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I'm a beginner in electronics, and started the 2nd chapter of AoE, regarding transistors. I've decided to design my own LED driver, as mentioned on page 76 (3rd ed.)


schematic

simulate this circuit – Schematic created using CircuitLab


Some specs:

  • V1 = +3.3 [V], V2_LOW = 0 [V] , and V2_HIGH = +3.3 [V].

  • The current (I1) through D1 should be 5 [mA].

  • Using a green LED (for example), the voltage drop V_D1 is 2 [V] at 5 [mA].

  • BETA of Q1 is at least 25.


Given the above, I need to figure out the values of R1 and R2. Here's my attempt:

To make sure that I1 = 5 [mA] when there is a signal, I'll need to find an appropriate R1:

  • V_D1 + V_R1 = V1, so the voltage drop across R1 (V_R1) is 3.3 - 2 = 1.3 [V].

  • V_R1 = I1*R1, so R1 should be 1.3[V]/5[mA] = 2.6 [k].

Now, I need to find R2 such that there is saturation when there is a signal:

  • I1 < BETA*I2 = BETA*(V2/R2), so R2 should be no more than BETA*V2/I1 = 16.5 [k].

In conclusion, R1 = 2.6 [k] and R2 = 16.5 [k].


Is my solution correct? If not, where and why is it faulty?

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  • \$\begingroup\$ Your calculation is wrong. You forgot Vce(sat). \$\endgroup\$ – Chupacabras Jul 5 '17 at 21:17
  • \$\begingroup\$ @Chupacabras -- What is that? \$\endgroup\$ – Fine Man Jul 5 '17 at 21:21
  • \$\begingroup\$ Vce(sat) is a voltage across your Q1 transistor in saturation between collector and emitter pins. You just assume there is zero volts, which is not true. You have to look into datasheet of your transistor. \$\endgroup\$ – Chupacabras Jul 5 '17 at 21:28
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You have made a small mistake when computing the \$I_2\$. You say \$I_2 = V_2/R_2\$, which is not correct.

You should say \$I_2 = (V_2-V_b)/R_2\$, since \$V_b\$ is not equal to 0.

Then, you can say \$V_b = V_{be}\$ and assuming you will have the transistor in saturation \$V_{be}\$ is actually known and is equal typically to 0.7 V.

Moreover, there is also a mistake in the collector side calculations. More particulary you have thought the voltage drop across \$R_1\$ wrong.

From Kirchoff's law you can write \$V_1 = V_{D1}+I_1 \cdot R_1 + V_{ce}\$. In this equation you are missing \$V_{ce}\$. If you assume the transistor will be in saturation it is typical to assume \$V_{{ce}_{sat}} = 0.2V\$ (this value is given at the datasheet of the transistor). Now you can calculate \$R_1\$! Note this is the complete formula. The way you did it is also correct, since \$V_{{ce}_{sat}}\$ is quite small. You just make some approximation by ignoring it.

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  • \$\begingroup\$ 0.7 [V] is awfully close to the the typical voltage drop of many diodes (0.6-ish [V]). Does this voltage drop come from the pn-half (which looks like a diode) of the transistor? \$\endgroup\$ – Fine Man Jul 5 '17 at 21:31
  • \$\begingroup\$ Also, what's the difference between V_b and V_be? \$\endgroup\$ – Fine Man Jul 5 '17 at 21:32
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    \$\begingroup\$ @Fine Man V_b it the voltage from the base of the transistor to the ground while V_be is the voltage from the base to the emitter. In your case the two are the same. But if there would be a resistor (R3) from the emitter to the ground then V_b would be V_be + V3. \$\endgroup\$ – Bence Kaulics Jul 5 '17 at 21:49
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    \$\begingroup\$ @FineMan Bence Kaulics said it all regarding \$V_b\$ and \$V_{be}\$. Regarding the 0.7V of \$V_{be}\$ you are correct. It is indeed coming from the voltage drop of the pn junction between base and emitter. Take a look at this question for more details. \$\endgroup\$ – nickagian Jul 5 '17 at 21:58
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    \$\begingroup\$ @FineMan Well, this magic smoke is normal, is what electronics are really made of! Since, when you blow them, it always comes out! :-) \$\endgroup\$ – nickagian Jul 5 '17 at 22:07

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