# BJT Transistor Biasing Problem

If I have the circuit where I have Vcc = 30V and Vee = -30V. Transistor's beta equals 300. I want my Ic to be around 10mA and Vce = 25V, Vrc = 25V, Vre = 10V. I want to bias the base of transistor from the common node of both positive and negative supply rails, that is GND.

• What will be the voltage drop across Rb (base resistor)?

I have no idea how to calculate it in such situation...

simulate this circuit – Schematic created using CircuitLab

• Did you do some calculations? could you share your calculations? – rsg1710 Nov 12 '17 at 19:00
• If you know Vre... knowing Vrb should be a no brainer.... – Trevor_G Nov 12 '17 at 19:01
• Re = 10V/10mA = 1k ohm. Rc = 25V/10mA = 2.5k and Rb = (30V - 10V -0.7V) /(10mA/300) = 580k Ohm. So wher is the problem? – G36 Nov 12 '17 at 19:18

You've said you want the voltage across $R_E$ to be $10\:\textrm{V}$ and you've specified your voltage rails. This means $V_E=-20\:\textrm{V}$, by definition. The only remaining question for you to ask yourself is "What is $V_{BE}$?" Because if you know that, you must then know the voltage across $R_B$. How can you not?

Seriously, this isn't so much a calculation as the fact that you've almost already specified the value, directly. $\beta$ doesn't even enter into the question, as it simply does not matter. You don't need to know the base current, as it simply does not matter. You've written the script and the answer is right there.

The only way to complicate this question would be to state, "Well, I don't know the value of $V_{BE}$ in this case." But that's probably not what you want to say.

• 30 - Vre - Vbe = Vrb -> I get that. But 30 - Vrc - Vbc ≠ Vrb -> Can you explain that?? – Keno Nov 13 '17 at 13:25
• @Keno Where did you get Vbc? – jonk Nov 13 '17 at 17:25
• Junction voltage - between collector and base, I measured it. According to KVL, the 30 - Vrc - Vbc should equal Vrb but it doesn't. – Keno Nov 13 '17 at 19:11
• @Keno In short, I can either help you understand a complete set of actual measurements, by applying theory to show you why they make sense; or else I can help you work out something from theory, and then explain measurements you make from there. But I cannot address myself to why one specific, actual measurement "here" doesn't add up to a list of theory-derived values "there." You have to keep apples with apples, so to speak. – jonk Nov 13 '17 at 19:21
• @Keno That's an odd question, now that you express it exactly that way. The answer is that you would not do that. The resistor divider technique allows you to set up ANY voltage you want (Thevenin) by just using two rails and the divider. Using ground with two other rails doesn't help because, once again, that means you are stuck with ground as your voltage. And you want the flexibility to select a voltage. Not the inflexibility of being stuck with just one, no matter which one it is. I think I see better where you were coming from, though. Thanks. – jonk Nov 13 '17 at 22:31

Maybe if you draw this circuit this way will help you find the voltage across Rb resistor.

simulate this circuit – Schematic created using CircuitLab

• There fore voltage drop across Rb = Ib*Rb = 10mA*580K/300 – rsg1710 Nov 12 '17 at 19:31
• Not really, because I already knew that before. But thank you for trying. – Keno Nov 12 '17 at 20:18
• 30 - Vre - Vbe = Vrb -> I get that. But 30 - Vrc - Vbc ≠ Vrb -> Can you explain that?? – Keno Nov 12 '17 at 20:52
• I don't understand your question. Can you elaborate? – G36 Nov 13 '17 at 13:26
• I think the problem here is that, due to the high resistance or Rb, the base voltage now is highly dependent on the transistor characteristics, which can vary greatly. In order to fulfill the requirement Vre=10V, you would want to have a fixed base potential of -19.3V. I would opt using a voltage divider to achieve this. – Bart Nov 13 '17 at 14:24