5V Step Up Converter Drops Voltage to 3.45 Volts when Connected to Load

Recently I combined two 18650 batteries in Parallel, and attached the output of the Paralleled 18650 batteries to a 5V Step UP Converter to convert my voltage up to a constant 5V and draw the 4A my load requires at 5V.

Here is a picture of the Step-Up Module I used, and link: EBAY
[size=small][img=250x250]https://i.ebayimg.com/images/g/q~EAAOSwjvJZXJT8/s-l1600.jpg[/img]
[/size]
- The Paralleled batteries input 3.45 Volts, with up to 15A constant power drain available.
- My load takes a required 5V 4A.
When I set the output on the adjustable converter to 5V and connect my load to the custom battery pack, whose input is connected to this 5V step up andI made from 2 18650 Samsung 30Q batteries,the output Voltage drops to the same voltage as the batteries. This problem did not occurwhen I used a 5V 1A load.

I am not sure why this is happening: immediately when I connect the battery powered device to the output with AWG Gauge wires, the Voltage I set is seemingly overriden.
Thanks!

The table on the listing specifies that the converter is tested to do 2a 5v with 3.7v in. This is often the case. Its maximum output is only achievable at higher voltages.

It would be better to use a buck converter and use 2s rather than 2p. This converter cannot be used as a buck converter unfortunately.

Also you posted this in the wrong section.

Usually always easier to use a buck converter than a boost converter. There's less amp draw on the source side, and the output voltage won't drop when the input voltage drops a little bit
I think also a buck converter is slightly more efficient than a boost converter

Geek said:

The table on the listing specifies that the converter is tested to do 2a 5v with 3.7v in. This is often the case. Its maximum output is only achievable at higher voltages.

It would be better to use a buck converter and use 2s rather than 2p. This converter cannot be used as a buck converter unfortunately.

Also you posted this in the wrong section.

Click to expand...

Hi Geek,

I wanted to use Parallel Batteries because I am afraid that if I put them in a series I won't have enough run time for the load.

In addition, I am scared that I am going to run into the same problem with the boost converter. If I use a buck converter from a 7.4Volt input, how can I guarantee that when it discharges that 5V 4A will be maintained? What happens when the battery pack drops down to 3.7Volts and I set the converter to buck down to 5V? It wouldn't work then, I assume.

Is there any way I can make it so that my boost converter will boost to a stable 5V 4A? I am afraid that I am going to run into all the same problems above and more using a buck converter.

In the end if nothing works, could I have two seperate boost converters and combine the wires to double the amperage? Essentially, two seperate batteries with two seperate converters?

---

Korishan said:

Usually always easier to use a buck converter than a boost converter. There's less amp draw on the source side, and the output voltage won't drop when the input voltage drops a little bit
I think also a buck converter is slightly more efficient than a boost converter

Click to expand...

Hi Korishan, I posted a follow up below if you could check it out!

It doesnt matter if you parallel or series cells. You still got the same amount of stored energy. Its more important to look at efficiency and other aspects like series require BMS or some sort of charging mechanicsm that ensures them being parallel.

Its not wise to parallel such converters to same output. Better getting a proper buck or boost converter.

All Buck or boost converters have feedback loop where they will ensure that the output is keept stable as long as it can. As long as input voltage is above or below the needed threshold

5V and 4A = 20w. thats quite alot and at 3.7V that would be 20/3,7 = 6A with some losses. So you basically need a boost that can to 6A or in china words 10A


a 2s battery should never be dropped below 6V due to cells then at 3V each.

I have used this one alot. This with a simple 2s bms you got a decent setup
Choose the BMS with balancer. Then you just need a powersupply giving out 8.4V and 1A for charging.

daromer said:

It doesnt matter if you parallel or series cells. You still got the same amount of stored energy. Its more important to look at efficiency and other aspects like series require BMS or some sort of charging mechanicsm that ensures them being parallel.

Its not wise to parallel such converters to same output. Better getting a proper buck or boost converter.

