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# Over Current Protection

I want Design a circuit for over current and short circuit protection,and also want

input voltage ~= output voltage

i have 24V/10A input power supply , i want its divided in ten parts as figure.

Kindly suggest me what i do for it.

• Hello,

I would recommend 2 x TPS4H160 and one TPS2H160. The on resistance upper limit is 280mOhm at 150C junction temperature. That makes the max voltage drop of 280mV between the 24V supply and the load. The device max power dissipation 4 x 0.28W = 1.12W. The junction temperature rise above ambient is 1.12W x 32.7C/W = 36.6C.

At 1 A load current the current limit accuracy is15% max and the device is fully protected and has full diagnostic features.

Regards

• Thanks sir

can you provide me circuit diagram ? plz.

• Hello,

Sure will do. I understood the design requires over current protection but I also need to know if the high side switch should communicate with MCU to report the faults on any single channel.

Regards

• Hello,

i'm not use any MCU or any External Power Supply.i want apply fixed Input Voltage (1~7)V   on Inputs Pins.

• 1A_10 Channels.pptxHello,

please check the attached circuit diagram and let me know for further assistance.

Regards

• over current protection possible in Version A.
• Hello,

Version A or Version B are both working is this application. They both have the same current protection function.

Regards
• thanks

how to calculate value of resistance? Kcl=?

• Hello,

Please refer to datasheet page 16. The formula relating external limiting resistor RcL, load current IOUT and current ratio KcL: RcL = VcL(th) x KcL/IOUT.

The highest variation is from current ratio KcL and KcL = KcL(typ) x (1 + dKcL/KcL(typ)). dKcL/KcL(typ) is the accuracy and it is +/-15% at 1A load current.

For making sure the load current does not hit the current limit during normal operation, RcL should be calculated with the lowest current ratio accuracy.

That means RcL = VcL(th) x KcL(typ) x (1 + dKcL/KcL(typ))/IOUT.

At 1A load current, RcL = 0.8V x 2500 x (1 - 0.15)/1A = 1700 Ohms

Regards

• hello

i'm trying this circuit but it's not working properly,kindly support me...

thanks

• Hello,

I do not see the actual circuit. In order to help we need schismatics at device level and details about your observation.

Regards

• Hello Verma,

You sent my drawing and I was hoping to receive your schematics and data from lab measurement. You have to check the following once you build the circuit:

1- Make sure the pins are connected correctly

2- After you power up the circuit, make sure the voltages are present

3- Make sure the input pins INx are high for enabling the channels

Regards

• Hello Sir,

I'm trying this circuit, there are some problems......

• Rcl Resister very heating, I'm currently use 0.5W.
• Output Voltage (24V) not same in each channels.
• During short circuit conditions my power supply also effective.

• Hello,

the circuit looks OK and should function correctly in term of power delivery and over current protection. With current limit resistor of 1.8K, the typical current limit is 0.8V x 2500/1.8K = 1.11A.

The short circuit current limit can be between 0.94A and 1.277A due to 15% tolerance.

The device max power dissipation is 4 x 200mOhm x 1.277^2 = 1.3W.

The current limit resistor max power dissipation is Vcl(H)^2/Rcl = 6.5V^2/1.8K = 23.5mW. There is no need for 0.5W power rating for limiting resistor.

If the limiting resistor is getting hot, can be from:

1- Lower resistor value

2- Higher voltage on CL pin and in this case the device may be damaged

Could you please double check the Rcl value and measure the voltage on CL pin? Is the power pad soldered to system ground?

Regards

• Hello sir

everything's OK in this circuits but currently i have only one problem during short circuit .........

when short circuit remove i want recover 24V/1A but circuit give me a output only 19V/250mA.
I want full recover without any replacement because when circuit in use no one present there for giving any external pulse.
i want solution for automatically recover my voltage/current
kindly provide me a circuit for this requirement

Thanks & regards

Sudesh verma
• Hello,

In your design the enable pin IN is not toggled. The over current trip is based on analog loop and pulls the FET gate low once the current hits the current limit and puts the FET in saturation region. When the over current is removed, the current loop suppose to release the FET gate and puts the FET in linear mode. Because it is an analog loop, the FET will go back to linear region only if the IN is toggled or the load current goes much lower value than the over current limit.

We recently changed the datasheet and added this sentence: "External current limit accuracy is only applicable to overload conditions greater than 1.5X the current limit setting"

The upper limit for this design should be 1.5 x 1A = 1.5A and with 15% accuracy the upper current limit is 1.15 x 1.5 = 1.725A. The external current limit resistor value should be 0.8V x 2500/1.725A = 1.159K.

Because the enable IN is not toggled, the solution is to lower the external current limiting resistor from 1.8K to 1.2K

Regards,

• hello Sir,

i'm trying many times but not got  a good response..so

Kindly provide me a circuit diagram,for full-fill my requirement....

• 1A over current Limmit
• Short circuit protection
• Auto recover power after overload or short circuit remove

• Current limit block diagram.pptxPlease use my previous explanation for this design and circuit diagram is attached.