Urgent! analysis of simple circuit diagram

Discussion in 'The Projects Forum' started by studystudystudy, Nov 17, 2010.

1. studystudystudy Thread Starter Member

May 21, 2010
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i attached a file. Need Pro in power electronic help me explain and analysis the circuit diagram.

this is my first design energy harvesting circuity. but teacher told me have some error, i spend a lot of time in searching the mistake i made. please help me to figure it out!

The voltage source is unstable can be ranging from 1 to 1000V
Input current is very small - about 25uA

1. C1 used to filter the DC voltage, but the value use is inappropriate. why? 1000uF with 10V

2. Zener diode connected series with diferrent polarity. used to regulated the voltage so that protect MemoryCap 5V? the Zener diode place in wrong way! why?

3. UF4007 is a ultra fast recovery rectifier. is it already rectifier? mean no need four UF4007 to make a rectifier?

These few question make me headache!!! your help deeply appreciated!! thank you thank you

Last edited: Nov 17, 2010
2. t06afre AAC Fanatic!

May 11, 2009
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Are you going to build this circuit, or is just some simulation project only? This setup look some kind of dodgy. What is it meant to do?
If you are simulating the negative end of C1 should be tied to the rectifier as you have done, but also to ground.
C1 must have a voltage rating so it can handle the input voltage. If the input is 500 volt AC. Is this peak or RMS. A 10 volt type cap will not do any good here in any case.
Your zener diode setup is more a clipper circuit. http://en.wikipedia.org/wiki/Clipper_(electronics) Using a 5.2 volt zener with a resistor in series is a more correct setup. The series resistor must be able to dissipate (VC1-Vzener)*I watt
UF4007 is a ultra fast recovery rectifier diode. You will still need 4 of them to make a bridge rectifier.

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3. wayneh Expert

Sep 9, 2010
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I think C1 is redundant, if you were to connect ground to the negative side of your rectifier. You only need one cap to store energy.
Well, ZD2 provides the protection by breaking down and passing current to ground whenever the voltage exceeds the threshold. ZD1 does nothing except drop voltage about 0.7v additional when current flows. effectively raising the threshold by that amount. Any diode would work in that regard.
No, it may be called a rectifier, but it's just a diode, and you need all 4 of them.

Last edited: Nov 18, 2010
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4. kubeek AAC Fanatic!

Sep 20, 2005
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Have you considered how long does it take to charge the cap, and its leakage current? With no leakage, 25uA will charge the 1F cap to 5V in whooping 55 hours.

Sorry, this is just not gonna work. That capacitor has leakage current of 315uA, and you neet to have available more than that just to get it charging. Datasheet vishay.

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5. studystudystudy Thread Starter Member

May 21, 2010
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to t06afre

i going to build this circuit

thanks.

i guess is RMS, i dont understand what is peak or RMS. then which voltage rating of cap will good in here? which should i used? i have no ideal.

can u provide example how to calculate?

6. studystudystudy Thread Starter Member

May 21, 2010
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the capacitor leakage current with 315uA is my Memory cap or C1?

7. Audioguru New Member

Dec 20, 2007
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The circuit takes the extremely low power from a piezo transducer in a shoe. It produces its tiny output for each step you take. It might charge the capacitor in a few years if you continuously run fast. Then you can use the charge in the capacitor to power something for a few seconds or a few minutes.
But the leakage current of the capacitor is higher than the output of the piezo transducer so nothing will happen.

There is a similar circuit on the web that powers a tiny low power watch. Its capacitor is much smaller so its leakage current is much less.

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8. studystudystudy Thread Starter Member

May 21, 2010
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to audioguru, thanks for your post

my Memory Capacitor have leakage current? or my C1

9. Audioguru New Member

Dec 20, 2007
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Both capacitors have leakage current.
The zener diodes also have leakage current.

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10. kubeek AAC Fanatic!

Sep 20, 2005
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As others said, C1 is pointless in that circuit, so it is leakeage of the memory cap. I even posted a datasheet for it, didn´t you notice?

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11. studystudystudy Thread Starter Member

May 21, 2010
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thanks kubeek

i need C1 because i need another same set of circuit connected parallel to the memory cap. mean i have another set of rectifier go through new capacitor like C1 and both set is connected with zener diode ZD1 and ZD2 and finally memory cap.

my memory cap have leakage current? i dont understand here, my memory capacitor is like supercapacitor, can stored the current without losing it.

thanks again for your post, really appreciated it

12. kubeek AAC Fanatic!

Sep 20, 2005
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As others said, in the original schematic you are missing a ground on the bottom side of C1, without that you will be only charging C1 and not the memory cap.
When you add this ground, you will notice that C1 becomes parallel with the memory cap, which makes it useless as it adds only a little to the 1F memory capacitor.

No capacitor is perfect, when you charge it to some voltage and just leave it unconnected it will slowly discharge. The rate of discharge is called leakage current, it can also be modelled as a parallel resistor with the capacitor.

Here is an updated schematic, this is what the circuit actually should look like. You can connect many more sources to the A-B points to overcome the leakage.

• supercap.png
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13. wayneh Expert

Sep 9, 2010
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I suppose this must be a silly question, but does D6 have any purpose other than dropping voltage?

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14. kubeek AAC Fanatic!

Sep 20, 2005
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No it doesn´t.

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15. Audioguru New Member

Dec 20, 2007
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D6 does not drop the voltage.
D6 is drawn as a zener diode but it is connected backwards to how a zener diode works so any low current silicon diode will work instead. Its function is to increase the effective clamping voltage of zener diode D5 by 0.7V. The two diodes clamp the voltage at about 5V to 5.2V to prevent the super-capacitor from blowing up.

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16. wayneh Expert

Sep 9, 2010
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Right, which it can only do by producing a DROP across itself.

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17. Audioguru New Member

Dec 20, 2007
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The circuit is silly to use a 4.7V zener diode in series with a 0.7V ordinary diode.
Why not use a single standard 5.1V zener diode instead?

What is the purpose of trying to charge a capacitor in weeks then having it discharge for only a few seconds?
What can be powered by the capacitor?

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18. wayneh Expert

Sep 9, 2010
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Maybe the rice vending machine?

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19. Audioguru New Member

Dec 20, 2007
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Why does anybody in the Orient need a rice vending machine?
Don't they have rice all over the place?

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20. studystudystudy Thread Starter Member

May 21, 2010
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i modify, i remove the C1 from original schematic
the new schematic can work? show in image thumbnail above

your help deeply appreciated. need comment and suggestion from both of you. thank you very much