18 Replies Latest reply on Sep 22, 2020 2:26 PM by jw0752

# 3.7 to 5v ?

Hello,

I have a device that has a power requirement of 5v, 1a. I wish to build a power source to make it portable.

I have several 3.7 lithium Ion 18650's. But I'm not sure if I should put them in series to make 11.4v then reduce down to 5v or paralle to 3.7 then boost up ?

Can anyone recommend a build sheet or video on this with parts list ?

Thank you. OT

• ###### Re: 3.7 to 5v ?

If you have an old auxiliary backup battery for a cell phone there is a circuit inside that takes the 3.7 volts of the battery and boosts it to 5 volts. This circuit will also allow you to charge the 3.7 Lithium battery with a standard phone charger. The only catch is that it may require a certain level of output current to remain on. In other words the 5 V device that you want to power with the battery might have to draw 20 or more mA to keep the circuit engaged.

John

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• ###### Re: 3.7 to 5v ?

Again, the device calls for "5V 1A". So if it requires "1A" and I have 11.1v - 3.5a and I step it down, or I have 3.7 with 10.5a with a step up, wouldnt that work ?

• ###### Re: 3.7 to 5v ?

Hi OT

That is a good question. If your device is designed to run on 5 Volts and it draws 1 Amp yes you can obtain this from either a step up converter (called a boost converter) or from a step down converter (called a buck converter). I will assume 100% efficiency for these conversions. In the case of the 3.7 volt boost conversion the converter will draw 1.35 Amps from the 3.7 volt battery in order to deliver the 1 Amp at 5 Volts. If the efficiency is less that 100% AND IT ALWAYS IS the current draw from the 3.7 volt battery will be more than 1.35 Amps. In the case of the 11.1 Volts and a buck converter the converter will draw at least 0.45 Amps from the 11.1 Volt battery in order to deliver 1 Amp at 5 Volts. There are DC to Dc converters on sale on ebay that will allow you to do either of these conversions.

If you want to build a converter yourself there are some integrated circuits available that allow you to build a converter by adding the correct resistors, capacitors, and inductors. Unless you have a good back ground in electronics it might be a fairly steep learning curve as a first project.

I want to make a clarification as I get the impression from your question that you may think that the battery or power supply decides the amount of current that it will supply to the device. The battery or converter has an internal resistance and provides a voltage. The device also has a resistance. When the voltage of the battery is put on the device a current flows through the resistance of the battery and through the resistance of the device. This current is equal to the Voltage of the battery divided by the added resistance of the battery and the device and is measured in Amps. I could have a 5 volt battery as big as a house and I would get the same 1 amp current when I hooked up your device that I would get if I hooked it up to a much smaller 5 Volt battery. For this reason batteries do not come with an amperage listed on them. What you may see however is an Amp Hour or milli-Amp hour rating. These ratings give us some idea how much power is in the battery. The device that we put on the battery will determine how much current in amps flow in the circuit and if we divide that measure of Amps into the Amp Hour rating of the battery we get a rough idea of how long the battery will run the device before it runs out of power.

If I have missed your question you will have to rephrase it and try again.

John

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• ###### Re: 3.7 to 5v ?

That is a good question. If your device is designed to run on 5 Volts and it draws 1 Amp yes you can obtain this from either a step up converter (called a boost converter) or from a step down converter (called a buck converter). I will assume 100% efficiency for these conversions. In the case of the 3.7 volt boost conversion the converter will draw 1.35 Amps from the 3.7 volt battery in order to deliver the 1 Amp at 5 Volts. If the efficiency is less that 100% AND IT ALWAYS IS the current draw from the 3.7 volt battery will be more than 1.35 Amps. In the case of the 11.1 Volts and a buck converter the converter will draw at least 0.45 Amps from the 11.1 Volt battery in order to deliver 1 Amp at 5 Volts. There are DC to Dc converters on sale on ebay that will allow you to do either of these conversions.

Hello John,

So would you say it would be better to go with a series circuit configuration for more efficiency and less heat ?

