Will this work?

Thread Starter

Lightfire

Joined Oct 5, 2010
690
Another "Will this work?" question asked by me.:D

The question is..

Will 75Ω resistor be fine for my experiment?:confused:

My radio's specifications are 3.0V, 0.14 A (based on my multimeter's reading) I guess it was 148 milliamps.

I used the formula given by SgtWookie. And found out that the resistor I need is 77.428571428 . I chose 75Ω because it is the closest among 68 and 82Ω.

Anyway, I am not about of wasting the power. I just want to see if it will work or not.

P.S. My power source will be 13.8 V .

Formula

Rlimit = Vsupply - (LED*numbers in series) / Ampere
13.8V 3.0V * 1 0.14
13.8V 3.0V 0.14

10.8V/0.14A

=77.428571428 OR 75Ω

Final question: Will radio work with 13.8 V battery with 75 ohms connected?

thanks and pls. answer if it will work.

any help will be appreciated.
 

Wendy

Joined Mar 24, 2008
21,929
When designing with electronics it is pretty common to select the closest value as you've done. It will probably work fine, but you could always do the math in reverse, and see what the part will do for your application.
 

Thread Starter

Lightfire

Joined Oct 5, 2010
690
When designing with electronics it is pretty common to select the closest value as you've done. It will probably work fine, but you could always do the math in reverse, and see what the part will do for your application.
But how could I do the math reverse?
 

Wendy

Joined Mar 24, 2008
21,929
You calculated the value you needed. Now calculate what the value you've chosen will do. If you are powering up an LED, for example, see what the current with the new value will do. It will be very, very close. This is more an exercise in math, the part will work fine.
 

Thread Starter

Lightfire

Joined Oct 5, 2010
690
You calculated the value you needed. Now calculate what the value you've chosen will do. If you are powering up an LED, for example, see what the current with the new value will do. It will be very, very close. This is more an exercise in math, the part will work fine.
I guess, I have done math reverse.;)

Wait, why when I get my multimeter and read the voltage of the battery (I put already the resistor), the value is still 13.06 something close to that. But I am wondering why when I connect it to the radio, the radio runs and it doesn't explode as what I expected because the volts is 13.06.

Also, the resistor is becoming hot as time runs, I guess I read it somewhere. But why?

Thank ya very much and I hope for any answers.
 

iONic

Joined Nov 16, 2007
1,650
I guess, I have done math reverse.;)

Wait, why when I get my multimeter and read the voltage of the battery (I put already the resistor), the value is still 13.06 something close to that. But I am wondering why when I connect it to the radio, the radio runs and it doesn't explode as what I expected because the volts is 13.06.

Also, the resistor is becoming hot as time runs, I guess I read it somewhere. But why?

Thank ya very much and I hope for any answers.
Resistors dissipate the power they consume as heat. If it is getting hot then you might want to increase the power rating of the resistor. Instead of a 1/4W resistor, use a 1/2W resistor.
 

praondevou

Joined Jul 9, 2011
2,942
I'd only use a resistor to step down voltage for a device whose power consumption is constant. A radio power consumption varies with volume.

Right now you have about 10mA. (0,74 V on 75 Ohm) If this radio is still working and you increase volume, voltage on the resistor should increase. (syntonized to a station of course)
 

Thread Starter

Lightfire

Joined Oct 5, 2010
690
Yes catapult, I think you may be going about this wrong. A resistor is not what you need. You should use a voltage regulator like this: http://parts.digikey.com/1/parts/612612-ic-regulator-3v-1a-to220fp-ba03t.html
Thanks for recommending the above link.:D

However, the resistor did not pop. I run my radio by 13 volts (with 75Ω resistor already). I run it for more than 15 minutes, or even 30 minutes. It just became hot, but not that popped.:D

So, why when I get my multimeter and check the readings, the readings was still 13.somethingV not 3.somethingV .

Pls. help...

Catapult
 

strantor

Joined Oct 3, 2010
5,195
Thanks for recommending the above link.:D

However, the resistor did not pop. I run my radio by 13 volts (with 75Ω resistor already). I run it for more than 15 minutes, or even 30 minutes. It just became hot, but not that popped.:D

So, why when I get my multimeter and check the readings, the readings was still 13.somethingV not 3.somethingV .

Pls. help...

Catapult

ok, lets clear this up. where are you measuring with your multimeter? between A & C? Between A& C you will always measure 13V because you are measuring the entire circuit (all the voltage drops). if your resistor is working properly, you should read 3V between B & C and 10V between A & B.

This setup is not proper. Radios are not LEDs, so you can't use the LED formula for this. If your radio has not burned up yet, then it probably has it's own voltage regulator, or it will burn up if you keep using it like that.
 

Attachments

Thread Starter

Lightfire

Joined Oct 5, 2010
690

ok, lets clear this up. where are you measuring with your multimeter? between A & C? Between A& C you will always measure 13V because you are measuring the entire circuit (all the voltage drops). if your resistor is working properly, you should read 3V between B & C and 10V between A & B.

This setup is not proper. Radios are not LEDs, so you can't use the LED formula for this. If your radio has not burned up yet, then it probably has it's own voltage regulator, or it will burn up if you keep using it like that.
Actually, I measure the readings on +13V and 0V .;)
 

wayneh

Joined Sep 9, 2010
16,162
.... it will burn up if you keep using it like that.
Maybe yes, maybe no. Definitely bad design. But it doesn't HAVE to fail, as long as it's always drawing enough juice such that the voltage at the radio stays within specs, less than say 4V. Data collected so far suggests it might be OK. (It ain't dead YET! :D) The B-to-C voltage measurement would help a lot.
 

strantor

Joined Oct 3, 2010
5,195
Actually, I measure the readings on +13V and 0V .;)
ok, then you are measuring between A & C.

Maybe yes, maybe no. Definitely bad design. But it doesn't HAVE to fail, as long as it's always drawing enough juice such that the voltage at the radio stays within specs, less than say 4V. Data collected so far suggests it might be OK. (It ain't dead YET! :D) The B-to-C voltage measurement would help a lot.
good point wayneh

As wayneh said, you should be measuring between B & C
 

SgtWookie

Joined Jul 17, 2007
22,210
If the load were also a resistor, you could use a resistor to drop the voltage.

However, your load is active, and current through it will change as the active circuit demands. As has already been mentioned, your voltage regulation will be very poor.

You really need to use a voltage regulator, with capacitors across both its' inputs and outputs.

If you know what the maximum radio current drain is, you can use a resistor between the battery and the regulator to take some of the heat off the regulator.
 

iONic

Joined Nov 16, 2007
1,650
What sort of radio is this, an mp3 player with radio? Do you know the make and model, a picture???

Are you trying to power the device with the adapter to avoid using batteries?

(Thinking to myself) Maybe a simple 5V cell-phone charger adapter and a T0-92. 3.3V fixed regulator.
 

radiohead

Joined May 28, 2009
508
Resistors resist circuit current. When they do, they develop a voltage drop across them. Resistors are NOT voltage regulators. Use a zener diode circuit or a voltage regulator as stated above. If you are dead-set on using resistors, try a voltage divider circuit.
 

wayneh

Joined Sep 9, 2010
16,162
Resistors are NOT voltage regulators.
And the OP did NOT ask for a voltage regulator, nor best practice. He asked whether something would work, and presented data supporting the answer that yes, it might. Only time will tell. Is it good practice or even recommended? No.
 
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