EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: rohan-04 on May 14, 2013, 07:38:03 pm
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Hey so I basically want to build my own bench power supply. I drew up a schematic (attached). Based of the LM317 I wanted the system to give out various preset voltages (i.e) 3V,5V,6V,9V,12V. and also power an Led for the respective voltage. Simple enough. I've attached the dip trace schematic below, would really appreciate it if some one would take a look at it for me.
Thank you
Rohan.
P.S - The correct resistor values are (in Ohms) :-
3.3V = 393
5V=720
6V=912
9V=1488
12V=2064
Cant find a proper 2P6P switch symbol for diptrace. Basically now the 6th position would need to switch the output off
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Please export it to PDF or something. Most of us don't have DipTrace installed to view the file.
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I'm a beginner so don't take my word as gold, but I think all your LED's are going to stay lit. You'll need to set that up a bit differently.
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I'm a beginner so don't take my word as gold, but I think all your LED's are going to stay lit. You'll need to set that up a bit differently.
Yeah I had my doubts about that too. Also Would I need to incress the values of the Resistors to accommodate the LED?
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It's not really going to work at all like that - none of the LEDs are going to light (because they are backwards), and even if you turn them around, they'll just all light. If you want individual selections like that with one LED each, you should use double-pole switches. You can get a 2P5T rotary switch like this (http://www.ebay.com/itm/2P5P-Double-Pole-Five-Position-Rotary-Wafer-Switch-/121106923634), and put the adjust resistors on one pole and the LEDs on the other. However, I'd recommend just using a potentiometer as the adjust resistor and a panel meter to show the output voltage. That way, you can get whatever voltage you want.
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You seem to think the switches and adjust resistors are going to have any effect on the LEDs at all. In case you didn't notice:
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The goal of this project is to have preset Voltages. Apart from the leds is everything else ok?
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As far as I can tell with the information I have, yes.
A couple requests:
- Input voltage?
- Output current?
Also, you have a bunch of connections that are not actually connected on that schematic, just so you know.
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you need resistor to limit the currents for LEDs ... and the switch for one value should connect the LED from VCC to ground in parallel with the adjustable resistor .. so its tow switch in one package .. not like your pictures or just remove all the LEDs
you need to add main switch off ....
and ofcource good heat sink .. and no more than 3 A loads
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haha, Sorry Im just getting my grips with the software.
Ah yes Input voltage is 18V. Output current the transformer is rated for 4A. From the system im looking to only get about 1.5A to 2A.
How can I connect one switch for both? Can I use a 2 pole, toggle switch to control both the resistor and the led?
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From the system im looking to only get about 1.5A to 2A.
LM317 only does 1.5A maximum, and that's with a large heat sink.
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you need resistor to limit the currents for LEDs ...
He has this.
you need to add main switch off ....
He has this.
and no more than 3 A loads
In what world do you get 3A from an LM317?
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From the system im looking to only get about 1.5A to 2A.
LM317 only does 1.5A maximum, and that's with a large heat sink.
Yeah sorry read the data sheet wrong.
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To elaborate on the current: LM317 has a maximum current of 1.5A. However, you need to watch out for heat. It will dissipate (Vin-Vout)*Iout watts of power. Multiply this by the total thermal resistance of whatever heat management system you are using to see how much hotter it will get. If it has no heat sink, the thermal resistance is 80°C/W. If it has a heat sink, you'll have to look up the thermal resistances of the heat sink and thermal compound, plus the LM317's resistance of 5°C/W. Of course, if you are using a heat sink you already have on hand, you probably won't know the thermal resistance; in this case, just run it dissipating a known amount of power and see how hot it gets. You need to keep it under 125°C, and preferably under 80°C.
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Under my full load of 12Vout thats only 9 watts but 3.3V is 22.05 the LM317 is rated for 20 right?
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You could do double pole double throw switches but you would need a few of them and you'd have to watch out for turning on more than one at a time. Would be much better to use the switch c4757p mentioned. Are you really going to need more than 1 amp? I'm just a hobbyist but so far I have not really used more than half that. And if you ever get into audio you may want to include negative voltages. And moving up to the 15 volt range.
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Under my full load of 12Vout thats only 9 watts but 3.3V is 22.05 the LM317 is rated for 20 right?
Yes, but "rated for" just means it can do it if you take care of the heat. With no heat sink, it's going to try to reach 720°C at 9W! Of course the thermal overload protection will kick in.
Are you really going to need more than 1 amp? I'm just a hobbyist but so far I have not really used more than half that.
Agreed.
And if you ever get into audio you may want to include negative voltages. And moving up to the 15 volt range.
One step at a time...
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You could do double pole double throw switches but you would need a few of them and you'd have to watch out for turning on more than one at a time. Would be much better to use the switch c4757p mentioned. Are you really going to need more than 1 amp? I'm just a hobbyist but so far I have not really used more than half that. And if you ever get into audio you may want to include negative voltages. And moving up to the 15 volt range.
Oh no never, I wanted to know the extream limits of the system..
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you need resistor to limit the currents for LEDs ...
He has this.
there is no limiting resistors for the LEDs in the attached picture more than that you should note that in his schematic all the LEDs will light ... but when he add every LED on its own switch ( with the adjustable resistor then he should add a limiting resistor for it for every LED due to the output voltage
you need to add main switch off ....
He has this.
there is no main switch in the attached picture
and no more than 3 A loads
In what world do you get 3A from an LM317?
you are right .. its 1.5 A .... party ..
so if you want higher current use the 3055 transistor it give u 5 A... just connect the transistor base to the LM317 output voltage ... c to + .. and E is your output again with good heat sink .. ( output value will be lower by 0.7 volt that will drown by BE junction )
and for limiting resistor value ..
R= ( Vcc - VLED ( 3.5 v) )/ I LED ( 0.02 A )
as 3.5 voltage on LED and 0.02 is LED current ..
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There is a switch, in the grey section next to F1, on the mains side of the transformer.
He doesn't really need separate resistors per LED. Yes, it will be dimmer at lower voltages, but newer LEDs are often so efficient that they will light sufficiently with even a small current, so not really a big deal. It could even be considered a feature that it's brighter at higher voltages... Either way, it's not going to blow anything up, and I'm sure he'd figure it out for himself when he sees the brightness changing.
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as 3.5 voltage on LED and 0.02 is LED current ..
If the LED is recent and didn't come out of an old junk box it's going to be blindingly bright at 20 mA.
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Is there any reason why i cant use toggle switch? Would the other switch be safer?
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What's the other switch? Just make sure it's rated for the proper voltage and current and you'll be fine.
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The 2P5P switch.
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What is the purpose of the five LED's on the right side of the schematic? Are you intending to sense the output voltage with those and light them up accordingly?
I presume that is the intention since they are marked with voltages.
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Yeah thats exactly it. The problem i have is in wiring them up correct. I think now I'll just use one switch ( rotatory) and connect the resistors aswell as the leds.