Oscar

• ###### Re: 3.7 to 5v ?

Hi Oscar,

As long as you use a switch mode converter, whether the batteries are in series or parallel  it will both work about the same for heat loss. If you are going to try to charge the batteries in the unit the Parallel would have some advantage as mentioned by a previous poster. Otherwise it is your choice and may also depend on whether you find a buck or a boost converter that you like.

John

1 of 1 people found this helpful
• ###### Re: 3.7 to 5v ?

I have reread your original question and I will add this. There is the same amount of energy in 3 batteries in parallel as there are in 3 batteries in series. If you had a 100% efficient switch mode converter to use in either case the device would run the same amount of time with the parallel as it does with the series configuration. For this to be true a switch mode converter would be required. There is also a buck converters called linear regulators which will also take 11.1 volts and change it to 5 volts. The linear regulators do not however do this efficiently and you would loose over 1/2 of your batteries energy to heat and your time of operation would likewise be cut in half. The linear converter will not work to boost from 3.7 Volts to 5 Volts.

John

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• ###### Re: 3.7 to 5v ?

Hi Again OT,

Here is a picture of one that I took apart to use in a similar fashion to what you describe.

John

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• ###### Re: 3.7 to 5v ?

Hello John,

I ordered the boost converter and I'm waiting for it to show up. But I have another question concerning a different device: if I have a larger capacity batt with the same voltage output as the device requires, do I necessarily need a protection board of any kind between the batt and the device ?  Thank you, Oscar.

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• ###### Re: 3.7 to 5v ?

Hi Oscar,

This is a great question.

There are two main dangers involved with Lithium Batteries. #1 is over charging them and #2 is allowing them to discharge below a certain point. These two conditions can damage the batteries permanently and in some cases cause them to overheat and catch fire. We have all seen the videos of devices exploding into fire. That is the purpose of the battery protection devices. They know when the battery charge is getting too low and they disconnect the battery from the load or they sense when the battery is fully charged and they stop more charge from going into the battery.

Now if you have a way for the your device to stop working before the battery gets too low and if you are removing the batteries to charge them on an approved charger then you probably do not need the protection circuit. I for example use lithium batteries in many of the LED flashlights that I have around the house. When the light beam from the flash light starts to get weak I remove the batteries from the flash light and put them into a special Lithium battery charge that has its own battery protection circuit and recharge them. I do not have separate protection circuits for these flashlight batteries.

We all like to use the Lithium batteries because they have what is called high density energy storage. Lots of energy in a small and light package. Unfortunately the side effect is the danger that all this energy presents if something goes wrong. Since the lithium battery is fussy about how it is charged and its intolerance of being fully discharged, battery protection circuits are important. All devices that use a lithium battery have the protection circuits built right into the battery packs.

Another way that protection circuits help the charging of lithium batteries is the situation where we want to charge two or more lithium batteries that are in series. Here the problem is called a balancing problem. If the batteries are not perfectly matched one battery may get more of the charge than the other. This is caused by a difference in the internal resistance of the batteries. This allows the charger to put a higher voltage across one battery than across another. Ultimately this damages the batteries and can cause failure of the battery pack. The protection circuit can sense the condition of each battery and treat each one as if it is the only battery that is being charged.

If you see the protection circuit as a way to protect the battery while it is being used and while it is being charged you will better be able to decide if a protection circuit is needed in your application.

John

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• ###### Re: 3.7 to 5v ?

The boost concept is a solid approach for generating 5.0V from a Li-Ion battery.  As John has pointed out there are a lot of available and cheap devices out there to do that.  I have chosen to build my own.  They are easy and affordable and you have a higher degree of flexibility in functionality build your own.  Here are a couple of links to some of my designs.

Good luck!

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• ###### Re: 3.7 to 5v ?

I agree with the boost converter scenario. Using the batteries in parallel implies self balancing and that could help with their life cycle, also you can upgrade to more batteries to achieve longer capacity or less batteries to achieve a more compact design.

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