PNP and NPN transistors in the circuit have identical Is values. Based on the PNP transistor is connected to a DC voltage source.

  • k = 1.38064852×10-23 J/K
  • qe = 1.60217662 ×10-19 C
  • Is = 0.69 fA at 17°C
  • VA = ?
  • β = 100 A/A (in Forward Active and Soft Saturation Regions)

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  1. Determine VOUT and IC when Vx=0.8V. Which region is the transistor operating at?

  2. Determine VOUT and IC when Vx=0.75 V. Which region is the transistor operating at?

  3. Determine VOUT and IC when Vx=0.85 V. Which region is the transistor operating at?


VT = kT/q, so VT = 25 mV

In PNP, IC = Is eVEB/VT, but how to determine VEB?


You have two active devices, so you're going to have to solve a system of simultaneous equations to resolve the unknowns.

It will probably help if you start by replacing the base bias network of Q1 with its Thévenin equivalent.


There's no requirement to solve simultaneous equations for this problem. It's not so complex as that.

The first step, though, is to Thevenize your resistor pair at the base of \$Q_1\$. This will be \$V_\text{TH}\approx 2.207\:\text{V}\$ and \$R_\text{TH}\approx 728\:\Omega\$. This means things will be saturated if the collector of \$Q_1\$ goes below about \$2.2\:\text{V}\$. Which means you don't have to worry about any collector current exceeding about \$1\:\text{mA}\$. This implies that the base current (for active mode) will be at or under \$\approx 10\:\mu\text{A} \$. The drop across \$R_\text{TH}\$ is therefore, again in active mode, around \$7\:\text{mV}\$ or less. So, given these details we can assume that when in active mode the base voltage will in all such cases be \$\approx 2.2\:\text{V}\$.

Now that you know the active mode base voltage, and since you know a-priori that because both saturation currents and beta values are the same in the two BJTs, it follows that \$V_\text{BE}\$ for each BJT is one-half of the difference between \$\approx 2.2\:\text{V}\$ and \$V_X\$, or \$V_\text{BE}=\frac{2.2\:\text{V}-V_X}{2}\$. Given your three values to test, this means \$V_\text{BE}=\left\{725\:\text{mV}, 700\:\text{mV}, 675\:\text{mV}\right\}\$ for \$V_X=\left\{750\:\text{mV}, 800\:\text{mV}, 850\:\text{mV}\right\}\$.

Keep clearly in mind, now, that we are talking only about the case where \$Q_1\$ is in active mode. This simply means that the collector voltage of \$Q_1\$ doesn't precede below the base voltage of \$\approx 2.2\:\text{V}\$ (where \$I_C\le\left(\frac{3.3\:\text{V}-2.2\:\text{V}}{1\:\text{k}\Omega}=1.1\:\text{mA}\right)\$.

From here, it's pretty easy to work out. You know the maximum value of \$I_C\$ when \$Q_1\$ is in active mode. So you know the worst-case values for \$V_\text{BE}\$: \$V_\text{BE}\le \left[V_T\cdot\operatorname{ln}\left(\frac{I_C}{I_\text{SAT}}+1\right)\approx 702.435\:\text{mV}\right]\$.

Therefore, it is immediately apparent which cases are active mode and which are not.

I've left some details for you to worry about (\$V_\text{OUT}\$ and \$I_C\$.)


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