24V-4A PSU design need your help and contribution

lm317 24v

The picture below is my PSU design. I want to assemble one PSU for my own LAB.

Specification: 1.25 ->24V, 4A with short circuit protection, overload and overvoltage protection also. The stability is 2% (for 4A sink, the output is decreased but not below 98% of max value)

- My transformer is capable of providing upto 30VAC and 10A
- Value of component:

* The DC Voltage input is 29VDC
* R1=22 Ohm, R3=220 Ohm and R4 is set so that the output voltage without load is 24V.

For load current below 1.5A, the PSU work properly. I mean that, the output voltage is not decreased (24V --> 23.96V) --> really good.

But when I decrease the load resistor so that it absorbs 2.5A (Load resistor of 10 Ohm).

The output Voltage decreased, from 24VDC to 22.5 VDC (1.5 Volt). My teacher said that, the stability is so bad and he ask me to re-design the PSU (this is also one part of my final project).

I use LM317T to produce the stablised voltage. I know that it use 1.25V Reference between Pin OUT and Pin ADJ. When the load absorb 2.5A, this Vref decreases to 1.148V. I think this is the reason which leads to the decrease of the output. But I don't know why? Can some one help me to fully understand how LM317T stablises its output and how to drive the Iadj at pin 1 of LM317T (so that I can drive the VRef 1.25V Between pin 1 and pin 2)

I intend to improve the quality of my PSU as below (some ways):

- to supply more current goes into the LM317 by using the circuit diagram as shown in file PSU1.JPG. R5=0.33Ohm so that when the current goes through 2N3055 reaches 2A, it leads to the voltage across R5 is 0.6. As a result, Q2 begins to conduct and the current Ic of Q2 supply more current for LM317T. I think that it would help to improve the Iadj of pin 1 of LM317T so that, Vref remains unchanged at 1.25V. But in fact, it does not. The Vref still remain decreased at 1.148V so the output Voltage is decreased.

Right now, I have no more idea to improve my design. I really need your help and contribution so that I can upgrade my design.

Thank you beforehand!

overvoltage protection lm338

Have you considered this (from the datasheet)?:

"The LM117 is capable of providing extremely good load
regulation but a few precautions are needed to obtain maximum
performance. The current set resistor connected between
the adjustment terminal and the output terminal (usually
240Ω) should be tied directly to the output (case) of the
regulator rather than near the load. This eliminates line
drops from appearing effectively in series with the reference
and degrading regulation. For example, a 15V regulator with
0.05Ω resistance between the regulator and load will have a
load regulation due to line resistance of 0.05Ω x IL. If the set
resistor is connected near the load the effective line resistance
will be 0.05Ω (1 + R2/R1) or in this case, 11.5 times
worse."

lm317 2n2955

Hi,

2N3055 is a NPN transistor and not a PNP as drawn on the schematic. Do you realy use a 2N3055 ?

24v lm317

I'm sorry, it's exactly 2N2955, not 2N3055 as shown in the picture.

Is there any more help????

+lm117 +oscillation

Change the 220R to 120R. The 220R is only valid for premium devices and not commercial ones. Look carefully at the datasheet and you will notice that the commercial device needs about twice the current into the ADJ terminal and hence 1/2 the resistance as generally quoted in the datasheets. With a 220R in place regulation will suffer.

Here is my the reasoning posted on this subject in the past: It is regarding a LM338 device but the principle is the same for the 317.
----------------------------------------------------
The adjustment pin current may vary anything from 45uA to 100uA
according to the datasheet. This output current combines with the
current through the voltage divider on the output formed by the 240
Ohm resistor and the adjustable resistor. This causes a error term in
the Vout formula .....+( IadjR2). To minimize this error the quiesent
operating current ( 1 to 3mA ) of the device is directed to the output
requiring a min. load current tot be present for the device to operate
within specs. If the load current is too small the output voltage will
increase.( Error term comes into play ). The required min load current
for the LM138 is 3.5mA to 5mA and for the LM338 3.5mA to 10mA.

At the min. voltage output of 1.25V the bottom resistor equals zero.
The top resistor is now responsible for the min. load current. Taking
the worst case, to get 10mA flowing through R1, R=V/I then
R1=1.25/10mA = 125 Ohm ( Closest standard value - 120 Ohm ). By using
a 330 Ohm the min load current of 3.8mA is lower than the worst case
of 10mA for the 338. For the 5mA of the 138/238 a value of 250 Ohm is
required ( Closest standard value - 240 Ohm ).
---------------------------------------------------------

lm317 and 2n2955

XNOX_Rambo said:

Have you considered this (from the datasheet)?:

"The LM117 is capable of providing extremely good load
regulation but a few precautions are needed to obtain maximum
performance. The current set resistor connected between
the adjustment terminal and the output terminal (usually
240Ω) should be tied directly to the output (case) of the
regulator rather than near the load. This eliminates line
drops from appearing effectively in series with the reference
and degrading regulation. For example, a 15V regulator with
0.05Ω resistance between the regulator and load will have a
load regulation due to line resistance of 0.05Ω x IL. If the set
resistor is connected near the load the effective line resistance
will be 0.05Ω (1 + R2/R1) or in this case, 11.5 times
worse."

