How much resistance value and watts of resistor need to lower the voltage output

Thread Starter

Kelvin Lee

Joined Oct 22, 2018
109
Dear Sir/Madam,

I only have a battery with fully charged at 28V and a 24V motor, I need going to join a motor car competition, the rule is the voltage should be limited below 22V, I plan to design a voltage divider to lower the voltage from the battery output. When I test with the battery directly connect to the motor, the current is 13.7A, in this case, how much resistance value I have to use and how many watts of the resistor need?

Best regards,

Kelvin.

Screenshot 2019-05-04 at 11.27.49 PM.png
 

ericgibbs

Joined Jan 29, 2010
8,755
hi Kelvin,
If you ignore R4, You need to use a (28v-22v)/13.7A = 0.43R , Rated at 82Watts, say 100Watts
Adding R4 to the circuit is going to increase the overall wattage and reduce the efficiency.
Consider an alternative method, say a SMPS Buck converter.
E
 

BobTPH

Joined Jun 5, 2013
2,029
Dropping 6 volts at 13.7 Amps is 82
Watts. You don’t want to do that.

Either use a buck converter or get a different battery. A 5 cell lithium battery is 21 V.

Bob
 

MrAl

Joined Jun 17, 2014
6,518
The overall efficiency when using a single resistor to drop the voltage is about 79 percent.
That's pretty good by old switching converter standards, but modern converters can do 90 percent.
The run time of the single resistor solution though is not as good as with the 90 percent efficiency buck converter.
With the buck you'll see about a 14 percent increase in run time. If that is significant then go with the buck, but if not then you only have to worry about what you are going to do with all that extra heat (82 watts resistor vs 33.5 watts buck).
 

shteii01

Joined Feb 19, 2010
4,667
Hi,

Sorry but he definitely does not want to use a voltage divider.
That was for educational purposes. The OP knew about voltage divider, but did not know how to calculate values of the resistors that form the voltage divider.
 

AnalogKid

Joined Aug 1, 2013
8,111
If the characteristics of the motor are well known, you do not need R4. The motor itself forms one leg of the voltage divider.

Surf around and look up a 0.43 or 0.47 ohm, 200 W resistor, then consider a buck regulator module from ebay, It will be smaller, lighter, cooler, and deliver a constant voltage to the motor as the battery discharges.

ak
 

Thread Starter

Kelvin Lee

Joined Oct 22, 2018
109
hi Kelvin,
If you ignore R4, You need to use a (28v-22v)/13.7A = 0.43R , Rated at 82Watts, say 100Watts
Adding R4 to the circuit is going to increase the overall wattage and reduce the efficiency.
Consider an alternative method, say a SMPS Buck converter.
E
Dear ericgibbs,

What is SMPS Buck converter? I have still a newbie to the electronic, I don't know whether voltage divider is the good solution, just some one told me to consider it.
 

Thread Starter

Kelvin Lee

Joined Oct 22, 2018
109
Dropping 6 volts at 13.7 Amps is 82
Watts. You don’t want to do that.

Either use a buck converter or get a different battery. A 5 cell lithium battery is 21 V.

Bob
Dear BobTPH,

By calculation, yes, 5 cell lithium battery is 21V, sorry that I forgot to mention the capacity is also the consideration, now my battery is 24v12ah.
 

Thread Starter

Kelvin Lee

Joined Oct 22, 2018
109
The overall efficiency when using a single resistor to drop the voltage is about 79 percent.
That's pretty good by old switching converter standards, but modern converters can do 90 percent.
The run time of the single resistor solution though is not as good as with the 90 percent efficiency buck converter.
With the buck you'll see about a 14 percent increase in run time. If that is significant then go with the buck, but if not then you only have to worry about what you are going to do with all that extra heat (82 watts resistor vs 33.5 watts buck).
Dear MrAI,

How is a modern converter? Is it a circuit design and individual electronic component?
 

Thread Starter

Kelvin Lee

Joined Oct 22, 2018
109
Hi,

Sorry but he definitely does not want to use a voltage divider.
Dear MrAI,

No, a voltage divider is not my only one solution, just someone told me about this but I don't know how to calculate the need. I know about the single resistance using Ohms Law calculation but don't know why the guy suggested me the voltage divider.
 

