So I've tried that one from analog alchemy, that one from AMZ website, first two from geofex ( NPN and Nfet ) and the one from general guitar website. Only the Nfet bypass versions did work, but they had some leakage ( in off position they would lit up in 5-10 seconds ). Maybe because I've used Bf 245 and 2sk30, but those are the fet's i had. What am I doing wrong? Any ideas?
I haven't tried millenium 2 ( the one with with mosfet ), it's on the menu today.
Go with the M2. I build loads and they're very reliable (thanks R.G.!). Just make sure your FX output has a route to ground (otherwise the LED will fade back in). BTW, I like to build mine on scraps of vero: I have 6x6 and 4x9 layouts for the offcuts I seem to generate...
I'm just about to solder the diodes and I'm not sure if they connect with input or go "above" it?
(http://www.geofex.com/article_folders/millenium/milckt3.gif)
Quote from: JebemMajke on September 24, 2012, 09:52:58 AM
I'm just about to solder the diodes and I'm not sure if they connect with input or go "above" it?
Input, gate, and both diodes connect together there.
That's a four-way connection. BTW, the "low leakage diode" is the C-B junction of an NPN transistor (the example from GEOFEX is 2N3904). Be sure to check out "The Millenium C(ontinues) (http://www.geofex.com/Article_Folders/Millenium/The%20Next%20Millenium.pdf)" at GEOFEX.
(Looks like R.G. got in there first while I was typing!)
FYI, here's what I do with my vero offcuts:
(http://www.bouron.org.uk/marc/M2boards.gif)
Q1 is BS170. If you substitute a 2N7000, face it the other way. Q2 is 2N3904. As R.G. suggests, season to taste with whatever resistor values suit.
Ok, here goes, it works! But I have some issues. I've placed 50k trimpot in place of resistor and bc550c as low leakage diode. Bs170 and green led. If resistance is zero led is at it's fullest but there is almost no sound. If I introduce some resistance the sound is back but the light is somewhat dimmed. I would like to be at it's shiniest and to have full volume. The circuit I've tested this with was and is Fetzer valve plus 100k resistor at the output.
Can you post schematics of those please, i don't know how to read vero :icon_redface:
Mosfet millenium from this pdf here ( http://www.geofex.com/Article_Folders/Millenium/The%20Next%20Millenium.pdf ) works great :)
Thanks R.G. for coming up with this :)
And thank you bluebunny for your help.
Ps i made it really small 1x2,3 cm. I guess I could make it smaller, but trimpot takes a lot of space.
Quote from: JebemMajke on September 24, 2012, 10:53:12 AM
Ok, here goes, it works! But I have some issues. I've placed 50k trimpot in place of resistor and bc550c as low leakage diode. Bs170 and green led. If resistance is zero led is at it's fullest but there is almost no sound. If I introduce some resistance the sound is back but the light is somewhat dimmed. I would like to be at it's shiniest and to have full volume. The circuit I've tested this with was and is Fetzer valve plus 100k resistor at the output.
Can you post schematics of those please, i don't know how to read vero :icon_redface:
I'm a little confused by what you've said here, and I'm not sure if you're using the M2 in the right way. First of all, if you're talking about using a trimmer in place of the LED resistor (R1 in the schematic below), then zero is a bad thing - you'll be supplying a full 9V to the LED, which will burn it out. :o Anything over 390R will work fine (for a bog-standard red LED) - my choices of around 1K5 give a balance between brightness and current consumption. BTW, the other resistor (10K in my layouts; R2 in the schematic below) is pretty much a random value according to R.G. and may even be omitted.
Anyway, to the second point. If you're finding a relationship between LED brightness and the volume of your circuit, then it's most likely wired up wrong! :) When your FX circuit is "on", it's
not connected to the M2. So one cannot affect the other. When your FX circuit is "off", its output is connected to the control input of the M2, providing a DC route to ground, and this switches the LED off. I routinely use the wiring diagram in the "offboard wiring (http://www.tonepad.com/project.asp?id=35)" article at Tonepad (click on the tonepad_offboardwiring.pdf link). Check out page 3 for wiring up a Millenium board.
BTW, this is the M2 schematic I'm working from:
(http://www.bouron.org.uk/marc/M2schematic.gif)
This is from the "Next Millenium" PDF on R.G.'s site which you've already found.
There weren't any issues relating led-killing-volume, the circuit I've used for testing wasn't biased properly. Hence the "volume issues". For that test I've used this schematic (http://www.geofex.com/article_folders/millenium/milckt3.gif), and the led survived with zero ohms xD.
And on the last post I've used that schematic you've posted in the previous post, and that exact wiring :) and it works since than.
Quote from: JebemMajke on September 26, 2012, 11:06:23 AM
... and it works ...
This is the most important detail! ;)
Having issue with Millenium 2 and fuzz face, LED won't turn off.
Here's schematic
(http://i1249.photobucket.com/albums/hh505/JebemMajke/BottomsUpFuzz-1_zps3d32fdd4.gif)
Any idea?
Ps, for low leakage diode I've used bf224. I have a lot of them and they are useless in any other project. Could that be the problem?
Edit
In order to replace bf224 with the usual bc548 I took it out and all of a sudden voila, it works??? I've placed bc548 and bc549 and other NPN transistors and it only works when there is nothing in place of low leakage diode.
Quote from: JebemMajke on October 24, 2012, 05:43:40 PM
Ps, for low leakage diode I've used bf224. I have a lot of them and they are useless in any other project. Could that be the problem?