Click to expand...

Dear sir,

As far as I know, I've tried this note in practice but no improve was seen.
Is this right to add a resistor between Pin2 (Vout) and Output of the PSU as indicated in the note??? (sorry if my english is too bad so maybe I misunderstand the meanings of the note).

I accompany here the modified schematic (This schematics introduce no improve compared with the original one!)

lm317t-4a

E-design said:

Change the 220R to 120R. The 220R is only valid for premium devices and not commercial ones. Look carefully at the datasheet and you will notice that the commercial device needs about twice the current into the ADJ terminal and hence 1/2 the resistance as generally quoted in the datasheets. With a 220R in place regulation will suffer.

Here is my the reasoning posted on this subject in the past: It is regarding a LM338 device but the principle is the same for the 317.
----------------------------------------------------
The adjustment pin current may vary anything from 45uA to 100uA
according to the datasheet. This output current combines with the
current through the voltage divider on the output formed by the 240
Ohm resistor and the adjustable resistor. This causes a error term in
the Vout formula .....+( IadjR2). To minimize this error the quiesent
operating current ( 1 to 3mA ) of the device is directed to the output
requiring a min. load current tot be present for the device to operate
within specs. If the load current is too small the output voltage will
increase.( Error term comes into play ). The required min load current
for the LM138 is 3.5mA to 5mA and for the LM338 3.5mA to 10mA.

At the min. voltage output of 1.25V the bottom resistor equals zero.
The top resistor is now responsible for the min. load current. Taking
the worst case, to get 10mA flowing through R1, R=V/I then
R1=1.25/10mA = 125 Ohm ( Closest standard value - 120 Ohm ). By using
a 330 Ohm the min load current of 3.8mA is lower than the worst case
of 10mA for the 338. For the 5mA of the 138/238 a value of 250 Ohm is
required ( Closest standard value - 240 Ohm ).
---------------------------------------------------------

Click to expand...

I haven't tried this proposal but I'll will try later. I'll post the result here as soon as possible.

By the way, I want to improve the Iadj (so that I can improve the voltage regulation capability) using this schematics. It works as follow:

If the load current does not excess 2A, Q2 is not conduct. When the load current excess 2A, Q2 begins to conduct, the colector current of Q2 will supply more current for LM317T, so the current Iadj will be increased either.

Is this right?

24v 10a by lm338

Hi,

Max current flowing through R1 = Vbe/R1 = 0.7/22 = 31.8mA. Hfe of 2N2955 is 70max @ Ic=4A so Ib (base current) must be equal to Ic/hfe = 4/70 = 57mA. Ib flows through R1. I think you need to reduce R1 value to 10ohms, allowing 2n2955 and LM317 to both work properly.

lm338 diagram

Why the 0.22R at the output?
Also I have seen these type of regulators go into some sort of burst-HF-oscillation under certain load conditions if you don't put some capacitance on the input Pin 3 when you have series resistance in this path, like you have. Put a scope on Pin 3 and monitor under various loads and you might see the oscillation. Putting a low value cap 0.1uF to 1uF directly on Pin 3 ->GND normally prevents this.

24v low voltage drop regulator

You understood the datasheet note backwards - sorry that I didn't clarify it.

You should reduce the resistance from pin 2 to the resistor, not
increase it. The 240 ohm resistor should be connected directly to the
case of the LM317, to eliminate any voltage drops caused by current
through PCB tracks.

/Rambo

24v/4a power regulator ic

E-design said:

Why the 0.22R at the output?
Also I have seen these type of regulators go into some sort of burst-HF-oscillation under certain load conditions if you don't put some capacitance on the input Pin 3 when you have series resistance in this path, like you have. Put a scope on Pin 3 and monitor under various loads and you might see the oscillation. Putting a low value cap 0.1uF to 1uF directly on Pin 3 ->GND normally prevents this.

Click to expand...

exactly, due to my misunderstanding of the LM317T note as quoted by NOX, I inserted this resistor. But I have corrected it.

Yeah, I'll put a capacitor of about 0.1 uF to ground all high-frequency component!

By the way, I do not understand the term "burst-HF oscillation". What is it?

lm338 24v

papyaki said:

Hi,

Max current flowing through R1 = Vbe/R1 = 0.7/22 = 31.8mA. Hfe of 2N2955 is 70max @ Ic=4A so Ib (base current) must be equal to Ic/hfe = 4/70 = 57mA. Ib flows through R1. I think you need to reduce R1 value to 10ohms, allowing 2n2955 and LM317 to both work properly.

Click to expand...

