How Current Limitation is happening in the circuit???

[xavier60]

That design has latching over current protection. The Cut switch triggers the protection circuit, disabling the PSU's output.
The Load switch resets the protection, enabling the output.

[ommsiva]

Sir,

how this circuit is going to work during short circuit condition?

[xavier60]

When the current gets too high, the voltage across the 0R22 triggers the SCR to turn on, applying voltage across the relay's coil.
The relay's NC contacts open, disconnecting the unregulated supply from the input to the regulation circuit.

[ommsiva]

Sir,

Whether this circuit is feasible for implementing short circuit protection?

if yes i have a transformer of 24-0-24v @ 10 amps, if i put rectifier and filter i would get 34V and nearly 10 Ampere, What should be the specification of the relay?

what is the specification of cut and load switch?

Thank you.

[xavier60]

Sir,

Whether this circuit is feasible for implementing short circuit protection?

if yes i have a transformer of 24-0-24v @ 10 amps, if i put rectifier and filter i would get 34V and nearly 10 Ampere, What should be the specification of the relay?

what is the specification of cut and load switch?

Thank you.

Because there is no instantaneous current limiting, short circuit current will be very high. How high? I dont know.
If the relay is too small, the contacts might weld. I'd use a 40A automotive relay.
Also because of the high instantaneous current, the output transistors might fail.

[ommsiva]

Sir,

What would be the best circuit to implement short circuit protection for LM317 or LM338?

[xavier60]

Sir,

What would be the best circuit to implement short circuit protection for LM317 or LM338?

I think something as simple as possible so it's easy to build and make work.
From a previous post with some corrections, https://www.eevblog.com/forum/beginners/how-current-limitation-is-happening-in-the-circuit/msg5375117/#msg5375117
The parts should be here soon so I can test it for myself.

[xavier60]

It definitely works but not in an altogether nice way.
Because the current limiting has control only of the power transistors but not the LM337, the LM337 goes to its full current when there is an overload. It can get hot enough to go into thermal limiting. Maybe a good thing in a way because it will reduce the dissipation at the power transistors.
It's also difficult to precisely set the current limit.
 The design is ok if brief overloads are expected.

[ommsiva]

Sir,

What is the function of Q1 transistor?

what should be the voltage across collector to emitter in normal operation and short circuit condition in Q1?

why there is 47 ohm resistance connected to the base of Q1?

Q2 and Q3 are current boost transistor

[xavier60]

Sir,

What is the function of Q1 transistor?

what should be the voltage across collector to emitter in normal operation and short circuit condition in Q1?

why there is 47 ohm resistance connected to the base of Q1?

Q2 and Q3 are current boost transistor

Q1 is for current limiting. R2 and R3 are for current sharing of the output transistors and also serve as Current Sensing. R6 and R7 combine the voltage from the CS resistors and apply it to the Base of Q1.
At 3.2A, the Vce of Q1 is 1.5V. It stays the same when the output is shorted because it was already limiting the current through the output transisitors, Q2 ans Q3.
 You don't need to use exactly the sane parts to test the design. If you have 0.47Ω resistors, use them in place of the 0.56Ω.

[ommsiva]

Sir,

I found this circuit on web. Whether this will perform as specified in the circuit.

please advice me.

[xavier60]

Sir,

I found this circuit on web. Whether this will perform as specified in the circuit.

please advice me.

I'm not certain. I'd need to build it but I don't have enough time.

[ommsiva]

ok sir

[ommsiva]

Sir,

Check this circuit and comment on the design.

Thank you.

[xavier60]

It will current limit so long as the output doesn't need to go below 1.25V such as when the output is short circuited.

[xavier60]

Correction. With a short on the output, the current can be no less than 5.7A. Does this suit your needs?

[ommsiva]

Sir,


At short circuit,  power dissipation = 5.7amps* 1.25V. And this will not be larger than 11 watts.


Which could be easily dissipated by MJ15004 with heatsink.

Why it is 5.7 amps at short at the output?

[xavier60]

You need to multiply the 5.7A by whatever the voltage is going to be across the power transistors.
Also, put  a resistor in series with the Base of the current limiting transistor to protect it, like 470Ω.

[xavier60]

The LM317 can regulate only down to 1.25V which will be applied across the 0.22Ω resistor when the output is short circuited.

[ommsiva]

Sir,

I found this circuit in web and calculated node voltage.

My doubt is

1)when i move the p1 pot up to top position my node voltage will be 0.79V , which is enough to turn on my internal current limiting transistor. Then why i need those parallel 0.22ohm (4 nos)?

2)which will be the negative terminal of power supply? either Right end or Left end(where filter capacitor were connected) of current sensing resistors(4x 0.22 ohm resistors). What is the technique employed?

3) what will happen if there is no 4x0.22 ohm resistor?


Thank you

[MrAl]

Sir,

I found this circuit in web and calculated node voltage.

