Measuring voltage of a power supply?

[DeusM]

Hey guys! So far you people have taught me a lot about electronics, but since I'm a beginner there are lots of basic things I don't understand so I will be bothering you again with some questions 

The way I measured the voltage of my power supply was from the positive of the power supply input jack (the other ring, its negative on the inside and positive on the outside) to the positive of the output guitar cable that goes to the other pedals or the amp. Now that I think of it there must be something wrong but the values I get was the same for all the pedals. 9.7 volts (I think that's too much). It didn't change now matter the pedals where on or of.

So my question is. Did I make it wright? How should I measure voltage? Is there a difference if the pedals are connected to the amp?

Hope you guys could help me 

[PRR]

Outside of a center-Negative power plug, to outside of an audio jack, is probably valid.

A regulated supply will hold its voltage for any load within its ability. It may deviate 0.1V, but probably less, often not enough to read.

9.7V is certainly "OK" for all "9V" pedals. A fully-fresh battery can be 9.36V, give or take a bit for chemistry details. 9.7V is only 4% "high", a typical tolerance for regulators. Certainly not enough to hurt anything.

[DeusM]

Thancks! I'll try to build the "ultra clean power supply" beacuse mine is a pretty cheap one and has some noise. Not a problem right now becxuse I'm not playing live but I'm gonna start a new band and need to get rid of all the noise for rehaersals.

[antonis]

Is there a difference if the pedals are connected to the amp?

Pedals are connected to the amp by a jack which transfers ONLY signal (or, at least, it should do it..)

So, there isn't any coming DC into amp - even for the in-brackets case there is a DC blocking capacitor right after amp's input jack..
(well, maybe not exactly after the jack in case of Hi-Low Input selection but let it be for the moment..) 

[Lizard King]

Thancks! I'll try to build the "ultra clean power supply" beacuse mine is a pretty cheap one and has some noise. Not a problem right now becxuse I'm not playing live but I'm gonna start a new band and need to get rid of all the noise for rehaersals.

Put a 1 - 2 mH choke (coil) in series with your + output and a 220 - 560 uF cap across + & -.  Careful with the polarity...caps make a nice popping sound when they explode. 

[antonis]

Chokes are forbitten words...!!! 

[DeusM]

Put a 1 - 2 mH choke (coil) in series with your + output and a 220 - 560 uF cap across + & -.  Careful with the polarity...caps make a nice popping sound when they explode. 

Is that so I dont have a voltage drop? I read about it somewhere. I know it's supouse to filter but it works with a resistance too
I took this from the guy thas has the DIY guitar pedals channel



Not shure why the diference in the coil

[antonis]

D2 doesn't need to be a Scottky (in fact, it doesn't need to be placed at all..)

Same for C1 despite of inductor or resistor existence..

[DeusM]

Scottky?

[thermionix]

Scottky?

Short for Scott County, Kentucky.

[DeusM]

So it's a  kentuchy cap? Still don't get it. I'm not from America you know   we no hablar ingles in my country señor

[thermionix]

He means Schottky diode.

[DeusM]

Oh! Sorry. I infused the C with the D. Don't know what a Schottky diode is anyway. Gonna google it right now. I guess I ruined the joke 

[DeusM]

Oh! It's the rectifier right? It's also called regulator or I'm wrong?

[Phoenix]

Oh! It's the rectifier right? It's also called regulator or I'm wrong?

A diode is a component that only allows current to flow in one direction, from anode to cathode (the cathode is the end marked with the bar). Diodes are used for rectifiers, but also for other purposes like blocking current, or their most common usuage in effects pedals, for clipping a signal (introducing distortion). All solid-state diodes have a forward voltage, that is the voltage that needs to be exceeded before they will conduct any current, but once they start conducting, the voltage across them stays almost exactly the same regardless of current (until they finally burn up). This is non-linear behaviour.
This is different to resistors which will conduct at any voltage, and as current through them increases, so does the voltage (or vice versa), in a linear manner.
There are different types of diodes, germanium which conducts about 0.3V, silicon (which are most common) conducts around 0.7V, LED's which start around 1.8V for red, and go up to 3V for white and blue, and there are schottky diodes, which conduct around 0.2-0.3V. Each has their own benefits and drawbacks depending on the application.

