Connecting different voltage rail in parallel

I am working on a buck-boost IC circuit design in which the load current requirement outweighs the capability of a single IC. Since I dont have the freedom for alternative, I'm thinking of connecting two IC circuits together in parallel to increase the current, which brings the following question into mind:

What would happen if there is a voltage mismatch? a few hundred milivolts? a few volt? So i did some simple simulation (see attachment) to see the effect of placing different batteries in parallel. By applying KVL and KCL, i got the current across R1 as 1.351A. This results in 2.5673V of effective voltage as seen in the attachment.


Going through similar thread like the following, i still could not find what i'm looking for, (or maybe its there, but i'm too dumb to figure it out):
https://www.edaboard.com/threads/71527/

My question is the following:
Is a simple KVL and KCL all it need for this kind of problem or am I overlooking the IC capabilities?
Is this practised in the real world? (connecting two IC circuit output in parallel to increase current)
In the case of car battery jumping, the battery is designed chemically to be rechargeable, would the IC really explode if i do the same?
Even if there is no voltage mismatch, will it still be safe?

I also did a simulation using the IC in question, but cannot put it here right now. it remains the similar however (Vout shares the same node, IC grounds share the same node)

Best regards,
Faizal

Connecting 2 batteries in parallel will always give voltage equal to the stronger battery.They will not add or average.
Connecting 2 output pins of an IC can also work if the output of both pins is same. This type of solution is also used in ULN where 2 output pins are connected together to get double the amount of current to drive heavy loads.

In general this does not end well with buck/boost UNLESS the part is a current mode controller, in which case you can tie the current control nodes together and the current loops will ensure even load sharing. Because of the voltage feedback loop, most regulators have very low output impedances, and particularly in the case of CCS sync converters you can end up with one running in buck and the other one moving power back the other way in boost....

Linear have a few parts that are explicitly designed for this kind of service, and even have some where there is an output that can be used to lock the slave device into a phase shifted mode of operation which will massively reduce the strain on the input caps. Have a poke around linears collection of external power switch buck/boost converters to see what I mean.

Regards, Dan.

Thank you for the replies.
As for the load in question, it is a 3g Module with following working current:


During GSM operation, there is a pulsed load period in which the current drawn would be up to 2A for a certain period. I believe it is in this period that I notice a significant voltage drop at the module input supply, which will reset the module causing a new device detection and initialization in Windows device manager.

The 2A which you are telling is called as burst current. Which will happen when the module is trying to register to its network. So ofcourse you need to able to supply upto 2A current. For solving this use one charge pump capacitor ( use electrolytic) with some 1000 or 2000uf of value. Place this capacitor nearer to the VCC and ground pins of your module supply.

Surendhar M said:

The 2A which you are telling is called as burst current. Which will happen when the module is trying to register to its network. So ofcourse you need to able to supply upto 2A current. For solving this use one charge pump capacitor ( use electrolytic) with some 1000 or 2000uf of value. Place this capacitor nearer to the VCC and ground pins of your module supply.

Click to expand...

The size of this capacitor violates the current design constraint . I did not use this capacitor type or supercap due to the ridiculously small height constraint(any component height above 26mm). I only used this one in the prototype evaluation and it work. But i dont have the freedom to use this right now.

Best regards,
Faizal

Either you need change IC part or add capacitor. Its better to do any one option. Give me the IC part number so that will verify anyother possiblities. What should be height of the capacitor which you need to use?

I could not edit the previous post anymore. Please note the maximum component height is 2.6mm, not 26mm as in the previous post.
The IC used for the working prototype is LTC3113EDHDPBF. The datasheet states that this IC should be able to handle the current requirement, but possibly due to my design and layout error, the supplied current is not enough. From my observation, the maximum supplied current is only 0.15A, which results in this thread.
I am now using MT3608 as a replacement (which works and the circuit component meets the constraint requirement) but is difficult to procure.