[Help] In the design of lithium battery power modules, why is every 1V of lithium battery voltage very valuable?

Redamancy785

[Help] In the design of lithium battery power modules, why is every 1V of lithium battery voltage very valuable? [Copy link]

 

When designing the reverse connection circuit of the lithium battery power module, the diode solution is often not used because the diode conduction will cause a 0.7V voltage drop, and the NMOS solution is often used.

Why is every 1V of lithium battery so precious? In other words, what will be the loss if the diode solution is adopted?

maychang

A 0.7V voltage drop means a power loss of 0.7W per ampere, or 0.7J of energy per second. This energy is supplied by the battery. In other words, the energy stored in the battery is not consumed by the load, but part of it is consumed by the diode. This obviously reduces efficiency.

qwqwqw2088

The internal resistance of the MOS tube is generally in the milliohm level, and the voltage drop is very small in the mV level. The smaller the voltage drop, the smaller the internal consumption.

The advantage of the diode solution is that it is simple and low-cost

chunyang

P=UI, the junction voltage drop of the diode is basically constant, so there is a power consumption of about 0.7I wasted, which is 0.7W at 1A, which is the biggest problem facing the diode solution. Now many MOS tubes have an on-resistance of less than 0.1 ohms, and the power consumption at the same 1A current is less than 0.1W.

Gen_X

There is nothing wrong with your analysis, but I think it is mainly a problem of "efficiency":

Assuming it is powered by a battery, the maximum voltage of a lithium battery is 4.2V, and a diode is 0.7.

For the load, 4.2-0.7 = 3.5V, which is the maximum voltage, so the minimum direct loss is 0.7/4.2=16.666%,

Therefore, the power loss is also the minimum 16.666%.

If you don't care about the ratio, you can use it.

soso

May I ask if the landlord's problem has been solved?