stuee Posted November 24, 2004 Report Share Posted November 24, 2004 Ok as you can see below is the diagram of my intentions.Ill explain through it.On the left side is fans, thats no problem. the right side, i want a main switch, when turned on it will send power to the ir switch and to the 12v xformer, i want the 12v to power a flashing led and a buzzer that buzzes quietlyin the background. These 2 are for safety so i know this is on and dont forget its on.The ir side is i want to set some sort of power switch for 3 settings. High, Med Low, some sort of rotory switch?If anyone can help it would be appreciated.This is for a strip of ir's and fans to dry the ink as its comming off a printer at work. Thanks Quote Link to comment Share on other sites More sharing options...
steven Posted November 24, 2004 Report Share Posted November 24, 2004 :) stuee theres no picture or jpg image for anyone to look at. :) Quote Link to comment Share on other sites More sharing options...
stuee Posted November 24, 2004 Author Report Share Posted November 24, 2004 umm odd. i will have to try again when i get home.thanks for letting me know :) Quote Link to comment Share on other sites More sharing options...
Guest Yevgenip Posted November 25, 2004 Report Share Posted November 25, 2004 I see it... Quote Link to comment Share on other sites More sharing options...
Guest Yevgenip Posted November 25, 2004 Report Share Posted November 25, 2004 Well, you can use a potentiometer, but if you want a switch:connect a 3 way switch. to each of the three switch settings, connect a different resistor to control the output current. then, connect all the resistors together and to the IRs.You should calculate the resistance you need depending on the power of the IRs. Quote Link to comment Share on other sites More sharing options...
stuee Posted November 25, 2004 Author Report Share Posted November 25, 2004 Im no good at these calculation things, still learning.Ive just pulled off a ir from the rack and the details are. 400w/230vI bacically want it so its full, med. low. not sure how though,. i can put on a 3 way dial switch.What resistors would i need too? i mean tolerence/watt?Thanks for your help Quote Link to comment Share on other sites More sharing options...
Guest Yevgenip Posted November 25, 2004 Report Share Posted November 25, 2004 hmm...I did some calculations but I don't know if I am right. Is there an Electronics god that can help?U = 230 VI = ?R = ?P = 400 WU = I*RP = U*I230 = I*R400 = 230*I230 = I*RI = 400/230 = 1.739 amp230 = 1.739*RR = 132.25 ohmI got these results, but I am not sure what they mean. Is R the IRs inner resistance? Ante, audioguru, MP, steven, someone - Please help. :-\ Quote Link to comment Share on other sites More sharing options...
steven Posted November 25, 2004 Report Share Posted November 25, 2004 :)r i think is the resistance you need to limit the current to the corect level for that infrared led , like a normal 5mm red led needs 390 ohms resister to run it off 9 volts, leave me a meesage if you need some help. although im not much good at technicle exsplanations ill help whenever i can or ill email you some handy information to assist you Quote Link to comment Share on other sites More sharing options...
stuee Posted November 25, 2004 Author Report Share Posted November 25, 2004 the Ir lamps are 7 tube lamps. They are not tanning ones though. just heat. Quote Link to comment Share on other sites More sharing options...
steven Posted November 25, 2004 Report Share Posted November 25, 2004 :)oh then disregard what i said Quote Link to comment Share on other sites More sharing options...
audioguru Posted November 25, 2004 Report Share Posted November 25, 2004 Hi Stuee,Don't use resistors to reduce the power to the heat lamps. The resistors would heat as much as the lamps! Hundreds of Watts.Just use a lamp dimmer. Make certain that it is rated for the load. Restaurants use high-power rated dimmers.What is the heat-lamp load? Your pic shows 8 X 80W but then you said 400W. Now you say 7 tubes, are they 400W each?The LED and buzzer can be connected to a wall-wart DC output power supply. Use a current-limiting resistor for the LED and maybe a wire-wound pot for the buzzer volume preset control. Quote Link to comment Share on other sites More sharing options...
stuee Posted November 25, 2004 Author Report Share Posted November 25, 2004 The diagram details are almost correct. its 8X 240v 400w units. the units each have 7 tubes in. the coil starts in one end, out the other then into the other one, so the ir only has 2 wires at the back. I only know this because one of the tubes broke on one so i had a look.what about this onehttp://www.rsaustralia.com/cgi-bin/bv/browse/Module.jsp?BV_SessionID=@@@@0196216012.1101423322@@@@&BV_EngineID=ccchaddddlllhkgcfngcfkmdgkldfih.0&cacheID=auie&3277248954=3277248954&catoid=-85980856 or this http://www.rsaustralia.com/cgi-bin/bv/browse/Module.jsp?BV_SessionID=@@@@0196216012.1101423322@@@@&BV_EngineID=ccchaddddlllhkgcfngcfkmdgkldfih.0&cacheID=auie&3278876357=3278876357&stockNo=4235660&prmstocknum=4235660&prodoid=260496Thanks Quote Link to comment Share on other sites More sharing options...
audioguru Posted November 26, 2004 Report Share Posted November 26, 2004 Hi Stuee,Your heat-lamp load is 3200W, correct?Sorry, I can't get your long links to work. Quote Link to comment Share on other sites More sharing options...
stuee Posted November 29, 2004 Author Report Share Posted November 29, 2004 Ok im back at work now, that link didnt work because my session expired, its pretty crap when i didnt even log in.Anyway, i was a way from work last week so going of what i remember, the proper details are 9X 230v - 400w lamps.I wired each lamp seperatly so i can put them all into a junction box or use all them seperately. pref want to use the whole altogether though.If all im doing is adding the watts toget to ge the total load well the total would be 3600w.thank for the help so far Quote Link to comment Share on other sites More sharing options...
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