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Topic: Charge capacitor using higher voltage (Read 12282 times) previous topic - next topic

Pavouk106

Have a look at the schematic for a possible way to consider doing this.


This looks interesting. I have some questions though:
1. Can I directly connect the 6,5V from solars to capacitor (provided that I cut it off before reaching capacitor max voltage)? I always thought that I shouldn't charge the capacitor with more than its maximum voltage. Well, my electronics experience is not really good...
2. Why is there resistor on gate (G)? To prevent frying my Arduino because of high current to G? Shouldn't I have one larger (ie. 22k) resistor connected between G and S?
3. Wouldn't there be a problem with zener? I always thought it would "eat up" all the voltage above its rating.

There are more :-) I'm measuring voltage on solars (source) with Arduino too. This is to prevent Arduino from dying if the voltage drops too much (it's powered from solars) = when the Arduino is low on voltage, it cuts capacitor from source, so it can regain power for running itself.

4. How would your circuit behave in the way of measuring source voltage? Wouldn't the source voltage be the same as capacitor voltage when MOSFET is turned on?
5. Wouldn't the zener limit the source voltage if the MOSFET is on and the capacitor is still charging?

I'm sorry, if these questions look dumb, but I don't know much about electronics, only basics.

I will try to write how I understand your schematics - Solars have enough power, Arduino is on and it opens the MOSFET, charging begins. Capacitor's voltage goes up, but solars voltage drops, because the capacitor is "hungry" at the first moments. Arduino keeps itself on, because it closes and reopens the MOSFET repeatedly keeping the solars voltage high enough to power Arduino. MOSFET open state gets prolonged as the capacitor voltage rises and at (ie.) 3V MOSFET stays on (because capacitor isn't that "hungry" and solars voltage doesn't drop that much) until capacitor reaches 5V. Then Arduino closes the MOSFET.

Sorry to write it as a story, but I think it's the most understandable description. Is it how it would work? My main fear is from frying the capacitor with too much voltage and from dropping the solars voltage too much (which should be prevented by software) when MOSFET is opened.

BillO


This looks interesting. I have some questions though:
1. Can I directly connect the 6,5V from solars to capacitor (provided that I cut it off before reaching capacitor max voltage)? I always thought that I shouldn't charge the capacitor with more than its maximum voltage. Well, my electronics experience is not really good...


The solar cell has an internal resistance that will prevent the open circuit voltage from appearing across the capacitor as it charges.  How long that takes depends on the current capability of the solar cells and the total resistance in the circuit, however the Arduino is capable of responding in a few micro-seconds so this should not be a problem.  Just to be on the safe side, do you have links to datasheets or specifications on both the capacitor and the solar cell?  That would help.  It may be that we need to add a resistor coming from the solar cell.

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2. Why is there resistor on gate (G)? To prevent frying my Arduino because of high current to G? Shouldn't I have one larger (ie. 22k) resistor connected between G and S?


Theoretically the gate resistor is there to limit the current sourced by the Arduino.  Again, I am not sure what size MOSFET you will need, nor how often you will need to turn on and off the MOSFET.  I'll have a better idea once you let me know the specs of the capacitor and the solar cell.  I do believe it is highly unlikely you will absolutely need one however, I was soundly castigated recently on this site for suggesting that there are cases where they are not needed.  So I promised to be a good boy and suggest them so as not to give other people heart trouble.  It's only a few cents, and having one certainly will not hurt in this application.  Use 75-150 ohms.

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3. Wouldn't there be a problem with zener? I always thought it would "eat up" all the voltage above its rating.
It is only there as a safety.  If for some reason the Arduino fails to shut off the MOSFET, the zener will prevent the capacitor from being over charged.  Since the zener is rated at 5.3V and the Arduio will shut off the MOSFET at 5V, it should never come into use under normal conditions.

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There are more :-) I'm measuring voltage on solars (source) with Arduino too. This is to prevent Arduino from dying if the voltage drops too much (it's powered from solars) = when the Arduino is low on voltage, it cuts capacitor from source, so it can regain power for running itself.

4. How would your circuit behave in the way of measuring source voltage? Wouldn't the source voltage be the same as capacitor voltage when MOSFET is turned on?


There is no need to do this.  If the Arduino is using all the solar power and all the capacitor power, disconnecting the capacitor will not help.  The capacitor is not like a battery and will only charge if there is excess power available and is only then a load on the circuit.  If there is only enough power to keep the Arduino going, the capacitor being connected will no hurt the situation.  I think you can skip doing this altogether.

