Shahriar Posted May 22, 2005 Report Share Posted May 22, 2005 HiI have made a power supply, the schematic is below.as you can see, i have put a surge resistor (R1 = 3.3ohm, 2watt). as You know this resistor controls the current at circuit start up. because all the caps act like short circuit at circuitstart up, it protects diode bridge from high current.but he problem with my circuit is that, the output votlage decreases when connected to load.you may say because the current passing R1, increases the voltage of "C" point and causes voltagedecreases in "A" point. but my theory is, the output voltage of LM317 proportion to its ground is alwaysconstant, so if the voltage of "C" point increases beacuse of load current, so LM317 should increase its outputto keep the voltage constant, BUT WHY THIS WOULD NEVER HAPPEN.I hope u understand what I meant.here another short circuit protected application from LM317 Datasheet. so output voltage in this application should decreaselike my circuit.ThanX in advanceShahriar Quote Link to comment Share on other sites More sharing options...
EdwardM Posted May 22, 2005 Report Share Posted May 22, 2005 Hi Shahriarthe first thing I can see is that R1 isn't really doing anything apart from being a current dump for the bridge rectifier. If you want to limit the inrush current to the reservoir capacitor then a resistor should be placed in series with the capacitor.It may be that most of the available current is being lost in R1, remove R1, and check the output voltage again and post itBest of LuckEd Quote Link to comment Share on other sites More sharing options...
Guest Alun Posted May 22, 2005 Report Share Posted May 22, 2005 I agree, and I'd remove the 200mH inductors on the output as well, I bet you that the problem is that they have a high internal resistance and they're not needed anyway as the LM317 already does a good enough job of removing any ripple. The only time I would use indutors is on a voltage regulator is if I'm powering a RF circuit or with a switching regulator and even then I'd use a small value like 4.7uH to 47uH maximum. Quote Link to comment Share on other sites More sharing options...
audioguru Posted May 23, 2005 Report Share Posted May 23, 2005 Hi Shahriar,1) You marked the wrong resistor in your schematic. the 3.3 ohm "anti-surge" resistor is on the left and is in series with the rectifier's negative terminal and the circuit.2) You don't need an anti-surge resistor. Rectifiers are designed to charge a filter cap with a very high current, and the resistance of the transformer will limit the current anyway. Read the very high max current spec's on the datasheet for your rectifier diodes.If you use a small-signal diode instead of a rectifier diode then an anti-surge resistor would probably be necessary.3) The output voltage drops when connected to a load because of the resistance of the 200mH coil that isn't needed and might continue dropping as the coil heats-up and its resistance increases due to the heat.To confirm it, simply measure the output voltage of the regulator IC then measure the voltage after the coil. ;DWhy are you using a coil? Quote Link to comment Share on other sites More sharing options...
Shahriar Posted May 23, 2005 Author Report Share Posted May 23, 2005 Thank you very much for all of your replies.As audioguru said i mark r1 by mistake. the surge resistor is r1 on the left side. when this happened tomy circuit, at first I short circuited the R1 and noticed that the voltage raised to the correct voltage (5.1V). I didn't get suspecious to the coil because I was sure that there is no coil in the world with so much resistance.So what I did? I raised up input voltage from 8V to 10V, and what happened!!!The voltage raised to the right volatge (5.1V) although the surge resistor were in the circuit.... ;DThis is related to the LM317 spec. if you take a look at it you will understand. the zener diode (6V at the bottom of the circuit) does not break correctly!!!!!ThanX a millionShahriar Quote Link to comment Share on other sites More sharing options...
audioguru Posted May 24, 2005 Report Share Posted May 24, 2005 The 6V zener diode in the LM317 isn't used for voltage regulation because with its ref terminal grounded for 1.25V output, the voltage is guaranteed with at least only 3V from input to output.Most of the LM317 spec's are made with 5V from input to output.Its typical dropout voltage is about 2V higher input voltage than its output voltage. When it has dropped-out, it is regulating poorly since its output voltage has dropped 0.1V. ;D Quote Link to comment Share on other sites More sharing options...
muckleluck Posted May 28, 2005 Report Share Posted May 28, 2005 Hi,I don't know if someone already covered my problem in a previous reply but I too am having a problem with a power supply that I built based on the LM317. The schematic can be found at:http://www.uoguelph.ca/~antoon/circ/vps.htmMy circuit is identical to this one except I haven't installed the voltmeter or the led branches. When I connect it to a 7408 TTL chip the voltage drops from 5V to 2.3V. In the construction section the author suggests adding two diodes to drop the min output voltage to zero. I found that the voltage went to 0 without them so maybe that is a problem? Quote Link to comment Share on other sites More sharing options...
