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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.

Last updated: 2026-10-28 Guitar effectsSwitchingMOSFETParasitic capacitance

An LED without a 3PDT

A footswitch has two jobs: take the effect out of the signal path when off (true bypass), and show its state. Wired the obvious way that needs a 3PDT (two poles for audio, one for the LED), which is expensive, stiff to press, and shorter-lived.

The Millennium bypass, published by R.G. Keen in 1999, keeps the DPDT and reads the state of a spare contact electrically to light the LED. In exchange, the circuit contains one node at around 1011 Ω.

A footswitch and the small-signal circuit that controls its indicator LED
A footswitch and the small-signal circuit that controls its indicator LED. AI-generated conceptual illustration; see the article’s diagrams and text for circuit details and conditions.

How the two poles are used

The pole assignments are in the figure legend. Get them straight, or the symptoms further down will not make sense.

The effect input stays wired to the input jack in both positions. Only the output side is switched, as in the original wiring diagram. It is called true bypass, but the input is never disconnected.

Millennium bypass schematicPole 1 of the DPDT ties the input jack to the output jack; pole 2 routes the effect output either to the sense line or to the output jack. The effect input remains permanently wired to the input jack. The sense line runs to the Millennium 2 board, to a MOSFET gate pulled up from +9V by the reverse leakage of a 1N914, with the LED and a current limiting resistor in the drain. IN Bypass path (closed when not engaged) Always connected. Input is never cut Effect circuit Always powered, always fed DPDT Bypassed position Open (unused) OUT This one when the effect is on How the poles are used Pole 1: input - output jack (closed in bypass, open when on) Pole 2: effect out - sense line (bypass) / output jack (on) Millennium 2 board Sense line +9V 1N914 rev. leak 10-15nA Low leakage (NPN C-B) BS170 1k-4.7k LED
Figure 1: the DPDT wiring and the Millennium 2 indicator circuit. The red trace is the permanent connection to the effect input.

All it detects is whether the sense line floats

Bypassed, pole 2 connects the sense line to the effect output. Past the output coupling capacitor, a volume pot or pulldown resistor is a DC path to ground and holds the MOSFET gate at 0 V. The MOSFET is off and the LED is dark.

With the effect on, pole 2 moves away and the sense line is left open. Nothing holds the gate down, so the reverse leakage of the 1N914 (10–15 nA) from +9 V pulls 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. A 10 MΩ pull-up passes 900 nA, which charges the effect's output capacitor while bypassed and becomes a pop when you switch.

The 10–15nA currents and parasitic capacitances below are illustrative assumptions. Reverse leakage depends on voltage, temperature and device variation. A datasheet maximum is neither a typical operating current nor a guaranteed minimum.

How the Millennium bypass switchesWhile bypassed, signal flows from the input jack straight to the output jack, and also into the effect circuit, which is never disconnected. The effect output feeds the sense line, the gate voltage is zero and the LED is dark. Pressing the switch moves both blades together: the effect output is routed to the output jack and the sense line is left floating, the gate voltage rises slowly and the LED fades in. Bypassed (not engaged) Input jack tied straight to the output jack; effect output goes to the sense line Effect on (engaged) Effect output goes to the output jack; the sense line is left floating IN Effect circuit 60dB gain, always running DPDT OUT Sense line Tied to the effect output, so the gate sits at 0V Floating, so leakage pulls the gate up Millennium 2 board Gate voltage V(GS) Example: 10nA into 60pF reaches 2V in 12ms Dark Lit (fades in)
Figure 2: the switching action. The input side never stops conducting; only the output and the sense line change.

What happens before the MOSFET turns on

A MOSFET gate is oxide-insulated, but its leakage is not zero. For this example, assume a net charging current of 10nA after other leakage paths and a total gate-plus-wiring capacitance of 60pF. Estimate the voltage change with ΔV = I × t / C.

With a constant net-current approximation, gate voltage rises linearly with time. Assuming 10nA and 60pF gives 0.17V/ms. Actual leakage varies with voltage and temperature, so this slope is not universal.

The BS170 is an N-channel enhancement-mode device: no drain current at VGS = 0 V. A depletion-mode part conducts at VGS = 0 V, so there would be no "off" state.

Under these assumptions the gate reaches 2V in 12ms and 3V in 18ms. Datasheet VGS(th) is specified at a small drain current, not at a guaranteed visible LED brightness. If the MOSFET is sufficiently on and its voltage drop is negligible, a red LED assumed to drop 2.0V with 2.2kΩ in series carries (9 − 2.0) / 2200 ≈ 3.2mA.

