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BJT biasing schematic

I got this question as a school assignment (1st year electrical engineering). I have to calculate the 3 unknown resistor values. The values which were given are in the schematic. Further \$\beta = 100\$ and the transistor is an NPN BC547B. The problem for me is the negative \$V_{EE}\$ voltage and the \$V_x = 0\$V. I don't really know where to start (already stuck for an hour). Can you give me some hints or an example so I know how to go further? I have already looked at a lot of BJT biasing examples but can't find one (I think) which really fits my problem.

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    \$\begingroup\$ Would it help to shift everthing up by 15 volts? Vee = 0 V, Vx = +15 V and Vcc = +30 V? The point we call "Zero Volts" is arbitrary. \$\endgroup\$ – Peter Bennett Jun 5 '15 at 17:23
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    \$\begingroup\$ Remember that absolute voltages are just a arbitrary choice for what to call 0. You can pick whatever node you want to call 0 V, as long as all the relative voltages everywhere are still as stated in the problem. \$\endgroup\$ – Olin Lathrop Jun 5 '15 at 17:25
  • \$\begingroup\$ Thank you for the clarification that I can change the point which I call zero volt. \$\endgroup\$ – 4lloyd Jun 5 '15 at 22:36
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Negative voltages are nothing to be afraid of. They work exactly like positive voltages, just with a minus sign. I'll walk you through the first part, which is to figure out \$R_1\$.

You know the voltage across \$R_1\$ is 15V. If you find the current through \$R_1\$, you can calculate the resistance with Ohm's Law. The current through \$R_1\$ is equal to the current through \$R_2\$ plus the base current. Ohm's Law gives you the current through \$R_2\$:

$$I_{R2} = \frac{0\ \mathrm V - -15\ \mathrm V}{5\ \mathrm{k\Omega}} = 3\ \mathrm{mA}$$

You're given the collector current and \$\beta\$, so you can easily calculate the base current:

$$I_B = \frac {I_C} {\beta} = \frac {1\ \mathrm{mA}} {100} = 0.01\ \mathrm{mA}$$

So the current through \$R_1\$ is:

$$I_{R1} = I_{R2} + I_B = 3\ \mathrm{mA} + 0.01\ \mathrm{mA} = 3.01\ \mathrm{mA}$$

This is almost (but not quite!) equal to \$I_{R2}\$ by itself. In real life, you would probably ignore \$I_B\$ unless \$\beta\$ were lower, but this is homework, so let's do it the hard way. :-)

Now you can calculate \$R_1\$:

$$R_1 = \frac {15\ \mathrm V - 0\ \mathrm V} {3.01\ \mathrm {mA}} = 4.98\ \mathrm{k\Omega}$$

I'll let you handle \$R_C\$ (trivial) and \$R_E\$ (harder) on your own. Please be sure to post the work you've done if you have any follow-up questions.

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  • \$\begingroup\$ Thank you very much for the help. +1 for the explanation on how to use the β value of a transitor. \$\endgroup\$ – 4lloyd Jun 5 '15 at 22:32
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In all cases it's not the "absolute" voltage which matters, it's the voltage across the resistors that matters.

\$R_1\$

\$R_1\$ and \$R_2\$ form a voltage divider with a small load \$I_B = I_C/\beta = 10\mu\$A. You're given \$V_x = 0\$, so the voltage across \$R_2\$ is \$0 - V_{EE} = 15\$V. By Ohm's Law the current through \$R_2\$ is \$15\text{V}/R_2\$, which is \$3\$mA. The current through the divider is much larger than the load current, so if you just need an approximation you can ignore \$I_B\$.1 In that case the voltage across and current through \$R_1\$ is the same as \$R_2\$ and therefore the two are equal. Alternatively, you can use the voltage divider equation, which can be simplified by noticing that the voltage divider output is halfway between \$V_{CC} = 15\$V and \$V_{EE} = -15\$V and therefore the voltage is divided in half. The simplified equation is

$$\frac{R_1}{R_1 + R_2} = \frac{1}{2}$$

which gives you the same result, \$R_1 = 5\text{k}\Omega\$.

If you can't approximate, note by KCL that the current through \$R_1\$ is the current through \$R_2\$ (\$3\$mA) plus \$I_B\$, or \$3.01\$mA. Now you have the voltage across \$R_1\$ (\$15\$V) and the current through it, so you can use Ohm's Law to calculate it exactly.

\$R_C\$

You're given \$I_C = 1\$mA and \$V_Y = 5\$V, so you know you have \$15-5=10\$V across \$R_C\$. Since you know the voltage across and current through the resistor, Ohm's Law gives you the required resistance.

\$R_E\$

The transistor is on since \$I_C = 1\$mA, so assume a \$0.6\$V (or \$0.7\$V) \$V_{BE}\$ drop. Since \$V_x = V_B = 0\$V, that means \$V_E = -0.6\$V. This gives you the voltage across the resistor: \$V_E - V_{EE} = -0.6 - (-15) = 14.4\$V. Remember

$$I_E = \frac{\beta + 1}{\beta}I_C$$

and both \$\beta\$ and \$I_C\$ are given, so you know \$I_E\$. Alternatively, you already know \$I_B\$ so you can also use the fact that

$$I_E = (\beta + 1)I_B$$

Again, you have the voltage across and current through the resistor so you can use Ohm's Law to calculate the required resistance.

1 Typically the rule of thumb is that if the unloaded divider current is 10 times the load current it is safe to make an approximation.

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  • \$\begingroup\$ (After Rc) "You're given IC=5mA and VY=5V, so..." You may want to edit that 5mA to 1mA \$\endgroup\$ – JIm Dearden Jun 5 '15 at 17:53
  • \$\begingroup\$ @JImDearden Thanks for the catch. I was thinking of the 5V I was about to type. For a minor correction like that feel free to edit my answer. \$\endgroup\$ – Null Jun 5 '15 at 17:56
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    \$\begingroup\$ @Null, Yes but this is a 1st year EE course. It takes some experience to understand when approximations can be safely used. Typically I figure that if the divider current is 10x the base current it is close enough. But that is in the real world, not in EE101. \$\endgroup\$ – mkeith Jun 5 '15 at 19:06
  • \$\begingroup\$ @mkeith Fair enough, for a 1st year EE course I should have been more explicit that I was making an approximation and why. \$\endgroup\$ – Null Jun 5 '15 at 19:09
  • \$\begingroup\$ @Null, I clicked the upvote because of your edit and request. \$\endgroup\$ – mkeith Jun 5 '15 at 19:10
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I will just give you some hints since it is course work. You can calculate Rc with the information provided, because you know the voltage on both sides of Rc, and you know the current flowing through it. You can calculate Ib with the information provided. After you do that, you can calculate the current flowing through R1, so then you can calculate R1 using the voltage and current. In order to calculate RE, you need to assume a certain voltage drop across base-emitter junction of Q1. You can pull that from the datasheet if you want, or you can use a typical value such as 0.6 or 0.7V. But since the type of transistor is actually specified in the problem, I would probably go look at the datasheet to see if it shows a graph of typical Vbe vs Ib.

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