LMR62421: How to get 2.1 amp maximum output current of the IC

Hi Zuhaib,

The 2.1 AMPS is the switch current limit, also the maximum peak inductor current ILpk. You may refer to the formula (6) and (7) in the datasheet for the relations between ILpk and input/output current.

As your second question, suppose no limitation on the Li-on battery, the maximum current is around 1A with a minimum 2.2uH inductor.

Regards

Lei

I have used 1uh 4 ampere inductor and 40v 3 ampere diode and I am getting 1 ampere 5v output do i still need to change the inductor to 2.2uh?

Hi Zuhaib,

I just calculated the minimum inductance based on a normal rule of 40% inductor current ripple. It's 1.4uH, so I recommended 2.2uH. If bigger inductor ripple of 60% can be accepted, you can keep the 1uH inductor.

Regards

Lei

okay thank you so much.

Actually I wanted 5v-2amp continuos  that is why I have selected this IC can you please suggest another part which can boost  battery li-ion 3.7v -5000MAH.

Hello can you please suggest some other part for 3.7v-4.2 volt input and 5v -2amp output.

Hi Zuhaib,

I suggest the device of TPS61033. Although it's in pre-release now, it will be formally released very soon in one month.

Regards

Lei 

Hello can you please also let me know for 5V 1 ampere output what will be the input current at 3V .? for LMR62421

Hi Zuhaib,

With the conditions you provided(1uH inductor), the average input current is around 2A, and the peak switching current is around 2.57A, which exceeds the guaranteed min switching current already.

Regards

Lei 

As you said It exceed guranteed minimum switching current that indicates I am using this IC in wrong way so please tell me the changes I have to make to smoothly run 5V 1 ampere at 3 V input as the lowest of lithium battery and goes upto 4.2 V

Also please let me know 5V *1amp=5 watt (output power) 

So how come input current is 2 amps at 3V it will become 6 watt 

and we know output must be equal to input 

so how you have calculated the input current

thanks and regards

ILpk = IIN + ΔIL

or

ILpk = IOUT / D' + ΔIL

please explain this formula as well 

Hi Zuhaib,

If the required peak current is out of the spec of the selected device, then you will have to select other devices with higher switching current.  

Following is a online tool, which will provide options based on your inputs of Vin, Vout and Iout.

https://webench.ti.com/power-designer/switching-regulator

If the efficiency of a boost device is 100%, then you are right output power = input power. Well, all the actual IC devices will have power loss. In the above calculation, I used a rough efficiency of 83%. That means, Input power * efficiency = output power.

In the boost application, the inductor current is not constant. This current increases during low-side MOS on and decreases during off in each cycle. So the inductor peak current = average input current(also equals to average inductor current) + ΔIL(half peak-to-peak ripple).

For boost, if ignore the efficiency,  VOUT = VIN/ D'. From VIN*IIN = VOUT*IOUT, you can get IIN = IOUT / D'.

Regards

Lei 

Can you please let me know what watt IC is lmr62421? is it a 5watt solution ?

Hi Zuhaib,

The maximum output power of LMR62421 can be larger than 5watt, but it doesn't mean that it can cover all the 5watt application. It heavily depends on the Vin, Vout and Iout.

In you application of 3V to 5V with 1A output, will TPS61022 or TPS61033 work for you?

Regards

Lei

Yes tps61022 and tps61033 are perfect but they are not available right now how will I test 

Hello what about TPS61288 at 3V input and 5v output how much current can U take from this device .

please let me know

Hi Zuhaib,

The max output will be larger than 5A. Here is the calculate method for your reference.

Will you consider TPS61033? It should be a better choice.

Regards

Lei