[SOLVED] Dual battery source to single circuit

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Dear All,

I have a circuit which uses 3.7v Li_Ion battery and it works for 5 days. Now I like to provide USB port to add external battery to the user to extend the circuit running period. User can just attach external battery on the USB port to extend the period. I need the circuit details. Kindly see the attachment and let me know your suggestions & recommendations.

thanks
pmk
 

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  • DualBattery_Supply.gif
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Hi,


Looks ok so far.
If you need less voltage drop then look for "ideal diode circuit"

Klaus
 

We can select suitable diode later. Now I am only bother about circuit. I want to know whether my circuit OK or do we consider any more issues while using 2 power supplies (batteries)?

thanks
pmk
 

The diodes block the passage of current from one source to the other while still letting both of them flow to your load. To minimize the voltage drop from the battery use a Schottky diode as Audioguru sugests (although they cost less than solar lights if you haven't got one to dismantle!) or investigate the ideal diode method suggested by Klaus.

For the USB power source you actually want to drop the voltage from 5V to 3.7V so you need to lose 1.3V along the way. I suggest using two normal silicon diodes (not Schottky types), maybe 1N4001 or similar in series. Each will drop about 0.65V at low currents so they do the blocking and voltage dropping at the same time. If you find the voltage is lower than 3.7V when fed from the USB supply, try using one normal and one Schottky instead of two normal diodes. Schottky diodes will drop slightly less voltage than normal ones.

Brian.
 

Hi,

You say USB port. I can't find it on your schematic.
Also i wonder if it is a host port or a device port.
Sending power into a host port may give problems.
Sending 3.7V minus diode voltage drop into a device port may alos give problems because USB voltage is specified with 5.0V +/- 0.5V if i remember right.

If you supply a resistive load , like in your schematic, it is
OK, but to supply electronic circuit you need capacitors, voltage regulators, protection circuits ...


Klaus
 

For the USB power source you actually want to drop the voltage from 5V to 3.7V so you need to lose 1.3V along the way....
Click to expand...

You say USB port. I can't find it on your schematic....
Click to expand...

Sorry. I am using USB port only to connect external battery (3.7v Li_Ion battery). Not power the circuit from PC. Both batteries have the same voltage range i.e. 3.7v.

any suggestions pl?

thanks
pmk
 

Most Lithium rechargeable battery cells (except the LiFePO4 ones) average 3.7V during a discharge. They are 4.2V when fully charged and are 3.2V when the load should be disconnected.
Does your load sense the battery voltage then disconnect it when it is 3.2V or less? Some battery cells have a "protection circuit" inside that disconnects the load when the battery voltage drops too low.

5 days is 120 hours. An 18650 Lithium cell (is that the size you are using?) is about 2600mAh so for it to operate a load for 120 hours then the current averages about 21.7mA which is very low. Then the series Schottky diode reduces the voltage about 0.2V. Will your load still work properly when the voltage drops to only 3.0V?

I recommend not using the USB connector because it is commonly used for 5V, not for 3.7V. Somebody WILL connect it to 5V then will your circuit survive?
 


Yes. My device has GSM Module, PIR Sensor, PIC MCU and buzzer only. It takes 3 to 4 mA (GSM in sleep mode). I am using 3.7v, 1450mA Li_Ion battery. I can use voltage only up to 3.8v due to PIR Sensor (through LDO - it requires at least 3.8v to run properly).

Does your load sense the battery voltage then disconnect it when it is 3.2V or less? Some battery cells have a "protection circuit" inside that disconnects the load when the battery voltage drops too low.
Click to expand...

Yes. I am checking periodically, once in hour, and if battery lower than 3.8v, it shuts down automatically and inform the user.

I recommend not using the USB connector because it is commonly used for 5V, not for 3.7V. Somebody WILL connect it to 5V then will your circuit survive?
Click to expand...

Why I preferred to use USB connector in my circuit is that readily available power bank (battery) uses USB port. Ok it is better to avoid USB as you said or we can add protection circuit to overcome this issue. Am I right?

thanks
pmk
 

Hi,
Why I preferred to use USB connector in my circuit is that readily available power bank (battery) uses USB port.
Click to expand...

I´m pretty sure that those power banks output 5V ( as specified with USB).

Klaus
 

I´m pretty sure that those power banks output 5V ( as specified with USB).
Click to expand...

let me check Klaus and update you with full details regarding power bank.

thanks
pmk
 

Yes. A power bank has one 3.2V to 4.2V lithium battery cell and has a charger circuit to charge its cell from 5V USB.
It has a voltage stepup circuit so its output is 5V USB to feed the charger circuit in a cell phone.

A 3.2V to 4.2V lithium cell in series with a Schottky diode has an output voltage that drops fairly quickly to 3.8V. Only 15% to 20% of the battery capacity is used.
 

I´m pretty sure that those power banks output 5V ( as specified with USB).

Klaus
Click to expand...

Yes. A power bank has one 3.2V to 4.2V lithium battery cell and has a charger circuit to charge its cell from 5V USB. It has a voltage stepup circuit so its output is 5V USB to feed the charger circuit in a cell phone....
Click to expand...

Yes. So, what we have to do is to use diode with forward voltage drop of 0.8v. So that we can get 4.2v output and circuit can work perfectly. Right?

A 3.2V to 4.2V lithium cell in series with a Schottky diode has an output voltage that drops fairly quickly to 3.8V. Only 15% to 20% of the battery capacity is used.
Click to expand...

GSM Module & PIR sensor have been connected at anode of the diode . i.e directly from (Internal) Battery. PIC MCU has only been connected at the cathode of the diode. So, we can only able to consume up to 3.8v due to circuit design. We have to accept that as of now. ***It can be improved in the next version.***

Suggestion required:

What happen if 2 batteries (one with 3.7V, 1500mAh Li_Ion & another one is also 3.7v, 2600mAh Li_Ion) connected parallel and feed the circuit without using any diode. Will both battery supply current to the circuit evenly? External Battery will be charged separately using external charger. Inner Battery will be charged using built-in charger circuit. Any other issues there to consider?

thanks
pmk
 

Hi,

in my eyes the best is to keep on the USB sepcifications.

I don´t know exactely, but if i remember right then it is 5V +/- 0.5V, @ 0..500mA load.


Klaus
 
Their charged voltages will probably not be the same so the moment they are connected together a very high current will flow between them and possibly start a fire. Maybe they can safely be connected together when both are almost dead and need a charge. When charged and when they have a load then the lower rated one will be charged by the higher rated one.
I have a battery from a cell phone that has two identical Li-Ion cells in parallel. They charge together and discharge together.
 
ok. thanks to both Audioguru and Klaus. We have to study in-depth on this issue before implementation.

thanks
pmk
 

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