3A adaptor for driving LCD

"chris_ivan" wrote in news:1139505907.756795.187500 @g43g2000cwa.googlegroups.com:

Please provide your schematic. Does it look anything like the figures in the datasheet, especially either Figure 1 or "1.2V?25V Adjustable Regulator" ?

What are the resistor values that you used?

Sounds like you are trying to use a voltage divider to directly drive your load. A voltage divider only provides the rated voltage if the load on it is small compared to the load of the resistors in the divider. When used in conjunction with a three terminal regulator (LM350T), the divider should be hooked up like Figure 1 in the datasheet.

Caution: It is risky to use your expensive LCD monitor for testing. It would be better to use a 5 ohm 50 watt (!) resistor.

Wrong. The datasheet guarantees 3.0 Amps. The 4.5 Amp value is what the "typical" part is capable of. Your part may or may not be typical. In any case, your design is marginal, you are on the edge of the guaranteed limit of the part. Oh, and this limit only applies if the difference between input voltage and output voltage is less than or equal to 10 volts. Otherwise, the limit is worse (lower current). More on this below.

Another concern: Your transformer puts out 18 VAC Root Mean Square. When rectified, this is about 25.5 volts peak. You are stressing your 4700 uF filter capacitor - verify by using a DC voltmeter to measure the voltage across it.

If your input voltage is 25.5 volts and your output is 14 volts, you are beyond the 10 difference that limits the output current. The guarantee of

3.0 amps no longer applies.

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You've said that the transformer secondary is 18 volts, but at what current? According to the National Semiconductor web site

the LM350T requires a _minmum_ of 10mA of load current. A LED with a series resistor, connected to the output and to ground, should satisfy that requirement. There is also the issue of the regulator's drop-out voltage. Again looking at the data sheet, the drop-out voltage is ~ 2V at 2A, and ~2.5v @ 3A. And don't forget that diodes have a voltage drop, too. It's possible that 4700uF cap is too small (I don't recall the formulas needed to size the cap), or that you'll need a transformer with a higher secondary voltage. Using an o'scope, look at the waveform at the capacitor. You will see hills and valleys. If the volatage at the valleys is less than 16.5 (2.5v +

14v), then the regulator won't work. A LDO (Low Drop Out) regulator may be a better choice. One of National's "Simple Switchers" would be another choice. BTW: if I recall. when using a bridge rectifier, the voltage at the cap is 1.4 x seconday volts. So 1.4 x 18v = 25.2 volts. 25 volts caps won't survive long. Try 35 volt caps, instead.

You may want to ask about this on sci.ectronics.design -you'll probably get more responses. Youc cold also google for linear power supply design. HTH

-Dave Polum

What no bypass or output filtering caps?

Also as stated in the paragraph below Figure 1, if insufficent load the output will rise. As someone else has said put an LED across the output with a resistor calculated on at least 21V at output for 5 to 10mA LED current.

I think he should scope his inputs and outputs as I doubt his levels being measured are truely DC (within normal noise parameters).

Another indication of marginal design is if he could get 14V out at 3A from

25.5V input that regulator is going to COOK without a very good heatsink.

(25.5 - 14) * 3 = 34.5W

Nice little heater he is designing, if the cap is still working.

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I'm quite sure that the LCD monitor draws current more than 1A. I've tried to use 1A adaptor to light up the LCD, and the LED indicator of the monitor flashes, then it goes off.

Will do. Thanks anyway for the response. Appreciated much.

ivan

Hi Ian,

Sorry for my late reply.I've got problems retriev> Please provide your schematic. Does it look anything like the figures in

I follow the schematic given at the LM350T datasheet. for the variable resistor I use 200ohm. Is it right to do that? As far as I know, the datasheet doesn't specify the specific value for the varistor. I'm gonna send you my schematic (JPG format) if you allow me, coz I don't know if I could post attachment to this group.

will do. thanks for the advice.

