Resistor Power rating

Thread Starter

Tan Kwan Yean

Joined Jun 15, 2015
7
Hi. If I have input voltage of 14.5 V. input current of 1 A and resistor of 0.25 ohm, how many power rating of resistor that needed so that the resistor won't burst in flame? Thx for advance.
 

Lestraveled

Joined May 19, 2014
1,946
14.5 volts divided by .25 ohms equals 58 amps. You say your input current is only 1 amp. So, your supply can not supply 58 amps. The .25 ohm resistor will be about the same as a dead short.

Edit: I assumed the .25 ohm resistor was the load. My mistake.
 

wayneh

Joined Sep 9, 2010
16,537
Power in watts is I^2•R with I in amps and R in ohms. In your case this is 0.25W. A resistor rated for 0.5W would be advised.
 

wayneh

Joined Sep 9, 2010
16,537
14.5 volts divided by .25 ohms equals 58 amps. You say your input current is only 1 amp. So, your supply can not supply 58 amps. The .25 ohm resistor will be about the same as a dead short.
I assumed the resistor was in series with whatever the REAL load is, not the load itself. But your assumption may be the right one.
 

R!f@@

Joined Apr 2, 2009
9,740
Too little info on how much current flows through the resistor.
Need to know if the R is the load or details on how the resistor is connected.
If connected, connected to what ?
 

Thread Starter

Tan Kwan Yean

Joined Jun 15, 2015
7
The 0.25 ohm resistor is the current limiter for a charger circuit. I have a output from buck converter of 14.5 V and 10 A. This output is split into 10 charging circuit as shown in the attachment. Ideally, after split, each charging circuit will has 14.5 V and 1 A as input. How much should the power rating of resistor Rs?
 

Attachments

john*michael

Joined Sep 18, 2014
43
Looks like the current limiter is set up so that the voltage across the sense resistor is 0.25 volts. Since you are using a 0.25-ohm resistor, you will have a maximum of 1 amp going to the battery, so the power is 0.25V x 1 amp or 0.25W.
 

Lestraveled

Joined May 19, 2014
1,946
Tan Kwan Yean
With only .25 ohms in series with your power supply and battery, small changes in battery voltage with cause large changes in charge current. May I suggest that you consider a higher power supply voltage and a higher resistance series resistor. The higher the voltage and resistances are, the less the charge current will change as the battery voltage changes. Calculate current at max and min battery voltage and you will see that this is true.
 

Lestraveled

Joined May 19, 2014
1,946
OK
Assume a battery low (discharged) voltage of 12.0V and a high (charged) voltage of 14.4V.

Case #1, power supply = 14.5V and Rseries = .25 ohms.

Vbat = 12.0V, Then, I = (14.5 - 12)/.25 = 10 amps
Vbat = 14.4V, Then, I = (14.5 - 14.4)/.25 = .4 amps



Case #2 power supply = 18V and Rseries = 6 ohms

Vbat - 12.0V, Then, I = (18 - 12)/6 = 1 amp
Vbat = 14.4V, Then, I = (18 - 14.4)/6 = .6 amps

OK, do you see the what I am talking about?
 
The UC3906 senses current through Rs in your charging circuit. The device uses the voltage drop across this resistor as feedback to set the maximum current that the charger will provide to the battery. The voltage across this resistor is compared to 0.25 volts and limited to this value by adjusting the base drive for the Darlington pair. So the voltage across the resistor is always 0.25 volts or less.

Your circuit has a 20 volt power supply. The sense current will have .25 volts at 1 amp or .25 watts, as explained above. The BUX10 will have the remaining voltage drop at 1 amp. If you are charging a 12-volt battery, and assuming a 0.7 volt drop across the MUR1540, you will have 20 -.25 (from Rs) - .7 (from MUR1540) -12 (from the battery) or a little over 7 volts from collector-to-emitter of the BUX10. So the BUX10 be the part dissipating your power, in this case 7 volts x 1 amp or 7 watts. It will need to be attached to a heat sink surface.
 
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