What happens before the MOSFET turns onOn the left, the circuit: about 10 nanoamps of reverse leakage from the +9V rail through a 1N914 flows into the gate and charges the gate and wiring capacitance, about 60 picofarads. Top right, a graph showing the gate voltage rising linearly with time and crossing the 2 volt threshold, at which point the LED begins to light. Bottom right, the transfer characteristic: an enhancement-mode device passes no drain current at zero gate voltage, whereas the depletion-mode curve drawn as a grey dashed line conducts at zero gate voltage and so can never be switched off. Current into the gate raises its voltage, and the LED lights once it passes the threshold Current flowing into the gate +9V 1N914 (reverse biased) Gate I(G) = 10nA C(iss) + wiring = 60pF N-ch enhancement mode I(D) +9V LED 2.2k Gate voltage now V(GS) = 0.4V V(GS) = 1.4V V(GS) = 2.0V = V(GS,th) V(GS) = 3.5V I(D) = 0, LED dark I(D) flows, LED lit 1. Gate voltage rising V(GS,th) = 2V (BS170: 0.8 to 3.0V) starts lighting here 4 2 0 V(GS) [V] constant current, so dV = I x t / C time 2. Transfer characteristic, I(D) against V(GS) 3mA I(D) 0 V(GS,th) V(GS) = 0 gives I(D) = 0 enhancement mode depletion mode The grey dashed curve is a depletion-mode part: it conducts at V(GS) = 0, so it can never be dark, and cannot be used here After turn-on the resistor sets the current: (9V - V(F)) / R = (9 - 2.0) / 2.2k = 3.2mA
Figure 3: current into the gate raises its voltage until the threshold is crossed and the LED lights. Bottom right is the transfer characteristic; the grey dashed curve is a depletion-mode part.

Pull-up current and the click

While bypassed, the pull-up current I runs into the effect's DC output resistance Rint and charges the output capacitor to V = I × Rint, which reaches the next stage as a click when you switch. A guitar puts out about 100 mV, so keep the step under 10 mV.

Pull-up method versus the largest effect output resistance that keeps the click under 10 mV
Pull-upCurrentAllowed RintUsable with
10 MΩ resistor (Rat)≈900 nA≤11 kΩonly low output-impedance circuits
22 MΩ resistor≈400 nA≤25 kΩstill restrictive
1N914 leakage10–15 nA0.7–1 MΩalmost anything

The third row is why the Millennium retrofits so easily: diode leakage instead of a resistor moved the usable range by two orders of magnitude.

Pull-up current and the clickLeft: pull-up from +9V through a 10 megohm resistor, about 900 nanoamps. Right: pull-up through the reverse leakage of a 1N914, 10 to 15 nanoamps. In both, the current runs down the sense line into the effect output and through the DC resistance Rint after the output capacitor, giving V = I x Rint. To keep the click under 10 millivolts, Rint may be up to 11 kilohms on the left and 0.7 to 1 megohm on the right. The pull-up current while bypassed makes the click The current through Rint gives V = I x Rint, which reaches the next stage when you switch. A guitar is about 100mV Pull-up: 10MΩ resistor +9V 10MΩ about 900nA Sense line Rint V = I x Rint Cout Effect output Click under 10mV: Rint up to 11kΩ Pull-up: 1N914 reverse leakage +9V 1N914 10-15nA Sense line Rint V = I x Rint Cout Effect output Click under 10mV: Rint 0.7-1MΩ
Figure 4: the pull-up current charges the output capacitor, and V = I × Rint becomes the click.

Why the LED comes up slowly

Charging time depends on the net current reaching the gate and its capacitance. If 60pF takes an assumed 0.5s to charge from 0V to 2V, I = C × ΔV / t gives a net current of 240pA. This is a calculation with assumed conditions, not a measurement by this site. Diode, MOSFET and board-surface leakage affect the net current, so timing alone does not establish the cause of a delay.

The node carries Ciss, wiring and two diode junctions: 30–80 pF in total.

Parasitics left in the Millennium bypassTop left, the open switch contact while bypassed still has 0.3 to 1 picofarad across it. Below it, the sense line and the bypass wire run in parallel and couple through wire-to-wire capacitance. Right, the sense node carries the MOSFET input capacitance plus wiring, 30 to 80 picofarad total. At the bottom, the equation for the time it takes a leakage current to charge that capacitance. 1. What stays connected in the signal path The contact is open, but not to a capacitor Effect output clipped, about 1V Open (unused) OUT Contact capacitance C(sw) 0.3-1pF Sense line (carrying full output) Bypass wire (the clean signal) Wire-to-wire capacitance 2. What hangs on the sense node A 10¹¹ ohm node charged by a few nanoamps +9V 1N914 (10-15nA) Sense node Low-leakage clamp C(iss) + wiring 30-80pF to the MOSFET gate Example: 60pF x 2V / 10nA = 12ms (assuming 0.5s implies 240pA net) What is left after the clamp, gate and board-surface leakage are subtracted from the pull-up
Figure 5: what is still connected while bypassed. Left, the signal path (the open contact, and the sense and bypass wires running together). Right, the capacitance of the sense node itself.

