Beginner doubt - Power section

[lepra85]

Hi,
i am new in the effects building art. Reading the "IC-Based Overdrive"  I realized that there is a resistor of 100 ohm in the power section.
In a lot of schematics this resistor is not included.
someone can help me to understand why is this resistor required?
Having a capacitor for filter the high frecuencies to Ground is not enought?

thanks for read!
lepra85

[amptramp]

The effect of a 100 ohm resistor and the 100 µF capacitor is to create a lowpass filter that has a corner frequency of 15.9 Hz or at least it would have with no load resistance (the IC power supply current and the half supply divider) since the amp is 3.2 Kohm minimum and the divider is 44 Kohm, not much changes.  The idea of the R-C network is to isolate the amplifier from the power supply.

If you are using a battery, it limits the current flow when the unit is turned on.

If you are running from a wall wart supply, it provides some filtering since these supplies usually do not have good filtering.  The resistor limits the pulse current flowing from the wall wart rectifiers and their usually inadequate filtering and prevents supply current changes (due to driving the clamping diodes into conduction) from going back to the wall wart.  If you think of the current through the rectifiers, they have to conduct the average current used by the amplifier but only conduct when the rectified voltage is greater then the capacitor voltage.  This results in large current pulses lasting a short time with a 120Hz repetition rate that would flow down the power lead and radiate to everything else.  The resistor limits this current and forces most of the rectifier current pulse to remain in the wall wart.

[Transmogrifox]

This also helps filter switch popping when you switch an LED (or latching relay) in/out:
http://www.muzique.com/news/connecting-an-led-to-an-effects-pedal/

Specifically,

[lepra85]

Thanks for your prompt answer. Following with the power section, I got another question.
I saw that in some designs the "protective diode" are in series (in the schematic you posted use to be in the R2 place) and other designs the "protective diode" is in parallel with the power jack.
Do you know what are the advantages/disadvantages of each configuration?

[amptramp]

Thanks for your prompt answer. Following with the power section, I got another question.
I saw that in some designs the "protective diode" are in series (in the schematic you posted use to be in the R2 place) and other designs the "protective diode" is in parallel with the power jack.
Do you know what are the advantages/disadvantages of each configuration?

Putting the protective diode in series gives excellent protection but the voltage drop of the diode is subtracted from the power supply voltage and it also has a series resistance in ohms of 26/I where I is the current in milliamps, so a circuit drawing 2 mA would see the diode as being 13 ohms.  With a 9 volt battery that can be assumed to discharge to 7.5 volts before it has to be replaced, the circuit would only get 6.8 volts which may not be enough.  This means the battery would have to be replaced at 8.2 volts, not much less than the new battery voltage.  As an added advantage, this connection is safe from accidental use of an AC supply.

A diode from input to ground connected so it does not conduct current during normal operation retains the full battery voltage.  But if the source impedance of the power supply is low enough, it can drive the diode current high enough to burn out the diode.  If the diode fails short, it still protects the rest of the circuit.  If it fails open, the rest of the circuit gets fried by reverse voltage.  But the battery voltage under normal operation is not affected.

[DiscoVlad]

You can also use an appropriately connected MOSFET as a replacement for series diode protection, which gives you the same benefits but also has a voltage drop proportional to the Rds(on) of whichever device you use. For most pedal loads, this is typically less than 10mV instead of the 0.6V (600mV) that the diode takes.

RG has an article about it here: http://www.geofex.com/Article_Folders/mosswitch/mosswitch.htm

[MrStab]

micro-hijack, something i've always been curious about: i always put the LED path after the filtering section, as i assume it might not get the full benefit of decoupling & it may be more likely to "contaminate" the signal later on. probably negligible if it's a matter of millimetres, but any truth to this concern?

[nocentelli]

i always put the LED path after the filtering section, as i assume it might not get the full benefit of decoupling & it may be more likely to "contaminate" the signal later on. probably negligible if it's a matter of millimetres, but any truth to this concern?

Most of the benefit of the decoupling cap is to remove AC ripple from a noisy DC adaptor, and stop it entering the audio circuit: Your LED will not need totally clean DC to light up. I also have the vague idea that placing the cap to ground between the LED and the audio power might help reduce the potential for LED-related switch pop, but I may have made this up.

[duck_arse]

any current the led draws when supplied through R2 creates an additional voltage drop at V+. when supplied via +9V instead, no drop.