nitindua Posted August 27, 2004 Report Share Posted August 27, 2004 all u geniuses tell me how to get this done requiement is of constant 12v and 500 mA current as the output with out the use of 7812 ics.Basically how to maintain the constant current with variable load Quote Link to comment Share on other sites More sharing options...
surajbarkale Posted August 27, 2004 Report Share Posted August 27, 2004 what do you want constant current or constant voltage? http://www.electronics-lab.com/projects/power/008/index.html i think this should help if you need constant voltage. but 7812 will do a better job. Quote Link to comment Share on other sites More sharing options...
ante Posted August 27, 2004 Report Share Posted August 27, 2004 Is the LM 317 allowed? Quote Link to comment Share on other sites More sharing options...
Ldanielrosa Posted September 2, 2004 Report Share Posted September 2, 2004 I've mentioned the fellow before, but here's a link to the circuit you describe.http://www.romanblack.com/a04.htmYou may have to change Rs to get the current you want. Quote Link to comment Share on other sites More sharing options...
ante Posted September 2, 2004 Report Share Posted September 2, 2004 Daniel,This circuit can Quote Link to comment Share on other sites More sharing options...
Kevin Weddle Posted September 6, 2004 Report Share Posted September 6, 2004 The basic requirements for a voltage regulator are feedback and a main current transistor that is controlled with the feedback. The feedback goes to the base of the control transistor. The collector of the control transistor goes to the base of the main current transistor and the input through a resistor. This is as simple as it gets. You have to set the main current right and you have to set the gain of the feedback transistor right. The gain is what can makes it difficult because the input voltage is normally not an ideal value and you have to use resistors to set the voltage at the collector of the feedback transistor. You also have to set the emitter voltage and that dictates the value of the emitter resistor. So it is imperative that you have the correct input voltage or use a high current feedback transistor that can have a variable amount of VCE. It's the base current of the main current transistor that hinders the gain setting. Quote Link to comment Share on other sites More sharing options...
Kevin Weddle Posted September 6, 2004 Report Share Posted September 6, 2004 I constructed this regulator. One of the problems is that the DC after rectifying and filtering is a little low. I should be getting something closer to the peak value. Has anybody experienced this? I know that the DC should be close to the peak value after filtering. Where is the loss? I think I would have this loss if my filtering were not good enough. But the filter capacitor is 2200uF +. Quote Link to comment Share on other sites More sharing options...
MP Posted September 6, 2004 Report Share Posted September 6, 2004 nitindua, what is the source voltage and what is the tolerance of error? Will the circuit be subjected to various temperatures or steady indoor temp?Is this a test for a course?MP Quote Link to comment Share on other sites More sharing options...
Kevin Weddle Posted September 9, 2004 Report Share Posted September 9, 2004 I found the reason for the drop in rectified voltage. It was the diodes. You must select diodes with a high enough impedance to filter. The peak to peak across the diode is high but the diode remains reverse biased because of the DC potential. This way you can have a high peak to peak across the diode which of course limits the peak to peak across the capacitor. Quote Link to comment Share on other sites More sharing options...
audioguru Posted September 9, 2004 Report Share Posted September 9, 2004 Hi Kevin,Of course your rectifiers cause a voltage loss in your power supply circuit. But it has nothing to do with high impedance nor peak-to-peak.The rectifiers conduct a very high current briefly while charging the filter capacitor. The rectifier current is limited only by transformer winding resistance and other wiring resistance. Since the rectifiers conduct only at the peak of the transformer's sine-wave, their current is 10 times or more the output current of the power supply.We tend to think that a silicon diode has a conducting voltage drop of about 0.7V. But that is only with a low current. A 1N400x rectifier is spec'd to have a maximum voltage drop of 1.1V at 1A. Its voltage drop is much more with a peak capacitor-charging-current of 10A.If your rectifier is a full-wave-bridge, then there are 2 rectifier voltage drops in series, causing double the voltage loss, which could total 2.6V!The datasheet for a 1N400x rectifier shows a curve for "typical" voltage drops at different currents. The guaranteed maximum is not shown. The datasheet is here:http://www.fairchildsemi.com/ds/1N/1N4007.pdf Quote Link to comment Share on other sites More sharing options...
Kevin Weddle Posted September 9, 2004 Report Share Posted September 9, 2004 If you have a low peak to peak at the capacitor, don't you still have the peak to peak at the input. This means there is a large peak to peak dropped by the diode. But it doesn't matter because the DC potential at the capacitor keeps it reverse biased. Just for you audioguru, I like to think of DC charging when it comes to sinusoidal waveforms and filters. The DC charging is an important aspect when considering voltage multipliers as well. Quote Link to comment Share on other sites More sharing options...
audioguru Posted September 10, 2004 Report Share Posted September 10, 2004 Hi Kevin,If you have a low peak-to-peak voltage (which is called ripple) at the capacitor, it shows that the capacitor is filtering well, and is large enough in value to supply current to the load without discharging too much before being charged again by the next rectifier conduction. If you have a high load current then you need a high-value capacitor. High-power car amplifiers that draw 100A use a filter capacitor of Farads, not just uF.The rectifiers have a voltage drop only when they conduct at the peak voltage of the sine-wave from the transformer. For most of the cycle when they are reversed-biased, the rectifiers are doing nothing. Quote Link to comment Share on other sites More sharing options...
absolution Posted September 23, 2004 Report Share Posted September 23, 2004 you can stack 7812 or lm317 for more current. there are regulated powers supplies on this site showing this that could produce up to 4A if necesary Quote Link to comment Share on other sites More sharing options...
audioguru Posted September 23, 2004 Report Share Posted September 23, 2004 Hi Absolution,The datasheets for the 78xx and LM317 show that a PNP power transistor can be used as a "booster" for more output current. Quote Link to comment Share on other sites More sharing options...
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