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I'm trying to design a supply circuit for a board that should deliver 3.3V using a LiPo battery, the battery should be that charged using a micro USB cable. here is what I've done so far. enter image description here

the LM3940 regulates the supply voltage to 3.3V and the MCP73831 is a battery charger. the Target voltage supplies the whole circuit.

after doing some research I figured out that need to take the load sharing in consideration, I found the following schematic for that :

enter image description here

My question is how big is the voltage on the MCU pin (my target voltage should be 3.3V) the way I see it , if **Vin ** is equal to 5V the output voltage is also 5V ? should I connect the the MCU pin to the voltage regulator ? thanks for any hint !

UPDATE made changes based on Andrew answer enter image description here

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    \$\begingroup\$ There is a mistake on your schematic, the Cin (4.7uF) capacitor should be between pin 4 (VDD) of the MCP7381 and GND. And the VCC pin of the USB should be connected directly to the VDD pin. Like Cout. Also the LiPo battery is connected into the 3.3 V node (Target Voltage) but the battery voltage can be anything from 4.2 V to around 3.2 V. \$\endgroup\$ – Bence Kaulics Aug 10 '16 at 11:29
  • \$\begingroup\$ @BenceKaulics thanks for your comment I'Ve correct it \$\endgroup\$ – Engine Aug 10 '16 at 11:32
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Your first circuit is a fine example of how to create a bad and potentially explosive circuit. If the battery is low it gets hit with 3.3V and as much current as the LDO can supply rather than being carefully charged. When the battery isn't low the charger is powering the circuit rather than the voltage regulator. The charger has no idea how much power is actually going into the battery and the circuit is being powered by up to 4.2V rather than 3.3V.

In the second circuit when VIN is high Q1 is off effectively isolating the battery from the MCU pin allowing the battery charger to do it's job. Power flows to the MCU pin via a diode, assuming 5V input with an LDO down to 3.3V the 0.2-0.3V drop in the diode shouldn't be an issue.

When VIN is off R2 causes Q1 to switch on and allows battery power to flow from the battery to the MCU pin.

Q1 needs to be sufficiently large to handle the current with appropriate voltage thresholds so that it is fully on and off under the normal operating conditions.

So yes, connect the MCU pin to the input of your regulator and you should get a clean regulated supply until your battery discharges to about 3.5V (assuming a 0.2V drop out LDO). Once your LDO drops out of regulation you should shut the system off since it's no longer reliable. You should definitely shut off before the battery gets below about 3.1V or you risk damage to the battery pack.

3.5V will normally be about 30% charge remaining but it is very temperature dependent. You could hit that voltage at 90% charged in low temperatures.

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  • \$\begingroup\$ thanks a lot for your help, I've change the schematic, would please take a look, so I could be sure that I did understand you right ! thanks again \$\endgroup\$ – Engine Aug 10 '16 at 12:42
  • \$\begingroup\$ It looks like C3 is shorted out. And you can't use an LM3940. That is designed for an input of 4.5 - 5.5V and has a dropout of 0.5V or more. A LiPo has a fully charged voltage of 4.2V. You need something with a lower dropout voltage. \$\endgroup\$ – Andrew Aug 10 '16 at 13:15
  • \$\begingroup\$ C3 should be the C2 in 2nd pic, and you're right about LM3940 \$\endgroup\$ – Engine Aug 10 '16 at 13:23
  • \$\begingroup\$ No, in the final picture C3 has a line through it and connection dots at the terminals. It looks like there is a line shorting it out. There is certainly something weird with the schematic there. \$\endgroup\$ – Andrew Aug 10 '16 at 14:59

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