Convert imput of 6s(22v-26v DC) To around 4s (14v-16.8v DC)

Thread Starter

Lbzcody

Joined Oct 6, 2021
2
Hello everyone, I'm an rc car enthusiast. I have many different cars and a fair bit of electronic knowledge. What I'm looking for help with should be easy for someone who understands resistors well. I have a cooling fan that runs on up to 4s lipo (4.2v per cell when full charged, 14.8v-16.8v) and I would like to run it on a 6s lipo (22.2v-25.2v). I don't want to seperate any cells because it will unbalance the battery. So what I was wondering is there a inline resistor I can run inline of the fan to bring the voltage down easily. The fan draws around .9amps to 1.3amps under different load situations. What size resistor do I need. I know there are other options but I prefer to do it this way if possible. I'm not interested in adjustable voltage regs..as they are big. So I'm hoping to leave the specifics up to you guys on size and type of resistor.

Thanks
 

Ian0

Joined Aug 7, 2020
4,169
You would need (25.2V-16.8V)/0.9A=9.3Ω for the first situation with a fully charged battery, or (22.2V-14.8V)/0.9A = 8.2Ω with a discharged battery. For the second situation you would need (25.2V-16.8V)/1.3A=6.4Ω with a fully charged battery, or (22.2V-14.8V)/1.3A = 5.6Ω with a discharged battery.
The resistor would dissipate 11W in the worst case, it would therefore be large and very hot.
A voltage regulator would avoid the need for different values, but it would still get just as hot.
Most engineers would use a switched-mode supply in those circumstances, the smarter ones would replace the fan by a higher voltage version.
 

Juhahoo

Joined Jun 3, 2019
257
Go with the:
LM2596s DC-DC step-down power supply module
or other similar buck regulator, they cost 1$ at cheapest and are: smaller, cheaper, energy efficient, than a big resistor.
 

Thread Starter

Lbzcody

Joined Oct 6, 2021
2
You would need (25.2V-16.8V)/0.9A=9.3Ω for the first situation with a fully charged battery, or (22.2V-14.8V)/0.9A = 8.2Ω with a discharged battery. For the second situation you would need (25.2V-16.8V)/1.3A=6.4Ω with a fully charged battery, or (22.2V-14.8V)/1.3A = 5.6Ω with a discharged battery.
The resistor would dissipate 11W in the worst case, it would therefore be large and very hot.
A voltage regulator would avoid the need for different values, but it would still get just as hot.
Most engineers would use a switched-mode supply in those circumstances, the smarter ones would replace the fan by a higher voltage version.
Thanks for the answers. I had a feeling a resistor would not be so easy. I don't know why I didn't think about ohms law to figure out the resistor needed. The DC to dc inverter is gonna have to work unfortunately.
 
Top