It's only for Ic=4A. But when my load absorbs only 2A, what will be????
I have done some measurement.
For example:

- Vin = 23.97V
- Load = 24 Ohm (I load = 1A)
- Ib of 2N2955 is 7.8mA (Hfe is aproximately 133)
- The output voltage is quite good, at 23.86V

But when I decrease the load resistor, the output voltage also decreases. And when I (Load) = 2.5A, LM317 produce only 22V instead of 24V ± 0.5V

lm138 dropout voltage definition

Burst meaning short random "sections" of High Frequency oscillation, not always present. More likely to show when you rapidly switch between low and high current conditions (Transient conditions).

I just thought of something that might be your problem. If you draw enough current in the GND or -ve line you will have a small voltage drop over the section between the bottom of the pot and the GND part to the load. This will make the regulator think that the voltage is too high and hence reducing the output voltage. Use "star" type GND connection by connecting the bot end of the pot with a separate wire to the GND side of the load. If you don't understand what I mean, I will draw a diagram Ideally the feedback resitor (220R/120R) should also sense as close to the load +ve to compensate for any voltage drops in that line

2n3055 pinout

Eh...

E-design - connecting the feedback resistor to the load is exactly what the datasheet
says not to do - please check the quote I hade in a previous post.
Otherwise the load current will generate a load-dependent voltage drop that
degrades the regulation.

Semiconductor - you say you have Vin = 23.97V and Vout = 23.86V.
In the datasheet they specify that Vin-Vout = 5V, i.e. you must have a 5V higher
input voltage than output voltage. All linear regulators "use" some voltage, the
"dropout voltage".

Generally I would recommend anyone to use the typical applications that are
given in datasheets, no more - no less. For starters al least.
On page 15 of this datasheet there is a high-current example. Why not try it?
**broken link removed**

Also be sure to use a proper heatsink, otherwise the thermal overload protection
will kick in and reduce your output voltage.

Good luck!
/Rambo

24v regulator 4a

XNOX_Rambo you are correct, I was thinking in general terms about remote sensing to eliminate drops and forgot about this problem. The ground point is still valid. You will need an additional opamp (as one application in the datasheet ) to sense the voltage drop in the wiring to the load if you can't connect directly to the tab which in most practical applications will be hard.

On the other hand when used with the current bypass, if it is possible to minimize the load current through the 317 (by using it as a reference only) and have most of the current pass through the external transistor the error term Rs X IL (from connecting the feedback resistor to the load side) may be much less than the voltage drop error at full current over the wiring. Just a thought.

2n2955 psu

E-design said:

Change the 220R to 120R. The 220R is only valid for premium devices and not commercial ones. Look carefully at the datasheet and you will notice that the commercial device needs about twice the current into the ADJ terminal and hence 1/2 the resistance as generally quoted in the datasheets. With a 220R in place regulation will suffer.

Click to expand...

I have done this but no improve can be seen. I reduce from 220R down to 100R and set the VR so that the output voltage is 24V. When I set the load to 12 Ohm (2A), the output falls to 23.8 and this result is quite good but if the load is 8Ohm (3A) the output voltage falls to 23V. I think that this result is so bad. So I do not try with 4A load.

How can I do now??? to improve my PSU (for 4A, the output falls at maximum of 23.5V)

2n2955 regulator shematic

XNOX_Rambo said:

Eh...
Semiconductor - you say you have Vin = 23.97V and Vout = 23.86V.
In the datasheet they specify that Vin-Vout = 5V, i.e. you must have a 5V higher
input voltage than output voltage. All linear regulators "use" some voltage, the
"dropout voltage".
Good luck!
/Rambo

Click to expand...

Oh, It's my mistake, exactly, Vout (without load) is 23.97V, not Vin!

In my opinion you have three choices:

1/ Use as is and live with the voltage drop with increased current. The configuration at present can't compensate for any voltage drop in the wiring after the connection point of the feedback resistor, unless you connect the tab of the 317 (which is also the sensing point connection for the feedback R according to datasheet recommendation) directly to the load connection. This might not be practical depending on your layout.

2/ Reduce the current through the 317 as in my previous post by making the external bypass transistors do 99.9% of the work by having them starting to turn on at very low load current ( you may need a darlington device here to provide the extra gain). By then connecting the feedback resistor to the load side (like the datasheet tells you NOT to do) you will be able to compensate for voltage drops in the wiring at high currents. By keeping the current through the 317 itself very low the load dependant reference error (Rs X IL) due to the resistance from the 317 wiring to the load may be insignificantly small (verify by testing) compared to the voltage drop you are having at the moment. It is very important that the bypass connection should feed in at a point, on or after the feedback connection point and NOT before. Also make sure that the GND reference is at the load itself, or as close as possible as in the diagram.

3/ Change your configuration like one example in the datasheet that includes an opamp to overcome the voltage drop problem by sensing the voltage at the load.