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As far as I know, to use transistor as switch we only need to apply voltage on the base to turn the transistor on. When this voltage is removed the transistor switches off. The idea is to make the base emitter junction conduct. All we need here is a base resistor to control base current.

However, in here the author is connecting a second resistor calling it R2 and giving value 10 times that of base resistor. Here is the picture:

enter image description here

It says "Resistor R2 is not essential to this circuit but is generally used for stability and to insure that the transistor switch is completely turned off. This resistor insures that the base of the transistor does not go slightly negative which would cause a very small amount of collector current to flow."

Questions are:

  1. What does author mean by stability, we are using transistor as a switch here after all.

  2. How does R2 help to maintain this so called stability

  3. How does one turn off this transistor? By applying 12V to R1?

  4. Why in the world would the base go ever slightly negative which I assume is <0V, as it explains near end of paragraph?

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  • \$\begingroup\$ \$R_2\$ is basically a "pull-up." It's providing a weak, but bi-directional, galvanic path to keep the base of \$Q_1\$ tied close to where the emitter is firmly tied. In thinking about it, imagine what might happen if you put this circuit on a protoboard and had the open node to \$R_1\$ as an unconnected, loose wire end. Suppose you touched it, on accident, or it touched something else? How might \$R_2\$ help in a case like this? \$\endgroup\$ – jonk Jan 17 '17 at 6:48
  • \$\begingroup\$ Are you referring to esd? \$\endgroup\$ – quantum231 Jan 17 '17 at 12:59
  • \$\begingroup\$ I'm referring to anything that might occur to a node that isn't connected to a low impedance driver. Often, I've added components to inputs that drive some circuit but where that input is at a connector pin or attachment point that may be left "open" at times. It's just wise precaution. Sometimes, that includes still more -- such as a pair of reverse-arranged diodes between both ground and Vcc, as well as a resistor (or two.) (It's possible for the \$V_{BE}\$ to be reverse-biased, with or without \$R_2\$, sufficient to zener and therefore damage it.) \$\endgroup\$ – jonk Jan 17 '17 at 19:27
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Adding R2 ensures that the base is biased firmly to turn the transistor off when the input is floating. Some transistors may leak collector-emitter current when the base is floating, some may not - transistors vary in characteristics quite a lot from device to device so it is wise to design your circuits with these large variations in mind. Assume you have the device with the worst characteristics in the datasheet, making sure you interpret the datasheet correctly when you do that.

In light of that, stability is referring to the stability of every circuit, be it made once or manufactured by the million. Keeping your circuit away from boundaries that can cause individual circuit behaviour to vary is a fundamental of design.

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3: How does one turn off this transistor? By applying 12V to R1?
Yes, that would be a good plan.

4: Why in the world would the base go ever slightly negative which I assume is <0V, as it explains near end of paragraph?
Your inference is in doubt, because this situation would indicate a dead transistor, whose base-emitter has become open-circuit. What was meant is that the base may become slightly negative with respect to the emitter. You'd not want the base approaching +11.4 V, where the transistor starts to conduct (emitter-to-collector). R2 pulls the base up, away from that point, ensuring that it is OFF.

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  • \$\begingroup\$ applying 12V to R1 would just put it in parallel to R2. Following the description, R2's job is already to be a pull-up resistor, so while applying 12V to R1 wouldn't hurt, letting it float would just be as good. \$\endgroup\$ – Marcus Müller Jan 17 '17 at 10:04
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Its similar to adding a pullup for a gpio in open collector/drain configuration.

1/4. Leaving the base floating may have some stray base current leading to unwanted biasing. 2/3. R2 is to limit the base current, otherwise here the point is to have Vbe = 0 (Base Emitter Vltage), R2 is chosen much higher than the actual base current limiting resistor R1 as R2 is only to compensate the stray base current.

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The PNP transistor doesn't conduct when the base is close to emitter voltage – but, when nothing is on the left-hand emitter, the base would "float" around, and to avoid that, you'd connect it (typically, using a relatively high-valued resistor) to supply voltage. If it floated, the transistor might suddenly change behaviour based on the load. That answers 1. and 2., and actually 3., because due to R2, you can simply leave R1 unconnected to turn the transistor "off".

Regarding 4.: Why shouldn't it? There's all sort of interference in the air, and if you look sharply, you'll notice that if the load has the chance to vary the voltage across it, you might get into situations where emitter voltage can simply rise above base voltage, even if that stays fixed.

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  • \$\begingroup\$ you seem to have the names "emitter" and "collector" reversed. The emitter is the lead with the arrowhead, whether the transistor is NPN or PNP. \$\endgroup\$ – Peter Bennett Jan 17 '17 at 0:19
  • \$\begingroup\$ @PeterBennett argh. Late night stackexchanging. Bad idea. \$\endgroup\$ – Marcus Müller Jan 17 '17 at 10:03

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