SatthisCy Posted September 18, 2006 Report Share Posted September 18, 2006 Hello to every one!! this is the first time firsttime im writting to forum.i found this website the most usefull for me since im a student in elect. eng.i hope that someone will help me with a problem i had building a LAB P.S.EI used a power supply from an old destroyed printer. I did some measurements and it was working JUST FINE!OUTPUT 1: 24v @ 2.5AOUTPUT 2: 5vSo, i decided to build a subcircuit using an LM317 regulator to adjust the 24v output, and a volt meter using an LCD. I was 100% sure that it will work!I built it, power it up, and .......... The voltmeter was indicating INFINITE VOLTS!I found out Quote Link to comment Share on other sites More sharing options...
MP Posted September 18, 2006 Report Share Posted September 18, 2006 Without looking at your diagram, I can only guess. But it sounds like you just need a split supply. You do not have to use the same common for two circuits. There are many designs where the common for one section of a circuit is floating at a different potential than the common of the other circuit.Looking forward to seeing your diagram.MP Quote Link to comment Share on other sites More sharing options...
SatthisCy Posted September 18, 2006 Author Report Share Posted September 18, 2006 Thanks Very much MP!!Here is the diagram... Quote Link to comment Share on other sites More sharing options...
MP Posted September 20, 2006 Report Share Posted September 20, 2006 With a quick look, C1 should be a much larger value. Also, the voltmeter has it's own setup. It must only use a samll voltage form the larger supply. For example, the voltmeter will have input resistors that scale down the voltage so that it is really only looking at approximately 200 mA. So if you have not set this up correctly, you will get a full scale.Hope it helps.MP Quote Link to comment Share on other sites More sharing options...
audioguru Posted September 20, 2006 Report Share Posted September 20, 2006 The - measuring input of the voltmeter circuit is not the same as its - power supply terminal. The voltmeter circuit probably needs a negative supply which can be made with a Cmos hex inverter package. Then maybe its positive supply voltage should be reduced. Quote Link to comment Share on other sites More sharing options...
SatthisCy Posted September 20, 2006 Author Report Share Posted September 20, 2006 HI.....MP, plase take a good look , the sample for the measuring point is taken on the middle point of a voltage divider of >680kohmsAnd i remind you that this proplem is solved then i supply the voltmeter circuit from a 9volts battery... This is the mystirious fact for me.??> The datasheet of the LCD panel says: " CAUTION! SUPPLY GROUND MUST BE ISOLATED FROM MEASURING GROUND"That's why there is no problem when i supply the cct from a battery; battery is a separate source from my cct!!!Thanks to MP, and audioguru. Quote Link to comment Share on other sites More sharing options...
MP Posted September 21, 2006 Report Share Posted September 21, 2006 Stath,I am not clear why you are using an adjustable regulator, then you have an additional adjustment pot after the regulator. This can cause some loading on the other circuit. Also, this depends a lot on the schematic of the voltmeter. If you must have an isolated ground, you can use a higher voltage regulator and lift the ground up from common with a resistor. I have also seen a post from ante where he coils a piece of wire and puts it on the transformer to make an isolated tap. Then rectifies it as an isolted small voltage supply.MP Quote Link to comment Share on other sites More sharing options...
audioguru Posted September 21, 2006 Report Share Posted September 21, 2006 You say that your meter circuit works fine when it is powered by a 9V battery.Our DMM project has an IC that makes its -5V from +5V. Maybe yours will work the same: Quote Link to comment Share on other sites More sharing options...
SatthisCy Posted September 21, 2006 Author Report Share Posted September 21, 2006 Stath,I am not clear why you are using an adjustable regulator, then you have an additional adjustment pot after the regulator.The second pot (POT2) is not part of the supply cct; it is part of the measuring cct.The resistence of the voltage divider (680kohms +the POT2) is very high so i think that its load is negligible . If you must have an isolated ground, you can use a higher voltage regulator and lift the ground up from common with a resistor. I've tried this with a simpler way ; i lifted the measuring ground from common by using a resistor. Same, bad resutls!I have also seen a post from ante where he coils a piece of wire and puts it on the transformer to make an isolated tap. Then rectifies it as an isolted small voltage supply.This is interesting!A Question...?? Do transformers can use for step-down in DC source as their are used in AC sources?Thanks MPaudioguru,I cant get the idea of using an inverter. what i can acheive with this? Quote Link to comment Share on other sites More sharing options...
audioguru Posted September 21, 2006 Report Share Posted September 21, 2006 audioguru,I cant get the idea of using an inverter. what i can acheive with this?The DMM IC that drives an LCD display and is the sister to the DMM IC that is used in our LED display project works from an isolated supply or a positive and negative supply. The inverter takes the positive 5V and makes a negative 5V. Then the IC can measure its own supply voltages. Quote Link to comment Share on other sites More sharing options...
SatthisCy Posted September 22, 2006 Author Report Share Posted September 22, 2006 Thanks audioguru,ill try this when i find the time, but i have a little problem..My LCD need at least 8volt supply . is there any IC that inverts this range of voltages? Quote Link to comment Share on other sites More sharing options...
audioguru Posted September 22, 2006 Report Share Posted September 22, 2006 My LCD need at least 8volt supply . is there any IC that inverts this range of voltages?A positive 5V supply and the negative 5V supply made by the IC circuit I showed add to 10V. 10V is more than the 8V minimum. Quote Link to comment Share on other sites More sharing options...
SatthisCy Posted September 22, 2006 Author Report Share Posted September 22, 2006 audioguru you a God!!!!Thank you.Ill post again after i try this. Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.