# Will this work?

Discussion in 'The Projects Forum' started by Lightfire, Aug 27, 2011.

1. ### Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
Another "Will this work?" question asked by me.

The question is..

Will 75Ω resistor be fine for my experiment?

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.

2. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
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.

3. ### Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
But how could I do the math reverse?

4. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
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.

5. ### Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
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.

6. ### iONic AAC Fanatic!

Nov 16, 2007
1,422
68
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.

7. ### JingleJoe Member

Jul 23, 2011
185
10
hold on hold on, you say your radio requires 3 volts ... but you are going to use 13V? Surely this will make it go pop.

8. ### praondevou AAC Fanatic!

Jul 9, 2011
2,936
489
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)

Oct 3, 2010
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10. ### Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
Thanks for recommending the above link.

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.

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

Pls. help...

Catapult

11. ### strantor AAC Fanatic!

Oct 3, 2010
4,302
1,989

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.

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12. ### Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
Actually, I measure the readings on +13V and 0V .

13. ### wayneh Expert

Sep 9, 2010
12,368
3,224
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! ) The B-to-C voltage measurement would help a lot.

strantor likes this.
14. ### strantor AAC Fanatic!

Oct 3, 2010
4,302
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ok, then you are measuring between A & C.

good point wayneh

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

15. ### SgtWookie Expert

Jul 17, 2007
22,183
1,728
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.

16. ### iONic AAC Fanatic!

Nov 16, 2007
1,422
68
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.