Is it better to first increase the voltage to 5V and then decrease it to 3.3V when using a 3.7V lithium battery? Or is it better to first decrease it to 3.3V and then increase it to 5V?

wangerxian

Is it better to first increase the voltage to 5V and then decrease it to 3.3V when using a 3.7V lithium battery? Or is it better to first decrease it to 3.3V and then increase it to 5V? [Copy link]

 

Recently, I have a project that requires peripherals with 5V and 3.3V. The main control is 3.3V, and there is a screen powered by 5V. The current is about 300+mA. The lithium battery is a single-cell battery with 3000mA. Should I increase 5V first and then reduce 3.3V? Or reduce 3.3V first and then increase 5V? Which method has the least loss? The power consumption of the main control is about 50mA, and the power consumption of other 3.3V peripherals is about tens of mA. I am also looking for domestically produced chips for 5V->3.3V and 3.3V->5V DCDC, with stable supply and suitable price~

chunyang

If both use switching power supplies, it doesn't matter which one is used first, but you need to weigh the cost. First step down the voltage, and both power chips must meet the maximum peak current. Whether this affects the cost needs to be noted.

mingplus

This is related to the discharge characteristics of lithium batteries. Generally, this type of lithium battery discharges to 3.6~3.5V at room temperature, and there is not much remaining capacity, so the lithium battery can directly step down to 3.3V and step up to 5V.

chunyang

In fact, there are three options: first increase and then decrease, first decrease and then increase, and two independent groups. It is really hard to say which one is better before making a specific choice, including efficiency calculation.

The first thing to consider is the input voltage range, which should preferably cover the final discharge voltage of the lithium battery in order to fully utilize the battery capacity. If both power supplies meet this requirement, two independent power supplies are most suitable. If the voltage is lowered first and then raised, or raised first and then lowered, the input voltage range requirement of the latter power supply can be greatly reduced, and whether it affects the cost must be considered. However, this is an extraordinary period, and chip prices may fluctuate greatly. If things go wrong, you may encounter a situation where you can use whatever you have, and you may not have much room for choice.

chunyang

Another option is to use a switching power supply for boosting and an LDO for bucking. Look at the manuals of the main control and peripherals. If you can tolerate the battery's full charge and discharge terminal voltage, you can even save the LDO and use the battery directly for power supply, which is the lowest cost.

If you are very sensitive to power consumption, you can lower the operating voltage on the 3.3V side appropriately. In this case, it is better to use a regulated power supply, such as 3V or even lower voltage.

Fred_1977

1. When the battery voltage is low, if the LDO is directly connected to the battery, the LDO is in an unstable state, which will affect the stability of your MCU.

qwqwqw2088

First use the switch chip to raise it, then use 1117 to supply 3.3V

Battery power supply generally requires low power consumption. If low power consumption is required, you need to choose a solution carefully.

wangerxian

mingplus published on 2022-5-6 11:52 This is related to the discharge characteristics of lithium batteries. Generally, this kind of lithium battery discharges to 3.6~3.5V at room temperature, and there is not much remaining capacity, so the lithium battery can be directly reduced to 3.3V...

In this way, first decreasing and then increasing, which one is higher, the loss or first increasing and then decreasing?

wangerxian

chunyang posted on 2022-5-6 12:20 In fact, there are three options, first increase and then decrease, first decrease and then increase, and two independent groups. Which one is better, it is really hard to say before making a specific choice, including efficiency calculation...

Indeed, choosing chips now makes people nervous, for fear of out-of-stock situations.

wangerxian

Fred_1977 posted on 2022-5-6 17:04 1. When the battery voltage is low, if the LDO is directly connected to the battery, the LDO is in an unstable state, which will affect the stability of your MCU.

I was also thinking about this. I thought the battery was too low and I shut it down directly.

wangerxian

qwqwqw2088 posted on 2022-5-6 17:28 First use the switch chip to rise, and then use 1117 to power the 3.3V battery. Generally, low power consumption is required. If low power consumption is required, you need to choose a solution carefully

Currently, we are considering this solution, which is to increase the voltage first and then decrease it. This will prevent the MCU from falling below 3.3V because it still has ADC collection.

se7ens

The operating voltage range of lithium batteries is about 2 (depending on the protection board) ~ 4.2V

From this point of view, the single topology does not meet the requirement of outputting 3.3V first. It is recommended to boost 5V first and then step down to 3.3V

Another advantage is that if the battery is charged with 5V, the product can be operated directly.

If you have any recommendations, you can go to LCSC to have a look. There are many domestically produced chips with such low current.

wangerxian

se7ens posted on 2022-5-7 13:47 The operating voltage range of lithium batteries is about 2 (depending on the protection board) ~ 4.2V. In this way, the single topology does not meet the requirement of outputting 3.3v first. It is recommended to upgrade first...

Got it! Thanks!

Gen_X

From a fail-safe perspective, doing things separately and running them in parallel and independently is efficient and reliable.

wangerxian

Gen_X posted on 2022-5-8 16:33 From a fail-safe perspective, doing things separately and running them in parallel and independently is efficient and reliable.

Because the two ends are independent, even if one end is broken, the other end can still continue to operate, right?

Gen_X

wangerxian posted on 2022-5-9 09:12 Because the two ends are independent, even if one end is broken, the other end can still continue to run, right?

Yes! It's the same principle.

Reliability is improved without increasing cost or circuit complexity.

H12315

Why does the battery only work up to 3.7V? The protection board I use lowers the voltage all the way to 2.7V. . . . .

wangerxian

H12315 posted on 2022-5-12 17:56 Why does the battery only work up to 3.7V? The protection board I use can lower the voltage all the way to 2.7V. . . . .

3.7V is the rated voltage, and the lowest can indeed reach 2.7V, but the general working voltage is around 3.7V.

H12315

wangerxian posted on 2022-5-13 09:22 3.7V is the rated voltage, and the lowest can indeed reach 2.7V, but the general working voltage is around 3.7V.

Since the working voltage will be lower than 3.3v, it can only be boosted first and then reduced.

chipsimi003

The common practice is to use two chips to control each channel separately: 1. Use LDO to stabilize the voltage to 3.3V, such as CSM5333BSC, with power consumption within 2.5uA; 2. Use boost DC-DC to increase to 5V, such as CSM9201SF.