LaVache Schematic Question

[MullisMan]

When I was looking at the schematic for the LaVache, I checked out the boost switch section of the circuit and I have a question.  It looks to me like the boost switch is just taking the 3.3M resistor out of the circuit, and the large cap is already in parallel with the emitter resistor, which is not really what's in the project documents.  I was thinking it would be more like the various BSIAB2 flavors where you actually switch the large cap in and out using the switch.

I haven't built it yet so I'm just going off my initial perusal of the schematic.  Any info?

Thanks.

[madbean]

In non-boost mode, the 3M3 essentially blocks the path to ground on the 47uF. In boost mode, the 3M3 is grounded out. With the 47uF connected directly to ground it acts as a bypass cap which increases the gain of the transistor.

[MullisMan]

Gotcha.  The way I'm looking at it, I just don't see the need for the 3.3M resistor.  Couldn't you just hook the bypass cap to the switch and have it bypassed that way? As such:



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[madbean]

You could do that, but there is a strong chance you will get a pop when switching it on and off. When you first connect the circuit to the power supply, that 47uF doesn't have a chance to charge. When the switch is engaged in your setup, current flows into the cap, it charges and begins shooting it to ground while continually recharging. When the switch is toggled off, the cap drains the remaining current to ground. If you switch back on before the current drains off completely, you will get a pop. This is very much the same as not having a pulldown to drain current stored in the input decoupling path.

[k.rock!]

You could do that, but there is a strong chance you will get a pop when switching it on and off. When you first connect the circuit to the power supply, that 47uF doesn't have a chance to charge. When the switch is engaged in your setup, current flows into the cap, it charges and begins shooting it to ground while continually recharging. When the switch is toggled off, the cap drains the remaining current to ground. If you switch back on before the current drains off completely, you will get a pop. This is very much the same as not having a pulldown to drain current stored in the input decoupling path.

Very interesting. I was actually thinking the same as  MullisMan regarding that 3M3 resistor, but after reading your explanation it makes a lot of sense. This is some good learnin' right here! hehe


-Kaleb