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Anik , 01-04-2024, 06:52 PM

This for change the line between USB and lipo 2s battery. The output its an input for 3.3V regulator of TPS63060DSCR . Will this circuit work properly??
Thank you.

QDrives , 01-05-2024, 12:23 AM

1) You do not need an opto-coupler (expensive), but can use a NPN+PNP too.
2) Is the 10k low enough to switch the FET off fast enough?
3) You can change the design for a lower current drawn from the battery.

Anik , 01-05-2024, 01:46 PM

The circuit look like thi is if i use optocoupler . And I can try to use transistor for the second version. I can make another circuit if optocoupler have any problem except the price.

Robert Feranec , 01-05-2024, 02:29 PM

maybe this can help you:

Robert Feranec , 01-05-2024, 02:29 PM

it is from our imx6 project here https://www.imx6rex.com/imx6-tiny-rex/baseboard-lite/ the full schematic is here https://www.imx6rex.com/wp-content/uploads/2016/04/iMX6-TinyRex-Baseboard-Lite-V1I1-Schematic.pdf

Anik , 01-05-2024, 04:32 PM

I have been checked this file clearly, I am unclear about the USB ID and also sir its little complex for my simple PCB.

QDrives , 01-05-2024, 08:44 PM

That is to switch the power on/off. It is not a switch between two sources as @Anik wants.
The schematic he shows will work. The FET is off in less than 20us.
And there is a voltage drop over D16 of course.

Anik , 01-05-2024, 08:51 PM

Thank you sir,
its maybe 0.3V will be drop on D16 but can I reduce it??

QDrives , 01-05-2024, 10:16 PM

You can by adding a mosfet in parallel to the diode. It will make your design a lot more complex as you need to disable Q3 before you enable the 'USB' mosfet. All the while keeping your battery drain low....

Anik , 01-06-2024, 02:02 AM

I understand,sir
thank you

Robert Feranec , 01-06-2024, 06:00 AM

is it ok if VUSB flows to VIN?

Anik , 01-06-2024, 11:32 AM

The battery is 7.4V but the USB is 5V and can I plug in the USB when the battery is supplied to VIN??

Anik , 01-06-2024, 01:31 PM

Do I use one more diode along the VIN??

Anik , 01-06-2024, 03:52 PM

Will this work?? Here USB is 5V and battery is 7.4V so it can burn the USB device means computer.
Please suggest me a circuit type and also is this ok?? little voltage drop across D19 is ok I think cause its for 3.3V regulator .

QDrives , 01-06-2024, 08:54 PM

Vusb can flow into the battery if it drained below 5V. As there is not cut-off in (under) voltage. You could add a zener, and transistor for R92 to Gnd.
Why not simulate this?

Anik , 01-06-2024, 09:01 PM

The lipo 2s has its own BMS so it will cut off around 6.4 or 6.5V. And with the diode D19 I have simulated on thinkercad its working .

Robert Feranec , 01-07-2024, 02:50 PM

diodes in power are not ideal for battery powered devices - unnecessary loss, but it is a simple solution

Anik , 01-07-2024, 03:36 PM

How can I solve this sir, your given circuit is little complex for me...I am unable to match with my need.

Anik , 01-07-2024, 06:04 PM

Hello sir, I need one more help about latcing switch. I found this on online but its not working. I need to make a latching switch like mobile phone power button.

Anik , 01-07-2024, 06:16 PM

I found this. I am thinking will it work like phone power button means I will press for 2 second then it will on and if I press for 2 second again it will off

QDrives , 01-07-2024, 07:13 PM

If everything would be easy, we would not have work.
Replace the diode with a P channel mosfet ('diode' in same direction) and the gates together, you are done (in a few us).
A phone switches off by software. Are you able to drive signal to do that?

Anik , 01-07-2024, 07:23 PM

I understand about the changover circuit.
No sir I dont have the access to drive the signal. But can I not use a zener diode and capacitor to make some delay to turn on?? If I do like this will it delay.
here my goal it to discharge the capacitor slowly until the IN pin LOW the switch wont turn on.