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5. Wouldn't the zener limit the source voltage if the MOSFET is on and the capacitor is still charging?


Only if the Arduino fails to turn off the MOSFET.

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I'm sorry, if these questions look dumb, but I don't know much about electronics, only basics.


Not everyone can be expected to know everything about electronics, and the only way we can learn is to seek knowledge.  There are no dumb questions.  The only dumb thing is not to ask.

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I will try to write how I understand your schematics - Solars have enough power, Arduino is on and it opens the MOSFET, charging begins. Capacitor's voltage goes up, but solars voltage drops, because the capacitor is "hungry" at the first moments. Arduino keeps itself on, because it closes and reopens the MOSFET repeatedly keeping the solars voltage high enough to power Arduino. MOSFET open state gets prolonged as the capacitor voltage rises and at (ie.) 3V MOSFET stays on (because capacitor isn't that "hungry" and solars voltage doesn't drop that much) until capacitor reaches 5V. Then Arduino closes the MOSFET.


Almost.  I would power the Arduino from after the MOSFET.  From directly across the capacitor.  Set the Arduino to turn on the MOSFET when the voltage reaches about 3.5V.  This will charge the capacitor and power the Arduino.  When the voltage reaches 5V the Arduino switches off the MOSFET and the capacitor runs the Arduino until the voltage drops to 3.5V, then the cycle repeats.

3.5V for the turn on point is only a guess.  If that is not appropriate for your application, choose a voltage that is (4V, 4.5V, what ever is best).

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Sorry to write it as a story, but I think it's the most understandable description. Is it how it would work? My main fear is from frying the capacitor with too much voltage and from dropping the solars voltage too much (which should be prevented by software) when MOSFET is opened.


I'll need the specs of the capacitor and the solar cell to help much further.  It would also help to know the current draw of the Arduno and any other circuitry while it's working.  Also helpful would be knowing what other circuitry is being used.

Pavouk106

The thing is, that Arduino should be there controlling the (dis)charging of the capacitor, not running from it, so it should be run right from the solars. I have to keep the Arduino alive by disconnecting capacitor when voltage drops, because I need it to hold SW variables.

Other things are really ok as you suggested them.

Solar cells are "china made", came without absolutely anything beside "5,5V/100mA" information, which is, to be honest, not the best info. Four of these in parallel runs Arduino at 6,7V at bright sunlight and above 5V when overshadowed from direct sunlight at around 27mA. I will measure current through and voltage on 56 ohm resistor tomorrow, supplying it with those 4 solars in parallel. It should give us some idea how much power these can do.

Capacitor's datasheet is here (I have 1F version), it's memory backup capacitor, I was running it at 50mA discharge many times, still going well (it is not probably designed for this task I know).

BillO

Okay, a few questions from my end.

1) What is the capacitor for?

2) Do you know the load (current draw) of your proposed circuit?

3) What does your whole design do?

Pavouk106


1) What is the capacitor for?

Driving motor (for a few seconds)

2) Do you know the load (current draw) of your proposed circuit?

Shouldn't be more than 50mA for those few seconds

3) What does your whole design do?

You didn't want to know that - solar tracker. The capacitor is there to power the turning motor, solars wouldn't have enough power to do it on their own. The main goal is to do it, not to use it for the rest of my life. I would like to make it on my own, not by "copy-paste" style of someone's work.

I still don't have all the parts, I'm missing the motor. But I still have troubles with capacitor, so who cares about missing motor at this moment :-) I know, You would like to know specs of it. "China made on 5V less than 25mA", no info about stall current. I will be pretty surprised, if it all works as it should at last.

BillO

No Problem.

Okay, then use the circuit I gave you for the charging the capacitor.  Forget running the Arduino off it.  Just add a resistor between the solar cell to the MOSFET.  Actual value is not critical, maybe 50 to 200 ohms.  It needs be be large enough that charging the capacitor does not steal too much from the Arduino supply but still charges the capacitor quickly enough for your use.  Experiment on the value for your particular needs.

Use another zener (maybe 5.7V to 6V) in an NPN transistor emitter-follower regulator to run the Arduino off the solar cells using it's own feed.  You can look up a single transistor emitter follower regulator on google.  They only require 1 resistor, 1 zener and 1 transistor.  It will only drop about .8V.  Less than most other regulators.

