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DrNomis wrote: Leds typically need a maximum of 20mA to light, the brightness of the led varies in proportion to the supply current, being a Diode, it will exhibit a constant voltage-drop across it, if you use a resistor-and-led to drop 9V down to 1.6V, and are worried about impedance effects, there's a simple solution, just bypass the led with a 470uF/16V electrolytic capacitor, you can use the led as a power on indicator too....

Not so sure -beware, i do not want to begin a war here- because, if you choose a low current for the led, a big cap will provide a low AC impedance at first, but on sustained playing it would be inefficient.
You cannot feed a 50 watts amp with a 10 watts power supply, no matter the filtering caps you put in it.
As an example, i was working on a comp this week, a transparent one with no delay, no squash etc, it's filtering is a 150 ohms resistor and a 100µf cap for a max 5ma consumption (not at all the same as the FZ-1A, it's an example not comparison) and it was squashy as hell, tiny lows etc.Took me a lot of time to figure out that my circuit was as good as my prototype, but i was using an old battery
So, in all cases (resistive divider or led regulator) the source must be feeded with 5 to 10 times the current that the circuit will draw.

kleuck wrote:

DrNomis wrote: Leds typically need a maximum of 20mA to light, the brightness of the led varies in proportion to the supply current, being a Diode, it will exhibit a constant voltage-drop across it, if you use a resistor-and-led to drop 9V down to 1.6V, and are worried about impedance effects, there's a simple solution, just bypass the led with a 470uF/16V electrolytic capacitor, you can use the led as a power on indicator too....

Not so sure -beware, i do not want to begin a war here- because, if you choose a low current for the led, a big cap will provide a low AC impedance at first, but on sustained playing it would be inefficient.
You cannot feed a 50 watts amp with a 10 watts power supply, no matter the filtering caps you put in it.
As an example, i was working on a comp this week, a transparent one with no delay, no squash etc, it's filtering is a 150 ohms resistor and a 100µf cap for a max 5ma consumption (not at all the same as the FZ-1A, it's an example not comparison) and it was squashy as hell, tiny lows etc.Took me a lot of time to figure out that my circuit was as good as my prototype, but i was using an old battery
So, in all cases (resistive divider or led regulator) the source must be feeded with 5 to 10 times the current that the circuit will draw.

Well, the basic function of a capacitor in a DC circuit is to store energy, when the circuit wants to draw more current from the power supply, the stored energy in the capacitor adds to the energy being drawn from the supply, so if the circuit is being powered from say a 9V battery, the capacitor acts like a second 9V battery that's connected in parallel with the one powering the circuit, this effectively doubles the output current capability of the supply, and therefore lowering the impedance of the supply, you will find that the supply doesn't sag as much as it would without the capacitor, , note that the capacitor only supplies it's stored energy to the circuit during peak current demands, the rest of the time it's being charged by the battery....

It stands to reason that the larger the capacitor is in value, the more energy it can store....

Note that this is assuming that the capacitor is being used to bypass the supply...

DrNomis wrote: note that the capacitor only supplies it's stored energy to the circuit during peak current demands, the rest of the time it's being charged by the battery....

Yes, providing the circuit is able to give the current consumed by the circuit, plus enough to recharge the cap.
If the impedance of the regulator is equivalent or higher than the impedance of the circuit, it's impossible (50 watts amp with a 10 watt power supply : only the first chord will be a 50 watts one -with chance-) so in the case of a led regulator or voltage divider, the reg/divider must be able to give more current than the circuit needs, whatever the cap you use.

kleuck wrote:

DrNomis wrote: note that the capacitor only supplies it's stored energy to the circuit during peak current demands, the rest of the time it's being charged by the battery....

Yes, providing the circuit is able to give the current consumed by the circuit, plus enough to recharge the cap.
If the impedance of the regulator is equivalent or higher than the impedance of the circuit, it's impossible (50 watts amp with a 10 watt power supply : only the first chord will be a 50 watts one -with chance-) so in the case of a led regulator or voltage divider, the reg/divider must be able to give more current than the circuit needs, whatever the cap you use.

That's true, the amplifier's power transformer should be capable of supplying the current during peak current demands, otherwise the supply will tend to sag, I'd recommend choosing a power transformer capable of delivering at least twice, or maybe two and a half times the current that the powered circuit is expected to draw from the supply, so that the amp can deliver it's rated power at peak demands....say for example the circuit is expected to draw 500mA from the supply, I would therefore use a 1.5A power transformer....

The thing we must not forget is that the current requirements of a 50 Watt amplifier, and an FZ-1 Fuzz circuit is vastly different....

Just an analogy

kleuck wrote:Just an analogy

Yeah I like using analogies too, they make explaining things alot easier, for example the flow of electricity can be likened to the flow of water through the plumbing system of a house, when I was studying Basic Electronics at Uni, we had a Lecturer who was like an ex-army Sergeant, he wanted assignments done in triplicate, I kid you not, anyway he used to use some pretty funny analogies in his lectures....

I used the same with young mens : height of the tap (is it the right word ?) figuring the voltage, and the water flow the current, the head of the kid under meausring the power