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Looking at the below schematic, with LED1 being a red LED (1.7V forward voltage) and LED2 being a green LED (3.2V forward voltage), I have a few questions that probably reflect lack of some fundamental principles of circuits, and I hope someone can get me into the clear :) [EDIT: When I say "Q1 is ON", I mean that I am supplying a 5V "high" signal to input IN1]

  1. if Q1 is ON, LED1 will shine and LED2 will be off. What are the voltages at V1 and V2? [My assumption is that they are both 1.7V as there is just one resistor (R2) in the path]
  2. if Q1 is OFF, LED1 will be off and LED2 is ON. The voltage at V2 is now 3.2V (the forward voltage of LED2). Correct?

Now imagine a scenario where LED1 is green and LED2 is red.

  1. if Q1 is ON, both LED1 and LED2 will shine (though LED1 only shines very faintly). Why is this? What is the voltage at V1 and V2? [From the observation that both LEDs are on I think V1=V2=3.2V, but I fail to see why since my first observation in question 1. above seems to say that V1=V2=1.7V - so I would have expected 1.7V here too and hence LED1 would not be lit].
  2. if Q1 is OFF, LED1 will be off and LED2 is ON. Voltage at V2 is now 1.7V (the foward voltage of LED2). Is this also correct?

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Why are you saying that when Q1 is on the voltage on LED1 and LED2 are 1.7V? They depend on the voltage drop over R2. The voltage drop over R2 is I*1k where I is a sum of currents through LED1 and LED2. Also V1 will be about 0.3V lower than V2. \$\endgroup\$
    – Alexxx
    Commented Apr 6, 2015 at 17:10
  • \$\begingroup\$ @Alex - I thought that to calculate resistances you needed to look at the required voltage drop over R2 (in this case 9V-1.7V = 7.3V) and therefore I = 7.3/1000 = 7.3mA. If I follow your logic, and assume here that my LED's have 25mA rating, the voltage drop would be 50V ((0.025+0.025)*1000). Or do I see this wrong? \$\endgroup\$
    – Phil B.
    Commented Apr 6, 2015 at 17:20
  • \$\begingroup\$ The LEDs are non linear. Therefore current drawn depends on the V/I curve for this particular LED. And your 25mA/1.7V is just a point on this curve for your red LED \$\endgroup\$
    – Alexxx
    Commented Apr 6, 2015 at 17:48
  • \$\begingroup\$ @PhilB. What was the value of IN1 you used when you obtained a faint glow (case 3)? \$\endgroup\$
    – nidhin
    Commented Apr 6, 2015 at 18:37
  • \$\begingroup\$ @nidhin 5V - I've added this as an edit to my question \$\endgroup\$
    – Phil B.
    Commented Apr 6, 2015 at 18:44

2 Answers 2

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I think that what is happening is that with LED1 green you are effectively powering it from the base as the voltage at at the collector will be 1.7V due to LED2 and this is too low to power LED1. If the base voltage (IN1) is 5V then this is powering LED1 through the 10k resistor and Q1's base-emitter junction, the current through LED1 would be (5-(0.7+3.2))/10000 = 0.11mA, hence a faint glow could be observed.

If LED1 is red(1.7V forward voltage) and LED2 is green (3.2V forward voltage) then when Q1 is ON the voltage across Q1 and LED1 will be (assuming Q1 is saturated) 0.2V + 1.7V =1.9V. This voltage will also be across LED2 (it is in parallel with Q1 and LED1) and since LED2 is green it requires 3.2V to turn it on therefore it will not light.

If however LED1 is green (3.2V forward voltage) and LED2 is red(1.7V forward voltage) then when Q1 is ON the voltage across Q1 and LED1 will be (again assuming Q1 is saturated) 0.2V + 3.2V =3.4V. This will be across LED2 and since it is much greater than the 1.2V it will turn LED2 On.

