bogdan Posted January 17, 2004 Report Share Posted January 17, 2004 i need to build a constant current load, to demonstrate that if a constant current passes trough a capacitor, then the voltage across it increases or decreases liniary. the problem is that i don't know how to build this. the caps will be charged to max 16V, and if i can get a constat load over an interval of about 5V then it is very good. but how can i build such a circuit? i know that if it connect a LED trough a BF256 transistor, and connect the gate with the source, then the current maintained trough the led is between 10 and 15mA for voltage from 5 to 30V. but how can i make a circuit that draws 1mA?? any help is much apreciated. Quote Link to comment Share on other sites More sharing options...
MP Posted January 17, 2004 Report Share Posted January 17, 2004 I hope I am not misunderstanding you, but ohm's law is the rule no matter what the circuit is. V/R=I. You must take into consideration the resistance of the total circuit. If the resistance of your circuit changes or the voltage changes, then the current will change according to the formula. In regards to capacitors, the energy (in joules) stored in a capacitor is given by the formula E= 1/2(C*(Vsquared)).Here is a link that goes into more detail about capacitance. Perhaps this will help you complete your project. http://en2.wikipedia.org/wiki/CapacitancePlease let us know about the results. :)MP Quote Link to comment Share on other sites More sharing options...
bogdan Posted January 18, 2004 Author Report Share Posted January 18, 2004 not quite like that. you see, considering the capacitor charged, it has a stored amount of electricity Q=C*U. and an energy of E=C^2*U/2 . now, Q=I*t, where t is time and I is the current trough the cap.if you discharge the cap trough a resisotor than I=U/Rbut U changes because the cap discharges. U=Q/C, but Q2=Q-I*tso U=(Q-I*t). if I is constant, than the voltage across the cap is decreasing in a liniar way.but if you discharge the cap trough a simple resistor, than I drops with the dropping of U. this way thec voltage on the cap decreases slower and slower. now....what i need is a circuit that can draw a constant current, even is the voltage applied on it is variable(of course, in limits). Quote Link to comment Share on other sites More sharing options...
MP Posted January 18, 2004 Report Share Posted January 18, 2004 Then you are probably looking for application using a Transconductance Operational Amplifier (OTA) such as the CA3080A or similar.This device can be used like a variable resistor by the control pin. It is used widely in music applications for effects filtering circuits and VCA applications.MP Quote Link to comment Share on other sites More sharing options...
bogdan Posted January 23, 2004 Author Report Share Posted January 23, 2004 well, i know i can do it that way, but what i want is a circuit that will not need external power. just from the cap. Quote Link to comment Share on other sites More sharing options...
Staigen Posted January 23, 2004 Report Share Posted January 23, 2004 A constant current diode maybe, but i dont think they can go all the way down to 0 Volt.//Staigen Quote Link to comment Share on other sites More sharing options...
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