# LPB1 circuit analysis

The LPB1 (Line Power Boost 1) is a boost pedal that was commercialized in 1968 by Electro Harmonix. It is the first boost pedal using a silicon transistor. Indeed, previous boosts like the Dallas Rangemaster Treble Booster were using germanium transistors. Another novelty with this pedal was that it was boosting all frequencies and not only trebles, making it the first clean boost available!

The circuit is really the most simple one you can imagine for a boost: First, there is a 0.1uF coupling capacitor that prevents parasitic DC current from the guitar to go in the circuit. With the R2 resistance, it forms a high pass filter : by changing its value, you can modulate the amount of bass going through the circuit. If you increase C1 value, more bass will go through, and vice versa.

Then, there are two resistors forming a voltage divider (R2 and R1), to provide a certain voltage to the base of the transistor. Here it is : R2/(R2+R3)x9V = 43/(43+430)x9=0,81V at the transistor's base.

The silicon transistor is a 2n5088 (originally a 2n5133 - same transistors that were used in the Big Muff later), wired as a common emitter. R4 and R3 will define the amount of amplification. If you increase R4, amplification will be larger. If you increase R4, there will be less gain.
A second 0.1uF coupling capacitor that prevents DC current from the battery to go out of the circuit. Finally, a 100k potentiometer wired as a variable resistor defines the final volume.

If you look carefuly, you can see that the last stage of the Big Muff circuit is exactly the same! A LPB1 circuit is used to increase the final volume.

Here is a global schematic for what does what: (click to enlarge)
##### 8 Comment Thank you for the schematic and analysis. As an ammateur electronic gadgets DIYer I tried to follow but hit a snag in the following two points:

i) from where the biasing of the transistor comes

and

ii) the function(s) of R3 and R4. Can you help me please?

In i) it appears that biasing comes from R1 and R2 but then later you say that R3 and R4 also bias Q1.

In ii) it appears that R3 and R4 work ..."against" each other!? To me it sounds like instead of complementing they "balance" each other. So what if I want to introduce a variable resistor in place of one of these resistors to provide a control for increasing/reducing gain? At the same time if R3 and R4 affect the biasing in conjunction with R1 and R2 then how introducing a pot will affect the entire design, thus final output? Thanks for your time and explanation. The transistor here is laid out in common emitter topology.

The bias voltage comes from the voltage divider formed with R1 and R2, here the bias voltage is simply VccR2/(R1+R2).

I would not say R4 and R3 are not bias resistors, that terminology is a bit misleading. From a small signal perspective, R3 and R4 are isolated from the base of the transistor. To reiterate, only R1, R2, and the voltage supply is setting the bias voltage for the transistor.

The resistors R3 and R4 set the gain. The gain of a common emitter amplifier will be approximately -(R4||Rpot)/R3, where Rpot is the resistance of the potentiometer, and the symbol || is to denote R4 and Rpot being in parallel: R4||Rpot = (R4*Rpot)/(R4 + Rpot).

Thus, the output signal will be phase shifted 180 degrees and be amplified by the ratio of R4*Rpot/( R3*(R4 + Rpot) ). Now, it is clear that the poteniometer is used to vary the volume of the amplitude/volume of the output.

If you would like to learn more about this circuit, I would suggest researching the common emitter transistor topology. If you want a more complete understanding of basic transistor topologies, there is also the common base and common collector.

This comment has been removed by the author. - Hapus What about this: "If you increase R4, amplification will be larger. If you increase R4, there will be less gain." I think that it wolud be: "If you decrease R4, amplification will be larger". What do you think? Hey man I was wondering if the second capacitor has an impact on tone or if it serves some other purpose, because in the screaming bird they change both the input cap and the second cap to .022. Sure does. If you change one or both it will decrease or increase bass accordingly. If you swapped out one capacitor on the Screaming Bird for an 0.1, you'd get a slightly thicker (but still Screaming Bird) sound and if you swapped them both out for 0.1s you would have an LPB-1. Hi everybody.
Firstly, keep in mind that every component you place can take part to the bias of a circuit, or to its small signal behaviour, or to both of them. In particular:
- Resistors and inductors are generally placed to set the bias of a circuit;
- Capacitors and inductors are placed to set the small signal behaviour of a circuit;
- The small signal behaviour is directly affected by the chosing of the bias point.
After this introduction, it holds that every resistor you place in a circuit will be used to set its bias.
A npn transistor is ON when its Vbe voltage (difference between base voltage and emitter voltage, in this order) is higher than a threshold, which generally is around 650/700 mV. A npn transistor works in linear region when it's ON, and when its Vce voltage (difference between collector and emitter voltages, in this order) is higher than a saturation voltage (generally around 100 mV). All the resistors we place around a transistor have a contribution in setting Vbe and Vce, and also setting the Ic and Ib currents, and so the beta factor (or hbe = Ic/Ib, which is the current gain of the transistor. Some people say is a constant, but this is actually not true).
This said, all R1, R2, R3 and R4 take part in the biasing process of the transistor. R1 and R2 voltage divider don't exactly set the base voltage: the entire circuit must be rearranged with a Thevenin equivalent, that results in a voltage generator who's value is Vt = Vcc*R2/(R1+R2), and a series resistance (between the Thevenin Generator and the base terminal of the transistor) which has to be properly calculated (in this case, both R1 and R2 take part, but also the npn rbe and R3 have a contribution. It should be the parallel between R1, R2 and input resistance of the bipolar Rin, which actually is Rin = rbe + hbe*R3).
Ic current is set by the chosing of all the placed resistors, and this happens because the transistor output response is directly affected by the input behaviour. This simply means that we cannot chose input R1 and R2 indipendently by R3 and R4, because the bias is related to all of them! Remember that rbe (and so the input resistance of the transistor) is a funcion of both the current Ic and the hbe.
It is, though, true that if we chose a correct bias poit that returns a quite elevated hbe (at least 100 or close to it), we can express the static gain of the Degenre Common Emitter topology (the one in the schematic above) APPROXIMATELY as Av = - R4/R3. REMEMBER: IT IS AN APPROXIMATION! So, increasing R4 will increase the gain, but lower the Vc, and thus also the Vce: so if we place a too big R4, the risk is that of pushing the transistor too close to the saturation voltage (so the transistor can go out of the linear region and part of the signal won't be amplified but cut, causing distortion). If we make R3 smaller we still increase the gain, but we increase Ve and so, again, the Vce becomes lower, and saturation may occur again; moreover, the Vbe can also decrease to the point the transistor goes completely OFF.
So, to answer to Δημήτριος comment: you can surely use a potentiometer in series to R3 in order to modify the gain of the transistor, but this will also modify the bias point of the transistor, and so also its input impedance. The potentiometer must be chosen carefully.
Just a comment to what Coda Effects said: remember that the input cutoff frequency of the filter is given by C1 and the resistance obtained by the result of the parallel R1||R2||Rin (where Rin = rbe + hbe*R3 again).
I hope all of this was helpful. 