The Millennium bypass, and the parasitics it leaves behind
It gets you an indicator LED without a 3PDT, which is why it turns up so often in home-built pedals. The circuit is simple, but it creates one node at a very high impedance, and the quirks all follow from that.
An LED without a 3PDT
A footswitch has two jobs: take the effect out of the signal path when you are not using it (true bypass), and show whether the effect is on.
Two poles for the audio, one for the LED. Wire it the obvious way and you need a 3PDT, which is expensive, stiff to press, and does not last as long.
The Millennium bypass, published by R.G. Keen in 1999, keeps the DPDT and reads the state of a spare contact electrically instead. You save a pole on the switch, and in exchange the circuit now contains one node sitting at around 1011 Ω.
How the two poles are used
Get the wiring straight first, or the symptoms further down will not make sense.
- Pole 1: input jack — output jack (closed when bypassed, open when the effect is on)
- Pole 2: effect output — sense circuit (bypassed) / output jack (effect on)
The effect input stays wired to the input jack in both positions. Only the output side is switched, and that is how the original wiring diagram has it. It is called true bypass, but the input is never disconnected.
All it detects is whether the sense line floats
The indicator circuit checks one thing: is the sense line open, or is it tied to ground through a low resistance?
Bypassed, pole 2 connects the sense line to the effect output. Past the output coupling capacitor there is a volume pot or a pulldown resistor, which is a DC path to ground. It holds the MOSFET gate at 0 V, so the MOSFET is off, the drain stays up at 9 V, and the LED is dark.
With the effect on, pole 2 moves away and the sense line is left open. Nothing is holding the gate down any more, so the reverse leakage of the 1N914 (10–15 nA) from the +9 V rail starts pulling it up. Past the threshold the MOSFET turns on, the drain is pulled down and the LED lights.
Using diode leakage rather than a resistor for the pull-up is the point of the circuit. A 10 MΩ pull-up passes 900 nA, and while bypassed that current charges the effect's output capacitor. The step it leaves behind is the pop you hear when you switch.
What happens before the MOSFET turns on
Worth going through "leakage pulls the gate up" in more detail.
The gate is insulated by an oxide, so almost no DC flows into it (the BS170 gate leakage is specified at 100 nA maximum, and is picoamps in practice). The 10 nA arriving from the 1N914 therefore does nothing but charge Ciss and the wiring capacitance. Call it 60 pF, and the voltage follows ΔV = I × t / C.
A resistor pull-up would give you an exponential. Diode leakage is nearly independent of voltage, so it is a constant current, and the gate voltage climbs in a straight line. At 10 nA into 60 pF that is 0.17 V/ms.
The BS170 is an N-channel enhancement-mode device. At VGS = 0 V there is no channel and no drain current. Current starts once VGS passes the threshold VGS(th). The circuit only works because of this: a depletion-mode part conducts at VGS = 0 V, so there would be no "off" state at all.
VGS(th) for a BS170 ranges from 0.8 to 3.0 V. A 2.0 V part takes 2.0 / 0.17 ≈ 12 ms; a 3.0 V part takes about 18 ms. The same circuit turns on at different times depending on which device you fitted.
Once past threshold, the drain current is set by the series resistor, not by the MOSFET. With a red LED at 2.0 V forward and 2.2 kΩ, (9 − 2.0) / 2200 = 3.2 mA. The MOSFET on-resistance is a few ohms and does not come into it.
The threshold is not a digital edge. In between, the MOSFET behaves as a variable resistor, and because the gate voltage rises so slowly here, you watch that whole region happen in the brightness of the LED.
Pull-up current and the click
While bypassed, the pull-up current I runs down the sense line into the effect's DC output resistance Rint and charges the output capacitor to V = I × Rint. Step on the switch and that voltage is handed to the next stage as a click.
A guitar puts out about 100 mV, so keep the step under 10 mV. That sets the ceiling on Rint:
| Pull-up | Current | Allowed Rint | Usable with |
|---|---|---|---|
| 10 MΩ resistor (Rat) | ≈900 nA | ≤11 kΩ | only low output-impedance circuits |
| 22 MΩ resistor | ≈400 nA | ≤25 kΩ | still restrictive |
| 1N914 leakage | 10–15 nA | 0.7–1 MΩ | almost anything |
The third row is why the Millennium can be retrofitted to nearly any pedal. Changing the pull-up from a resistor to a reverse-biased diode moved the usable range by two orders of magnitude.
Why the LED comes up slowly
Build it and the LED takes about half a second. The calculation in the previous section said 12 ms, so it is roughly 40× slower than that.