I agree.. I'm gonna check the input and output voltage of the regulator.

you are absolutely right. I'm gonna check it right away. thanks.

It does, using 10uF/16V tantalum cap, as stated in the datasheet.

hmm.. as hot as 30W solder gun! I also put 2 fans to cool it down using different transformer. is it helpful?

ivan

To be blunt, I think you're going about this the wrong way. For that kind of application, I would strongly recommend you use a switching regulator. However, with respect, if you're struggling with a linear regulator, you'll be in all kinds of trouble with a switcher. Why not just buy a switching module?

Steve

"chris_ivan" wrote in news: snipped-for-privacy@g44g2000cwa.googlegroups.com:

The data sheet assumes that you will calculate the value of the resistors based on your needs and the various datasheet parameters that appply. However, if you look at the other appplication circuits, you will see that R1 is typically 75 - 240 ohms. On page 5 under "Load Regulation" the data sheet says that R1 is usually 240 ohms.

If it were me, I would set up the simultaneous equations so that: Vout = Vref x (1 + R2/R1) I1 = 10 mA = Vref / R1 (see figure 1 for the definition of I1)

(I use 10 mA because the requirement is at least 5 mA and I like lots of margin. It is still small compared to the load of the monitor.)

Then choose the closest standard value resistors, substitute back in and check the results. If close enough, you are done. If not, then iterate.

By "done", I mean that you are now in a position to calculate the power dissipated by the resistors. If you don't like the answer, then you need to go back and iterate some more.

I'm

Send to izshefz at earthzlink dot znet

but first remove every z from the address. Sorry!

There are two resistors, and you gave me the value for one. As I understand it, you made R2 = 200 ohms. Then is your R1 something close to 19.6 ohms? The good news: If you did this, then you have a minimum load of about 64 mA, way beyond the requirement of 5 mA. The bad news: R2 is going to dissipate about 0.8 watts. I hope you are using a 1 watt (or more) resistor.

A better set of choices might be R2 = 2000 ohms, R1 = 196 ohms (or as close as you can get). Another choice would be R2 = 1224 ohms (or as close as you can get), R1 = 120 ohms.

--
Ian Shef     805/F6      *    These are my personal opinions    
Raytheon Company         *    and not those of my employer.
PO Box 11337             * 
Tucson, AZ 85734-1337    *

Hi Steve,

thanks for the recommendation. To be honest, I usually build power supply module for microcontroller system whose current need around 1A. So far I don't find any difficulties with linear regulator. This is the first time I build 3A adaptor. I'm gonna start learning the alternative using switching regulator. Ah, also in the city where I live in, electronic shops typically sells adaptor around 1A, if I need more current, I ought to build it by my own. I also looked for switching transformer to reduce the size of the stuff, but I hardly found it.

This may seems a stupid question, but I would be if I don't ask. I just wonder whether it is possible to build 3A adaptor using 1A linear regulators connected in parallel. In theory it's possible, isn't it?

Thank you for the suggestion, it really helps.

Ivan

Learning is great.

Kelly

It is possible in theory, but in practice you'll have problems with current sharing - i.e. it will almost certainly prove tricky to get the parallel regulators to share nicely. It can be done rather better with a single regulator reference/feedback/comparison circuit and a number of series-pass MOSFETs in parallel and bolted onto the same heatsink... but a switcher is a far better approach. A linear regulator providing 3A is, no matter which way you do it, going to waste an awful lot of energy as excess heat which will need dumping somehow.

Steve

You can easily build one for a few bucks. A 30V 7A MOSFET costs around $1. Unfortunately, most controllers are very small and difficult to build without a PCB. Everything else can be hand wired, includind the MOSFET. We are thinking about build an adaptor just for the controller (lm3485), expanding from MSOP-8 to DIP-8.

No, it's not. They all see overloading conditions and shut down itself. You can use by-pass MOSFET or IGBT. But why not using switchers with 80% to 90% efficiency?