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enter image description here

Following parameters are given for the above circuit:

R1=8.2 kΩ, R2=5.6kΩ, RE=2.7kΩ, VEB=Uj=0.7V, Vcc=10V, β=200

The question asks the maximum load resistance RL for the transistor in active mode.

I solve the question in the following way:

enter image description here

The maximum load resistance RL for the transistor in active mode means to me the transistor is approaching to saturation at that point.

So for that point, I take Vce=0 and set Vy=Vx+0.7V.

Since Vx=Vcc*R2/(R1+R2)

Vy=Vcc*R2/(R1+R2) + 0.7V

Vy=10*(5.6/13.8) + 0.7V = 4.76V

Now since Vce=0V, and Ie=(Vcc-Vy)/Re = 1.94mA

Ic = Ie approximately so

RL = Vy/Ic = 4.76V/1.94mA = 2.45kΩ

So I calculate the maximum RL in active region as 2.45kΩ, whereas the answer is 2.1kΩ.

Is my calculation wrong?

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  • \$\begingroup\$ Active mode =/= saturation; you will have some voltage drop from collector to emitter. \$\endgroup\$ – Don Joe Oct 3 '17 at 1:21
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    \$\begingroup\$ Vce=0 means the transistor is in hard saturation. Vce> 200 mV is more realistic. Solution likely uses Vbc =0 \$\endgroup\$ – sstobbe Oct 3 '17 at 1:21
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    \$\begingroup\$ I agree with @sstobbe. The answer uses \$V_{BC}=0\$, which is the same place I use to demark the end of the active region and the start of (very shallow) saturation. (You should get a base current of about \$9.6\:\mu\textrm{A}\$, which I think you do.) \$\endgroup\$ – jonk Oct 3 '17 at 1:24
  • \$\begingroup\$ Vce = Vbc+Vbe vectorially. When Vbc=0 since Vbe is always around 700mV, does that mean the question takes Vce=700mV instead of zero? \$\endgroup\$ – user1999 Oct 3 '17 at 1:43
  • \$\begingroup\$ Yes the point where Vce = 700 mV is a good point for separating the active mode and saturation regions. \$\endgroup\$ – Bimpelrekkie Oct 3 '17 at 6:42
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The following two schematics are equivalent:

schematic

simulate this circuit – Schematic created using CircuitLab

Where \$V_{TH}=V_{CC}\frac{R_2}{R_1+R_2}\$ and \$R_{TH}=\frac{R_1\cdot R_2}{R_1+R_2}\$.

From the above, and assuming \$I_B^{'}=\mid I_B\mid\$ and \$V_{BE}^{'}=\mid V_{BE}\mid\$, you can compute:

$$\begin{align*} V_{TH}+ I_B^{'}\cdot R_{TH}+ V_{BE}^{'} + I_B^{'}\cdot\left(\beta+1\right)\cdot R_E&=V_{CC}\\\\ \therefore\quad I_B^{'}=\frac{V_{CC}-V_{TH}-V_{BE}^{'}}{R_{TH}+\left(\beta+1\right)\cdot R_E} \end{align*}$$

Given your values, I get \$I_B^{'}\approx 9.6\:\mu\textrm{A}\$.

The start of shallow entry into saturation occurs right when \$V_{BC}=0\:\textrm{V}\$ or when \$V_B=V_C\$:

$$\begin{align*} V_C&=V_B\\\\ I_C\cdot R_L &= V_{TH}+I_B^{'}\cdot R_{TH}\\\\ \beta\: I_B^{'}\cdot R_L &=V_{TH}+I_B^{'}\cdot R_{TH}\\\\ \therefore \quad R_L &=\frac{V_{TH}+I_B^{'}\cdot R_{TH}}{\beta\: I_B^{'}}=\frac{1}{\beta}\left(R_{TH}+\frac{V_{TH}}{I_B^{'}}\right) \end{align*}$$

From which I get \$R_L\approx 2130\:\Omega\$.

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Maximum gain for CE stage is VDD/0.026; the current is irrelevant (within reason).

Consider a 26 volt battery. You can operate at 1mA (thus 24Kohm allows 2 volts across the Vce, so bipolar is just barely out of saturation), or 10mA or 100mA, or 1uA (with 24,000,000 Ohm Rcollector)

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    \$\begingroup\$ Yeah. It maxes out at \$40\cdot V_{CC}\$. But I think the OP had a different question. Or did I miss something? \$\endgroup\$ – jonk Oct 3 '17 at 4:23
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    \$\begingroup\$ Indeed, the question was a very different one. \$\endgroup\$ – Bimpelrekkie Oct 3 '17 at 6:44
  • \$\begingroup\$ Then I apologize for providing insight to the OP. \$\endgroup\$ – analogsystemsrf Oct 3 '17 at 16:28

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