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Middle of the Pacific
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Not sure where to start this thread so I figured Bar Sport would be safe.

I am looking into creating a educational display for teaching kids about capacitance.  I was think about creating a display to measure the capacitance of the human body.  

I found something similar here:

http://web.mit.edu/Edgerton/www/Capacitance.html

from the explanation they first charge the person with 600V then discharge it into a .06 mF capacitor.  Then the voltage across this capacitor in milivolts is the body's capacitance in pico farads.

How accurate is the Arduino at measuring voltage?  Will it be able to pick up a few mili-volts of difference.  It will not be interesting if it reads the same value every time.

Any safety issues with shocking kids?

Any thoughts would be greatly appreciated.
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Denver
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Any safety issues with shocking kids?

Having just made a  Geiger counter that cranks out about 500V at very low amperage, I can offer a few things to consider . . .

It's impressive to see how well a voltage like that can travel through the body. Holding a wire in each hand or even 2 people each holding a wire and touching fingers - would give me a nice "click" on my pickup circuit.

No shock was felt, but this made me wonder about any affect the HV might have on a pacemaker. I don't know but, I'd check that out.

Also note that due to the very low current (I hope!) the 600V will be impossible to measure with a standard DMM. You will get a lower reading due to the loading. However, it looks like the op-amp takes care of that problem for you. So adjust your HV via the op-amp.

Interesting project!
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Middle of the Pacific
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I am having problems understanding the explanation for the HV circuit:

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The display uses an adjustable, regulated low voltage power supply (ranging from 1 to 12 volts at up to 1.5 amperes), which powers an emitter follower connected to the primary of a transformer made by eight turns on a 5 mH ferrite core inductor. The secondary of the transformer is tuned to about 220 kHz by a capacitative divider which provides positive feedback to the transistor base. A half wave voltage doubler then yields DC outputs between 100 volts and 1200 volts at less than one milliampere

Can someone help me understand what is happening here?
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They are creating a switching DC/DC converter that boosts an input voltage
from 1-12V up to 100-1200V.

Another way to demonstrate capacitance would be a Theremin. There
are no high voltages involved and the output is a musical tone.
One of my favorite performances --


(* jcl *)

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Middle of the Pacific
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They are creating a switching DC/DC converter that boosts an input voltage from 1-12V up to 100-1200V.

I was able to gather that much from reading it but I do not get this:

Quote
The secondary of the transformer is tuned to about 220 kHz by a capacitative divider which provides positive feedback to the transistor base

Can anyone clarify?
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SW Scotland
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Without seeing the circuit it's impossible to be specific but here goes.

By installing a capacitor across the secondary winding, which is inductive, there is an LC combination which has a natural resonant frequency dependent upon the values of C and L.  In this case that is 220kHz (which seems rather high - it being in the radio frequency band)

The oscillator drive component (transistor) requires feedback in order to maintain a stable oscillation.  Unlike an amplifier, which uses negative feedback to maintain stability free from oscillation, an oscillator requires positive feedback to initiate oscillation.

It is normal to take the output signal and feed it back, in the correct phase, to the active device.  However, because the output is several hundred volts, it would be much too high for the transistor base drive.   An attenuator is required to reduce this signal down to a few hundred millivolts.    Normally, one might use a resistor divider but this is inefficient in an oscillator circuit so in this case they have made a voltage divider by arranging the C capacitor from a combination of 2 capacitors in series and tapped the feedback from the smaller one.  

Capacitors in series are calculated using a formula similar to that used for resistors in parallel  viz  C= (c1xc2)/(c1+c2)

Trust this helps
jack
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