1
\$\begingroup\$

Say I have this circuit:

enter image description here

I am confused as to how using Thevenin's theorem yields correct results. Say I choose a bias point on an I-V characteristic curve at the IC = 16mA. If this transistor has a beta of 200 then that means a IB = 80uA. Lets also say I want 1 volt across the emitter resistor R2. Now lets take the resitive divider biasing scheme and create its thevenin equivalent:

Step 1:

enter image description here

Step 2:

Shorting Voltage Source and opening Current Source.

Step 3:

enter image description here

$$V_{\text{TH}} = V_1 \times \frac{R_3}{R_4+R_3}$$ and $$R_{\text{TH}} = R_3 \times \frac{R_4}{R_3+R_4}$$

So here is the problem, I know I want the base current to be 80 uA, VBE = 0.7 V and VE = 1 V. Which means I want VB = 1.7 V which is what I would have chosen for VTH as well. As we can see though, VTH needs to be greater than 1.7 V to give me the desired voltage at VB. Where am I going wrong in my thinking?

\$\endgroup\$
1
  • \$\begingroup\$ Alex, are you trying to figure out how to work out the resistor values using only the Thevenin equivalent process for the base biasing? Or, do you just want to work out the same resistor values, but with an easier method that doesn't involve Thevenin equivalents but achieves the same results? Do you care about BJT variations in beta and saturation current? Do you care about the Early Effect? Etc. What are you looking to achieve here? How much do you want to bite off? \$\endgroup\$
    – jonk
    Mar 29, 2020 at 6:56

4 Answers 4

0
\$\begingroup\$

(EDIT: I've completely rewritten my answer to hopefully provide a better answer.)

Consider the application of Kirchoff's Current Law for the two cases where the transistor's base is and is not connected to the midpoint of voltage divider R3 and R4.

Let \$V_X\$ be the voltage at the midpoint of voltage divider R3 and R4.


CASE 1 - The Transistor's Base Is Connected (\$V_X=V_B\$)

$$ I_{R4} = I_{R3} + I_B\\ \rightarrow \frac{V_1-V_X}{R_4} = \frac{V_X}{R_3} + I_B\\ $$

$$ \begin{align*} \rightarrow V_X &= \frac{R_3\,V_1 - I_B\,R_3\,R_4}{R_3+R_4}\\ &= \frac{R_3\,V_1}{R_3+R_4} - \frac{I_B\,R_3\,R_4}{R_3+R_4}\\ &= V_{TH} - I_B\,R_{TH}\\ &= V_B \end{align*} $$


CASE 2 - The Transistor's Base Is Not Connected (\$V_X=V_{TH}\$)

$$ I_{R4} = I_{R3}\\ \rightarrow \frac{V_1-V_X}{R_4} = \frac{V_X}{R_3}\\ \rightarrow V_X = \frac{R_3\,V_1}{R_3 + R_4} = V_{TH} $$


Case 1 has an additional term—the transistor's base current \$I_B\$—that changes the currents in resistors R3 and R4 compared to case 2 which does not have the \$I_B\$ term. Consequently, the voltage drops across R3 and R4 are different for the two cases, and therefore different voltages \$V_X\$ are present at the voltage divider's midpoint for the two cases. For case 1 (the transistor is connected) the voltage divider's midpoint voltage \$V_X\$ is the transistor's base voltage VB. For case 2 (the transistor is not connected) the voltage divider's midpoint voltage \$V_X\$ is the Thevenin voltage VTH. Because VB's value is influenced by the \$I_B\$ term and VTH's value is not, it's evident that for \$I_B \ne 0\$, VB and VTH cannot be equal; they must be different values.

In case 1 you can make the substitution

$$ I_C / \beta = I_B $$

if you'd like to calculate VB's value based on the collector current \$I_C\$ and your chosen value for β (e.g., \$\beta=200\$).

