Zener wattage for effects?

[Laika]

Sorry if this is obvious but what wattage zener should I be using in general for the power side of things when a layout doesn't specify.. Is 400mw good for everything in a 9V/12V stompbox? And can higher wattage (1W) be used/interchanged the same as different wattage resistors can be?

Thanks   

[defaced]

It depends, 400mw wouldn't do squat if not protected against a 9v battery, but it may be fine depending on how it's used.  Got a schematic?  Going up in power is always permissible, going down is risky unless you've done the math and understand the circuit well enough to understand what was intended, and what you're doing by changing it. 

[Laika]

Effects like the Madbean Firebomb or Macheen.. they ask for 9.1v zeners but no wattage mentioned.

[R.G.]

Mike is correct.

Power dissipated is always P = I * V, where I is the current through the device and V is the voltage across it. V is the zener voltage, because zener action will limit the voltage to this amount.

You need to know the current through the zener to tell how much power it will generate. For anything but very brief peaks, the zener rating must be larger than the power it generates.

You can use the P = I * V equation backwards to find the current. If your zener is 400mW and (for example) 5.1V, then the maximum average current it can stand without burning out is I = P/V = 0.4V/5.1V = 0.0784A, or 78.4ma.

If you subbed in a 1W, 5.1V zener, then the max current is I = 1W/5.1V = 196ma.

Where you get into trouble is where you can't figure out the current through it. If you hook a 5.1V zener across a 9V battery, what happens? The 9V battery is determined to make its output voltage be 9V, the zener is determined to make it 5.1V. The only thing that limits the current at all is the internal resistance of the battery and the wires. 9V batteries can put out an amp or more when shorted, so the internal resistance is less than about 9 ohms. There is a 9V - 5.1V = 3.9V difference between what the battery wants and the zener wants, so that voltage has to appear across the internal resistance of the battery, so that's at least 3.9V/9 = 433ma, nearly half an amp, so the zener dissipates 2.21W. If it's a 5W zener, the battery fries. If it's a 400mW zener, the zener fries.

It is critical to know how much current flows through the zener. If there's a resistor in series with it, you can calculate the currents. If there are other parts, you can usually calculate it. If you can't calculate it, you don't know how hot the zener will get.

It all depends on where it is in the circuit.