only_lonely Posted August 20, 2004 Report Posted August 20, 2004 111.GIFi want to replace the 4.5v supply with 12v...am i need to change the value of the resistor and capacitor????how to calculate the value??? Quote
Kevin Weddle Posted August 20, 2004 Report Posted August 20, 2004 This isn't at all biased right. What transistor are you using? There isn't any current because you don't have a potential difference between the base and emitter. Keep the emitter positive and use a resistor to ground on the base. I don't see a signal input so I'll assume that this circuit is about constant current adding or subtracting. R1 is just for bias and C3 creates the impedance as seen by the signal through the diode. Quote
audioguru Posted August 20, 2004 Report Posted August 20, 2004 Hi Ray Orbison wannabe or fan,If you want the rise-time of the voltage across C3 to be the same, then you simply multiply the value of R1 by the ratio of supply voltages, which is 12 divided by 4.5, so it will be about 12.5K to 12.7K.If the circuit is activated by the junction of R3 and R5 going close to ground, and you want the same low LED current, then multiply the value of R5 by the different voltage ratio that will be across it (subtracting the voltage across the LED). Quote
ante Posted August 21, 2004 Report Posted August 21, 2004 Audioguru,His name was ROY 8) and the song was Quote
audioguru Posted August 21, 2004 Report Posted August 21, 2004 Yeah, I agree Ante,Roy was a cool dude. May he rest in peace.I think he stole that song from Frank Sinatra. Quote
MP Posted August 23, 2004 Report Posted August 23, 2004 A good rule of thumb....wait until they post the rest of the schematic when they ask such questions. See his other post (with the complete schematic). He is about to fry a PIC. :oMP Quote
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