Robert Feranec , 01-08-2024, 06:42 AM

@Anik I would recommend you to build these simple circuits by yourself and play with them. You may be surprised how these simple circuits behave in real life - you may find out it is not what you expected. That is much better way to learn than asking questions - when you ask questions, you will not really learn what is happening in your circuits. We do it exactly that way, even the circuit I posted above, we built multiple versions of them because the first version didn't work as we expected.

Robert Feranec , 01-08-2024, 06:43 AM

Anik , 01-08-2024, 07:01 AM

I have some components in my workshop, I do always when I have them. In our country maximum of the components are unavailable. If I try to order from aboard supplier like mouser, digikey it will charge me $80 per ship then with all Customs charge around $100 its depends on my components quantity. All around cost if I buy $100 of components I need to expense total $280 Its very high for me each time. Also I have financial problem to do invest more and buy more components at a time.
I will arrange everything. But I need time.

QDrives , 01-08-2024, 10:12 PM

Then do simulations. Use correct models of the components. Add parasitics, and check that it works.
However, there are still plenty of times that simulation alone on going to help you, as Robert already mentioned.

Anik , 08-21-2024, 08:01 PM

I have an LED which consume 3.3V max and 3A. My power supply is 9V. I have calculated the series resistor but the value is 2ohm and 18 watts. If I gor implement this, the size is going too large. So that I have added a 5A 3.3V LDO for implement this in a small area. Will this work?
I dont have of these components in my hand. So that I can not test.

Mini , 08-21-2024, 08:20 PM

Just curious. What kind of LED it is? 3A is a lot of current. Is it some kind of LED module where lots of LED in paralleel?

Mini , 08-21-2024, 08:22 PM

But to answer to your question. Your calculations are kind of correct. (9V-3.3V)*3A= 17.1W. Meaning your series resistor should dissipate about 17W.

Mini , 08-21-2024, 08:32 PM

But since your load is LED you need something to keep it's current under control. You need constant current. I would search for LED drivers.

Mini , 08-21-2024, 08:34 PM

I would also mention that even if you would use this then your linear regulator has to still dissipate 17W. And you need to give heat somehow away as well. TO-252 gives heat to PCB. But how you get that 17W away? Also your regulator would like 130 degrees instantly and probably more since there is also thermal resistance to the board. You would need big heatsink and perhaps some fan for active cooling. Anyway this way it would not work anyway.

Mini , 08-21-2024, 08:35 PM

Also it would be really inefficient. I would use step down converter. But since your load is LED you can't simply connect it directly to LED. You need constant current driver.

Anik , 08-22-2024, 02:59 AM

https://www.digikey.com/en/products/detail/creeled-inc/XMLBWT-00-0000-0000U2051/3778206

This is the LED, regular rating is 2.72V and 700mA.
But for max, 3.2V@3A

Mini , 08-22-2024, 11:27 AM

It's about LUX. More current more light basically. So depends on how bright you want to drive it. 3A is a lot of current and also you realize this LED needs heatsink. I would look for LED driver I think. Maybe someone else got better idea.

Mini , 08-22-2024, 11:31 AM

I googled 3A led driver. Check these out. I haven't opened datasheet or anything, but search something like this:

Mini , 08-22-2024, 11:31 AM

https://www.st.com/en/power-management/aled6000.html

Mini , 08-22-2024, 11:32 AM

https://www.monolithicpower.com/en/mp2480.html?srsltid=AfmBOoormmIvUTqguYdg1mBdnCttp5kldPaSRjab9Iy4GeT_Ky0R6mci

Mini , 08-22-2024, 11:33 AM

Or maybe look even higher current LED driver. But then you gotta be sure not to overdrive your LED.

QDrives , 08-22-2024, 08:00 PM

Use a LED driver.
Pick one of the over 6000 here: https://www.digikey.nl/en/products/filter/power-management-pmic/led-drivers/745
Most LED drivers are like switch mode regulators and require an inductor, no so much a capacitor.
There are one that buck, boost and buck-boost.

Mini , 08-22-2024, 08:13 PM

Buck type led driver is needed here. Don't pick linear one.