Guest Alun Posted May 29, 2005 Report Share Posted May 29, 2005 What current are you atempting to draw?What's the volatage and current rating of the transformer?If you want to power a TTL chip then use a LM7805 to provide a fixed 5V output. Quote Link to comment Share on other sites More sharing options...
audioguru Posted May 29, 2005 Report Share Posted May 29, 2005 My LM317 project has a minimum voltage of 1.2V like it is supposed to, and powers things with a very well-regulated voltage just fine without the voltage dropping. I usually use a 12VDC/1A wall-wart transformer to power it then its max regulated voltage is 9V.Without a load I set its output voltage to 5.00V. With a 1A load, its output voltage drops in the wires to the load but at the LM317 it measures 4.98V.Instead of the 240 ohm resistor, I used 120 ohms so that it has enough idle current for all LM317's at any output voltage. The 240 ohm resistor is used with the more expensive and less idle current LM117, as shown on the 1st page of the datasheet.Using 120 ohms instead of 240 ohms the value of the pot must also be halved. With a 30V output from the project, the pot must be rated at 300mW or more. With a voltage between its input and its output greater than 15V an LM317 will reduce its output current to protect itself. With 15V or less between its input and output, an LM317T is guaranteed to produce at least 1.5A. With 40V between its input and output it is guaranteed to produce only 150mA. The guaranteed output current spec's apply when its junction temperature is 25 degrees C. When it heats-up, it is less. ;D Quote Link to comment Share on other sites More sharing options...
Guest Alun Posted May 29, 2005 Report Share Posted May 29, 2005 I always thought the output current is inversly proportional to the junction temperature. A LM7805 3V difference without a heatsink might only be able to output 300mA continious but might be able to handle 1.5A surges if the junction is cool enough, isn't this the same for a LM317 with a 40V difference with a good heatsink? 1.5A surges might be OK but it will 150mA to if the load is left connected. Quote Link to comment Share on other sites More sharing options...
muckleluck Posted May 29, 2005 Report Share Posted May 29, 2005 If someone has a similar supply taht works I can only assume that I have assembled something incorrectly. I'll take a careful look at it and if all else fails start again. Thanks.Muckleluck Quote Link to comment Share on other sites More sharing options...
muckleluck Posted May 29, 2005 Report Share Posted May 29, 2005 I bought the transformer specifically for the project. It is rated 24 VA so that is more than enough for this circuit I think. The only current that I am trying to draw is whatever the 7408 requires with just the power connected. Quote Link to comment Share on other sites More sharing options...
Guest Alun Posted May 29, 2005 Report Share Posted May 29, 2005 That's a fairly low current.What's the secondary voltage?Are you planning to use this to power anything else? Does it really need to be variable? I recommend you use a 9V 500mA transformer with a fixed LM7805 regulator and a heatsink won't be needed. Quote Link to comment Share on other sites More sharing options...
audioguru Posted May 29, 2005 Report Share Posted May 29, 2005 Hi Alun,The LM317 and 78xx regulators reduce their output current with more than 15V across them to protect their pass transistor and keep it in its Safe Operating Area. Die temp affects only the typical 2.4A max current with less than 15V across it a little, down to about 1.6A typical. Quote Link to comment Share on other sites More sharing options...
muckleluck Posted May 29, 2005 Report Share Posted May 29, 2005 Ok I got it to work now on my breadboard. When I eventually put it together permanently I am going to connect a 7805 so that I have a fixed supply and a variable all in one unit. What should I use for a heat sink on the 7805 and the LM317?Muckleluck Quote Link to comment Share on other sites More sharing options...
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