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Power Supply Design that gives variable voltage 0-20V and 3A current

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mengghee

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hiye,

i would like to build a power supply that can give a variable voltage of around 0 to 20v and current of around 3A. does anybody know if there is any reference i can look into ? thank you


regards,
mengghee
 


    mengghee

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variable power supply using lm350k circuit

Well first I must know the kind of application
ripple and transient response.
There are three kinds of power supplies: Linear, LDO and Switched power supplies.
If you need an easy one with low noise, Linear one.
Better efficience LDO is the selection.
for the best efficiency Switched mode power supply is the best but it is the most complex of all.
 

    mengghee

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50v 3a power supply filtered dc power schematic

well, i just need a bench power supply because it is too expensive to buy one. ldo ? what is ldo actually ? thank you



regards,
mengghee
 

power supply lm350t circuit

Once I built a very similar power supply. You can use the LM350T as a variable regulator. It can regulate from 1.25V to around 30V output voltage.

I would design as follows:

one tranformer with 24Vac x 3A secondary
a rectifier bridge for 50V x 3A (four 1N5404 diodes also do the work)
one 10,000uF x 35V electrolitic capacitor (maybe 4700uF also works)
the LM350T with a huge heatsink
a resistor of 270 ohms and a potentiometer of 4k7 as voltage adjustment (see LM150/350 datasheet to calculate the resistors)
and a output capacitor (maybe 100uF x 25V)

To improve your design, attach an analog voltmeter (0-30V) and an ammeter.
 

    mengghee

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power supply variable with lm350

hiye,

i am just wondering, for a power supply as such, do we have to worry about our load, such as inductive, capacitive etc. and do you have a schematic diagram ? from the first post from ianp, i noticed in the schematic diagram, it doesn't uses regular like lm350t. what kind of power supply is that ? thank you.

regards,
mengghee
 

power supply design -12v, 3a

Hi Mengghee !

Don´t worry about the kind of the load. It is a DC power supply with output protection (short circuit, thermal, etc). Of course you should read the LM350 datasheet (National Semiconductors) to design properly your power supply. But it is very simple. With the components I listed, I think the only difficult is to know how to connect the resistor and the potentiometer.

Before that, the transformer secondary wires connected to the rectifier bridge (tilde terminals), the positive terminal connected to the positive of big capacitor and the negative (groud/earth0 connected to the negative of the same big capacitor (the filter) . Basic unregulated power supply. It should provide a 24 to 28Vdc with ripple.

Ok, the LM350T (TO220 package) has 3 pins: looking at it front view: pin 1 is the adjust, pin 2 is the output voltage (regulated) and pin 3 (the rightmost) is the input (from the positive of the filter capacitor).
Connect the resistor between Voutput and Adjust pins (2 and 1) and the potentiometer (the center terminal and one of the other terminal) between the pin Adjust (2 of the LM350T) and the ground (the negative terminal of the big capacitor). The remaining capacitor (smaller) you just connect between output pin
(positive of the power supply) and ground (negative of the power supply).

Sorry for not drawing the schematic, but I think the directions are enough. If you have any doubt, ask again. Remember this power supply will not give you output voltages below 1.25V (because it is used as reference voltage at the LM350). But the construction is very simple and all components are easy to find and buy.

Special attention to the heatsink. LM350T has a thermal resistance junction to case of 3K/W and the maximum junction temperature is 125C. So at 1.25V output and 24V input and 3A current, you have (24-1.25)V x 3A = 68W (impossible to dissipate)

The LM350T with a 4K/W heatsink (a very common size) and the thermal compound could dissipate only 125C/ (3+4+1)K/W = 15W.

One alternative is to use "K" package (LM350K) which has a lower thermal resistance (1.2K/W) together with a larger heatsink (maybe 2K/W) to obtain 125/(1.2+2+1) = 30W power dissipated (or 10V difference between input and output at 3A current), below this output voltage, your output current must be below 3A. This is another disadvantage: you are restricted to some voltage and currents, or in the other words, you cannot use it for the full range of voltages with 3A consumption.
 

    mengghee

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0-30 vdc variable power supply drawing

thank you very much. as stated, the ic has it's own protection. i am just wondering what would happen when the protection is exceeded ? thank you.



regards,
mengghee

Added after 2 minutes:

and also ... by using the ic, will it give a very good dc output ? with very less ripple ? thank you
 

a 0-30 vdc 0.002-3a stabilized power supply.

Hi !

As stated in the datasheet, the ripple rejection is 65dB (without output capacitor) and can be improved to 86 dB by using a bypass capacitor, It means that at 65dB if you have a ripple of 1Vpp @ 120Hz (voltage oscillates from 24 to 25 V for example) in the unregulated input voltage (after the filter capacitor), the LM350 will attenuate it by 1778, resulting in a 1/1778 = 560uV.
With the capacitor, the attenuation turns to 20000 times, the output ondulation becomes 50uV. I think at UK, the frequency is 50Hz, so the ripple frequency will be 100Hz, but this frequency is too close to 120Hz and the resultant output ripple very similar.

Concerning the protection: I think that if the temperature is exceeded, the IC turn itself off, protecting the power supply. If happens overcurrent, the IC limits the output current to the 3A protecting again the circuit.
 

    mengghee

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