High Current Power Supply - voltage collapses under load

Thread Starter

JeremyCondie

Joined Jan 5, 2015
12
Hi Guys
I found this circuit online. I'm driving it with a Honda EU2200i generator [12v - 8.3A unregulated DC output]. Being unregulated, the voltage changes as the load on the generator changes so I'm wanting to regulate the output to a consistent 12V DC. The generator output is a sine wave that cycles between 0V to 18V.

I used a 680 microfarad capacitor from Vout to ground to smooth (I hope that is correct).

The 4k7 in diagram was replaced with a variable resistor. I did not use the 6a4 diode.

I built circuit and it works great with low amp loads (e.g. a few LED lights).

I then tried a 50W light bulb but the output voltage dropped from 12V to around 1V. The voltages at "1" and "2" also dropped significantly.
I was expecting the 2N3055 to isolate the LM317 from the higher current draw, so why does the voltage drop so dramatically?

high current power supply.jpg Thanks
 

crutschow

Joined Mar 14, 2008
23,814
If the generator output is a sinewave then you need to a add a rectifier and filter capacitor to generate a steady DC voltage to the LM317 input.
It's not designed to operate with a pulsating input.

Also the base drive to the transistors should come from the LM317 pin2, not pin 1.
Pin 1 is the regulation reference terminal.
 

Ylli

Joined Nov 13, 2015
854
Somewhat non-conventional configuration, but should work (poorly) as shown. The diode is not really needed at all, the output of the LM317 is normally from pin 2, but will work from pin 1. The 2N3055's have no ballasting resistors, so there is no guarantee they will share the current equally.

That said, with your voltmeter set to read DC voltage, what do you see at the input with no load? What do you see at the input when under load?
 

AlbertHall

Joined Jun 4, 2014
8,847
The diode is not really needed at all, the output of the LM317 is normally from pin 2, but will work from pin 1.
The base current of the 3055s when the supply is loaded must flow through the 220Ω resistor and this will pull down the output voltage as the load is increased as crutschow says.
 

ebp

Joined Feb 8, 2018
2,332
No, it won't, not ever. The max current at the adjust pin is 100 uA.
But in the circuit shown, about 5.7 mA will flow through the 220 ohm resistor. The output of the 317 will rise as high as is necessary to try to maintain 1.25 V across that resistor. This is the basis of a current source made with a 317. Most of the current will go into the bases of the transistors in the circuit as drawn. A little will go into the "bottom" resistor that would normally adjust voltage.

===
In general, you can't directly parallel bipolar transistors very successfully. There will be variation in the base-emitter voltages between individual transistors, even from the same production batch. The transistor with the lowest Vbe and/or highest gain will tend to hog most of the current and heat up the most. In consequence, its Vbe will drop further increasing the base current, plus the gain, unless it is very hot, will also rise and it will hog even more current. The usual approach is to introduce some negative feedback for each transistor. This is done by adding a small "ballast" resistance between the emitter of each and the common connection to the load. With all transistors sharing a common voltage at the base, if the current in one transistor rises, its base-emitter voltage will be reduced, producing the desired negative feedback. BJTs are usually thought of as current operated devices, but they also can be thought of as voltage operated. The relationship is logarithmic and a small change in base-emitter voltage will have a large effect on collector current.

A light bulb, if you are talking about an incandescent lamp, is a not a constant resistance load. A general rule of thumb is that the resistance of the filament when at room temperature will be about a tenth of what it is at operating temperature, but the ratio can be higher. That means that your 50 W lamp would take 500 W at the instant of turn on. Because the LM317 is actually acting as a current source, trying to deliver about 5.7 mA and the gain of 2N3055s isn't very high, the transistors simply won't be able to supply anywhere near enough current.
 

AnalogKid

Joined Aug 1, 2013
8,231
But in the circuit shown, about 5.7 mA will flow through the 220 ohm resistor. The output of the 317 will rise as high as is necessary to try to maintain 1.25 V across that resistor. This is the basis of a current source made with a 317. Most of the current will go into the bases of the transistors in the circuit as drawn. A little will go into the "bottom" resistor that would normally adjust voltage.
All very true, but a) 220 ohms is way too high a value to drive three paralleled power transistors to any reasonable output current, such as an output current that is greater than the 317 can provide on its own; b) the intent of the circuit is a regulated output voltage, not current; and

c) the statement I was responding to referred to pin 1 as an optional output pin. Also, I think an inexperienced user such as the TS will interpret that as an optional output pin that is a regulated output voltage (since that is the context of this thread) and that is not ever true.

ak
 

ebp

Joined Feb 8, 2018
2,332
Doh! for the second time in two days.

I'd actually written the first paragraph quite differently, didn't like it, went back and changed it and took out a sentence I intended to leave in about the output voltage being pretty much indeterminate. For me, writing sensibly seems to be incompatible with my experiments in drinking myself to death. The time to chose one or the other (or something stronger) may be nigh.
 

Picbuster

Joined Dec 2, 2013
983
lets assume average gain 3055 = 50 you need 50/12 4Amps approx. ideal split over 3 3055 = 3/4 amps
base current per 3055 = 0.75/50 = 15mA to deliver by 317 3 x 15 = 45mA
45 mA in 220 ohm = 9.9V there is also a emitter collector voltage ( Vec)
output will be 12-(9.9 +Vec)

resume: Crutchow & AlbertHall are absolutely correct.

Picbuster
 

BobaMosfet

Joined Jul 1, 2009
893
I think it should be noted that having 12V to start with, in a sinusoidal wave form means you will never have 12V output (after regulation) (unless you boost it at the expense of current). Regulation will use some of it, and you will wind up with an RMS value at best.

Frankly, if it were me, I'd use a cap to drop it to A/C then put a transformer on it to boost it to a higher voltage (like 18 or 24V) and then rectify that to the 12VDC if that's what I needed.
 

AlbertHall

Joined Jun 4, 2014
8,847
I think it should be noted that having 12V to start with, in a sinusoidal wave form means you will never have 12V output (after regulation) (unless you boost it at the expense of current). Regulation will use some of it, and you will wind up with an RMS value at best.

Frankly, if it were me, I'd use a cap to drop it to A/C then put a transformer on it to boost it to a higher voltage (like 18 or 24V) and then rectify that to the 12VDC if that's what I needed.
The generator output is a sine wave that cycles between 0V to 18V.

I used a 680 microfarad capacitor from Vout to ground to smooth (I hope that is correct).
 

Thread Starter

JeremyCondie

Joined Jan 5, 2015
12
If the generator output is a sinewave then you need to a add a rectifier and filter capacitor to generate a steady DC voltage to the LM317 input.
It's not designed to operate with a pulsating input.

Also the base drive to the transistors should come from the LM317 pin2, not pin 1.
Pin 1 is the regulation reference terminal.
Thanks Crutschow (and others). I sorted the problem with your advice. Moved the large cap (680 microfarad) to the input to smooth the sine wave, then moved the base drive to pin 2. Worked a treat. Thanks.
 
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