All Buck or boost converters have feedback loop where they will ensure that the output is keept stable as long as it can. As long as input voltage is above or below the needed threshold

5V and 4A = 20w. thats quite alot and at 3.7V that would be 20/3,7 = 6A with some losses. So you basically need a boost that can to 6A or in china words 10A


a 2s battery should never be dropped below 6V due to cells then at 3V each.

I have used this one alot. This with a simple 2s bms you got a decent setup
Choose the BMS with balancer. Then you just need a powersupply giving out 8.4V and 1A for charging.

Click to expand...

This is immensely helpful, but I am at work now. Please expect a ton of questions to come! You guys are the best.

OmarSumadi said:

daromer said:

It doesnt matter if you parallel or series cells. You still got the same amount of stored energy. Its more important to look at efficiency and other aspects like series require BMS or some sort of charging mechanicsm that ensures them being parallel.

Its not wise to parallel such converters to same output. Better getting a proper buck or boost converter.

All Buck or boost converters have feedback loop where they will ensure that the output is keept stable as long as it can. As long as input voltage is above or below the needed threshold

5V and 4A = 20w. thats quite alot and at 3.7V that would be 20/3,7 = 6A with some losses. So you basically need a boost that can to 6A or in china words 10A


a 2s battery should never be dropped below 6V due to cells then at 3V each.

I have used this one alot. This with a simple 2s bms you got a decent setup
Choose the BMS with balancer. Then you just need a powersupply giving out 8.4V and 1A for charging.

Click to expand...

This is immensely helpful, but I am at work now. Please expect a ton of questions to come! You guys are the best.

Click to expand...

The one thing that has constantly confused me when I use a boost converter is the information I
received on stack exchange:

According to a user over there, if I bought a boost converter that could handle ~10A input that I
should be able to boost to a stable 5V 4A; however, what you said makes sense when I look at
the test data. Often times it states that at an input of 3.7V, for instance, it will boost to 5V 2A.
It seems so counter intuitive - why isn't it boosting to the load of 5V 4A?
- View StackExchange Link: Here

Anways, it makes a lot more sense now to go for a series - I kind of feel like an idiot.

I was hoping you could give me clarificaiton/tell me if I am right about a couple of concepts:

1) Series vs Parallel run times will be the same since in a Series less amperage will be be
drawn, and mah is based on amperage drawn.

2) Since my boost converter projects have failed to output 5V 4A, if I put two batteries in a
series and use the buck converter you mentioned, will I finally get 5V 4A output?
- However, I see the buck converter minimum input is 8v, what if I get something like this:
EBAY LINK

Would these work to finally give me 5V 4A, given that the voltage will drop to a low of 6V since I will have 2 batteries in series?
I would put the batteries in series to make 7.4V, then I would use the buck to step down to 5V in which my load would be drawing 4A, and since buck input can go down to 5V, when voltage decreases amperage of 10A cap is more than enough? BMS would be wired accordingly.


3) Given the BMS, I would charge by soldering to P+ and P- and connecting my power source?
What happens if I don't have a 8.4V 1A power supply? Can I use a 12V 2A power supply?

Off-topic:
4) For boost converters, what about something like this?
https://www.ebay.com/itm/DC-DC-Step...516268?hash=item4b151923ac:g:P6IAAOSwdd9ae7C-
It seems like it's designed for 18650s, and can output 3a (even though it's not what I want, I just wanted your opinion).
Could this small circuit even handle a 5V 3A output when the battery drains to 3.2 V and input amperage will be around 8 amps? Just wondering
for more own information.

Omar Sumadi said:

According to a user over there, if I bought a boost converter that could handle ~10A input that I
should be able to boost to a stable 5V 4A; however, what you said makes sense when I look at
the test data. Often times it states that at an input of 3.7V, for instance, it will boost to 5V 2A.
It seems so counter intuitive - why isn't it boosting to the load of 5V 4A?

Click to expand...

