Dual Power Supply possible with Non CT Trans?

I have a simple doubt.. Is this circuit possible to work out?? Or will it simply short out the transformer?? I am putting this possibility hoping that the bridge wont allow interconnections b/n the 2 ckts.. Help me out. Please dont take it serious if i am wrong.

Hello,

The circuit you described will give trouble at the transformer.
You could make it with two seperate transformer windings.
Note: The voltage is rather high (30 Volts) 15 to 18 Volts will be enough.

Greetings,
Bertus

No this is not the way to accomplish this.

The bridge rectifiers work on both +ve and -ve half cycles so you will be cross connecting if you do this. Follow the connections from one side of the transformer through a diode in each bridge to see this.

There are several ways to do this.

Firstly using one of the transformer terminals as a reference you can generate simultaneous + and - supplies from the other using alternate half cycles, by way of a single rectifier for half cycle and thus each supply. You will obviously have to size components accordingly to overcome the massive ripple this will engender.

Secondly you can use a single bridge / reservoir /regulator, connected across the whole secondary, and split the result using complementary emitter followers to create an artificial zero point. This can be made variable and tracking.

Thirdly you can interpose a series capacitor between one of your bridges and a transformer terminal in your circuit. This has to be of sufficient current rating to take the whole supply. Such capacitors are hard to obtain and expensive for any significant current.

Each of the circuits, including your own are probably more costly than simply obtaining a second transformer.

Well Thanks for all your replies. I just wanted to clear a doubt that came to my mind. I normally use centre tap transformers so couldn't make out what would be the result of such a circuit. Just wanted to learn, thats all. Anyway thanks. Now I have a few more doubts regarding variable RPS.
I'm building a Variable RPS using LM317 for a max voltage around 18vdc..

1. If I use a 100 ohm resistor between adj and out of LM317 what would be the variable resistor I can use for getting a peak 18v and also market available variable resistor?
2. I plan to add a fine adjustment to the circuit. so I guess I need to use another low value variable resistor in series with the first one. Will 100 Ohms be adequate for a fine control of 1volt?
3. How can I add a simple overload LED indicator?
4. And finally I would also like a current control of the output. I know LM317 can be controlled that way but dont know how to wire it out. Please help me on that also.

Hello,

On this page of the EDUCYPEDIA there are some idea's for making powersupplies.
http://www.educypedia.be/electronics/circuitssupplies.htm

Greetings,
Bertus

See if this doesn't help...

http://forum.allaboutcircuits.com/showthread.php?p=84026#post84026

Thanks for all that Bill,but your thread focusses more on dual supplies. I actually am NOT looking to build a dual polarity supply.. But I appreciate the Info I got from your work... Nice work Bill..
And the other links to educypedia led to the place from where I got the same circuit.. Anyway let me have a dive in it .. I just need to clear some minor doubts.. Thats where I need your help guys. I

With regard to the 18 volt PS with the LM317, refer to the data sheet for it to get min/max component values and how to sense current. I don't believe the 100 ohm resistor is quite right.

By the way, if you want to make a negative regulator, use an LM337.

The LM317 requires a minimum 10mA load to provide guaranteed regulation.
Vref is the voltage drop between the OUT terminal and the ADJ terminal; it is nominally 1.25V, but may be as low as 1.2V or as high as 1.3V.

Since it may be as low as 1.2V, using a 120 Ohm resistor (R1) from OUT to ADJ will result in a minimum 10mA current that will satisfy the minimum load requirement.

One COULD use a 100 Ohm resistor, but that would result in somewhere between 12mA and 13mA current. Why waste the extra 2mA-3mA if you don't have to?

So, for now let's use 120 Ohms for R1. Since the nominal Vref is 1.25v, you'll have a nominal 10.417mA current flowing through R1.
(I = 1.25/120 = 10.417mA)

If the ADJ pin is then grounded, the nominal output of the LM317 will be 1.25V, with 10.417mA flowing through R1 to GND.
Q: What if we want a higher output voltage?
A: We'll need an R2 to go from the ADJ terminal to ground.

