# resistor + db question



## Etac (Aug 26, 2008)

edit: let me reword this

.. ok i need to bring a speaker with 4 ohm resistance down 9 db's..

so if im doing it correct
the total amount of resistance i would need to add would be 12 or 28.. (without including speakers resistance) ..?

im confused on whether i need to double the inline restistors as resistance overall is increased...?


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## Stephen_Sawyer (Jan 10, 2010)

I'll give this a shot-
If you simply place a resistor in series with the 4 ohm speaker, you'll need the resistor to be 7.5 ohms to get a -9.1dB change.

Here's some math behind it:
Vspkr = Rspkr * Vamp/(Rser+Rspkr)
dB = 20*log(Vspkr/Vamp)

You'll need to watch the power rating of the series resistor (Rser). For example, if the amplifier output voltage is 5v, there will be ~1.4W dissipation in Rser.
Pser = (Vamp/(Rser+Rspkr))^2 * Rser

hope this helps...


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## Etac (Aug 26, 2008)

i guess a better question for me would be

ok say you had a 4 ohm speaker and a 4 ohm resistor wired in series..

now i know the amp would see an 8 ohm load.. to simplify things.. lets say we were giving it 100 watts at 4 ohms.. at 8 ohms the amp would put out 50 watts 

now take the 4 ohm speaker and 4 ohm resistor... would the speaker see all 50 watts, or would the resistor eat half the power, causing the speaker to see only 25?


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## Stephen_Sawyer (Jan 10, 2010)

Yes, if the amp is putting out 50W then the power will be split between the two equal loads in series- 25W a piece. That's a lot of power for a resistor.


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## zoomer (Aug 2, 2009)

Etac said:


> edit: let me reword this
> 
> .. ok i need to bring a speaker with 4 ohm resistance down 9 db's..
> 
> ...


You have not mentioned what configuration you have. If this speaker is connected to a crossover, then adding a resistor in series will also change the crossover frequency! so plase be carefull.
If that is the case then you need an Lpad which is a constant resistance attenuator.


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## Etac (Aug 26, 2008)

yeah i know... the crossovers are being made by me too.. i was gonna ptut the resistor e before it.. but i always thought an l-pad was more complicated, but apparently its rather simple lol, so ill be using that..

but now im just wondering... 

consider my question before.. i understand that the power is split 50/50 when the resistance is the same.. but now say we had a 4 ohm speaker and an 8 ohm resistor.. giving an overall resistance of 12.. the speaker would carry 1/3 the resistance while the resistor has 2/3...

now.. would the speaker see more power or the resistor? i know in parralel the speaker would, but the series part confuses me... i know its a noob question.. but part of me thinks the speaker would cause it obviously has less resistance, but then again since we are talking about series.. it has to go through the resistor.. so would the speaker see 2/3 of the generated power or 1/3? (simplified, i dont care about heat loss.. imp rise.. or whatever else you ppl wanna throw in the equation =P)


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## zoomer (Aug 2, 2009)

Etac said:


> yeah i know... the crossovers are being made by me too.. i was gonna ptut the resistor e before it.. but i always thought an l-pad was more complicated, but apparently its rather simple lol, so ill be using that..
> 
> but now im just wondering...
> 
> ...


What you are saying then is that you are using woofer/mid/tweeters that vary widely in sensitivity and therefore you have to *pad* one or more of the drivers. I hope it is not the woofer!. 

Still, even if you put a resistor in series between the amp and the crossover you are still changing the xover frequency! That is ok if you are doing the design yourself and can adjust L and C values accordingly 

here is a link to pie or T attenuation networks. 
attenuators


as for power thru the speaker or resistor. 

lets assume Rs is the series resistor and Rd is the driver resistance

Starting with ohms law of V=IR or I= V/R
and then Power= I^2 x R or Power= (V^2)/R

The current going thru both resistors is the same, and is =I 
therefor the power dissipatd by the series resistor is I^2 x Rs
and the power dissipated by the driver is I^2 x Rd

The power dissipated by the resistors , with I^2 cancelling out, proportional to the resistance ratio. In other words if Rs is half of Rd then the power dissipated in Rd is half of what is dissipated in Rs. 

Imagine 3 equal 4 ohm resistors in series.. each will dissipate 1/3 of the power. Combine 2 of the resistors onto one 8 ohm resistor and it will dissipate 2/3 of the power, with the remaining 4 ohm resistor disspating 1/3.

You can use the same logic with parallel resistors. Use 3 8ohm resistors in parallel. Each will dissipate the same power. Combine 2 of the resistors together to get a 4 ohm resistor. It will dissipate 2x the power of the remaining 8 ohm resistor!.

The beauty of ohms law is you can use it for series or parallel.
In Parallel resistors both see the same voltage and you use P=(V^2)/R 
This is an inverse relationship and a smaller resistor will dissipate more power because if R gets smaller then P gets bigger. The 4 ohm resisstor will dissipate twice as much power as the 8 ohm resistor.

In Series resistors both see the same current and you use P=(I^2) x R
This is a direct relationship and the smaller resistor will dissipate less power because when R gets smaller, so does P. 
The 4 ohm resistor will dissipate half the power of the 8 ohm resistor. 

Hope this straightens things out. 
Remember Ohms law V=IR and the P=IV equations are simple High School math and are your friend either for DC or for AC calculations thru resistors.


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## Etac (Aug 26, 2008)

i would liek to thank you both for your time and responses

also no this is not for a woofer, its for a tweeter that wont be seeing that much power..

and i will be using an L-pad, the last question was just sheer curosity =]


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## zoomer (Aug 2, 2009)

Etac said:


> i would liek to thank you both for your time and responses
> 
> also no this is not for a woofer, its for a tweeter that wont be seeing that much power..
> 
> and i will be using an L-pad, the last question was just sheer curosity =]


9 db is quite a difference in sensitivity! That is like 10x power! Are these drivers you picked by choice or stuff you just got your hands on? 
Please dont tell me you are using a very inneficient car subwoofer with an efficient horn tweeter!


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## Etac (Aug 26, 2008)

lol its from a collection of various parts i've had over the years, i know they arent an ideal match, but i may as well put em to use.

the sub in it, is the 6.5 tang band sub (20-600hz).. TB 3 inch mids (600 - 6khz) and a Vifa tweeter.. (6k-20k)

the vifa tweeter is slightly horn loaded and has a high sensativity (around 93 @ 1w).. the tb sub is very low (only around 83 db's @ 1watt).. i got them on sale awhile ago for half price from pe.. even though they are kind of ineffcient, they are impressive little guys... they can take about 125 watts, free air down to 20hz, and not bottom out (12 mm xmax one way).. but i put the sub in a .75 cubes enclosure ported to around 40 hz... 

and another note i gotta add, i've used a couple $10-$20 tweeters from vifa, and for their price, they are pretty nice.. i also hear people sometimes say bad things about TB.. now im not saying they are the best.. and some of their stuff can be a *little* pricey, but honestly, the few parts i've gotten from them, i have nothing but good things to say.


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