It's not that it "can't" do it, it's that the difference between the input/output is to close. Try setting it to 10V with a 3.7V (or 4.2V) and see what kind of sustained amps you can get.

Omar Sumadi said:

1) Series vs Parallel run times will be the same since in a Series less amperage will be be
drawn, and mah is based on amperage drawn.

Click to expand...

The run time will be about the same, yes. mAh is based on Amps "and" voltage, but they are proportional. If one goes up, the other must go down to equal the same mAh rating

Omar Sumadi said:

2) Since my boost converter projects have failed to output 5V 4A, if I put two batteries in a
series and use the buck converter you mentioned, will I finally get 5V 4A output?
- However, I see the buck converter minimum input is 8v, what if I get something like this:
EBAY LINK

Click to expand...

That should work, and it has two trim pots, which is good. You can do current limiting with it. So you can set it to 4A max

Omar Sumadi said:

Would these work to finally give me 5V 4A, given that the voltage will drop to a low of 6V since I will have 2 batteries in series?
I would put the batteries in series to make 7.4V, then I would use the buck to step down to 5V in which my load would be drawing 4A, and since buck input can go down to 5V, when voltage decreases amperage of 10A cap is more than enough? BMS would be wired accordingly.

Click to expand...

Ok, first off, it won't be using 7.4V, unless your upper limit on the SoC of the cells is 3.7V, and you wouldn't drain to 5V (2.5V/cell) either. You'll have a range, from 8.4V (4.2V/cell) to 6V (3V/cell). I would actually recommend going with 2s2p if you can (not sure of your application/space), or even go with 3s if possible. The buck will take what ever voltage input within it's operating range and then output a specific voltage. So no matter what the input is the output will remain stable (within reason)


Another to note, for a good buck, or a boost, converter, the minimum voltage difference should be 1V, if not 1.5V. Meaning if the desired output is 5V, then the input should either be 6V for a buck, or 4V for a boost. If you start to get smaller than that, you could start to have issues with the logic of the converters.

Korishan said:

.....

Another to note, for a good buck, or a boost, converter, the minimum voltage difference should be 1V, if not 1.5V. Meaning if the desired output is 5V, then the input should either be 6V for a buck, or 4V for a boost. If you start to get smaller than that, you could start to have issues with the logic of the converters.

Click to expand...

That should be fine with an end voltage of 3v per cell (2S will give 6v - you may see problems if you go any lower but). That buck converter should do the job no problems.

Also, with any of those adjustable converters, I get the settings I want, then remove the pot and replace it with a fixed value resistor. No good when they fail!

Korishan said:

OmarSumadi said:

According to a user over there, if I bought a boost converter that could handle ~10A input that I
should be able to boost to a stable 5V 4A; however, what you said makes sense when I look at
the test data. Often times it states that at an input of 3.7V, for instance, it will boost to 5V 2A.
It seems so counter intuitive - why isn't it boosting to the load of 5V 4A?

Click to expand...

It's not that it "can't" do it, it's that the difference between the input/output is to close. Try setting it to 10V with a 3.7V (or 4.2V) and see what kind of sustained amps you can get.

OmarSumadi said:

1) Series vs Parallel run times will be the same since in a Series less amperage will be be
drawn, and mah is based on amperage drawn.

Click to expand...

The run time will be about the same, yes. mAh is based on Amps "and" voltage, but they are proportional. If one goes up, the other must go down to equal the same mAh rating

OmarSumadi said:

2) Since my boost converter projects have failed to output 5V 4A, if I put two batteries in a
series and use the buck converter you mentioned, will I finally get 5V 4A output?
- However, I see the buck converter minimum input is 8v, what if I get something like this:
EBAY LINK

Click to expand...

That should work, and it has two trim pots, which is good. You can do current limiting with it. So you can set it to 4A max

OmarSumadi said:

Would these work to finally give me 5V 4A, given that the voltage will drop to a low of 6V since I will have 2 batteries in series?
I would put the batteries in series to make 7.4V, then I would use the buck to step down to 5V in which my load would be drawing 4A, and since buck input can go down to 5V, when voltage decreases amperage of 10A cap is more than enough? BMS would be wired accordingly.