Q: So, what value of resistor?
A: Since R=E/I, and we know that the current is 10.417mA, and our minimum output voltage is 1.25(nominal)...
R2 = (DesiredOutputVoltage-1.25) / 10.417mA
So, if you wanted 18V out:
R2 = (18V - 1.25V) / 10.417mA = 16.75 V / 0.010417 A = 1608 Ohms (rounded off)

Or set the output with a 2K trimmer configured as a variable resistor.

jj_alukkas said:

Thanks for all that Bill,but your thread focusses more on dual supplies. I actually am NOT looking to build a dual polarity supply.. But I appreciate the Info I got from your work... Nice work Bill..
And the other links to educypedia led to the place from where I got the same circuit.. Anyway let me have a dive in it .. I just need to clear some minor doubts.. Thats where I need your help guys. I

Click to expand...

If you really read the article it most assuredly is not. It is how to create a pseudo ground with a single source supply, which is what you asked, no?

If the single source is not grounded you can ground the pseudo ground and it will resemble a dual tracking power supply. It is meant as a how to design, not as a design in and of itself.

Ya Bill, that was my first question and all of you said its not possible with my circuit and you provided me with a solution for that. Ok I understand that. Then I had my next few questions regarding single polarity power supply. Please dont mistake me. You gave me the correct answer for my first question. you are never wrong.

With regard to the 18 volt PS with the LM317, refer to the data sheet for it to get min/max component values and how to sense current. I don't believe the 100 ohm resistor is quite right.

Click to expand...

Somewhere which shows out power supplies with LM317 states using a 100Ω resistor to ensure the minimum current is more than 10mA. Thats why I went for a 100Ω. If 120 gives a better efficiency why not try that then. Anyway thanks with the values SgtWookie... And I appreciate everyone's help..

What they were saying is you're original schematic wouldn't work as it was drawn. This was why I showed the alternate approach, long before you asked. There is an old saying, there is more than one way to skin a cat.

There are other ways to make a power supply also, without using a LM317. The LM317 is a good part, quick and easy, but the 10ma can be a problem sometimes. This thread worked through an alternate method. You need to define your specs, then design for them.

I chose that due to simple design and fewer components.. Thats all.. My specs include a variable regulated power supply anywhere near 18 volts single polarity and 1.5A.. thats all... WHich other chip would you reccommend??

jj_alukkas said:

My specs include a variable regulated power supply anywhere near 18 volts single polarity and 1.5A

Click to expand...

18V? How much above and below 18v do you need to be able to adjust it?

Will there always be a minimum current load on the supply when your circuit is operating?

An LM117/LM317 will put out up to 1.5A current, but that is it's maximum rating.
An LM150/LM350 is a similar adjustable regulator with a 3A output. Besides having twice the current rating, the maximum voltage difference from input to output is 35v, where the LM117/LM317 can handle up to 40v - but this does not matter in your application.

Whatever voltage regulator IC you wind up using, you will need a heatsink on it.

Anyway I'm now happy with my final decisions. Im going to use a 120ohm for R1 and R2 would be a 2K pot with a 100 ohm for fine adjust in series. No current control as its not very necessary for me. And If I need the voltage to start at 0v, I will use 2 diodes in series. Hows that Idea Bill?? Pretty fine?? One more help from you. Any idea for a simple overload indicator? Not compulsory but just asking for knowledge.

"18V? How much above and below 18v do you need to be able to adjust it?

Will there always be a minimum current load on the supply when your circuit is operating?

An LM117/LM317 will put out up to 1.5A current, but that is it's maximum rating.
An LM150/LM350 is a similar adjustable regulator with a 3A output. Besides having twice the current rating, the maximum voltage difference from input to output is 35v, where the LM117/LM317 can handle up to 40v - but this does not matter in your application.

Whatever voltage regulator IC you wind up using, you will need a heatsink on it.

Click to expand...

Ohh sorry.. its 0-18 to use as a lab VRPS.. And always there wont be a load as it is a lab power supply. And I definitely have to use a heatsink. I know I cant otherwise.Ahh... By the way, can i bolt the LM317 to the metal case? will it short the chip?? Assume as if my case is ground.

Let's see what you have so far. An overload indicator is pretty easy, but I've got to work with what you have. The two diodes will get you pretty close, but not exact. You'll have to build it to see where it goes, varibility is always with electronics.

If you are using a LM317 in a TO3 case style just use the same kit you would use for a transistor in a TO3 case, including heat sink compound.

Circuit is ready on breadboard.. just got to replace that 100 with 120

How about a schematic?