Hello everyone,

I am building a power supply with 24v 5A transformer to compensate the voltages dropped by diodes and other components. I have posted a design earlier however it had too many conflicts. I redesigned it again without current limiting, but I'd like to add an adjustable current regulator. I am not sure of where to put it! The design is made to decrease the ripples [http://imgur.com/QmGobR8]. Please do criticize, and tell me what can I make to make it better or fix it.



Notes:

-I only found a 24v 5A transformer in our country (others are 12v 1A max)... I wanted a power supply that runs max V: 14-15v and a MAX current of 3A (adjustable)

- Circuit done in KiCad

- I know that the lm338 dissipates alot of heat so i will use a radio heatsink laying around with a fan [https://farm6.staticflickr.com/5641/22741611580_ed371c13ca_b.jpg]

A couple of questions:

How can i determine the best values for caps?
How can i improve the circuit?
Where can i find Vripple in datasheets?
Is it possible to add an adjustable current regulator/limiter?

Sincerely,

teklucifer

 

Capacitor size:
√2 C Er(p-p) F = I
You need to reduce R to 120 ohms to comply with the minimum load on the chip.
You can add a current limiter with a few more components. People usually try to put a resistor after the chip to run the current limiter, but the chip only uses about 80 microamps so you can put the current sensing resistor before the chip if that's close enough to suit you.
There are many ways to do this. I prefer the LM723 chip because it has, "foldback" current limiting. It just about shuts down when an overload occurs. Other people want to drive a constant current, adjustable, and the 723 chip is not the best for that job.

 

• think about wiring the pot properly - what if the wiper goes open?
• read the datasheet about the size of the capacitor on the output.
• a crude current limiter can be built into the circuit by placing a power resistor between the regulaotr's output pin and that 270ohm resistor.
• do number your parts.

 

think about wiring the pot properly - what if the wiper goes open?

dannyf has a good point. It is always good practice to wire the wiper to one end of the pot if it is used in a variable resistance application.

 

You need a transistor T1 in the negative line, with a low value series resistor R1, R4 is the base drive limiter, when 0.7V appears across R1, transistor T1 turns on and pulls the output voltage down, altering R1 controls the current limit.

circuit

 

Over what current range do you want to adjust?
That will determine the circuit required.
The wider the range, the more complex the circuit.

 

@crutschow I was looking for an output from 0 to 3A max i couldn't find any adjustable current circuit for the regulator. I was thhinking to add another lm338 regulator with a variable resistor before the voltage regulator circuit.

Thanks guys for the quick replies I'll redesign the circuit again with labels. I havent umderstood what you meant by "wiper". Also, is there another regulator rated for max of 3A?

 

I havent umderstood what you meant by "wiper". Also, is there another regulator rated for max of 3A?

LM350
The wiper of a pot is the middle terminal.

 

Look at the data sheet for the LM1084 adjustable in a TO-220 package.
Essentially an updated version of the old 317.

 

A reverse biased diode between pins 2 and 3 of the LM338 if this is a bench supply.

 

I have re-designed the circuit again with an LM1084 TO-202, with fixes as well as minor changes. I have added a transistor current-limiting circuit rather than another voltage regulator. I also added extra capacitors to make sure the ripple get as low as possible. P.S: I hope there is minimal Inrush current that can harm the circuit. I need your approval stamp so that I can start finding values for caps and resistors.

Circuit:

A few questions again:

-I still haven't understood the Caps calculation if possible please re-explain

- How can I find the resistor values for the current limiting circuit?
- What wattage should i choose for the resistors?
- Is the current limiting circuit in the schematic better than using a regulator circuit?

Please do criticize as well as tell me how to improve it better!


Best regards,

Teklucifer


P.S: I may not be able to reply fast so I apologize in advance

 

The minimum current limit value is ≈ 0.7V / R2.
Make sure you consider the power dissipation (I²R) in R2 at the highest current level.
The other resistors can be ¼ W types.

 

Q1,Q3,D1,,Rv1 are useless, put the current limit in the negative line with Q2, like my previous post#5, or use an op amp.

 

Here's the LTspice simulation of a current limit circuit using a comparator that can be adjusted down to near zero amps, as determined by the minimum output voltage of 1.25V.
You can substitute the LM1084 for the LT1084.

 

Here is a combination of several online schematics. Note that this is a concept schematic, and some of the component values might need adjustment for a specific application. For example, the schematic shows an LM317, which cannot handle 3 A, and the shunt resistor says 1 W because that's what's in my design library. The real shunt resistor would have to be 5 W min.. The current shunt is set for 2 A minimum.

Placing a pot in parallel with the shunt resister gives this circuit some adjustability, but it does not have a very wide range because the current sense amplifier Q1 has a relatively high threshold of 0.6 V and no gain. Still, if you need only a limited adjustment range it is a fairly simple circuit.

ak

 

Hello,

@AnalogKid , the current limitation in your circuit will be 2.3 A or higher.
The resistor will dissipate about 1.6 watts at the 2.3 A.
The LM317 will most likely limit the current sooner.

Bertus

 

I have re-designed the circuit again with an LM1084 TO-202, with fixes as well as minor changes.

@crutschowI havent umderstood what you meant by "wiper".

You still do not have the wiper of the pot, RV2, connected correctly. The center terminal should be connected as shown in post #14 by crutschow.

 

the current limitation in your circuit will be 2.3 A or higher.

Actually, a 4401 starts conducting at around Vbe = 0.5 V, so the minimum limited current will be less than 2 A.

The resistor will dissipate about 1.6 watts at the 2.3 A.
The LM317 will most likely limit the current sooner.

Yes, I pointed out those limitations in the text.

Another issue with the circuit is that the current limit set point will drift as Q1 warms up. Overall it is not a perfect circuit, but does add some adjustable limiting with very few components.

ak

 

First of all, thank you for your replies,

I am not familiar with op-amps and how they work @crutschow , @Dodgydave. I have been searching all the time for a current limiter using an op-amp. Moreover, I have checked the data sheet for the LM393 and it should me a small input current; How can it handle the 5A input. I may add @crutschow 's current limiter design to the circuit I have built if it's going to adjust it to 3A without many conflicts.

@AnalogKid How can i determine the resistor values as well as the caps P.S: I used a calculator to find the R3 as well as the RV2. Another thing is what you meant by

that the current limit set point will drift as Q1 warms up.

and how can i make it almost perfect [using LM1084].

Thank you all again,

Teklucifer

 

First of all, thank you for your replies,

I am not familiar with op-amps and how they work @crutschow , @Dodgydave. I have been searching all the time for a current limiter using an op-amp. Moreover, I have checked the data sheet for the LM393 and it should me a small input current; How can it handle the 5A input. I may add @crutschow 's current limiter design to the circuit I have built if it's going to adjust it to 3A without many conflicts.

@AnalogKid How can i determine the resistor values as well as the caps P.S: I used a calculator to find the R3 as well as the RV2. Another thing is what you meant by


and how can i make it almost perfect [using LM1084].

Thank you all again,

Teklucifer

Just a couple of questions:
The voltage on the regulators will not go to zero volts - only to 1.2 volts. Is that okay?
The same is true of the current limit. It will only reduce the voltage to 1.2 volts so if the output is shorted the current will still be pretty high. Is this okay?
Your voltage is to high for the LM1084. When you add filter capacitors to the bridge rectifier the voltage will be somewhere around 33 volts. The maximum voltage from input to output is 29 volts.
So stick with the 338 or similar.