adjustable current limit control

Thread Starter

mrh586

Joined Dec 5, 2021
90
Hello friends
How can I add an adjustable current limit control to the circuit below? Please explain with schematic.

The output value doesn't matter to me; it is important for me to learn how to change and control the current with this UC3842AD8 IC.

The input current is 15 amps and the output is suitable for me from 400 mA to 15
 

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crutschow

Joined Mar 14, 2008
31,123
One way:
You add a small shunt resistor in series with the output and measure the voltage across that to determine the output current.
(There are high-side current-measuring ICs that can do that.)
Then you compare that output with an adjustable voltage reference for the desired current limit to generate a limit signal.
That limit signal is then summed with the voltage feedback signal into the VFB input to reduce the output voltage and thus the current.
 

Thread Starter

mrh586

Joined Dec 5, 2021
90
One way:
You add a small shunt resistor in series with the output and measure the voltage across that to determine the output current.
(There are high-side current-measuring ICs that can do that.)
Then you compare that output with an adjustable voltage reference for the desired current limit to generate a limit signal.
That limit signal is then summed with the voltage feedback signal into the VFB input to reduce the output voltage and thus the current.
Thank you
I am a beginner
Could you please explain these contents with schematics?
I saw somewhere that pin 3 of the IC was connected to pin 8 with a potentiometer
But this method was not accurate
 

crutschow

Joined Mar 14, 2008
31,123
Another simpler way is to amplify the voltage from the peak current-sense resistor R7 to the current-limit input R7, but that will require a high-speed, variable gain amplifier as determined by the switching frequency of the converter.

Sorry, but I'm not able to generate a schematic for your circuit at this time..
 

Thread Starter

mrh586

Joined Dec 5, 2021
90
Sorry, but I'm not able to generate a schematic for your circuit at this time..
my God
It was a pity
Thanks for your help, but I'm a beginner and need to be careful
Otherwise, the circuit may be damaged
Please post the schematic whenever you can
 

Dodgydave

Joined Jun 22, 2012
10,591
R7 is the current sense resistor, increasing this will lower the output current draw, .
The Voltage feedback is wrong as it should have an optocoupler.
 

crutschow

Joined Mar 14, 2008
31,123
Please post the schematic whenever you can
I generally don't post schematics of circuits I can't verify with simulation, and I don't have a simulation model for your converter.
And since you are beginner I can't use you to test the circuit and make any tweaks or modifications that may be required to get it to operate properly, since switching converters are tricky beasts even for experienced designers, and not something that a beginner normally builds.
 

Thread Starter

mrh586

Joined Dec 5, 2021
90
R7 is the current sense resistor, increasing this will lower the output current draw, .
Grateful
According to my studies, this is not a standard task, because this resistor is actually a filter
And in reality, this can only be effective to a very imprecise extent (tested).
The Voltage feedback is wrong as it should have an optocoupler.
Optocoupler is usually used for non-isolated circuits, this IC is accurate and advanced enough
 

Thread Starter

mrh586

Joined Dec 5, 2021
90
I generally don't post schematics of circuits I can't verify with simulation, and I don't have a simulation model for your converter.
And since you are beginner I can't use you to test the circuit and make any tweaks or modifications that may be required to get it to operate properly, since switching converters are tricky beasts even for experienced designers, and not something that a beginner normally builds.
Your opinion is respected and shows your humanity.
But really, I experiment for fun and experience and learning
And I think what makes me stronger are complications and problems
Thankful
 

Ian0

Joined Aug 7, 2020
6,707
You could add a resistor from the junction of R5 and C9 to the 5V reference.
That increase the voltage at the current sense terminal and makes the IC limit the MOSFET at a lower value.
 

Thread Starter

mrh586

Joined Dec 5, 2021
90
You could add a resistor from the junction of R5 and C9 to the 5V reference.
That increase the voltage at the current sense terminal and makes the IC limit the MOSFET at a lower value.
My experience showed that pin 3 of this IC has mostly 0 and 1 states and tampering with it makes the output of the circuit unstable. Probably changing the amp should be through feedback or an external factor.
 

Ian0

Joined Aug 7, 2020
6,707
My experience showed that pin 3 of this IC has mostly 0 and 1 states and tampering with it makes the output of the circuit unstable. Probably changing the amp should be through feedback or an external factor.
pin 3 is the most important feedback in the circuit, and it is not a digital input. It is the feedback of the primary current sense. The waveform on it should be a sawtooth. If it isn’t there’s something wrong.
 

Thread Starter

mrh586

Joined Dec 5, 2021
90
pin 3 is the most important feedback in the circuit, and it is not a digital input. It is the feedback of the primary current sense. The waveform on it should be a sawtooth. If it isn’t there’s something wrong.
No, I didn't mean that pin 3 was digital, I meant that trying to control the current through pin 3 would make the output unstable.
Its performance is almost similar to digital (not exactly).
Its waveform is sawtooth, but the changes in it will mess up the waveform
 

Ian0

Joined Aug 7, 2020
6,707
No, I didn't mean that pin 3 was digital, I meant that trying to control the current through pin 3 would make the output unstable.
Its performance is almost similar to digital (not exactly).
Its waveform is sawtooth, but the changes in it will mess up the waveform
On the contrary - a standard technique to reduce the voltage across the sense resistor.
 

Dodgydave

Joined Jun 22, 2012
10,591
Here is a simple method to limit the output current by putting a series resistor in the output side, as the voltage gets to 0.7 V across the resistor the transistor turns on and pulls the voltage sense pin up thus lowering the output voltage. Current out is selected by choosing the resistor size . I = 0.7/ R

IMG_20221202_133412.jpg
 
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