Society of Robots - Robot Forum
Electronics => Electronics => Topic started by: cyborgemu on April 09, 2010, 10:58:13 AM
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I'm trying to integrate an oscillator into a circuit powered by a 4V, 100uA PV cell. I built the right hand circuit here (http://www.discovercircuits.com/DJ-Circuits/OSC4.htm), which claims to draw only 1uA. But it caused the circuit voltage to drop to 0.5V, obviously pulling far more than 1uA. Why might this be? I triple-checked the circuit: no short circuits or crossed leads. Is there a reason why using a PV cell as power source would affect this? Thanks!
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Did you actually measure the current of that circuit to see if the claim is true?
PV specs are at two extremes:
100uA shorted, ~0 Ohm, ~zero V
4V open, infinite Ohm, zero current
They don't do both under the same conditions.
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Hi,
The current drain of 1µA is the average drain of the circuit.
A capacitor over the oscillators supply terminals is needed in your case.
The formula for capacitance is:
C = As/V [Farad]
Where:
A = Ampere
s = seconds
V = voltage drop allowed
If we assume each pulse is eg. 1ms in the 500ms period, the current drain will be:
500ms/1ms*1µA = 500µA.
500µA in 1ms assuming an allowed voltage drop of 0.5V, will take a capacitor of:
0.0005 x 0.001 / 0.5 = 1µF
A tantalum capacitor will be best, since it has lower ESR than an electrolyte. You might need a bit higher value than calculated, as they have a large tolerance on the marked capacity.