The use of the ESP8266 in the world of IoT

User avatar
By to1140
#13477 Thanks for sharing this. I was thinking to try TI LMR14006 which does auto-sleep, depending on load current.

What's the input current consumption with no output load? -- one post mentioned 1~2 mA, not sure if it is measured # and if the no load input current depends on input voltage. Some folks recommend < 30 mA total consumption even with car battery to be feasible for use in a car.


ArnieO wrote:If you're not very pressed on time or project volume, you get 20 of these for 10.8$, free shipment (54 cents each!);
http://www.aliexpress.com/item/20-pcs-Ultra-Small-Size-DC-DC-Step-Down-Power-Supply-Module-3A-Adjustable-Step-Down/32262311443.html
It only converts down, but that what I mostly need.
User avatar
By trackerj
#14154 A quick setup for a rechargeable battery power solution:

- LGDB 318650 - Rechargeable 3,7V/2600mA Li-ion Cell
- 318650 Cell holder
- Li-ion Charger/Monitor with protection

It is based on TP4056 1A Standalone Linear Li-lon Battery Charger with Thermal Regulation chip

The TP4056 is a complete constant-current/constant-voltage linear charger for single cell
lithium-ion batteries. Its SOP package and low external component count make the TP4056
ideally suited for portable applications. Furthermore, the TP4056 can work within USB and wall
adapter.

Li-ion Cell Protection is done with dedicated DW01-G battery protection IC.

DW01-G is designed to protect lithium-ion/polymer battery from damage or degrading the lifetime due to overcharge,overdischarge, and/or overcurrent for one-cell lithium-ion/polymer battery powered systems, such as cellular phones.

Details: http://www.esp8266-projects.com/2015/03/mailbag-arrival-new-battery-solution.html

ADC - 0-2 - battery - 0_2.jpg


ADC - 0-3 - battery charger - 0_3.jpg
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User avatar
By ArnieO
#14187
to1140 wrote:What's the input current consumption with no output load? -- one post mentioned 1~2 mA, not sure if it is measured # and if the no load input current depends on input voltage.


Good question! So I measured one.
Vin = 12V
Vout = 5,0 V, I_out = 0,000 mA: I_in =0,23 mA => Pin = 2,8 mW, Pout = 0,0 mW => Loss 2,8 mW
Vout = 5,0 V, I_out = 6,5 mA: I_in =4,03 mA => Pin = 48,4 mW, Pout = 32,5 mW => Loss 15,9 mW