Charge Pump

Audioguru

Joined Dec 20, 2007
11,248
Some of the other kids in your class are posting the same assignment on other websites. Maybe they will post here too.

A zener diode voltage regulator wastes a lot of power. But you have very low power and cannot afford to waste any.
 

Bernard

Joined Aug 7, 2008
5,784
Expect wild variations in output V. Might use a series string of red LED's to give 2 V steps. What is required accuracy of 2 V steps? LED's measured 1.93 to 2V @ 20 mA. Are caps to be charged sequentially or all at once?
I vote for current driver & step-up transformer.
 

hgmjr

Joined Jan 28, 2005
9,027
What are the rules of the competition exactly? You mentioned the voltages but what are the values of the capacitors to be charged? And, can you skip the intermediate voltages and go straight to 20V?

hgmjr
 
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Thread Starter

falconite

Joined Nov 6, 2010
23
@Bernard....

I did a bit of study on charge pumps.
Using charge pumps seems to be a very bad idea now.:(
Thanks for the info about the LEDs though.:)

@hgmjr....

The values of capacitors has not been mentioned.
Only the supply voltage of 2V and the clock amplitude of 2 V(peak-to-peak) is given.
Even the frequency is not given.

The intermediate voltages have to be generated as well.
But since generating 20V seems to be the trickiest part,I thought I would concentrate on it first.
If nothing else, the intermediate voltages may be generated using resistors.
 

Thread Starter

falconite

Joined Nov 6, 2010
23
Okk got another stupid question.
Most probably it wont work but still would like to ask the experienced people.

Suppose I have a voltage source Vs providing very low o/p current.
Can i use a buffer op-amp to increase my o/p current,considering I bias my op-amp with the same voltage Vs???

Of course both the i/p voltage and the bias voltage of the op-amp in this case will be the same voltage Vs.
 

Audioguru

Joined Dec 20, 2007
11,248
An opamp needs to have a power supply. The output current of the opamp comes from its power supply. If you use charge-pump ICs to increase your 2VDC input to 8V then the current is too low for an opamp to do anything.

You might use the 8VDC to power a low current oscillator that drives a voltage multiplier made with diodes and capacitors.
 

ifixit

Joined Nov 20, 2008
652
Hi falconite,

Are you allowed to use an inductor? If so then a simple SMPS would work.

I have attached an example that uses back emf to get 20V from 2V. The Energy that goes into the inductor is equal to what you will get out. The magnetic field is built up when the transistor is on using the 2V source. When the transistor turns off the collapsing magnetic field reverses the voltage on the inductor. The voltage will go as high as needed to get rid of the energy that is stored.

In the example the 10K resistor limits that voltage to 20V or so. This energy conversion repeats once every millisecond so smoothing to a DC voltage is easy with a capacitor.

Read up on this subject and design your own SMPS to suit your needs.

Regards,
Ifixit
 

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Thread Starter

falconite

Joined Nov 6, 2010
23
Hello ifixit...

I did think about using inductors once.
But i wanted to avoid any designs with inductors.
Seems like the only option now.
thnx for the suggestion.:)
 

Thread Starter

falconite

Joined Nov 6, 2010
23
Aren't inductors supposed to be bulky and all????
Plus can the winding resistance associated with an inductor be determined and controlled accurately?????
 

Audioguru

Joined Dec 20, 2007
11,248
My solar garden lights use an oscillator made with two transistors and an inductor that looks like a 1/2W resistor. Its battery voltage is only 1.2V and the circuit increases the voltage to 3.5V or more so it can light a white or color-changing LED.
 

Thread Starter

falconite

Joined Nov 6, 2010
23
Sounds encouraging.:)
Well i was reading up a bit on SMPS and it looks like the o/p voltage is controlled by changing the duty cycle of the clock signal.
There should be some other way as well to change the o/p DC voltage, like by changing the value of the capacitor, isnt there????
 

shortbus

Joined Sep 30, 2009
10,045
Why not get a disposable flash camera, and remove secondary windings from the transformer until you get to 20 V ? The used cameras are free from most drug stores, so you could get several and experiment till you get correct voltage. They work with a AA cell so the 2 V is no problem.

After getting the 20 V you can drop it to the other voltages you need.
 

ifixit

Joined Nov 20, 2008
652
Sounds encouraging.:)
Well i was reading up a bit on SMPS and it looks like the o/p voltage is controlled by changing the duty cycle of the clock signal.
There should be some other way as well to change the o/p DC voltage, like by changing the value of the capacitor, isnt there????
Hi falconite,

In the example I gave the output is controlled mostly by the load resistor (10K). Lowering the resistance will lower the output voltage.

Since you require very little output power i.e. <100mW, then you don't need any bulky, expensive, difficult to make, inductors. A small one from Digikey would do (811-1295-ND).

http://parts.digikey.com/1/parts/1591944-inductor-radial-3-3mh-0-1a-22r335c.html

This one is; small, 3.3mH, 0.1Amp max, Rser=10Ω. The "charge" current will be limited to approximately (2V-Vce)/10Ω = 0.17Amp max. (289mW)

You can also limit the output by replacing the load resistor with a zener of 20 Volts because the output power is very limited. Then use a resistive divider to get the 2V steps.

Learning is fun,
Ifixit
 

Thread Starter

falconite

Joined Nov 6, 2010
23
Hey ifixit,

Sorry to bother you again.
But i cant find any material which explains how V_out depends on R. :(

I checked out on some simulations on net where the V_out varies with R, but no explanation is provided.
Could someone refer some material pleaseeeee??

I dont need the actual component values.
But i do need to know how this stuff works.
 

ifixit

Joined Nov 20, 2008
652
Ohm's law is the formula you are looking for. My power supply example has an unregulated output, so it will vary with load just like any unregulated power supply would.

I believe your requirements call for a fixed resistive divider load so regulation is not required... is it?

Regards,
Ifixit
 

Thread Starter

falconite

Joined Nov 6, 2010
23
Ya it does.
The voltages tapped out will be connected to capacitors.
Hope that doesnt change the output voltage considerably.
 
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