Simple power supply?

[Mike Taylor]

12v, 3A transformer, LM350T VR, 5A diode bridge, 22000 microfarad cap., 220 & 1K control resistors, 1mf cap. Large cap is on output side of bridge, small cap on out side of VR. Standard circuit diagram.

Volts on out side of bridge is 17. Volts on out side of VR is 7.1 under no load. Load is an 8' piece of nichrome wire, 1 ohm per foot, plus a large ohmite ceramic resistor. VR is on a large heat sink with fan.

Problem is the output voltage starts dropping once the current gets over 0.5 amps. At 0.65 amps the out volts is about 5.1 volts. The volts on the out side of the bridge only drops to about 16.5. If I decrease the load resistance more the out volts will go down to 2.5.

What am I doing wrong? I've run it for an hour at 7v and 0.5 amps. The heat sink stays cool. I need 7v and 2 amps. Same results with a 4700mf cap.

Mike

 

[Minder]

Basically you are using the wrong type pf control, it not very efficient even if it works.
Many use a light dimmer on the primary side of the transformer, I prefer to use one on the secondary, for that you have to use a home built one intended for the lower voltage using a Triac and a couple of components.
It will still operate OK on AC.
No bridge, no electrolytics.
M.

 

[dorke]

Mike,
The thing you do wrong is running the LM350 at a Vi-Vo > 10Volts.
In this case the current limit is no longer guaranteed to be 3A .
There is a de-rating factor ,only 0.25A is guaranteed (for vi-vo=30v).
see here: from LM350 data sheet





If you only need a "fixed 7Volts output", the simple solution is:
Use a lower value transformer (9V will work fine,you will get Vi=12.6V).


BTW,
how did you come to 22,000uF cap?

I highly recommend you use protection diodes and caps (close to Reg.)
See here

 

[Alec_t]

Perhaps the VR is oscillating? Try connecting a 0.1uF cap directly between the VR input pin and ground, as close to the input pin as possible (as per the datasheet).

 

[Old Steve]

Mike,
The thing you do wrong is running the LM350 at a Vi-Vo > 10Volts.
In this case the current limit is no longer guaranteed to be 3A .
There is a de-rating factor ,only 0.25A is guaranteed (for vi-vo=30v).
see here: from LM350 data sheet

He's actually below the Vin-Vout = 10V, at which a potential 3A out is possible. His is 9.9V. It may not be the most efficient way to do this, but should work at the test current of 0.5A if the heatsink is big enough to effectively dissipate 5W. (He says the heatsink stays cool, so the internal thermal protection isn't pulling the voltage down.) He might have problems in that regard when he steps up to 2A though.
The fault must be something else. And you're right, 22000uF is huge.

 

[Mike Taylor]

Dorke nailed it. I switched to the 6 v center tap and was able to get 1.5 amps at 5v. The v out of the bridge was about 8.3. Wonder if I can stage 2 VR's?

 

[(*steve*)]

Wonder if I can stage 2 VR's?

You can, but a switchmode regulator will be way more efficient. I would use one unless there are other considerations which prohibit it.

 

[Mike Taylor]

I just happen to have a dozen 22000mf only one 4700 but it's axial leads.

 

[Minder]

If this is a Nichrome foam cutter etc, the Triac is the simplest way to go IMO.
M..

 

[rickselectricalprojects]

If this is a Nichrome foam cutter etc, the Triac is the simplest way to go IMO.
M..

you always say IMO, what does that mean?

 

[Mike Taylor]

I'm just using the nichrome wire as a test load. I have a cell phone antenna and amplifier at my place in the country. The power supply for it quit so I'm attempting to build one. The label for the amplifier says 6v & 2 amps. I am retired and like to tinker. I've make lots of VR's for DC voltage reduction for game cameras and to drop voltage from solar panels again for game cameras. None of that had the >= 10 volt differential thing. Not high tech stuff but it keeps me off the streets and out of beer halls.

 

[Mike Taylor]

Let me ask another question. What other component would drop 18v down to 6v and give 2A? I don't have a 9v transformer. Got many LM350's. That's why I thought about putting two in series. I was originally thinking of over-driving the amp at 7v. It seems to tolerate 6.5v fine. Or what if I just did half wave rectification. What would be the ripple voltage at 2A with a 22000 cap?

 

[Mike Taylor]

Oh shucks I get a delta-V of only 0.75 volts. If I used the 4700 cap and one diode that would get the voltage down even more.

 

[dorke]

The other side of the LM350 limit is Vi-Vo>3V.

So if you have another LM350 use them in series to do this:

Use the 12V of the transformer!
12VAC -->retifier+filter cap-->LM350T output set to about at 11V-->LM350T output set to 6V.
important:;
use a full rectifier ( bridge or dual diode) not a single diode!
also use the protection diodes for the LM350s.

If you desire the lowest ripple possible use Cadj=10uF
it will be something so low you can't measure...

On the inputs of the LM350s use a 0.1uF to ground,
very close to the inputs ans very short wires.
Between the two just a 0.1uf (no electrolyte needed).

 

[Old Steve]

you always say IMO, what does that mean?

In My Opinion

 

[Mike Taylor]

Roger the full wave bridge. Wouldn't get the amperage I need with half wave.

 

[Mike Taylor]

Dorke, where does Cadj go? Same as C2 in the above diagram? Last sentenance of your last note - between the two inputs of the LM350's?

 

[Old Steve]

Dorke, where does Cadj go? Same as C2 in the above diagram? Last sentenance of your last note - between the two inputs of the LM350's?

Even I can answer that - you're right, Cadj = C2 in the diagram in post #3, (from the adjust pin to ground).
And for two series LM350s, the 0.1uF dorke mentioned would go from the output of the first reg, (which is also the input of the second reg), to ground.
And the same diodes as in the diagram in post #3, fitted to both regulators.

 

[(*steve*)]

Dorke, where does Cadj go? Same as C2 in the above diagram? Last sentenance of your last note - between the two inputs of the LM350's?

That, and the DC produced has far greater ripple under load.

The last sentence means between the output of the first and the input of the second. I would place it as close to the input of the second (the same as you have been advised earlier).

 

[Mike Taylor]

Got the circuit pattern printed on paper. Next to transfer it to film and iron it to a board. I counted like 55 holes to drill. Not sure when I can populate the board. Taking a trip in a couple of days. I do have the design in a PDF file that I can put online.