\$\endgroup\$
6
  • \$\begingroup\$ Here is the problem I am facing with equation 2, when I find the VTH it is equal to VB because VB and VTH are in parallel but after I find RTH and complete the transformation, now VTH is not equal to VB. \$\endgroup\$
    – alex
    Mar 29, 2020 at 2:48
  • \$\begingroup\$ @alex, I've completely rewritten my answer to hopefully provide a better answer. Does this description help to answer your question? \$\endgroup\$ Mar 29, 2020 at 7:38
  • 1
    \$\begingroup\$ Thank you, I now see where I was mistaken. I kept thinking R3 should be equal to VB since it was in paralell but I was neglecting the effect of the open circuit. I tried again by choosing what currents I want through R3 and R4 and choosing what voltages across them which gave me their vaules through ohms law. Then I took the transformation and it everything worked out. Thank you! \$\endgroup\$
    – alex
    Mar 29, 2020 at 18:20
  • \$\begingroup\$ @alex, If you aren't already familiar with the concept of "voltage divider stiffness" you might look into that also. For voltage divider bias, the resistance values chosen for resistors R3 and R4 determine the divider's stiffness. Various authors who discuss this topic recommend various degrees of stiffness ranging from 10% (10:1 firm), 5% (20:1) to 1% (100:1, stiff). Solutions that use the STIFF (100:1) option tend to require very low resistance values for R3 and R4, which is often undesirable. So I suggest starting with the 5% (20:1) option. \$\endgroup\$ Mar 29, 2020 at 20:02
  • \$\begingroup\$ From what I undersrand the "stiffness" is about making RTH small compared to (beta+1)*RE such that IC is less dependent on beta correct? \$\endgroup\$
    – alex
    Mar 29, 2020 at 20:51
1
\$\begingroup\$

Usually the current in the voltage divider is 10 times the base current because a transistor part number has a range of beta. Your transistor might have a beta that is 100 minimum, 200 typical and 300 maximum. You get whatever is available.

\$\endgroup\$
0
\$\begingroup\$

VTH is a little greater than 1.7V.

R3 & R4 values would be chosen to give 1.7V at the base of Q1 taking into account the base current which is drawn through R4. This base current increases the voltage drop across R4.

Now if we disconnect the transistor, the voltage at the junction of R3 & R4 will rise to the value of VTH which is a slightly higher voltage than 1.7V.

\$\endgroup\$
1
  • \$\begingroup\$ So this mean there Will be consequent deviation from the required q-point and does it creates a problem for the bias? \$\endgroup\$
    – imran muhd
    Oct 11, 2023 at 21:26
0
\$\begingroup\$

Say I choose a bias point on an I-V characteristic curve at the IC = 16mA. If this transistor has a beta of 200 then that means a IB = 80uA. Lets also say I want 1 volt across the emitter resistor R2.

So first what we do is to select R2 resistor value:

$$R_2 = \frac{1V}{16mA} = 62\Omega $$

Now we want a "stiff" voltage divider which means that \$R_{TH} < \frac{1.7V}{80\mu A}\$ or \$R_{TH} < \beta R_2\$

So, if I choose \$R_{TH}\$ one-tenth of \$ \frac{1.7V}{80\mu A} = 21.25k \Omega\$ we get \$R_{TH} = 2.2k\Omega\$

Therefore the \$V_{TH}\$ value is \$V_{TH} = R_{TH}\cdot I_B + 1.7V = 1.876V\$

Additional if the supply voltage is \$V_{CC} = 10V\$

And from the voltage divider equation, we know that:

\$\frac{R_4}{R_3} = (\frac{V_{CC}}{V_{TH}} - 1)\$ and \$R_{TH} = \frac{R_3 \cdot R_4}{R_3 + R_4}\$ we can solve it and get this result:

$$R_3 = R_{TH} \cdot \frac{V_{CC}}{V_{CC} - V_{TH}} = 2.7k\Omega $$

$$R_4 = R_{TH} \cdot \frac{V_{CC}}{V_{TH}} = 12k\Omega$$

And we are done.

But normally when finding the voltage divider resistor values we do not use Thevenin's theorem.

All we need is a KCL and Ohm's law.

Again to get a stiff voltage divider we assume that the voltage divider current is much larger than the base current. So, for example:

\$I_B = 80\mu A \$ and divider currently should be larger than this say ten times, therefore:

$$R_4 = \frac{V_{CC} - V_B}{11 \cdot I_B} = \frac{10V - 1.7V}{11 \cdot 80 \mu A} = 9.4k\Omega = 10k\Omega $$

$$R3 = \frac{V_B}{10 \cdot I_B} = \frac{1.7}{10 \cdot 80\mu A} = 2.2k\Omega $$

Now we can recalculate the circuit and check if we meet our requirements.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.