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  • \$\begingroup\$ I understand this for the LED1 red and LED2 green scenario. Since the path over LED1 has a lower forward voltage, it will be used instead of the higher voltage LED2 route. No issue there. I'm still unclear on the LED1 is green scenario. What I observe is that LED1 barely lights (a faint glow). This would either mean a) my LED is not a real diode, there is actually some current flowing at lower voltages, and since the voltage over LED2 is 1.7V I'm seeing the same voltage over LED1 and the non-diode behavior explains the faint glow, or b) the voltage is 3.2V(or 3.4V) but that leaves me confused. \$\endgroup\$
    – Phil B.
    Commented Apr 6, 2015 at 17:25
  • \$\begingroup\$ You're seeing a faint glow when LED1 is green but Q1 is off? \$\endgroup\$
    – dwightreid
    Commented Apr 6, 2015 at 17:30
  • \$\begingroup\$ No, when Q1 is ON, that's when I see the faint glow. When Q1 is OFF, no matter whether LED1 is green or red, the LED is not shining at all (as expected). \$\endgroup\$
    – Phil B.
    Commented Apr 6, 2015 at 17:33
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    \$\begingroup\$ Ok, I'm just realizing that there is no resistor in LED2, so if LED1 is green then theoretically it doesn't come on at all, whether Q1 is On or OFF. But you are experiencing a faint glow when Q1 is on. So yes I would say the voltage across Q1 and and LED1 would be 1.7V when LED2 is red, so LED1 shouldn't come on. Faint glow could be due to base-emitter circuit if IN1 is greater than (3.2V+0.7V). Is this the case? \$\endgroup\$
    – dwightreid
    Commented Apr 6, 2015 at 17:54
  • \$\begingroup\$ Yes, IN1 is 5V which is > 3.9V - so you say that I'm actually powering the green LED1 from the base ? \$\endgroup\$
    – Phil B.
    Commented Apr 6, 2015 at 17:55
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  1. If Q1 is ON, LED1 will shine and LED2 will be off. V1=1.7V, the forward voltage of LED1 and \$V_2 = V_1+V_{CEsat}\$. Where \$V_{CEsat}\$ is the collector-emitter voltage at saturation whose value will be in the range 0.2-0.3V. (assumed that IN1 can bring the transistor into saturation)

  2. Yes. if Q1 is OFF, LED1 will be off and LED2 is ON. The voltage at V2 is now 3.2V

  3. LED2 will be ON. So that makes V2 = 1.7V. So the voltage that can appear at base will be around 2.4V. So if the emitter junction is forward biased (since you said LED1 glows), the voltage at emitter will also be around 1.7V.
    The forward voltage 3.2 is for a specific current. Current (less than the rated) can start flowing before reaching the forward voltage. And I think your LED is glowing faintly because of this small current flowing.

  4. If Q1 is OFF, LED1 will be off and LED2 is ON. Voltage at V2 is now 1.7V

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  • \$\begingroup\$ This is getting real close to answering all my questions :) Only question mark is your answer to 3 ... I have empirical evidence that LED1 faintly glows, so there is current. And where does the 0.7V (2.4V-1.7V forward voltage from LED2) come from in your answer? Does Q1 supply extra voltage due to the voltage at base perhaps? Can you explain? \$\endgroup\$
    – Phil B.
    Commented Apr 6, 2015 at 17:28
  • \$\begingroup\$ @PhilB. the collector-base junction will be forward biased. hence the 0.7V extra at base than collector. \$\endgroup\$
    – nidhin
    Commented Apr 6, 2015 at 17:35
  • \$\begingroup\$ ok - but isn't the fact that I make IN1 "high" (e.g. 5V) enough to make transistor Q1 ON? I'm not sure I understand what the 2.4V at base has to do with the functioning of the LED ... I can understand that 2.4V at the EMITTER would not be enough for LED1, but that doesn't seem to be what you are saying. \$\endgroup\$
    – Phil B.
    Commented Apr 6, 2015 at 17:40
  • \$\begingroup\$ @PhilB. For an npn transistor to become ON, voltage at base should be greater than the voltage at emitter. So 2.4 at base means that the transistor will be ON only if the voltage at emitter is 1.7V. \$\endgroup\$
    – nidhin
    Commented Apr 6, 2015 at 17:43
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    \$\begingroup\$ @PhilB. The forward voltage is for a specific current. The current can start flowing before reaching the forward voltage. In this case, the LED may emit light but at a reduced intensity. I think that is what is happening in case 3. \$\endgroup\$
    – nidhin
    Commented Apr 6, 2015 at 17:49

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