NPN with no pullup: understanding?

[davens]

Hi. I'm using a 2N3904 NPN transistor to ground out an audio signal (from a guitar pickup). It does kill the audio signal, but pulls it to a non zero DC voltage (3mV in simulation, 18mV in practice). This is causing a click when the audio signal is small and the transistor is switched. An equivalent circuit is attached (the pickup is replaced with R13, assuming no output from the coils, and there is a sensor circuit instead of S1). I imagine this is some characteristic of transistors I don't understand? I've only really used them where the collector is pulled high.

I think what I'm asking is: could someone please provide insight into why an NPN with saturated I(BE) has a non zero V(CE) when the collector is pulled to GND?

Thanks.

 

[IanP]

When you "turn" a BJT on, a voltage of roughly 0.7V has to be applied to the base (Ube), so the non-zero collector voltage can be explained as follows:
- current through the emitter-base junction allows current through the reverse-biased base-collector junction ..
- the action of base current can be thought of as “opening a gate” for current through the collector, more specifically, any given amount of emitter-to-base current permits a limited amount of base-to-collector current; for every electron that passes through the emitter-base junction and on through the base wire, a certain number of electrons pass through the base-collector junction ..
It's almost as if you connected two diodes back-to-back (it's just a model) .. see attached picture ..

Rgds,
IanP

 

[dick_freebird]

Vce(sat) is the difference between a hard driven Vbe (highly
doped junction) and a soft driven Vbc (lightly doped junction)
that eventually steals the excess base drive; current density
wise the b-c is always less than e-b in a vertical NPN that
is drawn right-side-up.

Since lighter doped (B-C) junctions always have a lower Vf
than heavier (E-B) there is always a net nonzero C-E offset.

To eliminate the click, make your turnoff less abrupt.
Perhaps a simple base-emitter capacitor would do, if you
sized it right.

 

[davens]

Thanks a bunch, both these posts were insightful. I do have a B-E capacitor (since i was trying to soften the transition). I will tune C4 and R12 to try and make the turnoff indistinguishable.
Another idea this sparked would be to use a voltage divider to the collector (in the 500kohm range) and decouple the collector from the output with a suitably large cap. This way the collector would already be biased to Vce(sat)? Just a thought, I'll have to think about this more and simulate this evening. Thanks.

 

[FvM]

By operating a BJT transistor in inverse mode (exchanging emitter and collector), you can considerably reduce the DC offset.
It's a standard solution for chopper circuits with BJT. Unfortunately the current gain drops to about unity, so the base current
must be increased.

But why don't you use a JFET or MOSFET? It's the commonly used analog switch (apart from LDRs) in most guitar amps and
effects. A P-channel JFET (e.g. J175) can be used as normal closed switch, opened with a positive gate voltage, or a
N-channel MOSFET (e.g. BSS138) as normally open switch. Unlike a BJT, both can operate at effectively zero bias current
and have no DC output offset.

 

[davens]

I ended up swapping the collector and emitter, and boosted the cap a bit, and it softened the edge well and dropped the offset by a good order of magnitude. I simulated with an NMOSFET and this looks significantly better too. Depending on how it plays out when the guitar is done, I may switch to that.
Thanks a bunch.