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schematic

simulate this circuit – Schematic created using CircuitLab

For this circuit, I am trying to figure out the value of \$V_1\$ DC which will turn on the LED, D1.
Since the forward voltage drop of the LED is 1.8V,

$$ I_{R_3} = \frac{5-1.8}{220} = 14.5\ mA $$

To find the required \$V_1\$, I would use the gain of the transistor to find \$I_{b}\$, where \$I_b = I_c \beta_f\$. Then,

$$ V_1 = R_1 I_b + 0.7\ V $$

(Am I right in ignoring \$R_2\$, since it is just connected to ground?)

The problem is that the gain of the transistor is not given. Is there another way to work out the value of \$V_1\$ required to turn on the LED, without using the gain of the transistor? Any help would be much appreciated.

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  • \$\begingroup\$ For switch operations (seems that's what you are discussing), it's common practice to use \$\beta=10\$. You could choose another value. But if it's not specified, then that's taken by default to be the intended value. \$\endgroup\$ – jonk Feb 13 '17 at 8:05
  • \$\begingroup\$ When you say, the gain is not given, the transistor in the schematic (2N3904) does not resemble the real transistor, right? \$\endgroup\$ – Ariser Feb 13 '17 at 8:12
  • \$\begingroup\$ @Ariser yes, the original circuit diagram does not have the transistor model given \$\endgroup\$ – abruzzi26 Feb 13 '17 at 8:14
  • \$\begingroup\$ I thought you were looking at this as a reliable switch design. Are you asking, instead, about trying to figure out the moment at which you might just barely see the LED as you raise the voltage and are NOT looking for this as a reliable switch design? \$\endgroup\$ – jonk Feb 13 '17 at 8:41
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If gain \$\beta\$ is unknown, you should make a worst case assumption about it. As others have said, 10 is a good guess. You should also allow some margin (+20% is commonly used) for collector current in your calculations.

Usually the problem is formulated the other way round: which base resistor should I choose in order to make sure that the base current saturates a given transistor for a given input voltage, even in worst case (lowest \$\beta\$) conditions. You may want to do your calculations using this approach.

Also, beware of \$R_2\$. Usually, a much higher resistance is chosen in order to be able to make the assumption that all the the current through \$R_1\$ goes to the base. If \$\dfrac{V_{be}}{R_2}\$ is comparable to the base current, then you can't ignore the effect of R2.

EDIT:

To be more specific, the effect of R2 on V1 will be as follows:

$$ V_1=R_1I_{base}+\left(1+\frac{R_1}{R2}\right)0.7 $$

So, assuming \$\beta=10\$:

$$ V_1 = \frac{1\ k\Omega·14.5\ mA}{10} + 2·0.7 = 2.85\ V $$

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  • \$\begingroup\$ In my case, both R1 and R2 are 1K. Does that mean I will have to consider the parallel combination of R1 and R2 i.e V1 = ((R1||R2)*(Ibase)) + 0.7V ? \$\endgroup\$ – abruzzi26 Feb 13 '17 at 8:46
  • \$\begingroup\$ No, they're not in parallel. At most, you may be led to think that they are forming a voltage divider, but that's not true either. The voltage drop in R2 is tied to Vbe. \$\endgroup\$ – Enric Blanco Feb 13 '17 at 8:53
  • \$\begingroup\$ The formula you're looking for is V1 = R1*Ibase + (1+R1/R2)*0.7 \$\endgroup\$ – Enric Blanco Feb 13 '17 at 9:02
  • \$\begingroup\$ Thanks, but how do you get the (1+R1/R2)*0.7 part? \$\endgroup\$ – abruzzi26 Feb 13 '17 at 9:07
  • \$\begingroup\$ Let I1 and I2 be the currents through R1 and R2 respectively. Then, V1-0.7 = R1*I1 and I1=Ibase+I2. You substitute I1 into the first equation, then isolate V1. As I said in the answer, you can only safely ignore R2 if its value it's much higher than R1, because then you have that aprox V1 = R1*Ibase + 0.7. Otherwise, you can't ignore R2. \$\endgroup\$ – Enric Blanco Feb 13 '17 at 9:13
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I am trying to figure out the value of V1 DC which will turn on the LED, D1.

about 1.4v: the voltage drop on R2 is 0.7v, so the minimum amount of current going through R1 is 0.7v/R2. As R1 = R2, the minimum amount of voltage drop over R1 is 0.7v / R2 * R1 = 0.7v. thus the answer.

the key here is that the transistor has sufficient current gain so that its base current can be ignored. if not, you just need to add that to the current going through R1 and re-calculate.

But conceptually, it is the same.

edit: here is your circuit in spice.

enter image description here

the current through the led goes from 1ma@V1=1.3v to 14ma@V1=1.5v.

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With Q1's emitter tied to ground you can assume that the base voltage needs to be about 0.7 volts to turn on the LED. This means about 1.4 volts applied to R1. Don't worry about thinking in terms of hFE or current gain for this type of circuit. Having said that, to lightly turn on the LED, only about 1 volt will be needed because, under these conditions hFE will be large but start to fall as more collector current is taken. There is no black and white answer but a range of uncertainty.

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  • \$\begingroup\$ That does make sense, thanks! In the case where the gain is given, was I right to ignore R2 while calculating V1 using Ibase? \$\endgroup\$ – abruzzi26 Feb 13 '17 at 8:22
  • \$\begingroup\$ Did you forget \$R_2\$, which is going to sink about \$750\:\mu\textrm{A}\$? \$\endgroup\$ – jonk Feb 13 '17 at 8:25
  • \$\begingroup\$ @abruzzi26 No, you can't ignore \$R_2\$. It's sinking a lot of the needed current. \$\endgroup\$ – jonk Feb 13 '17 at 8:26
  • \$\begingroup\$ No, you cannot ignore R2 because the base-emitter junction is a forward biased diode and R1 and R2 form a potential divider - you might need 0.5 volts on the base to create a little bit of light - base current will be tiny compared to R2 current so R2 is relevant. \$\endgroup\$ – Andy aka Feb 13 '17 at 8:26
  • \$\begingroup\$ @jonk "Did you forget..." was that for me or the OP? \$\endgroup\$ – Andy aka Feb 13 '17 at 8:27
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For switch operations (seems that's what you are discussing), it's common practice to use \$\beta=10\$. You could choose another value. But if it's not specified, then that's often taken by default to be the intended value.

So, you have this equivalent circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

Given the high current, I'd estimate \$V_{BE}=700\:\textrm{mV}+\operatorname{ln}\left(\frac{14.5\:\textrm{mA}}{4\:\textrm{mA}}\right)\approx 740 \:\textrm{mV}\$. From this, I'd suggest \$\frac{V_1}{2}=740\:\textrm{mV}+\frac{14.5\:\textrm{mA}}{10}\cdot 500\:\Omega\approx 1.47\:\textrm{V}\$. That value must then be multiplied by 2 to get \$V_1\$.

You can tinker around with that to get whatever you want, regarding precision. But given the variability in everything I think \$V_1=3\:\textrm{V}\$ would be the right value to use for an assumed saturated switch BJT with \$\beta=10\$.

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