YAY!!!

...are you going to do as Steve suggested and put a flyback diode across the motor?

Last edited: May 27, 2013
11. ### henrib736

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May 25, 2013
Do you think its necessary to use a flyback diode? The voltage isn't much. I have two solar panels which have a combined voltage of 2.5 volts. Are voltage spikes still bad at that level?

I have two solar panels in series. 1 inch from my desk lamp, one makes 1 volts and the other makes 1.6 volts. In normal room light, one makes 0.2 volts and the other makes 0.5 volts. Are you familiar enough with solar panels to know how many volts a 3.5 volt solar panel will have in room light? I've seen videos on youtube of the same circuit but with solar cells half the size of mine, and they are under normal room light.

12. ### henrib736

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May 25, 2013
Do you think the flyback diode is necessary? Combined, my two solar cells produce 2.5 volts. Are voltage spikes still dangerous at that level?

Also, are you familiar enough with solar cells to know how many volts a 4.5 volt solar cell would produce under normal room light? In normal room light, one of my solar cells produces 0.3 volts, and the other produces 0.5 volts, but I need at least 2.5 volts for the motor to run. To get my solar cells to produce enough voltage, I need to put my desk lamp 1 inch from the cells. I've seen Youtube videos of the same circuit as this but the solar cell is half the size of mine, and the light conditions are normal room light conditions. Do you know how much voltage a 4.5 solar cell would produce under room light?

13. ### poor mystic

1,067
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Apr 8, 2011

You could research "switched inductors" and would then begin to get an inkling why the 2.7V supply can still produce high voltage spikes.

As for how well your solar cells will perform, you'll have to try using them to find out.

PS Steve's resistor is a good idea too. Excepting myself, everybody who has answered you is a well-regarded expert. Listen to them!

Last edited: May 27, 2013
14. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
Yes. The power is coming from the capacitor at a significant current. You could very easily get spikes up to 50V or more. -- in fact the voltage will pretty much rise to the value needed to cause avalanche breakdown of the transistor.

Well, that says you need 5.

I would recommend you connect the solar cell to a capacitor and check it every minute or so to see how high (and how fast) you can charge it.

More panels (in series) will increase the voltage. Larger panels (or panels in parallel) will charge the capacitor faster.

The charge time will determine how active your device is in lower light conditions.

That is something to experiment with.

Last edited: May 27, 2013
15. ### henrib736

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May 25, 2013
Ok, I'll definitely put a flyback diode across the motor after learning how it works.

Sorry for posting twice about the same thing. I didn't see there was a second page for this thread, so I thought my post got deleted.

16. ### KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
You should also take Steve's advice about putting a resistor between the PNP's collector and the NPN's base. This was my first reaction on seeing that circuit.

When the transistors turn ON, a significant amount of energy will be wasted due to the heavy current flow into the NPN's base. Since this circuit is solar powered, I assume you want to waste as little energy as possible.

Start with 100 ohms as Steve suggested, but you might have to adjust it. Too much resistance will cause the NPN to turn off unnecessarily early, especially if the motor draws a lot of current. Too little resistance (including no resistance) wastes energy.

17. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
One of the inefficiencies of this circuit is that it drains the capacitor too much. Ideally you want the circuit to turn on at the point there is "enough" voltage on the capacitor (which this does), but also to turn it off again when there is insufficient to power the motor any longer. This circuit doesn't do that very well.

Let's explain a few things as background.

The motor requires more power (voltage and current) to start than it does to keep running. This means that the capacitor needs to have a higher voltage to start the motor than it does to keep it running.

However, at some point there is insufficient voltage across the capacitor to keep the motor running (the actual voltages depend on the motor and the load it is under).

The ideal circuit would turn off the power to the motor at the point just before the motor stopped. This means that the capacitor retains some charge. If it keeps allowing the capacitor to discharge through the stalled motor, you just waste power for no good reason. You then need to gather that wasted energy from the solar cell in addition to the power that will actually be used to turn the motor.

So, the drawback that Kris mentioned (turning off unnecessarily early) can be exploited to increase efficiency. You should pick a resistor value that provides the best balance of performance.

18. ### KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
Good thinking, Steve.