# Circuitry Beginner needs help

Discussion in 'The Projects Forum' started by Deado10, Aug 11, 2012.

Aug 11, 2012
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Morning Folks,

At the moment i am working on a circuit design that i wish to build but there is a few questions i would like to ask that may clear up some aspects of the design. Below is a simulated diagram of the circuit.
http://imgur.com/g88mF

(the blue line is just a border in diagram)

My goal is to have 6*1.5V batteries connected in series to create a 9v charge, this charge is to then be regulated by the 7805 to output a 5V charge. Both capacitors are used to remove any oscillations from the circuit, and finally the USB port will deliver a 5V current to an electronic device. My concerns are as follows:

- I am unsure as to whether or not i have correctly placed by regulator.
- My simulation shows an input of 4.45V and an output of 0.54V on the USB port, does this mean the port is able to supply a solid 5V or only 4.45V?
- I plan to charge my 6 batteries with a small solar panel that i have, in optimal conditions it can output 6.5V, i am however unsure where to connect it to circuit (as i dont want it powering the circuit, only the batteries). I was considering using a switch between the V2 source and the C4 capacitor, but once again not entirely sure.
- Finally, with regards to the solar panel and its 6.5V output, is it alright to directly connect the panel to the batteries for charging, or will it require a regulator as well? from what ive heard, batteries can handle quite a bit of damage and since the solar panel output may vary significantly depending on the amount of light its subjected to, i wasnt sure if it was safe to connect the panel to the batteries directly.

Thanks again,

James

2. ### BSomer Member

Dec 28, 2011
433
107
First thing to change is the capacitors. They should be tied Vin to GND and Vout to GND, not in series to the input and output as you show in the schematic.

I think that to charge your batteries you will need more than 6.5V from the solar panel. The solar panel output would be tied in parallel with the batteries.

Placing a switch on the output of the solar panel could isolate the batteries when you do not need to charge them.

As for the charging of the batteries I think you will need a circuit of some sort so as to not damage the batteries. This circuit will depend on the type of batteries used. I personally do not have much experience with solar panels and charging circuitry so someone else will hopefully clarify this part.

3. ### wayneh Expert

Sep 9, 2010
16,102
6,219
Most solar battery chargers have an unloaded panel voltage about double the nominal voltage of the battery. This allows charging at less than full sun, and gives some room for the voltage drop across the blocking diode.

You cannot directly charge a 9V battery with a 6.5V panel. The battery would have to be at 6.5-0.7 = 5.8V to accept any charge, and this would be too low for it.

You said your panel is small. It's unlikely to give enough juice for what you are expecting to accomplish, even if you got past the voltage issues.

Aug 11, 2012
9
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@BSomer, is this what you were suggesting? the voltages into the usb port seem to have dropped even lower...
http://imgur.com/PYLDi

@Wayneh, i understand your point in charging the 9v battery, but if its 6 individual AA (1.5v) batteries, and i charge them in some sort of parallel configuration (perhaps 2 parallel chargings of 3 batteries in series (4.5V)), would that work? or is there another way to do it? (i do have 2 solar panels) i also have 2 of each capacitor and 2 7805's. could you think of a combination?

5. ### DerStrom8 Well-Known Member

Feb 20, 2011
2,373
1,355
That is how the capacitors should be connected, yes.

What voltage are you getting at the USB port?

Aug 11, 2012
9
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an input (VCC) of 1.1V and an output (GND) of 3.89V

7. ### bretm Member

Feb 6, 2012
152
24
He's getting 5V now. It just looks funny because GND isn't at 0V so it shows +1.1V and -3.9V at the two terminals.

8. ### WBahn Moderator

Mar 31, 2012
24,559
7,696
Why are you using six batteries (9V nominal)? Why not 6V or 7.5V?

I can think of some possible reasons and I'm not saying that you should use four batteries, but I'd like to know YOUR reasoning.

Aug 11, 2012
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@bretm, so disregarding the solar charging section, this circuit is now correct?

Aug 11, 2012
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@WBahn, initially i was going to opt for 8 AA's and have 2 lots of 4 (series) as parallel so i could get a longer charge time out of them, but i went for the 6@9V because i wanted to keep the design smaller, but still large enough that i could produce enough power for a fast charge.

Aug 11, 2012
9
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This is the design with the solar panel integrated.
its kind of hard to see but the panel powers two sections of batteries (that are in parallel). these parallel sections are in series with the panel. any suggestions on this design? as mentioned above they are AA's with 1.5V output.
http://imgur.com/NjzbI

12. ### absf AAC Fanatic!

Dec 29, 2010
1,930
549
The moment the switch is ON, your solar panel is shorted at V19 + terminal and V14 - terminal. You need to rearrange the battery connections not to destroy the solar panel.

BTW putting a fuse and a diode in series is a must in your circuit after the switch, just in case something goes wrong.

Allen

Last edited: Aug 12, 2012

Aug 11, 2012
9
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do you have any suggestions?
what if i was to put a second switch on the wire that connects terminal 19 and 14, that would be switched off first, the turning the solar switch on?

14. ### absf AAC Fanatic!

Dec 29, 2010
1,930
549
I dont have much experience with solar panels. The ones that I saw are large ones made by BP for charging 2x 12V Lead Acid Battery in series for the "M.A.R.S." communication systems made by Philips. It was used in remote unmanned stations and I only visit it once a month for other maintenance works.

To charge 9V battery with only 6.5V supply is hard to design. You need a circuit to step the 6.5V to something like 11 or 12V in order to charge your battery. That's why WBahn was asking whether you can reduce the number of battery to 5 or 4. That would make the charger easier to manage the charging current and voltage.

I have a simple sketch attached, but I am not sure how the charging circuit block would look like but that's the idea I have.

Allen

• ###### SOLAR.PNG
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Last edited: Aug 12, 2012

Aug 11, 2012
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Okay i believe im just going to ignore the solar panel situation for now and get started on the circuit, as a final request for the time being can somebody please help explain why the input on the USB is 1.1V and the output is -3.89V. As it was previously commented, im still getting the 5V through my USB port, i dont understand how?
http://imgur.com/6HSLk

Aug 11, 2012
9
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As seen in the above image, J3(VCC) now displays a voltage reading of -0.36V and J3(GND) now displays a voltage reading of -5.36V. It seems that the VCC and GND should be the other way around, and also positive... can anyone see what the problem may be?

17. ### WBahn Moderator

Mar 31, 2012
24,559
7,696
I'm not familiar with your simulator, but all voltage measurements, whether simulated or real, are taken between two points (because that is how the very concept of 'voltage' is defined). So when you put a probe on some node and it tells you a voltage, that is the voltage relative to some reference point. That reference point is arbitrary and if you don't specify one then the simulator has to. If you want to know the voltage between two nodes in your circuit, you need to subtract the two node voltages. So if you want the voltage of J3(VCC) relative to J3(GND), you have (-0.36V)-(-5.36V) which is +5V.

Every simulator I have worked with, when left the task of assigning a reference, arbitrarily picks on of the nodes in the circuit and calls it Node 0 (or whatever it uses to indicate the ground reference). It doesn't appear that this is happening in your simulator; whatever method it is using isn't obvious to me. But there should be some kind of global ground symbol that you can attach to a node that will mark it as being electrically tied to the ground reference.