Pavouk106

I don't get the point in using emitter-follower regulator. Should I supply Arduino through it - connect it between solars and Arduino? Because I run Arduino from solars directly now, they are connected to Vin on Arduino board and everything is going just fine like that. I hope I don't do something wrong...

Thanks for replies and advices, I'm a little more experienced now ;-)

BillO

Sorry, I thought you asked where you connect the Arduino.  I also thought that you felt the regulator on the Arduino dropped the voltage too much.  If you are using Vin and are happy with that, then there is no issue.

Pavouk106

Arduino will be connected directly.

I was asking for some kind of regulator to drop voltage on capacitor, but you suggested I don't need one.

To be sure, once more - Can I connect 6.5V supply to 5.5V (not fully charged) capacitor without damaging it if I disconnect the supply before capacitor reaches its maximum voltage? If I can do this, there is absolutely no need to use any kind of regulator ;-)

dc42


To be sure, once more - Can I connect 6.5V supply to 5.5V (not fully charged) capacitor without damaging it if I disconnect the supply before capacitor reaches its maximum voltage? If I can do this, there is absolutely no need to use any kind of regulator ;-)


Yes if you have sufficient series resistance between the 6.5V supply and the capacitor. "Sufficient" here means much higher than the internal resistance of the capacitor (the ESR measured at DC).
Formal verification of safety-critical software, software development, and electronic design and prototyping. See http://www.eschertech.com. Please do not ask for unpaid help via PM, use the forum.

Pavouk106

dc42: Much higher... Ri (internal resistance) is rated at 30 ohm by the datasheet. Would much higher mean 100, 330, 1k ohm or even more? Forgive me, but I'm learning electronics along the way of my "projects".

dc42

I would go for 10x higher (so about 330 ohms) if that gives you a high enough charge rate (for a 1F capacitor it will be 330 seconds to 63% charge, 660 seconds to 86% charge). The danger with using too low a resistor is that when you turn the mosfet on, you will get a voltage step on the capacitor, which may be large enough to damage or and/or cause the charging circuit to keep switching on and off. With the resistor 10x the internal resistance, when you turn the mosfet on the voltage step will be no more than 1/10 of the difference between the current capacitor voltage and the supply voltage. In fact it won't be a step, more like a sawtooth, because the internal resistance is distributed through the capacitor.

If you need to charge it faster, you could use two mosfets, switching in a low value resistor (e.g. 100 ohms) for initial charging and a higher value resistor (e.g. 470 ohms) when charging with the lower value resistor has been completed but the voltage has dropped a little below target. This may be overkill - it's hard to be precise because it depends on the characteristics of the capacitor.

If the internal resistance of the capacitor is 30 ohms, then it will drop 1.5V when supplying 50mA to your motor. Have you checked that that capacitor is rated for a 50mA discharge current?
Formal verification of safety-critical software, software development, and electronic design and prototyping. See http://www.eschertech.com. Please do not ask for unpaid help via PM, use the forum.

Pavouk106

dc42: The voltage drop you mention looks believable, I have tried some kinds of load and the drop is there (depending on the load).

I haven't found information about current that capacitor could accept/give out. I know it can be charged from 5V source directly (without series resistor) and that it can discharge at somewhere between 50 to 70 mA. Did a few dozens cycles this way and still wokring. Maybe it will die soon by this usage, but still keeps going.

Final application of this capacitor will be probably something like "keep charging/stay charged forever and the circuit will sometimes need ~50% of your charge". There shouldn't be problem with long charging time.

BillO

#28
Jul 10, 2013, 03:57 pm Last Edit: Jul 10, 2013, 06:08 pm by BillO Reason: 1
As I said before, the series resistor will depend on some factors around your desired charge rate and indeed the turn-on voltage.  AS dc42 suggests, there will be a voltage dip once the MOSFET is turned off, but since the turn off voltage is very close to the peak voltage of the solar cells the dip should not be too much and can be compensated for depending on the choice of series capacitor resistor.

For the following calculations, we define:
Vc = Final voltage across capacitor (to be calculated).  
Vo = Voltage across capacitor at cutoff
Vi = Open circuit voltage of solar cells
Io = Current at cutoff (estimated)
Rt = Total circuit resistance
Rc = internal resistance of capacitor
Rs = Added series resistance
Ri = Internal resistance of solar cells

The specs on the solar cells are scant, but if we take the specs we do know (6.7V open, 5.5V @100ma) then we get an internal resistance per cell of about 12 ohms.  4 in parallel will give you Ri = 3 ohms.  Not much, but we can add it into the calculations.