Edit
In order to replace bf224 with the usual bc548 I took it out and all of a sudden voila, it works??? I've placed bc548 and bc549 and other NPN transistors and it only works when there is nothing in place of low leakage diode.
You don't actually need the low-leakage diode- it doesn't do anything useful. I think RG put it there to supposedly protect against gain-source overvoltage, but that simply can't happen in a 9V circuit. The low leakage diode is redundant.
Quote from: merlinb on October 25, 2012, 05:49:35 PM
You don't actually need the low-leakage diode- it doesn't do anything useful. I think RG put it there to supposedly protect against gain-source overvoltage, but that simply can't happen in a 9V circuit. The low leakage diode is redundant.
I would say that another way. The low leakage device is not active in normal circuit operations. It is a protective device against electrostatic transients. It is only active during installation if you put the Millenium Bypass in as a retrofit, or if your circuit has some hefty electrostatic discharges during operation. It is redundant and does nothing useful in the same way that a fire extinguisher does.
Quote from: R.G. on October 25, 2012, 07:14:04 PM
It is only active during installation if you put the Millenium Bypass in as a retrofit, or if your circuit has some hefty electrostatic discharges during operation.
As long as the circuit runs off less than about 20V, you can't get gate-source overvoltage anyway, since the gate will be clamped to the supply voltage by the other diode. So it's redundant in the same way as a fire extinguisher in a room that can't catch fire!
electrostatic at the gate doesn't really care what the supply rails are doing, if it's below the potential of the 9V rail then hundreds or thousands of volts will go through the mosfet if it's without protection diodes. normally this is a zener series pair to ground rated above the gate voltages expected, but that doesn't really apply in the M2.
rooms that don't catch fire are like people that don't walk across carpets prior to picking up their pedals. not that common.
Can I use multiple LEDs? Like 3 or 4?
Quote from: merlinb on October 26, 2012, 08:27:13 AM
Quote from: R.G. on October 25, 2012, 07:14:04 PM
It is only active during installation if you put the Millenium Bypass in as a retrofit, or if your circuit has some hefty electrostatic discharges during operation.
As long as the circuit runs off less than about 20V, you can't get gate-source overvoltage anyway, since the gate will be clamped to the supply voltage by the other diode. So it's redundant in the same way as a fire extinguisher in a room that can't catch fire!
You are absolutely correct, in theory. As Yogi Berra said, "in theory, theory is the same as practice; in practice, it isn't". In a properly constructed circuit, once the parts are all soldered in, there isn't any predictable way that the voltage on the gate could get outside the power supplies. In a properly constructed fireproof room, you don't need a fire extinguisher - until someone walks in with a can of gasoline, or even a can of paint. Probably - probably - you don't even need the extinguisher then.
However, Mother Nature has taught me that the world can be a complex place, which is simply another way of saying that theories are abstracts, simplified models of how things work when there are no other confounding or hidden factors.
The original version of that circuit did not have the low leakage diode in it, and the circuit works fine without it. It's the single most confusing part in the circuit, and the subject of probably 75% of the questions I get on the circuit. So I go through the explanation that it's not needed for operations, but is a protection device only, another time for every few beginners that try it. As a side light, about 20% of the questions go "my Millenium 2 LED is always on - what's wrong? I did every thing exactly correct." In that case, it is almost always getting the MOSFET turned around backwards.
By the way, I'm sure you're familiar with the fact that 86% of all statistics are made up on the spot, right? :icon_lol:
So here's the pronouncement I generally make - Hey! To everyone making a Millenium 2: the low leakage diode is a protection device. It's mostly never needed. I advise you to put it in there, or I would not have put it into the schematic. But you're free to leave it out, and probably - probably - your Millenium 2 will work fine without it.
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. Likewise, be careful about plugging in daisy chains, as this exposes the power and ground lines to electrostatic arcs and also disconnects the battery, so that electrostatic sparks can get at your circuits, including any Millenium 2s. This is very unlikely, of course, so if you just want to take responsibility for debugging and replacing parts in the unlikely event it ever does happen, you can choose to leave the low leakage diode out.
Quote from: JebemMajke on October 26, 2012, 08:51:56 AM
Can I use multiple LEDs? Like 3 or 4?
For the LED it lights? Depends on how you hook them up. LEDs take 1.5 to 3V mostly, so if you want to run from 9V and have 3V LEDs and want to stack them in series, the best you can do is two - there has to be some voltage left over from the power supply for the series current limiting resistor. If you hook up a bunch of LED+resistor sets in parallel, you can drive lots of LEDs. The common MOSFETs for this are the BS170 and 2N7000, which are good up to ... 600ma, I think. I didn't look it up. At 10ma per LED+resistor, that's a whole bunch of LEDs you can light.
Did you mean can you use LEDs for the low leakage diode??
QuoteDid you mean can you use LEDs for the low leakage diode??
Nope
QuoteIf you hook up a bunch of LED+resistor sets in parallel, you can drive lots of LEDs.
This is what I wanted to know. So can you explain how? What would the values of resistors be?
Quote from: JebemMajke on October 26, 2012, 11:05:47 AM
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?
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?
500mA for BS170.
By the way there's this great thing called datasheets:
http://www.fairchildsemi.com/ds/BS/BS170.pdf
Quote from: R.G. on October 26, 2012, 10:38:10 AM
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?
the other diode takes care of that.