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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?

Solution:

VT = kT/q, so VT = 25 mV

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

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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.

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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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