Rc = 30 ohms, from the capacitor datasheet.
Rt = Rc + Ri + Rs
Vi = 6.7V, measured
Vo = 5V, initial arbitrary value
Io = (Vi - Vo) / Rt
And finally Vc = Vo - (Rc * Io)


Case 1:  High charge rate -> Rs = 50 ohms.
Then:
Rt = 83
Io = (6.7 - 5) / 83 = 20ma
Vc = 5 - (30 * 0.02) = 4.4V

4.4V is not enough for the final capacitor voltage. Using a cutoff voltage of 5.4 will yield:
Io = (6.7 - 5.4) / 83 = 15.5ma
Vc = 5.4 - (30 * 0.0155) = 4.93V

So, for a quick charge rate use a 50 ohm series resistor, stop charging at 5.4V and use a 5.6V protection zener


Case 2:  Medium charge rate -> Rs = 150 ohms.
Then:
Rt = 183
Io = (6.7 - 5) / 183 = 9.3ma
Vc = 5 - (30 * 0.0093) = 4.7V

4.7V is not enough for the final capacitor voltage. Using a cutoff voltage of 5.2 will yield:
Io = (6.7 - 5.2) / 183 = 8.2ma
Vc = 5.2 - (30 * 0.0082) = 4.95V

So, for a medium charge rate use a 150 ohm series resistor, stop charging at 5.2V and use a 5.6V protection zener


Case 3:  Low charge rate -> Rs = 330 ohms.
Then:
Rt = 363
Io = (6.7 - 5) / 363 = 4.7ma
Vc = 5 - (30 * 0.0047) = 4.85V

if 4.85V is not enough for the final capacitor voltage. Using a cutoff voltage of 5.1 will yield:
Io = (6.7 - 5.1) / 363 = 4.4ma
Vc = 5.1 - (30 * 0.0044) = 4.97V

So, for a low charge rate use a 330 ohm series resistor, stop charging at 5.1V and use a 5.3V protection zener

A turn on voltage of 4.2V or less would eliminate any cycling in any of the scenarios.

Keep in mind that Rc = 30 ohms is quoted at 1kHz.  The steady state resistance is likely to be a bit different and it might be a good idea to check out what it is.  It the difference is significant ( > 10 ohms), it might be a good idea to readdress these calculations although the differences will be well withing tolerance.

Also keep in mind that due to that internal resistance Vc is the actual voltage appearing across the capacitor (not Vo, the cutoff voltage) and neither ever exceeds the rated voltage (5.5V).  Given that the maximum surge voltage is 6.3 volts, there is a good margin of safety.

BillO

#29
Jul 10, 2013, 04:13 pm Last Edit: Jul 10, 2013, 05:29 pm by BillO Reason: 1
In fact it won't be a step, more like a sawtooth, because the internal resistance is distributed through the capacitor.


I'm not sure about this.  Wouldn't the vast bulk of that resistance be in the conductive material rather than in the electrolyte?

Is the model more like:

----/\/\/\/\---||-----

Or:

----/\/\/\---||---/\/\/\---||---/\/\/\---||-----...

Or:

Code: [Select]

-----------------------------------------...
 |          |         |         |
  \          \         \         \
  /          /         /         /
  \          \         \         \
  /          /         /         /
 |          |         |         |
 =          =         =         =
 |          |         |         |
-----------------------------------------...


Sorry about that last one.

or maybe this:

Code: [Select]

  ------------||----------------------------------
--|                                               |--
  ----/\/\/\---||---/\/\/\---||---/\/\/\---||-----



Update:  Okay see the attached.  I found this model of the double layer super capacitor on Wikipedia.  It's like the 2nd one above with the leakage capacitance across it.  I think a capacitor adhering to this model would not have a relaxation, but rather a step.  What do you think?

Update to update:  It seems there is another model, but I am not sure if it applies to the double layer technology.  That model is of series connected RC elements.  See the Wiki article: http://en.wikipedia.org/wiki/Supercapacitor and read the sections on capacitance and internal resistance.  A model of this type would relax not step once the charge voltage was removed.  It also mentions that the DC inrush resistance can be much higher than the ESR (as quoted in the datasheet) and increases as the capacitor charges.  The ESR of this capacitor is already quite high so I'm unsure how much higher the DC resistance would be.

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