Millenium bypass issues

[R.G.]

This is what I wanted to know. So can you explain how? What would the values of resistors be?

The MOSFET will act like a switch, either completely interrupting the current from the power supply, or holding its drain near ground. Within certain limits, you can connect any number of things between +9V and the MOSFET drain.

For each LED you want to light, you need one resistor in series with it. You can then parallel up these resistor+LED units between power supply and MOSFET drain.

To select resistors for any LED,  there are two main ways. One is to just guess, with the knowledge that on a 9V supply 3.3K to 4.7K is generally going to get you some light, and lowering the resistance makes it brighter until it burns out if you get too low and the current goes over 20ma or so for normal LEDs.

The second way is to calculate the resistor to get a specific current. LEDs in general have a light output that is nearly linear with current. Zero current gives zero light. They burn up at much over 20ma for the standard 5mm T 1-3/4 LED package, so 20ma is as bright as they can do continuously. So you guess at a current you want for how bright the LED should be. Ordinary LEDs are kinda dim at low currents, and "super-ultra-hyper-mega-bright" LEDs can be blinding at 20ma.

Then to calculate, do the following:
1. Figure out how much voltage you have across the LED+resistor. In this case, it's 9V (if you're using a 9V supply) minus any voltage across the MOSFET when it's turned on. This time I did go look at the 2N7000 datasheet for values. With the gate higher than the source by 4.5V, the 2N7000 will pull 75ma down to less than 0.4V drain to source. So we have at least 9V minus 0.4V, or 8.6 V to use for the LED+resistor.
2. Look up or measure the forward voltage of your LED. It's easiest to look at the LED datasheet if you have it for "Vf". This will be between 1.5V and 3V, usually.
3. We're trying to figure out what voltage is across the series resistor, so subtract the LED voltage from the previous 9V  minus the MOSFET. Let's say we have a 2V LED; that gives us 9V - 0.4V - 2.0V = 6.6V.
4. Now you pick an LED current. Let's arbitrarily pick 10ma, half as bright as it can get. We know the current through the resistor will be 10ma. We know from the voltage handwaving that the voltage across it is about 6.6V. Ohm's law tells us that the resistor must be 6.6V / 0.01 = 660 ohms. 680 is the nearest standard resistor value. 1K would be dimmer, 470 would be brighter.

Do this once for each LED. If they're all the same kind and color, you can just pick a resistor value once, and use that value on the resistor each LED has.

You can also use a 5K pot instead of a resistor for a test-LED, turn it to the brightness you like, then measure the pot resistance and pick the nearest standard value. If you turn it too low, the LED might die.

Once you know (or have decided) a resistor value for each LED, you hook up each LED with its resistor, and put all of those LED+resistor groups in parallel between +9V and the MOSFET drain.

Since I noticed that the 2N7000 is specified for a maximum of 200ma, and that they only talk about performance with 4.5V and 10V on the gate, it is possible to put too many LED+resistor groups in parallel for the MOSFET to handle. At 5ma, you can put 75/5  = 13 LED+resistor groups in parallel with no problem. At 10ma, you are PROBABLY OK up to 8 groups.

I'm sure I cloudied that more than I cleared it. What did I confuse you with?

[JebemMajke]

Actually it's pretty clear explanation. Thanks. Ps I'm using Bs170. Is Bs170 specified for max 200mA as 2n7000 is or 600mA as you previously mentioned?

[Kesh]

500mA for BS170.

By the way there's this great thing called datasheets:

http://www.fairchildsemi.com/ds/BS/BS170.pdf

[merlinb]

If you leave it out, it would make sense to take electrostatic precautions when you're constructing the circuit and installing it in the pedal, and also whenever you change the battery in the circuit, as when the battery is removed there is not a good clamping element in the circuit to clamp the static sparks under +/-20V on the gate.

Come to think of it, how is the low-leakage diode supposed to stop the gate going 20V above the source?

[Kesh]

the other diode takes care of that.