0-50V 1A PSU schematics

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neazoi

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Hi, I need a simple 1A, 0-50V adjustable PSU for the lab.
I have found this which looks ok for it's simplicity.
https://www.electroschematics.com/945/01v-50v-power-supply

However I wonder is a single regulator can do this, like the LM317 which is used up to 30V.

- - - Updated - - -

I know the thread is for 0-50v, but I see that the LM317 can go as high as 37V. Can you help me design a PSU with this one for 1-37V or so?
I am not sure about the transformer and bridge/capacitor values so as not to exceed 40V (max input for LM317).
 


Hi neazoi
LM317 is dealing with a lot of dissipation and it's regulation isn't so precise ! furthermore , if you need 50 volt you should have more than 50 volt as input to have a regulated output ! LM317 can't handle it however with some specific tricks you can optimize it but i've a better suggestion ! less dissipation and more quality ! try LM2576 ADJ HV read it's datasheet . i'll bet you'll like it !

Best wishes
Goldsmith
 
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    neazoi

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Thnks a lot!
However the LM2576 is a switching regulator. I do not want to use these in my RF circuit.
 

Hi,

50W of power dissipation is a lot of heat....
Big heatsink with fan..

Klaus
 

However the LM2576 is a switching regulator. I do not want to use these in my RF circuit.
Click to expand...
RF circuit ? i though you're looking for a lab PSU !

Anywhere what's the point ? why not a switching regulator in RF circuit ? if you use a good filter as out put filter you won't have any problem and interference !

BTW as KlausST has mentioned it can has at least 50 w power loss ! ( if inlet voltage be more than 50 volt and as it should be ! it would be more dissipation ! )
And increasing the temperature will deal with thermal noise !

However if you're interested in a linear regulator i can help you too .

Best Luck
Goldsmith
 
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    neazoi

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What I mean is a power supply capable of powering an RF circuit. I did not know that the LM317 can go as high as 37V, that is why I mentioned the circuit on post #1. I aimed for a 50V PSU as the title states, but the simplicity of the LM317 circuit made me compromize these 13 more volts for the shake of simplicity.
I apologize for the confusion.
Do you think the LM317 will be heated too much at an output voltage of 25-37V with a 40V transformer?
 

Heat produced = (Vout - Vin) * Iout. Obviously it will vary according to the output voltage you set it to and the current you are pasing through it. All you can do is base calculations on a worst case scenario, which is probably lowest output voltage at full current. That gives the dissipation figure in Watts. You can then decide on how hot you are prepared to let it get by choosing a suitable heat sink. Taking into account the thermal resistance in the device package itself and any insulating washers you might place beneath it, pick a heatsink that has suitable C/W rating. It's the number of degrees temperature increase per Watt of power.

Example, (ignoring thermal resistance of the device mounting itself) if it drops 30V at 1A it dissipates 30W of heat. A 1C/W heatsink will therefore increase in temperature by 30C. The conditions defined by the heatsink manufacturer have to be observed, for example mounting in in a small enclosed space will reduce the quoted rating because it will be unable to radiate/convect the heat away sufficiently.

Brian.
 
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    neazoi

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Linear regulation will be very inefficient at 1A for mid range voltage where load and regulator share ~equal voltage drop and power dissipation of 25 W or more.

I would modify an existing universal Lap top charger to change the feedback voltage to adjust the output, as a preregulator then an LDO to filter out noise which tracks the input voltage with the minimum drop for efficiency such that Vio is fixed at 1V or less. I.e. Vout = Vin -1V but filtered.
Laptop chargers are typically fixed 65W and many have stepped outputs from 12 to 24V which you can modify to make linear control from low to 50V with suitable Cap V rating checks. I have done this. But designing your own SMPS may be a better solution using ANalog devices Linear Tech or TI design tools
 
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    neazoi

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My power supply has a 12V 3A transformer, and a 2N3055 which provides resistive drop.

Suppose I draw 6V 2A. Then the transistor must dissipate 12 watts. It gets hot enough to burn my finger. I mounted the transistor to the steel enclosure for a heat sink, with a plastic film insulator and thermal grease. It is just about adequate.
 

Hi,

This Intersil datasheet for the CA3130 op amp has a 0.1 - 50V supply, it's the same circuit as the schematic you posted, in case you ever want to refer to it out of curiosity:

https://www.intersil.com/content/dam/intersil/documents/ca31/ca3130-a.pdf

I like Intersil datasheets, they often have additional helpful info. This datasheet - I don't know if it's of use to you as I was only browsing to compare the CA3130 to something similar and see if they were compatible.
 

Indeed the old datasheets had lots and lots of very useful circuits and applications.

The CA3130 comes all the way from RCA Semiconductor in the mid 70s, which was later along with the original Intersil purchased by General Electric, who in the Jack-I-hate-electronics-Welch years was sold to Harris, who again spun part of it again as Intersil, with the logic families sold to Texas Instruments.

All told a very storied brand with lots of unique ICs.
 
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    d123

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neazoi said:
Thnks a lot!
However the LM2576 is a switching regulator. I do not want to use these in my RF circuit.
Click to expand...


Which do you think would be easier or cheap to filter ripple including cost of 50W heatsink to <10mV? or <1mV?

The line frequency transformed and rectified will result in very narrow pulsed current , i.e. narrow duty factor, % df. at 2x line f, such that the peak current is inverse to the %ripple which is equal to % df. When RC=T the pulse interval= 2/f the % d.f. =%Vp-p =40%.

For 50Hz in and 100Hz pulse out @ 1Adc average, this RC=0.01s and for 50V C will be huge... Can you define Vpp ripple and value of C yet? You ought to be able now.

To reduce ripple from 40% to 1% the cap. ripple current increases to 40xIavg=40Ap-p or ripple the cap size increases from C=T/R to 50x as big. , resulting is ripple of 40% if load R and storage cap RC=1/2f. To get ripple down from 40% to 5% you need C=5x bigger such that RC= 5 times the pulse rate and to get down to 1% ripple or Ipk/Iavg=40 then you need 40x the size compared to C=1/(R*2f)

You may be compromise sag with a series R or add a choke that now at 100Hz has to be 1000x bigger L than at 100kHz, as a LPF at 1A.

Note that ac line transformers are only 60~80% efficient in this bulk peak current pulse charge rectifier mode and then Linear Regulator efficiency ranges from 90% at 50V to 2% efficient at 1V from a 55V supply.

Some clever designers have used the LM317 for high voltage buck operation by using series bypass MOSFET and only letting the LM317 conduct low current with a large drop zener in front, or have a simple buck and LDO follower with only a 1V drop or so.

Now you do the math and redefine your requirements..please. Vpp, cost , V,I and resolution.
 
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