My doubt is

1)when i move the p1 pot up to top position my node voltage will be 0.79V , which is enough to turn on my internal current limiting transistor. Then why i need those parallel 0.22ohm (4 nos)?

2)which will be the negative terminal of power supply? either Right end or Left end(where filter capacitor were connected) of current sensing resistors(4x 0.22 ohm resistors). What is the technique employed?

3) what will happen if there is no 4x0.22 ohm resistor?


Thank you

[1]  The four 0.22 Ohm resistors in parallel create a resistor of value 0.055 Ohms, which senses current.  You need that to sense current.
[2]  The output of the power supply is on the right (-) terminal.  The input is on the left.  The two can not be made into a single common.
[3]   With no 0.22 Ohm resistors there is no current sense, hence there will be no current limit at all.

You can analyze this by looking at the voltage divider created by the pot and the associated resistors, along with R5 the current sense resistors.
If you look at all the top side resistors in series and call them RA, and all the bottom side resistors and call them RB, then the voltage at the base of the lower NPN (voltage referenced to the output common terminal) would be:
Vbase=Vref*RB/(RA+RB)

but it is also biased by the voltage across R5 and the output current because the emitter is tied to the input common not the output common. This means the output current affects Vbe of the lower NPN as:
Vbe=Vref*RB/(RA+RB)+Io*R5

where Io is the output current.

Replacing RA and RB with the actual resistors we end up with:
Vbe=(R5*(Io*R6+Io*R2+Io*Rp)+Vref*(R6+Rd))/(R6+R2+Rp)

and here Rp is the total value of the pot (500 Ohms in the schematic) and Rd is the lower resistance part of the pot after adjustment.
With the values substituted in for the variables and setting Vbe=0.7 volts for an approximation, we get current limit as follows.
With the pot set all the way down, the current limit is around 11 amps.
With the pot set all the way up, the theoretical current limit is actually about -1 amps, which is probably not possible, so the output current would probably be zero at that point.  If we adjust it slightly down from top, we get a theoretical zero amps output.

Note Vbe may be slightly different than 0.7 volts, and changes with temperature (heating up of the IC chip and ambient temperature).

The main point though is that R5 is definitely needed to sense current or you can not get current limit.  If you short it out, no current limit, if you open it up, no output at all except maybe some leakage current.  Either of these extremes is not an option that will work.

[ommsiva]

Sir,

thank you , what i understand is

V_BE=V_B-V_E

V_B=V_ref * RB/(RA+RB)

V_E =I_oR₅

R_A =R2+Partial resistance from p1

R_B=R1+R3+R6+Partial resistance from p1

I cannot correleate and understand .these equations

Vbe=Vref*RB/(RA+RB)+Io*R5

Rectify me.

[Zero999]

Since this topic has drifted.

Please start from the beginning. What are you trying to do?

[MrAl]

Sir,

thank you , what i understand is

V_BE=V_B-V_E

V_B=V_ref * RB/(RA+RB)

V_E =I_oR₅

R_A =R2+Partial resistance from p1

R_B=R1+R3+R6+Partial resistance from p1

I cannot correleate and understand .these equations

Vbe=Vref*RB/(RA+RB)+Io*R5

Rectify me.

The total resistance of the pot P1 is Rp, and the resistance of the upper part of the pot is Ru, the lower part is Rd.

The voltage divider formula plus the voltage due to the output current times the sense resistor R5:
Vbe=(Vref*RB)/(RB+RA)+Io*R5

Since RA=R2+Ru we have:
Vbe=(Vref*RB)/(RB+R2+Ru)+Io*R5

and since RB=Rd+R6 we have:
Vbe=(Vref*(R6+Rd))/(R6+R2+Ru+Rd)+Io*R5

and since Ru=Rp-Rd we have:
Vbe=(Vref*(R6+Rd))/(R6+R2+Rp)+Io*R5

This last result is the same as that given previously just factored differently.
The formula shown on the updated schematic is the same also, also factored differently.

Note R1 and R3 are not included in the calculation because the additional current Ix through R5 is so small.
Ix=Vref/(R1+R3)
Vx=R5*Ix
If you like you can add that voltage to the calculation for Vbe also, although since Ix is so small (less than 700ua) the additional voltage will be very, very small (less than 40uv).
There is also a small contribution from P2 and R10 which will also be small.

The updated schematic is attached.

[ommsiva]

Sir,

1)Normally Base emitter voltage V_BE=V_B-V_E , But you have taken V_BE=V_B+V_E.
What is the reason for that?

2)What is the contribution of these terms interms of current limiting from these equation, Vbe=(Vref*(R6+Rd))/(R6+R2+Rp)-Io*R5 ?

for example
when load current is 10 amps,V_R5=10*0.055=0.55V, then what is the contribution of Vref*(R6+Rd))/(R6+R2+Rp) term?

                           the first term should be 1.25V (0.7V+0.55V) and second term should be 0.55V. is that correct?

Thank you