Click to expand...

Ok, first off, it won't be using 7.4V, unless your upper limit on the SoC of the cells is 3.7V, and you wouldn't drain to 5V (2.5V/cell) either. You'll have a range, from 8.4V (4.2V/cell) to 6V (3V/cell). I would actually recommend going with 2s2p if you can (not sure of your application/space), or even go with 3s if possible. The buck will take what ever voltage input within it's operating range and then output a specific voltage. So no matter what the input is the output will remain stable (within reason)


Another to note, for a good buck, or a boost, converter, the minimum voltage difference should be 1V, if not 1.5V. Meaning if the desired output is 5V, then the input should either be 6V for a buck, or 4V for a boost. If you start to get smaller than that, you could start to have issues with the logic of the converters.

Click to expand...

You guys are the best here, everything makes so much damn sense when I post here.

So in the end, just to confirm, given a 2S battery configuration, that buck converter should be able to output 5V 4A if my load is 5V 4A?

I just want to clarify because I am trying to think of what actually goes on with the battery.

Tell me if i'm right:
1) Battery starts at 8.4V fully charged.
2) Load calls for 5V 4A
3) Buck booster converts down to 5V but here is where I get a little confused, tell me if I am right:

My total draw @ 5V 4A is 20 Watts: so, the boost converter will signal to pull from 8.4V (20Watts/8.4V) = 2.38 A. Then, when the battery discharges to say 7.4V, it signals to the batteries to pull (20/7.4) = 2.7 A.

But aren't we setting it to a constant 5V, so wouldn't that mean: the buck converter is signaling the battery to always pull 5V from the battery and thus (20/5) = 4A?

I guess my question is: how is the buck converter actually outputting the correct voltage when the voltage of the battery is different than 5V? In addition, how does it pull 4 amps as well while maintaining 5 Volt output and adjusting amperage to match given voltage?

Thanks,
I don't know how to thank you!

---

Geek said:

Korishan said:

.....

Another to note, for a good buck, or a boost, converter, the minimum voltage difference should be 1V, if not 1.5V. Meaning if the desired output is 5V, then the input should either be 6V for a buck, or 4V for a boost. If you start to get smaller than that, you could start to have issues with the logic of the converters.

Click to expand...

That should be fine with an end voltage of 3v per cell (2S will give 6v - you may see problems if you go any lower but). That buck converter should do the job no problems.

Also, with any of those adjustable converters, I get the settings I want, then remove the pot and replace it with a fixed value resistor. No good when they fail!

Click to expand...

Thank you Geek - I don't know how to thank you and Korishan. I posted my last question below - endless thanks!

This video may clear things up on how a buck and a boost converter works:

btw, there is 3 types: 1) buck converter; 2) boost converter; 3) buck/boost converter
1 & 2 are built differently, but "almost" mirror each other. 3 is a combination of both

Korishan said:

This video may clear things up on how a buck and a boost converter works:

btw, there is 3 types: 1) buck converter; 2) boost converter; 3) buck/boost converter
1 & 2 are built differently, but "almost" mirror each other. 3 is a combination of both

Click to expand...

Hey Korishan - great news!

Everything worked out as expected: the BMS was perfect and the buck converter worked perfect - beyond my expectations.

I have a quick question about reducing size though:

Say I want to output 30 Watts from the Batteries, the batteries are rated at the lowest point at 8.8-9V using 18650s (over-discharge protection at 3V). Since I have the batteries set up in a series, that means the lowest voltage the battery can hit is 9V.

In that case, 30/9 = 3.333, so theoretically, the maximum currentinput is 3.33A - however, the buck converter will convert that to 5V 6A meaning that my limiting factor in purchasing a buck converter is not the input current and voltage, but the output current and voltage.

Is that correct? I ask this in light of input current problems with boost converters.

Thanks