9v wall wart question.

[ljoe1969]

my 9v 200ma wall wart puts out 16.70 volts.

is there a way i can add a resistor or something to make it be actually be 9v to run my pedals?

joe

[Andr3as]

When you put it under a load (connect a pedal) it will drop to around 9V.
You could make a voltage divider or put in a voltage regulator if it still is high when under a load.

[Mark Hammer]

I concur.

[GGBB]

To elaborate a little (because this is one of the few things I actually know a little about) - your power supply is unregulated and therefore relies on load to bring its voltage down.  As already mentioned - probably nothing to worry about since it is "rated" for 9V.  There's also a chance that your pedals have regulation/over-voltage protection built in.  If you're really worried, search on power supply regulation (100ohm resistor + 9.1V Zener diode), or buy a regulated power supply.

[Mark Hammer]

If it's unregulated (or more likely, just poorly regulated), the bigger concern is not what the meter reads, but rather the hum you will likely hear when powering your effects with it.  You'll want to smooth that power out more.

[ljoe1969]

this wall wart originally was for a baby monitoring system.

joe

[asatbluesboy]

Is it rated for 16V? It should say it somewhere.

[LucifersTrip]

a bunch of my 9V adapters that are un-regulated measure ~ 14.x  volts under no load

also, note that most normal adapters (probably the one you have for the baby monitor) are center (+) and the ones used for most pedals use center (-)

[ljoe1969]

it says on the casing 9v 200mA

wiring it for center positive is already done.

when tested with the volt meter it reads 16.70 volts or there abouts.

i was told earlier to use a 9.1V zener diode and a 100 ohm resister.

i just don't know how to connect them.

joe

[R.G.]

it says on the casing 9v 200mA

What that actually means is "it will drop to no lower than 9V when there is a 200ma load on it; at less load, it may be higher".

wiring it for center positive is already done.

What he was telling you is that this is *backwards* from the way almost all pedal power jacks do it. If you don't rewire the adapter to be center negative, then it will reverse power every pedal you plug it into. If you rewire your pedals so they take this adapter OK, then your pedals will be reverse powered by any normal/other adapter.

when tested with the volt meter it reads 16.70 volts or there abouts.

See the answer to the first item.

i was told earlier to use a 9.1V zener diode and a 100 ohm resister.
i just don't know how to connect them.

There's actually a problem there. Your adapter provides 16.7Vdc open circuit and 9V (we think) when the load is 200ma. That means it acts like a 16.7V voltage source running through a resistor already. The resistance can be approximated by using Ohm's law: R = V/I = (16.7-9)/0.2 = 38.5 ohms.

If you put a 100 ohm resistor in series with the output and a zener to ground (that's how to connect them, BTW) and take the voltage off the resistor and zener junction, then the zener will conduct and limit the current voltage to 9.1V. The current that flows is I = (16.7-9.1)/(100+38.5) = 54.8ma. And you can then power any pedal up to a maximum of 54.8ma. If your pedal(s) use more, the voltage will sag lower than 9.1

[GGBB]

There's actually a problem there. Your adapter provides 16.7Vdc open circuit and 9V (we think) when the load is 200ma. That means it acts like a 16.7V voltage source running through a resistor already. The resistance can be approximated by using Ohm's law: R = V/I = (16.7-9)/0.2 = 38.5 ohms.

If you put a 100 ohm resistor in series with the output and a zener to ground (that's how to connect them, BTW) and take the voltage off the resistor and zener junction, then the zener will conduct and limit the current voltage to 9.1V. The current that flows is I = (16.7-9.1)/(100+38.5) = 54.8ma. And you can then power any pedal up to a maximum of 54.8ma. If your pedal(s) use more, the voltage will sag lower than 9.1

R.G. - in that case, if he needs more current, could he not get away with just the zener in reverse across the PS?  That would give him up to I = (16.7-9.1)/(38.5) = 197.4ma.  Or does he need a real resistor of some sort otherwise something will go wrong.  How's about just a 10ohm to give 156.7ma?

[R.G.]

R.G. - in that case, if he needs more current, could he not get away with just the zener in reverse across the PS?  That would give him up to I = (16.7-9.1)/(38.5) = 197.4ma.  Or does he need a real resistor of some sort otherwise something will go wrong.  How's about just a 10ohm to give 156.7ma?

That works fine - as long as the zener can take the heat.

If there is no load, then the power supply still looks like 16.7V in series with 38.5 ohms. The current that flows is (16.7-9.1)/38.5 = 197.4 ma. And the zener has to eat all of that. So the power dissipated in the zener is 9.1V * 0.1974 = 1.8W. So you need a zener rated at least 2W, probably 3 to 5W to survive with some degree of reliability.  The no-load power waste in the zener is much worse than when there is a load to eat some of the power.

To use a 9.1V 1W zener, you have to limit the current to 1W/9.1V = 110ma, and so you have to add a resistor to the internal impedance of the power supply to do this. The necessary resistor is ((16.7V - 9.1V)/0.11A) - 38.5 = 30.6 ohms. And with that, the 1W zener will heat up to it's full 1W temperature, which is probably 85-100C, when there is no load.

It's possible, but using a shunt regulator like a zener requires just as much thought as a series regulated supply, just different thoughts.

[GGBB]

Thanks.