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I am pretty new to the world of electronics so apologize yet again for a very basic question.

I am trying to understand the functioning of a transistor, but my confusion may underpin some more fundamental confusion about voltage. I think understanding this circuit will help me grok the fundamentals.

enter image description here

When the button is closed like in picture, the light is off. I actually built it and verified that when it's closed, the voltage between base and emitter is actually zero, and the voltage across the 1k resistor is around 0.8V. I also found that when increasing this resistance to around 5k or more, the transistor is on whether the button is pressed or not.

I do not understand why is the voltage 0 at the base and why it's always on when the bottom resistor increases. The explanation in the text doesn't make sense. Could anyone help me understanding this? Thanks

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3 Answers 3

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Maybe it helps if I re-draw the circuit and annotate some values of voltages and currents.

The circuit on the left is the situation when the switch is open and the LED is on.

The circuit on the right is the situation when the switch is closed and the LED is off.

schematic

simulate this circuit – Schematic created using CircuitLab

Note how adding R5, the 1 k resistor, makes the voltage at the base of Q2 drop so low that there's only 0.8 V left for both the Base-Emitter of Q2 + LED D2. That 0.8 V would be enough for only the NPN (an NPN needs about \$V_{BE}\$ = 0.6 V to do anything but there's also the LED D2. A LED needs (depending on the type) at least about 2 V do start conducting. So that 0.8 V is not going to make any current flow into the base of Q2. So Q2 stays off and so does the LED.

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  • \$\begingroup\$ Great explanation. I was under the impression that the current would pass through the diode, and then split between the negative side of the battery and R5. Instead from your graph it looks like the splitting happens before the transistor, and the current goes in the opposite direction that I thought. So why doesn't the current split after the diode and goes to R5? \$\endgroup\$
    – Vaaal88
    Jan 27, 2021 at 10:13
  • \$\begingroup\$ Where did the 0.2V on the emitter in the right pic come from? \$\endgroup\$
    – Lundin
    Jan 27, 2021 at 10:18
  • \$\begingroup\$ @Lundin I just subtracted 0.6 V from the 0.8 V at the base. If the LED has a large leakage current that could happen. Maybe it is better to remove the 0.2 V as that voltage isn't well defined. I removed it. \$\endgroup\$ Jan 27, 2021 at 10:21
  • \$\begingroup\$ So this 0.6V is the same as Base-Emitter Saturation Voltage? The part specifies it between 0.6V min to 2.6V max. \$\endgroup\$
    – Lundin
    Jan 27, 2021 at 10:23
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    \$\begingroup\$ @Lundin So this 0.6V is the same as Base-Emitter Saturation Voltage? Yes. Do realize that the "Base-Emitter Saturation Voltage" might be specified in the datasheet for very large values of \$I_B\$ (base current) and \$I_C\$ (collector current). In practice, at low currents ( < 10 mA) 0.6 V is a good enough approximation. \$\endgroup\$ Jan 27, 2021 at 10:29
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// why it's always on when the bottom resistor increases.//

Bimpelrekkie gives the right answer. There is one more point to be told. As per the second image in his answer, when you increase the R5 value and put a switch between R5 and R3/R5 junction point, the following thing occurs.

Until the resistor R5 value is less than 4 kΩ, the circuit will act like this:

  1. When the switch is opened, Transistor will turn on and so the LED.
  2. When the switch is closed, Transistor will turn off and so the LED.

Why is this happening because when the switch is closed, R3 and R5 make the divider circuit and so the voltage is divided before applied to the base of the transistor.

Consider that to turn on the transistor, the applied voltage should be greater than 2.7V (0.7V of base + 2V of the LED). until the R5 reaches around 4 kΩ, the applied voltage via the divider network when the switch is closed will be less than 2.7V.

Say R5 is 3.6kΩ, then the applied voltage to the bias is, (9V * 3.6kΩ)/13.6 kΩ = 2.38V.

To turn on the transistor here, the applied voltage should be >2.7V.

When the resistor is increased more than 4 kΩ, say 4.7kΩ, the applied voltage via the divider network will be

(9v*4.7k) / (10k+4.7k) = 2.88V. This voltage is enough to turn on the base of the transistor and it conducts a signal.

This is why, when you increase the resistor value and close the switch, it will turn on the transistor instead of turn off.

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  • \$\begingroup\$ Great answer. This part is still not clear to me 'turn on the transistor, the applied voltage should be greater than 2.7V (0.7V of base + 2V of the LED).' why does the voltage need to be greater than 0. 7+2V? I thought it just needed to be higher than 0.7v is it because otherwise the current can't pass through the led and the circuit is' open'? \$\endgroup\$
    – Vaaal88
    Jan 27, 2021 at 11:18
  • \$\begingroup\$ That is because the forward voltage barrier created by a diode and LED itself. For more info: electronics.stackexchange.com/a/286872/154096 \$\endgroup\$
    – CNA
    Jan 27, 2021 at 11:25
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    \$\begingroup\$ The base-emitter junction is acting as a diode and an LED is connected in series with the base-emitter junction. So, even if you apply more than 0.7V supply, it is not sufficient for the current flow in the transistor. Because an additional 2V barrier is created by LED and it restricts the current flow between base and emitter after that only it eventually starts to conduct current from collector to emitter. \$\endgroup\$
    – CNA
    Jan 27, 2021 at 11:32
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When you press the button, you create a current path 9V->10k->1k->GND, where transistor base is attached to the middle of it. Of this VOLTAGE DIVIDER. In other words, when the button is pressed, you drop 10k/(10k+1k)~10/11 of the voltage on the 10k (because it passed 10/11 of resistance on its path to ground), which leaves you with 1/11 of 9V, which is roughly 0.8, transistor gate and circuit may distort that thing, it's just a bad circuit for demonstrating this phenomenon. So as you increase the bottom resistor, the proportion of upper (10k) resistor in the voltage divider decreases (imagine bottom resistor as another 10k - you would halve the 9V voltage if not for transistor gate), so the voltage divider outputs more and more voltage until the gate is at enough voltage level to have current and also open the LED for conducting.

This is, however, a weird and overcomplicated circuit. Find some simpler examples on the internet for "transistor as a switch", they will explain stuff better with fewer distracting and imo completely unnecessary components around. Take it slower.

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  • \$\begingroup\$ Thanks for your reply. But when the button is open, the voltage drop on the resistance should be 10k/10k and thus there should still be no voltage across the base and emitter. Yet in this case the led works. Why? \$\endgroup\$
    – Vaaal88
    Jan 27, 2021 at 9:25
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    \$\begingroup\$ That is wrong. When the switch is open, there is no path to ground through 10k via the switch. There is no voltage divider. So the voltage on both sides of 10k would have been 9V. No current = no voltage difference. But since we have transistor base (which is close to diode) followed by another diode, there is a path through the transistor base, voltage at the 10k resistor (9V on both sides) is enough to open all diodes, so I would expect voltage of Vled + Vbe on the 10k resistor. Again, this is A BAD CIRCUIT to learn from. It's very confusing and overcomplicated for beginners. \$\endgroup\$
    – Ilya
    Jan 27, 2021 at 9:32
  • \$\begingroup\$ It's a diode circuit parallel to resistor circuit, it's NOT something I would use to show how transistors work. Not initially. Srsly, drop this circuit for a week and find simpler examples. After a week you will probably get it. It's just not a fitting example. It's like teaching arduino by showing spaceship circuits. \$\endgroup\$
    – Ilya
    Jan 27, 2021 at 9:34

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