Best way to power an ESP8266 (3,3V) from AA batteries with LDO

Question

My latest project uses an ESP8266 (ESP-01) to transfer data when a button is pressed. The ESP is always connected to the AP. No deep sleep. Right now I am powering it with an USB-charger which is directly connected to an AMS1117 LDO to regulate the voltage to 3,3V. This works fine so far.

Now I need to power the ESP with AA Nimh rechargeable batteries. I thought of putting 4 AA in a battery tray which should result in nearly ~5V. Because of the high quiescent current (>1mA) of the AMS1117 I am searching for a 'better' LDO. I see a lot of projects using the MCP1700 LDO in combination with the ESP. This LDO has a very low quiescent current (<2 µA), but the datasheet says the max output current is 200mA, from what I've read the ESP can have peaks up to 250+ mA. Isn't this a problem? I also found the MCP1825s which has an output current up to 500mA, but a higher quietscent current of <120µA.

1) Which LDO should I use to have a stable but power saving setup?

2) How can I calculate the time this setup will work until the batteries voltage drops below their minimum? Using 4 AA (each 1900 mAh) starting at about 5V until 3,6V. 3,6V because the dropout of the LDO is about 0,3V and the ESP needs 3,3V. Assuming an average current draw of about 100mA. Is it just simple 1900mAh/(100mA+LDO current) = ~19h ? How fast will the voltage drop from 5V to 3,6V?

3) Which caps should I use for decoupling the LDO? Just the ones suggested in the datasheet? Is it important if the capacitors are ceramic or electrolytic?

Answer 1

I see a lot of projects using the MCP1700 LDO in combination with the ESP. This LDO has a very low quiescent current (<2 µA), but the datasheet says the max output current is 200mA, from what I've read the ESP can have peaks up to 250+ mA. Isn't this a problem?

It may not be a problem if the output capacitor on the regulator can provide the 50 mA extra over a short period of time without drooping the output voltage too much. If the capacitor is (say) 100 uF and the current demand is for (say) 0.1ms you can use the formula I = C dv/dt to calculate the voltage droop (dv) in the 1ms (dt).

So 0.05 amps x 0.1ms = 100 uF x dv, hence dv = 50 mV i.e. not much of a droop. Having said all this I don't know how long the peak of 250 mA will last for.

Which LDO should I use to have a stable but power saving setup?

I have no idea what you ask?

How can I calculate the time this setup will work until the batteries voltage drops below their minimum?

If each battery is rated at 1900 mAh, the total series connection will still be 1900 mAh but that rating may be a full depletion down to 2.5 volts (just an example) so play safe and assume you only have about 1000 mAh. Alternatively read the data sheet for the battery type and see what graphs it shows. If it's a decent battery it will have data sheets. If you bought it from ebay or aliexpress then don't count it it.

Which caps should I use for decoupling the LDO? Just the ones suggested in the datasheet? Is it important if the capacitors are ceramic or electrolytic?

Use the types in the data sheet but bear in mind you should try and use low ESR electrolytics unless the DS specifies otherwise.

Answer 2

It is more efficient to use distributor search filters to determine if a better choice exists. You are looking for 5.5V in max 3.1 to 3.3V out, Iq<10uA and Iout=250mA min. Vdrop at 250mA=0.3V max or similar ESR < 1 Ohm i.e. 300mV/300mA drop.

Change any assumptions I made to suit yours. e.g.

https://www.digikey.com/products/en/integrated-circuits-ics/pmic-voltage-regulators-linear/699?k=&pkeyword=&pv2065=198&pv2065=948&pv2065=7&pv2065=259&pv2065=1231&pv2065=1232&FV=204001b%2C204003c%2C2040043%2C204004c%2C2040051%2C2040061%2C1bcc018b%2C1bcc018d%2C1bcc0032%2Cffe002bb%2Cmu1.5%C2%B5A%7C477%2Cmu1%C2%B5A%7C477%2Cmu2.5%C2%B5A%7C477%2Cmu3%C2%B5A%7C477%2Cmu4%C2%B5A%7C477%2Cmu5%C2%B5A%7C477%2Cmu6%C2%B5A%7C477%2Cmu680nA%7C477%2Cmu8.5%C2%B5A%7C477%2Cmu8%C2%B5A%7C477&quantity=0&ColumnSort=0&page=1&stock=1&pageSize=250