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In every Boss compact pedal there’s one part of the circuit that is identical and that’s the flipflop. The flipflop is the part of the pedal that switches the effect on or off. Our discussions about problems or modifications often concern the flipflop so I decided to explain how this circuit works. I think I got it pretty well figured out but if something is incorrect or unclear, please reply to this post. The explanation is quite detailed so minimal knowledge of electronics are needed in order to understand the document.
The flipflop is a standard circuit that we can find in many devices. It is also called a bistable multivibrator. Bistable means that it has two stable states. In our case this means that the pedal can be either on or off. Below you can see the schematic for the flipflop as it is implemented in most Boss pedals. Here the component numbering matches the TW-1 schematic.
A transistor has 3 legs named collector, base and emitter. They can be used as an amplifier or as a switch. In the flipflop circuit we use both transistors as switches.
When the base is applied a voltage of around 0.6V higher than emitter, the internal resistance between collector and emitter will decrease. The transistor will start conducting and a current will flow from collector to emitter. In this case we say that the transistor is on. When the base voltage is further increased the transistor will become saturated. If the base-emitter voltage falls below 0.6V, the internal resistance between collector and emitter will be very high. No current flows from collector to emitter and in this state the transistor is off. We say that the transistor is in the cut-off state.
Until we apply a 9V voltage to the circuit both transistors are off. Their internal resistance is very high and no current flows through them. There is not an infinite resistance to ground so there will flow a current through the circuit. One current goes through R41, R37 and R38. Another current goes through R40, R36 and R39. We can ignore the capacitors as these will act as infinite resistances to DC currents. Because the resistor values in each of the two paths are the same, the same current will flow through each path. This current will be
I = V/R = 9/(56k+100k+56k) = 42uA.
Both transistors’ emitters are connected together and grounded at 0V potential. It is the voltage at the base of each transistor that determines whether it turns on or remains off. The base voltage of Q1 is the same as the voltage across R38 and the base voltage of Q2 is the same as the voltage across R39.
V(R38) = V (39) = I * R = 42uA * 56k = 2.3V
That more than enough to bring the transistors into saturation. There will always be tiny differences in the characteristics of the transistors and also some difference between the resistors in the circuit. This will cause one of the transistors to turn on a little before the other one. Let’s see what happens if Q1 starts conducting a current before Q2 does.
The base voltage of Q1 reaches 0.6V first and a current will flow through R40 and continue from Q1’s collector to emitter. The resistance between collector and emitter will be small compared to R40 and this will lead to the collector voltage dropping. The base voltage of Q2 will be about 1/3 of the collector voltage of Q1 so this will drive Q2 further into cut-off. The datasheet for the 2SC945 transistor that is used here tells us that the typical collector-emitter voltage when saturated is 0.15V. That means that the rest of the 9V will be across R40. The voltage across R39 will be around 0.05V and Q2 will be in cut-off while Q1 is saturated.
When Q2 is in cut-off, its collector voltage will be high because there is a high resistance between collector and emitter. The voltage is 9V minus the voltage across R41.
Vc = 9V – V(R42) = 9V – (I * R42) = 9V – (42uA * 56k) = 6.7V
There will be 2.3V (9V- 6.7V) over R41, This voltage will also be across D2 and the LED. D2 is a zener diode rated at 5.1V. Until the voltage across it is higher than 5.1V, no current will flow through it and this means that the LED will in this case remain dark.
S1 is the switch that we can use to turn the pedal on. When S1 is pressed, Q1 will stop conducting, Q2 will start conducting and the flipflop changes from one stable state to another. Let’s examine more closely what causes this to happen.
When S1 is pressed the potential at the bottom of C18 and C19 will quickly go from 9V to almost 0V. Keep in mind that a capacitor acts as a shortcut to AC voltages and as an isolator to DC voltages. The switch creates what is effectively an AC pulse when the positive voltage suddently goes to zero. Consequently the voltage drop on C18 and C19 will lower the voltage on the transistors base terminals as well. The Q1 will immediately stop conducting and it’s collector voltage will rise. The quickly increasing collector voltage on Q1 is transferred directly to Q2’s base through C16. This voltage will be way above the 0.6V needed to get Q2 conducting and Q2 is quickly brought to saturation.
As with Q1 earlier, the collector voltage of Q2 should now be around 0.15V. If our supply voltage is exactly 9V, V(R41) will be 8.85V and this voltage will also rest across D2 and the LED. D2 will attempt to maintain a stable voltage around 5.1V across itself and the LED will have the remaining 3.75V. The LED should in this case light up and the pedal will have turned on.
The explanation above only told us how the LED was turned on or off. In addition to the LED, the collector voltage of one or both transistors will be used to control the gate voltage of an n-channel JFET transistor. Here’s a simplified look at how this is implemented on the HM-3.
The FET transistor is similar to a bipolar transistor but its characteristics with a very high input resistance and low noise makes it ideal for use as a switch. The connections are named drain, gate and source compared to the bipolar transistors collector, base and emitter. Similar to a normal transistor, the current between drain and collector is controlled by the voltage applied between gate and source. The JFET will however conduct fully when V(G-S) is 0V and not at all when V(G-S) is around -4V. It is not completely correct to say that a JFET is in cut-off mode or saturation like a bipolar transistor. I will still use this terminology as it explains well how the circuit works and it’s close enough to the truth.
The source terminal of the JFET is connected to the +4V supply through a 220k resistor. When the flipflop’s Q1 collector is low, there will be no voltage on the gate and V(G-S) ~ -4V. In this state the JFET is cut-off and no signal passes through it. When the flipflop switches, Q1’s collector will increase to about 6.7V. This will give us a positive V(G-S) and the JFET is brought to saturation. The signal will then flow freely from drain to source. Typically there will be a similar circuit connected to the collector of Q2 so that one FET will always conduct while one is always cut off.
Some people favour pedals with true bypass. In these pedals the flipflop and JFET switch is replaced with a mechanical switch. This switch will be in the signal path so the disadvantage is that a loud click can often be heard when the pedal is turned on or off. On the upside it means that the true bypass pedal can be designed so that it doesn’t draw any current and the signal doesn’t pass through any components when the pedal is turned off.