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, the 10 nA goes there. Do not skip cleaning.
  • Touching the node changes the behaviour. Against 1011 Ω, a fingertip or a dirty joint counts as a conductor.
  • A high-impedance node picks up hum. Run the sense wire past the supply and the LED can flicker.
  • The slow turn-on is the specification, and leaky gold-doped diodes (1N914, 1N4148) suit it. That is why R.G. Keen's 2004 note warns against low-leakage diodes.

Bleed-through while bypassed

The most common complaint is hearing fuzz or distortion faintly while bypassed. The effect input is not disconnected, so a 60 dB gain stage keeps producing a clipped full-level output, and 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 the next stage:

Contact capacitance 0.5 pF, 1 MΩ load, effect output 1 Vrms (clipping hard)
Frequency|ZC|AttenuationBleed at the output jack
100 Hz3.2 GΩ−70 dB0.31 mV
1 kHz318 MΩ−50 dB3.1 mV
5 kHz63.7 MΩ−36 dB15.7 mV
10 kHz31.8 MΩ−30 dB31 mV

The clean signal is about 100 mV, so 15.7 mV at 5 kHz is −16 dB against it, plainly audible in a quiet passage. The coupling rises at 20 dB/decade, so only the top end gets through, as fizz.

Bleed-through at high gainBelow the clean bypass path from input jack to output jack runs the leakage path: from the input jack into the permanently connected effect input, through a 60 decibel gain stage to about 1 volt, then across 0.5 picofarad of open switch contact back to the output jack. At 5 kilohertz the budget is 15.7 millivolts of bleed against 100 millivolts of clean signal, or minus 16 decibels. How sound leaks through while bypassed The input is never cut, so the gain stage keeps producing full output IN Clean bypass signal, about 100mV OUT Gain stage (60dB) runs while bypassed, output about 1V Open (unused) Open contact 0.5pF 63.7 megohm at 5kHz Level budget at 5kHz 100mV guitar output (0dB reference) about 1V gain stage output (+20dB) -36dB 0.5pF against 1 megohm 15.7mV bleed at the output jack (-16dB) Coupling rises 20dB per decade, so what gets through is the top end, not the body of the sound
Figure 6: the bleed path and the level budget. The input is never cut, so the gain stage produces full output the whole time you are not using it.

Reducing the bleed

It is a product of the uncut input, the gain, and the contact capacitance, so a fix has to attack one of them.

  1. Reduce the contact capacitance. Use the outer contacts of the switch and leave the neighbours empty; that alone is worth several times.
  2. Shorten the sense wire, or shield it. It carries the clipped full-level output, and alongside the bypass wire the coupling can exceed the contact capacitance.
  3. Add a pulldown at the effect output. Less bleeds, but too low brings the pop back, so keep Rint to a few hundred kΩ.
  4. Cut the input as well. This works best, but 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, and for anything with real gain, switching both ends is safer. It is still worth reading as an example of the trade: one fewer contact, at the price of a very high-impedance node, where parasitic capacitance and surface leakage become design values.

Frequently asked questions

Is the Millennium bypass really true bypass?
The output is disconnected, so the effect does not colour the signal. But the input stays connected, so the effect's input impedance (usually 0.5–1 MΩ) still sits across the guitar while bypassed.
Does the MOSFET have to be enhancement mode?
Yes. A depletion-mode part passes drain current at VGS = 0 V, so the LED would not go dark while bypassed. The BS170 and 2N7000 are both N-channel enhancement mode.
My LED will not light. What should I check?
Clean the board first: flux residue forming a few hundred MΩ path swallows a 10 nA pull-up. Then check the diode type; a low-leakage part takes seconds or never lights. Finally, confirm pole 2 really leaves the sense line open.
Can I make the LED come on faster?
From ΔV = I × t / C, reduce C or raise I. Raising I (a leakier diode) works better than shortening the sense wire. If you need under 0.1 s, use a 3PDT or a relay.

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 limits and test conditions (page 2); these do not guarantee 10–15nA at 9V.

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