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I have 3.3v regulated coming out of pmic but one of my chips has a weird requirement of 3.6-4.2v in its vbat pins. Now I don't work with electronics at all except for stuff I did in school a long time ago. This datasheet says it can take my input voltage and convert it to between 3.6-4.2 depending on the resistors I choose. I remember reading somewhere that it you can only step down with an LDO, if that's true then I can't use this. Do I use one of those buck boost converters instead? I really don't know what I'm supposed to do since I can't test any of this and have 0 experience.

EDIT: Here are the datasheets of the chips I am interfacing.

Wifi: - on page 10 it shows the voltage specs and current consumption for the vbat pin? For current, on page 11 it says typical current consumption, does this mean vbat has to meet typical current consumption needs for it to work? I really don't know.

wifi datasheet

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  • \$\begingroup\$ What voltage do you have going into your pmic? It must be higher than 3.3v so can you use that for Vbatt? \$\endgroup\$ – Steve G Mar 30 '16 at 0:45
  • \$\begingroup\$ @SteveG the input for pmic is 5v from usb power, I need to convert 3.3 coming from a dcdc pin to 3.6-4.2v to power vbat for another chip \$\endgroup\$ – jack sexton Mar 30 '16 at 0:47
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    \$\begingroup\$ @jack or use a ldo from the 5V usb, avoiding the double efficiency loss of 5 -> 3.3 -> 3.8 \$\endgroup\$ – Passerby Mar 30 '16 at 2:05
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    \$\begingroup\$ @jacksexton, this is why it's helpful to give more details in your question up front. Since you have a 5V source readily available, it sounds like boosting the 3.3V line is overly complex. Just use another LDO that gets you between 3.6V-4.2V. \$\endgroup\$ – Dan Laks Mar 30 '16 at 2:43
  • \$\begingroup\$ What chip is it? Do you want battery backup on that chip? What's the operating voltage of that chip (I think it's 5V)? In a few cases, if you don't need battery backup, then connecting Vbat to GND may work. for ex - in case of DS1307 RTC chip. In such cases Vcc should be a bit higher than Vbat for normal operation and connecting Vbat to GND certainly ensures that. Again we won't know whether this is what you need because you haven't disclosed all necessary details. \$\endgroup\$ – Whiskeyjack Mar 30 '16 at 6:42
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No, a Low Drop Out Linear Regulator cannot be used to boost a voltage up. That specific one has a Drop out of 0.1 ~ 0.25 Volts, so the maximum V OUT is VIN - 0.25 Volts.

You need a Boost circuit. Depending on your specific current needs, there are different types of boost regulators that will work. A charge pump uses a few capacitors and will be the simplest for under 200 mA. Otherwise there are plenty of normal regulators that use an inductor and diode pair, with internal mosfet, that can be used.

TI and every other manufacturer have tools that basically design it for you.

The three other options, since you have a 5V power source in the first place, are:

  • Parallel the 5V to 3.3V PMIC with a 5V to 3.8V~4.0V LDO. Less complex, and no need to double up on 5V down to 3.3V, then up to 3.8V.

  • Change the PMIC to be 3.8V output, and use an LDO to get 3.3V from that.

  • See if you can just change everything to 3.6~3.8V. No more multiple regulators.

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  • \$\begingroup\$ If you don't mind can you give me a link to the TI tools, the datasheet for the chip says I can expect 350ma on typical usage. This is a wifi chip btw. \$\endgroup\$ – jack sexton Mar 30 '16 at 0:59
  • \$\begingroup\$ ti.com/lsds/ti/analog/webench/… the single power first option \$\endgroup\$ – Passerby Mar 30 '16 at 1:14
  • \$\begingroup\$ thanks for those suggestions, I'll try to look and see if my other chips can do option 3. I still have one doubt though. I've updated my question to include the chip I'm working with. On page 11 it has typical current consumption for the chip, does this mean I provide the max of those values on the vbat line to ensure it works? \$\endgroup\$ – jack sexton Mar 30 '16 at 14:09
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You are correct that linear regulators (including LDOs) cannot raise voltage. The output voltage is limited to Vin minus Vdo (dropout voltage).

In this case, you can use the resistors in the voltage divider to set the output voltage, assuming the input voltage is high enough.

You do need a boost converter to raise voltage.

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