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Electronics => Electronics => Topic started by: Melcin Powell on August 05, 2010, 06:51:32 PM
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I've been learning about different parts used in circuits ,and i don't fully understand how the capacitor releases its energy (discharges). What i am getting at, is the only time it discharges is when the power source is removed and replaced by a some sort of connection creating a complete circuit please try to explain how it works.
Thanks - Melvin Powell*
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also i understand it will discharge only if there is another electric component that needs the stored electricity, but how do i use it as a timer? Lets say there is a 6V buzzer i want to go off in 10 seconds im using 9v battery to supply electricity and the capacitor is a 15-uF capacitor. How do i do it?
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Good question that has a long answer.
Google the following terms to find some explainations:
RC time constant
RC circuit
voltage comparator
555 timer IC
Yes, there needs to be a completed circuit to discharge the cap.
For your 9V and 15uF cap you need some voltage level detector. This could be the biased transistor base or a comparator IC. Lets say you use an NPN transistor with a grounded emitter. When the base is above ~0.6 volts current flows from base to emitter and the collector to emitter path conducts.
The equation:
V(t)=V(0) e^(-t/RC) [Wiki RC time constant]
gives the voltage at time t. since you want t = 10sec, and the C = 15uF with a starting voltage V(0) = 9V we solve for R. So:
1/R = -(C/t) * ln (V(t) /V(0)) or
1/R = -(15e-6/10) *ln (0.6/9) = 1/246179 Ohm or R ~250 kOhm
Grab some caps and resistors, plug them into a protoboard, grab your voltmeter and give it a try.
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I've been learning about different parts used in circuits ,and i don't fully understand how the capacitor releases its energy (discharges). What i am getting at, is the only time it discharges is when the power source is removed and replaced by a some sort of connection creating a complete circuit please try to explain how it works.
Thanks - Melvin Powell*
Apologies if the following is TOO basic.
Think of electronics => plumbing of water pipes.
ie bigger resistance = thinner pipes - so you can get less water (current) through it than with thicker pipes (less resistance).
A capacitor is a bucket. You fill it with water from some water (current) source until it is full. But if it is connected to something else then water is also draining out. How quickly this happens depends on how thick the pipes are in and out.
Also: all capacitors (buckets) have a leak in them. So the water will drain away eventually all by itself. ie turn off the power switch and the caps will discharge by themselves over time.
As per 'walt' post then one resistor constrains the pipe (ie how quickly water flows into the bucket) and the size of the capacitor is the size of the bucket. So the two in combination dictate how quickly the bucket becomes 'full'.
The same is true for draining the bucket via another circuit.
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Hi,
For your 9V and 15uF cap you need some voltage level detector. This could be the biased transistor base or a comparator IC. Lets say you use an NPN transistor with a grounded emitter. When the base is above ~0.6 volts current flows from base to emitter and the collector to emitter path conducts.
The equation:
V(t)=V(0) e^(-t/RC) [Wiki RC time constant]
gives the voltage at time t. since you want t = 10sec, and the C = 15uF with a starting voltage V(0) = 9V
That will have the transistor open in the almost 10s and then it will close!
There needs to be a resistor from the base to the 9V as well, so this will change the timing, acting as a voltage divider.
Melcin <- I know it's just an example, but the way to go about it would be to first spec what time you need, what voltage you have available etc. and then select a capacitor that's reasonable.
If you need to drive a 6V buzzer from 9V, you need to drop the extra 3V somehow.
A 9V (PP3) haven't got much oomph, so if the buzzer is mechanical, you will get a very short battery life.