Thread Starter

Kelvin Lee

Joined Oct 22, 2018
109
If the characteristics of the motor are well known, you do not need R4. The motor itself forms one leg of the voltage divider.

Surf around and look up a 0.43 or 0.47 ohm, 200 W resistor, then consider a buck regulator module from ebay, It will be smaller, lighter, cooler, and deliver a constant voltage to the motor as the battery discharges.

ak
Dear AnalogKid,

Yes, I can use Arduino by changing the pin value to control the motor speed, is it that you said the motor itself already forms one leg of the voltage divider?
 

shteii01

Joined Feb 19, 2010
4,667
Dear AnalogKid,

Yes, I can use Arduino by changing the pin value to control the motor speed, is it that you said the motor itself already forms one leg of the voltage divider?
Motor has some electrical resistance. Therefore you can represent motor by its resistance. Then you have a simple circuit where you have three parts: battery, resistor one, resistor that represents motor.

For example. This motor: https://www.digikey.com/product-detail/en/crouzet/82740402/966-1741-ND/3190266 has resistance of 4.2 Ohm. Then your circuit becomes: battery, resistor one, 4.2 Ohm.
Since you want 24V across your 4.2 Ohm resistor, the voltage divider formula becomes:
\(
24V=28V\frac{4.2\Omega}{R1+4.2\Omega}
\)

Solve for R1.
 

mvas

Joined Jun 19, 2017
538
Dear Sir/Madam,

I only have a battery with fully charged at 28V and a 24V motor, I need going to join a motor car competition, the rule is the voltage should be limited below 22V, I plan to design a voltage divider to lower the voltage from the battery output. When I test with the battery directly connect to the motor, the current is 13.7A, in this case, how much resistance value I have to use and how many watts of the resistor need?

Best regards,

Kelvin
what does " ... should be limited below 22 volst " mean
Does that mean 21 Volts or 21.9999 Volts?

Here is a 28 Volt to 22 Volt Buck-Inverter
20 Amps Continuous
30 Amps Peak
https://www.amazon.com/General-Step-down-Adjustable-Voltage-Regulator/dp/B01MQ045LQ

The Buck Converter needs to supply the Locked Rotor amps of your motor.
 

mvas

Joined Jun 19, 2017
538
Motor has some electrical resistance. Therefore you can represent motor by its resistance. Then you have a simple circuit where you have three parts: battery, resistor one, resistor that represents motor.

For example. This motor: https://www.digikey.com/product-detail/en/crouzet/82740402/966-1741-ND/3190266 has resistance of 4.2 Ohm. Then your circuit becomes: battery, resistor one, 4.2 Ohm.
Since you want 24V across your 4.2 Ohm resistor, the voltage divider formula becomes:
\(
24V=28V\frac{4.2\Omega}{R1+4.2\Omega}
\)

Solve for R1.
The Volts ( and Amps ) required for "Constant Velocity" is only one part of the project.
The Amps required for hard acceleration, must also be considered.
A resistor will reduce the Amps during hard acceleration, which will result in slower acceleration.
 
Last edited:

Thread Starter

Kelvin Lee

Joined Oct 22, 2018
109
Motor has some electrical resistance. Therefore you can represent motor by its resistance. Then you have a simple circuit where you have three parts: battery, resistor one, resistor that represents motor.

For example. This motor: https://www.digikey.com/product-detail/en/crouzet/82740402/966-1741-ND/3190266 has resistance of 4.2 Ohm. Then your circuit becomes: battery, resistor one, 4.2 Ohm.
Since you want 24V across your 4.2 Ohm resistor, the voltage divider formula becomes:
\(
24V=28V\frac{4.2\Omega}{R1+4.2\Omega}
\)Solve for R1.
Dear shteii01,

Sorry I don't understand what the Omega mean.

How can you find the 4.2 Ohm from the datasheet? I can't find it, is it by calculation? How? Do you mind teaching me?

By the way, Solve for R1, how about R2? Remove it?

Best regards,

Kelvin.
 

Thread Starter

Kelvin Lee

Joined Oct 22, 2018
109
The Volts ( and Amps ) required for "Constant Velocity" is only one part of the project.
The Amps required for hard acceleration, must also be considered.
A resistor will reduce the Amps during hard acceleration, which will result in slower acceleration.
What is the hard acceleration mean?
 
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