Author Topic: Need help converting a circuit to schematic please  (Read 604 times)

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gotwind2

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Need help converting a circuit to schematic please
« on: May 20, 2007, 05:34:06 PM »
I am making a 1 foot diameter wind powered mobile phone charger, 1 watt max I suspect.


I have one of those hand cranked mobile phone chargers, which has the correct circuit for my needs, however I need to replicate the circuit.


I am having trouble visualizing the circuit - I struggle with electronics...


Can anyone convert the circuit below into a schematic please, so I can mock it up on a test board. Using a multimeter I get no more than 9 volts DC, so I am guessing that diode is a zenner?


R1 is 200 ohms

R2 is 100 ohms


Much appreciated

Ben








« Last Edit: May 20, 2007, 05:34:06 PM by (unknown) »

(unknown)

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Schematic
« Reply #1 on: May 20, 2007, 12:54:09 PM »
Hi there,

try this one

« Last Edit: May 20, 2007, 12:54:09 PM by (unknown) »

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Re: Schematic
« Reply #2 on: May 20, 2007, 12:58:26 PM »
 Sorry I forgot the condensator C1 between D1 (cathode) and ground or neutral on  the schematic but I think you can figure that
« Last Edit: May 20, 2007, 12:58:26 PM by (unknown) »

gotwind2

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Re: Schematic
« Reply #3 on: May 20, 2007, 01:01:11 PM »
Thank you mike1965


very quick response, I assume the extra 'tail' on D2 denotes a Zenner diode?


Cheers

Ben.

« Last Edit: May 20, 2007, 01:01:11 PM by (unknown) »

gotwind2

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Re: Schematic
« Reply #4 on: May 20, 2007, 01:14:32 PM »
That's o.k, I can see that.

condensator = Capacitor in English I assume.


Thanks Mike


Ben.

« Last Edit: May 20, 2007, 01:14:32 PM by (unknown) »

alancorey

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Re: Schematic
« Reply #5 on: May 21, 2007, 09:14:15 AM »
capacitor = condensor, same thing.


BTW the C3067 is an abbreviation of 2SC3067. Datasheet: http://pdf1.alldatasheet.com/datasheet-pdf/view/108588/SANYO/2SC3067.html


   Alan

« Last Edit: May 21, 2007, 09:14:15 AM by (unknown) »

gotwind2

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Re: Schematic
« Reply #6 on: May 21, 2007, 11:52:50 AM »
Thank you alancorey.

Can you or others, explain what this circuit is actually doing please.

I understand the components, but not fully in this application.


Cheers

Ben

« Last Edit: May 21, 2007, 11:52:50 AM by (unknown) »

jimovonz

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Re: Schematic
« Reply #7 on: May 21, 2007, 12:06:43 PM »
D1 is a half wave rectifier and provides a single positive pulse per cycle from the generator (a full wave bridge would give you two pulses). Q1, R1 and D2 form a simple series regulator as described here: http://en.wikipedia.org/wiki/Linear_regulator.  This is not particularly efficient as excess voltage (above the point set by the zenner D2) is dissapated across Q1 (i.e. lost as heat). R1 is there to limit the current through D3 (LED) which is there simply to show you a pretty light.

Much room for improvement.
« Last Edit: May 21, 2007, 12:06:43 PM by (unknown) »

gotwind2

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Re: Schematic
« Reply #8 on: May 21, 2007, 12:33:01 PM »
Thanks jimovonz

You say that D1 is a half wave rectifier.


The motor is DC, so rectification maybe not needed?


Maybe D1 is just a reverse current blocking soltion.


Cheers


Ben

« Last Edit: May 21, 2007, 12:33:01 PM by (unknown) »

jimovonz

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Re: Schematic
« Reply #9 on: May 21, 2007, 12:49:25 PM »
If the generator does supply DC then yes, D1 would just serve to block reverse current. Sorry if you had already made this clear.
« Last Edit: May 21, 2007, 12:49:25 PM by (unknown) »

jimovonz

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Re: Schematic
« Reply #10 on: May 21, 2007, 01:28:04 PM »
"Much room for improvement." - On second thoughts, loosing heat due to excess voltage is exactly what we normally do with a dump load. There might be something to gain if you were to use the likes of an LM2575 switching regulator as this would help minimize the losses in the generator/rectifier but may not be worth the effort for what you want.
« Last Edit: May 21, 2007, 01:28:04 PM by (unknown) »

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Re: Schematic
« Reply #11 on: May 22, 2007, 01:41:59 AM »
Sorry for the Term `Condensator´ but I´m an Romanian working in Germany so the English isn´t much spoken around.

For your Apliccation don´t need such an Voltage stabilizer, in high winds the load will be smaller and the turbine will overspeed. I suggest an simplier Circuit made from 1 Rectifier -D1 in your Schematic-  and a Zenner Diode parallel with the output a value between 5V1 and 6V2 should be OK, depending of your phone charge voltage. The power of your diode should be enough to dissipate the maximum power generated from motor, normally something between standard 1,3W and 3W will do the job, so the turbine will stall easier in high winds

Best regards

Michael
« Last Edit: May 22, 2007, 01:41:59 AM by (unknown) »

Ungrounded Lightning Rod

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Re: Schematic
« Reply #12 on: May 22, 2007, 11:25:48 AM »
The diode is to prevent backward voltage on the circuit - particular the electrolytic capacitor - if the motor is cranked backward.


Back-biasing an electroloytic will break down the electrochemical plating that forms the insulation between the + plate and the - solution.  It will fail shorted, or its voltage rating will be reduced, causing it to break down and short later when attempting to operate normally.  It may then outgass and explode or jet some foul-smelling crud.

« Last Edit: May 22, 2007, 11:25:48 AM by (unknown) »

Ungrounded Lightning Rod

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Re: Schematic
« Reply #13 on: May 22, 2007, 11:28:02 AM »
Reverse voltage on this circuit would also put a high current through the zener and the base-collector junction of the transistor.
« Last Edit: May 22, 2007, 11:28:02 AM by (unknown) »

Ungrounded Lightning Rod

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Re: Schematic
« Reply #14 on: May 22, 2007, 11:43:42 AM »
In particular:


R2 and D2 form a basic reference voltage source.  They pull a small amount of current to provide this reference.


T1 is an emitter-follower amplifier.  It:


 - provides a higher-current copy of the reference voltage (minus the base-emitter drop),

 - dissipates the excess voltage times the output current as heat in the transistor, and


 - only loads the basic power supply with the required output current (rather than the MAXIMUM output current, which is what would happen if you tried to just shunt-regulate with the zener and a lower voltage resistor).  (It also transfers some of the regulator's idling current to the load.)


It's actually a pretty good circuit for the original application.  To get more efficiency you'd need to go to a switching regulator, which would have been overkill, AND have a lot of overvoltage from high RPM to pay for the power consumed by the regulator's internals.  Hand-cranking would be at some narrow range of comfortable speeds, which can be shifted to something appropriate by the fixed gear ratio, so large overvoltage to be salvaged wouldn't have been an issue.

« Last Edit: May 22, 2007, 11:43:42 AM by (unknown) »