Here, those diodes are being used for polarity protection. D1 is a series protection diode - it takes away some voltage from your circuit, but it makes sure that if the battery or a reverse polarity power supply is accidentally connected, no reverse voltage can get through to the pedal circuit and damage it. Because it's taking away some voltage and we normally only have 9V to work with, we want to minimize the amount of voltage we're losing, so we wouldn't use an LED. Instead here a 1N5817 schottky diode is recommended, which will only lose about 0.2V, so our pedal still gets 8.8V.
D2 is actually doing absolutely nothing in that circuit. A diode placed in the same way is very often seen in effects pedals though, but only when D1 is not also present. If D1 is not present, then when the correct voltage is connected, D2 is reverse biased (remember current only flows in one direction through diodes), so it does nothing, and the effect gets the whole 9V. However, if the power supply is reversed, the diode will "clamp" the voltage coming from the battery or power supply to its forward voltage (0.3-1V depending on the type of diode used), and will eat up all the current the battery or power supply can put out. This means that the effect only sees ~1V of reverse voltage, which is usually not enough to damage the effect itself, but the diode and the battery or power supply are now in a fight to the death. The diode might beat a battery, but it will almost always lose against a power supply unless the power supply is smart enough to detect that there's a problem and turn itself off - otherwise the diode will burn up (possibly catch on fire), and the circuit will then have no protection and be destroyed anyway. So, it will protect against clumsy hands accidentally putting a battery in backwards for a few seconds, but there are some people that feel it's a way for pedal manufacturers to easily tell if you've been using a power supply other than the one they sell so they can regect warranty claims - "unathorised power supply used, warranty void".
Anyway, as I mentioned, D2 does nothing when D1 is also in place, because the way they are connected, they conduct in opposite directions, so if a battery or power supply is connected backwards, there is nowhere for the current to flow, so it's pointless to add D2 if D1 is fitted.

Now a regulator is a completely different thing. A regulator takes some varying voltage in, and does its best to put out a smooth, stable voltage out. There are a few different types of regulators, but the most common type you'll see in pedals are so-called linear regulators. These usually have three pins, in, ground and out. They have what is called a dropout voltage - if the input to output voltage difference falls below this, then they stop regulating and you get noisy output voltage instead. This is usually 2V for linear regulators - so to get 5V out, you need at least 7V in. One of the most common places you'll see linear regulators used is in smoothing voltage that has come from a transformer, through a rectifier and into a filter capacitor. The filter capacitor does most of the work, but will have ripple voltage at twice the mains frequency (100 or 120 hertz). The regulator then smooths this to nearly perfectly stable DC voltage. But this ripple voltage must always stay above the dropout voltage or the regulator will stop regulating. If you want 5V out of the regulator, and there's 2V of ripple on the main filter cap, then the peak voltage on the filter cap must be at least 9V (5V out + 2V dropout + 2V ripple). You also want some extra margin in there for variation in wall voltage, so 10V would be a better bet. If you can't change the voltage of the transformer but you still have too much ripple voltage, then you'd need to increase the value of the capacitor instead to get less ripple voltage.

Ok, that got off topic fast. Hope it was useful anyway....

[DeusM]

Thanks for the explanation. I already know what a diode is, just dint't know about schottky and regulators. Some interesting stuff. I also know they are used for reverse polarity protection and as clipping. I'm not sure if I'm going to make that power supply anymore. Just found out my power supply is regulated and filtered. I guess I hear noises because other reasons.

but the diode and the battery or power supply are now in a fight to the death

Pretty funny explanation 

[bluebunny]

A Schottky diode is formed by a metal-semiconductor junction.  Compare with a "regular" diode which is semiconductor-semiconductor.  Schottkys are faster and have lower V_F.

[antonis]

If I only knew that a forgotten "h" should cause so much trouble... 

[DeusM]

If I only knew that a forgotten "h" should cause so much trouble... 

Hey! No trouble. Now I know a little more. I read about it some time ago but I forgot. I will nerver forget now