3.7 V - 9V Booster Circuit

Hello everyone

I'm a bit stuck. I've just been browsing RS Components' website for boost converters over the past few days to try and find the correct one that will suite my needs.

I've had a look at the TPS61175 from Texas Instruments which states that it has got a 3A, 40V internal switch. When I look at the max ratings then it shows that the switch can handle 3.8A Nominal and 5A max.

Since I need my output to be at 9V and to be 2A or 2.5A I've calculated accordingly and found that I should use a 0.47 uH inductor to be able to get 2.6954A out at 9V but further down in the data sheet they say that you shouldn't use an inductor with a range below 4.7 uH. So it means that this device wont support the output current that is required.

I've browsed now to try and find boost converters that will be able to deliver a 2A output but with a 3.7V input it doesn't seem that I'm having any joy.

I'm trying to boost a 3.7 V li-ion battery voltage to 9V to drive a solenoid, the solenoid is the BERMAD Model S-402-3W-BB, it is a 9 - 40 V dc latching solenoid, the coil resistance is 6 ohm meaning if I supply it with 9V then the solenoid will draw 1.5A for a period of 20 - 100ms. I must also use a 4700 uF cap to handle the current spike according to the solenoid datasheet.

Any suggestions from anyone as to how I can accomplish a 2A / 2.5A output from a boost converter boosting from such a low voltage?

Use a boost converter or a 555 tripler to charge a supercap to perhaps 10 or 12V,size it to deliver the minimum voltage u require based on its discharge time constant (100ms of 2A @ 9V min). They can deliver very high currents happily. Use a freewheel diode for the current spike.

@Dirka

My first impression it that you may have made an error in the calculation. The TI/National application notes are usually pretty accurate. If you will share the exact numbers you used, we can recheck them for you.

Another approach is to use Webench. Here are links to the starting page and a worked example. I entered Vin = 3.7 to 4.2V, Vout = 9V, and I = 2A.

Home page:
https://www.ti.com/ww/en/simple_swi...er&DCMP=A_SVA_General&CMP=KNC-GoogleTI&247SEM

Example:
https://webench.ti.com/webench5/pow...=9&O1I=2.0&op_TA=30&simpleswitcherpreferred=Y

Finally, one of the icons of switcher power supplies was Jim Williams. He also had a sense of humor. Here is an often referenced application note from him. Be sure to check page 22 on inductor selection:

https://www.electro-tech-online.com/custompdfs/2013/01/an35f.pdf

That original article has been updated here, but some of the earlier originality has been lost:

https://www.electro-tech-online.com/custompdfs/2013/01/an73f.pdf

In summary, inductor selection is not that critical. I would suggest getting a couple in the range recommended by TI and trying them. The inductor you use must be able to handle the current without saturation.

John

0.47 uH sounds way too small for a boost switcher. Something is not right.

10uH for example gives 0.5 amp delta I for the inductor current.
20uH then would be around 0.25 amp delta I.

@ Mosaic

My knowledge about supercaps / sizing caps / caps in general is a bit vague so please bear with me.

If I need to charge it to to say 12V then I need to apply 12V to it so that it will be charged to that voltage correct? Then according to what your saying is that I must size it so that over a period of 100ms the capacitor will only discharge to 9V minimum when supplying 2A. I've got no idea how to size a capacitor is there a specific formula that relates to input/charging voltage, current spike and time or something like that so that I can calculate the value that is required?

And the freewheel diode is to block the current spike when the charge is released from the solenoid?

@ John

Thanks for the links John, I had a look and configured the calculator to give what I need, I will have a look through some of the other available boosters.

According to my calculations for the TPS61175 there is not much that I can change except the inductor, unless I'm mistaking Iout(crit) with Iout going to my load. I've changed the calculations to yield an inductor value by inserting a wanted current into Iout(crit), have a look though my calculations, it's not much but as I said, I'm a bit stuck. Once again thank you for the info.

Dirk

You want continuous mode. See page 12 and the equations that follow. The very first sentence:

Unless otherwise stated, the design equations that follow assume that the converter is running in continuous mode.

Click to expand...

And as confirmed by MrAI, you want to define the ripple that is acceptable to you. His calculations are much more in accord with TI's suggestions and the Williams' ap note.

John

Hello everyone

Thank you all for your responses, obviously I'm doing something wrong, I'm going to struggle to find out where I'm going wrong with my calculations, will adhere to the rules for design when running the converter in continuous mode.

I've just registered to be able to utilize webench and wow what a tool. I will read through the document of Jim Williams a bit later, I did however scan through it and I must agree, everyone is always in a rush to get a prototype running asap , way of the world I suppose. Thank you once again.

Will revert back to you guys when I've found my error.

Dirk

The incomparable Jim Williams. A genius with the gift of explaining difficult subject matters in a simple, humorous way.
His passing away was a great loss to the engineering community.