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Here's an image from another question:

pull down

I get what the pull down resistor is there to do, but I don't really understand exactly how it works.

When there is no voltage applied through the base wire, why does any induced current then go through R2, but when a voltage is applied, it goes to the transistor? My understanding is that voltage always follows the path of least resistance, so if applying say, 5v turns the transistor on, that means that the transistor has a lower resistance than R2, so I would also assume any induced current would also flow through the resistor - so why does it then go through R2?

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    \$\begingroup\$ Your understanding that "current always follows the path of least resistance" is incorrect, or at least misleading. Current will follow all possible paths - the greatest current will flow in the path of least resistance, but current will also flow in all other possible paths, in inverse proportion to their resistances. \$\endgroup\$ – Peter Bennett Jan 10 '14 at 1:47
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I've seen some comments here stating that the pulldown is needed to keep the transistor turned off, instead of floating, or for noise reasons. The base pulldown does in fact help keep the transistor off, but no one has answered why. This is why I chose to answer, rather than extend the comments. User alexan_e is thinking correctly about it, so I'll elaborate here.

There is a Miller capacitance between the collector and the base of all BJTs. Most designers know all too well about the MOSFET's Miller Capacitance and forget that the BJT has some as well. The BJTs Miller capacitance provides a leakage path from collector to base, injecting charge carriers into the base region, which can be amplified by the BJT's Hfe (gain). This allows noise current to flow from collector to emitter. The inclusion of the base pulldown will provide a path to ground to discharge the Miller capacitance and keep the BJT hard off and noise free.

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When the left side of R1 gets disconnected then there is no current flow through R1 or R2 and that means that there will be no voltage drop across it.
If there is no voltage drop across R2 the base level becomes 0v and the transistor is held in an off state.

On the other hand when there is a voltage applied to the left side of R1 then R1 and R2 form a voltage divider that drives the base and with R1 being much smaller than R2 the voltage drop across R1 is fairly small.

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  • \$\begingroup\$ I don't think that answers the question. Isn't the point of R2 being there to keep the signal wire in a determinate state? Based on what you're saying, if R2 was removed, the operation of the circuit would remain more or less the same, but I know you need pull down resistors to keep everything in a known state when no voltage is applied - how does it do this? \$\endgroup\$ – Cameron Ball Jan 10 '14 at 1:48
  • \$\begingroup\$ @CameronBall When there is no voltage applied to R1 the only way for the transistor base to become positive is to pick up some kind of noise but R2 being connected across the base and emitter lowers the impedance and prevents this from happening. \$\endgroup\$ – alexan_e Jan 10 '14 at 1:52
  • \$\begingroup\$ But when it picks up current from noise, how is that different to regular voltage? IE why does it pass through the resistor to ground instead of through the transistor as a regular 5v signal would? \$\endgroup\$ – Cameron Ball Jan 10 '14 at 1:56
  • \$\begingroup\$ @CameronBall A floating base has a high resistance to the ground so it can pick up noise easier. Having R2 in place introduces a low resistance from the base to the grounded emitter (much lower that the base floating) which offers immune from picking up noise. \$\endgroup\$ – alexan_e Jan 10 '14 at 2:01
  • \$\begingroup\$ @CameronBall Note that in order to turn the transistor on, noise has to reach a voltage of about 0.6v across base-emitter but with a "low" resistance path from base to to ground (R2 that can be considered as load) it is very hard to reach that level. \$\endgroup\$ – alexan_e Jan 10 '14 at 2:18

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