Zobel network - resistor wattage

[activexp]

Hi all,

Just wondering how you go about calculating the correct wattage for a resistor in a Zobel network on a woofer.

I've used a 10 watt on my economy 2-way boxes:
http://sites.google.com/site/stevess...peaker-project

At the moment they are connected to a Denon DM37DAB rated at 30W/ch RMS into 8 ohms but could be connected to amps of higher power. No idea if 10W resistors are sufficient

Any advice greatly appreciated

[MaxS]

Hello,



To my mind, a simple approach is possible here. Voltage across resistor is equal to output amp (voltage across capacitor is nile due to sinusoidal current), so power is defined by:

Pr = E²/R. (1)

E = sqrt(P*R) (2)
E ~ 16V

(1)+(2)=> Pr = 256 / R

Let you show us your ohmic value



MaxS

[pacificblue]

Capacitor and resistor form a frequency-dependent voltage divider. Xzobel = Xc + R, with Xc = 1 / (2 * PI * f * C). At DC all voltage will be across the capacitor. The higher the frequency, the more voltage will drop across the resistor, but at the same time the cross-over will reduce the total voltage across the Zobel as frequency increases.

10 W will probably be generous for your speakers.

[tinitus]

I have used 5watt cement many times without problems
However I do use 10watt in some places, just to be sure
My tweeter series resistor is 5watt, and have never failed
For tweeter series resistor 5watt sounds better(0R8)
For paralel resistors 10watt seem to sound slightly better

btw, a stripped cement sounds good
If a lead has broken I have just soldered copper leads on it, and it works too
But there are different kinds of cement, when you look at how the ends/leads are made

[Inductor]

Originally posted by tinitus
I have used 5watt cement many times without problems
However I do use 10watt in some places, just to be sure
My tweeter series resistor is 5watt, and have never failed
For tweeter series resistor 5watt sounds better(0R8)
For paralel resistors 10watt seem to sound slightly better

btw, a stripped cement sounds good
If a lead has broken I have just soldered copper leads on it, and it works too
But there are different kinds of cement, when you look at how the ends/leads are made

Hey tinitus, activexp did mention a woofer (>more energy).

[activexp]

Originally posted by MaxS
Hello,



To my mind, a simple approach is possible here. Voltage across resistor is equal to output amp (voltage across capacitor is nile due to sinusoidal current), so power is defined by:

Pr = E²/R. (1)

E = sqrt(P*R) (2)
E ~ 16V

(1)+(2)=> Pr = 256 / R

Let you show us your ohmic value



MaxS

It's an 8.2 ohm resistor. I'll leave you to do the maths as my calculator is broken!


Many thanks to everyone for all the helpful replies

Regards,
Steve

[MaxS]

Hello all,



Interesting things all around there : practical experiences, full frequencies study and so on.


So, R = 8,2 [ohm].

Pr = 256 / 8,2 ~ 24 [W]

This value seems high. HOWEVER, but this is sized for constant full power.
Virtually not happened except if you are listening as a deaf or you nut

Pr = sqrt(P*Z)^2/R => (P*Z)/R
Pr = 10

=> (Pa*Z)/R = 10 <=> 10 R / Z = Pa.

R ~ Z
=> Pa = 10 W

Your ten wattage rated resistor seems to be designed for a ten watt constant value. You'll see, is it enough ! (for your neignbours, at least)



MaxS

PS : if your calculator is broken, Google should calculate it for you

[pacificblue]

Originally posted by MaxS
Your ten wattage rated resistor seems to be designed for a ten watt constant value.

What a surprise.


Why so complicated? An amplifier that is rated as 30 W into 8 Ohm will deliver ~30 W into an 8,2 Ohm resistor connected directly across the speaker terminals. No need for a calculator.

But if the resistor was connected like that, the speaker and the resistor would lead to a ~4 Ohm load for the amplifier, which is not the case. You have to include the capacitor in your calculations.

With the information in post #3, you can derive the impedance curve of the Zobel circuit. Then you need to add the cross-over to see, how the voltage is reduced at higher frequencies. A 5 W resistor would be fine in that place for normal use. A 10 W resistor is enough for worst case scenarios and/or more powerful amplifiers.

[Inductor]

Originally posted by MaxS


Pr = E²/R. (1)

Following the diagram for the First Order (6db/octave) xover here http://sites.google.com/site/stevess...t/xoverdiagram I get a simulation like this pic.

For a voltage that is approximately equal to the full 7V peak, or 5VRMS (for ~30W on 8ohm res. load).

v15 (blue) is the voltage in respect to frequency. Max. voltage (peak) below xover freq. at ~5.2 volts/3.7 vRMS.

with equation (1) you have 5.2x5.2/8=3.38 watts

(v6 - woofer simulat.)
(v1 - tweeter simulat.)
(v15 - zobel R)

[activexp]

Perhaps I should add that I'm now using a 2nd order XO here:
http://sites.google.com/site/stevess...oved-crossover

But in any event it seems that a 10 watt resistor is more than adequate.



Regards
Steve