The capacitance is not the problem. The current that reaches the gate is two orders of magnitude below the datasheet leakage. What is left after subtracting the clamp diode's reverse leakage, the MOSFET gate leakage and the leakage across the board surface is all you get. Half a second works out to about 240 pA.
On the capacitance side, the node carries Ciss, the board trace and the wire to the switch, and the junction capacitance of the two diodes: 30–80 pF in total.
Things to watch when building it
- Flux residue or humidity will stop the LED lighting. Once the board surface falls to a few hundred MΩ, that is where the 10 nA goes. This is not a circuit where you can skip cleaning.
- Touching the node changes the behaviour. Against 1011 Ω, a fingertip or a dirty joint counts as a conductor.
- It picks up hum. A high-impedance node couples capacitively to everything near it. Run the sense wire past the supply and the LED can flicker.
- The slow turn-on is the specification. To speed it up you want a leakier diode — a gold-doped 1N914 or 1N4148. Fit a good low-leakage switching diode and it will take many seconds.
This is why R.G. Keen added a 2004 note telling people not to use just any diode. Here the leakage is the spec, which puts "better part" the wrong way round.
Bleed-through while bypassed
The most common complaint about this scheme is hearing the fuzz or distortion faintly while it is bypassed. The asymmetry in the wiring is the reason.
As we saw at the start, the effect input is not disconnected while bypassed. The guitar signal keeps running into the circuit, and a 60 dB gain stage keeps producing a fully clipped output. One open switch contact is all that stands between it and the output jack.
An open contact still has 0.3–1 pF across it. Take 0.5 pF and work out the divider against the 1 MΩ input of whatever comes next:
| Frequency | |ZC| | Attenuation | Bleed at the output jack |
|---|---|---|---|
| 100 Hz | 3.2 GΩ | −70 dB | 0.31 mV |
| 1 kHz | 318 MΩ | −50 dB | 3.1 mV |
| 5 kHz | 63.7 MΩ | −36 dB | 15.7 mV |
| 10 kHz | 31.8 MΩ | −30 dB | 31 mV |
The clean signal going to the output jack is about 100 mV. 15.7 mV at 5 kHz is −16 dB against it, which is plainly audible in a quiet passage.
The coupling rises at 20 dB/decade, so the bottom end never gets through and the top end does. That is the fizz people describe, rather than a quiet version of the effect.
Reducing the bleed
It is a product of three things — the input not being cut, the gain, and the contact capacitance — so a fix has to attack one of them.
- Reduce the contact capacitance. Use the outer contacts of the switch and leave the neighbouring ones empty. That alone is worth several times.
- Shorten the sense wire, or shield it. While bypassed it carries the clipped full-level output across the enclosure. Run it alongside the bypass wire and the coupling can exceed the contact capacitance.
- Add a pulldown at the effect output. A lower impedance on that node while bypassed means less couples off it. As the click section showed, going too low brings the pop back, so stop at a few hundred kΩ.
- Cut the input as well. This is what actually works, and it costs a pole, which means going back to a 3PDT.
So the higher the gain, the worse the fit. For a buffer, a boost or a modulation circuit whose output stays in the hundreds of millivolts, the bleed lands below −50 dB and never becomes a problem.
If you were building it today
3PDT switches are cheap now. For anything with real gain, switching both ends with a 3PDT is the safer choice.
It is still worth reading as an example of the trade it makes: one fewer mechanical contact, at the price of a node at a very high impedance. Soft-touch bypass built from a momentary switch, a microcontroller and a relay has the same shape, and so does a reset circuit that holds its state on leakage current.
Once you have a node like that anywhere, the parasitic capacitance and the surface leakage become numbers you have to design around.
Frequently asked questions
Is the Millennium bypass really true bypass?
Does the MOSFET have to be enhancement mode?
My LED will not light. What should I check?
Can I make the LED come on faster?
Standards and references
- R.G. Keen, "The Millenium Bypass", GEOFEX (1999, updated 2004) — The original. Schematics for Millennium 1 (JFET) and 2 (MOSFET), the comparison with the Rat bypass, and the note on low-leakage diodes
- R.G. Keen, "The Technology of Bypasses", GEOFEX — The lineage of bypass schemes and the loss and noise of each
- onsemi BS170 datasheet — VGS(th) range, Ciss and the gate leakage spec, used for the estimates above
- onsemi 1N914 / 1N4148 datasheet — Reverse leakage specification; gold doping is what makes it leaky