# Analyzing PNP BJT Homework: Active/Sat Mode & V_EC Calculation

• STEMucator
In summary: I_C$$In summary, the homework statement is that a transistor is in active mode if the emitter voltage is equal to the base voltage. If the transistor is in active mode, the collector current is proportional to the beta value. STEMucator Homework Helper ## Homework Statement ## Homework Equations ## The Attempt at a Solution I have several thoughts about this circuit. My first thought was to decide whether or not the transistor is in active or saturation mode. Assuming active mode operation and neglecting the base current, I see ##V_B ≈ 0V##. So the emitter voltage will be ##V_E = V_{EB} ≈ 0.7V##. Hence the emitter current ##I_E = \frac{5V - 0.7V}{2.2k} ≈ 2mA##. Since ##R_C = R_E##, we know ##I_C ≈ I_E##. So:$$2 mA = \frac{V_C + 5V}{2.2k} \Rightarrow V_C = (2 mA)(2.2k) - 5V = - 0.6V$$Hence ##V_{EC} = V_E - V_C = 0.7V + 0.6V = 1.3V## So ##V_{EC} > 0.3V## and I would think the transistor is in active mode. Does this seem reasonable? Am I approaching this correctly? It`s an approximate solution - and I cannot detect any fundamental error. However, as you certainly know - the correct solution requires a finite base current (other than 0). By the way: We can set IC≈IE (if beta is large enough) - indpendent on the condition RC=RE. I think the idea with this assignment is to not assume VB = 0. You can set up a system of equations to give you the device currents as a first step. Okay so assuming the transistor is active, we have:$$I_B = \frac{V_B - 0V}{R_B} = \frac{V_B}{20k}V_E = V_B + V_{EB} = V_B + 0.7VI_E = \frac{5V - V_E}{R_E} = \frac{5V - (V_B + 0.7V)}{2.2 k} = \frac{4.3V - V_B}{2.2 k}$$From here I am not entirely sure what the next step should be. Should I use:$$V_C = V_E - V_{EC} = V_B + 0.7V - V_{EC}$$Then I figure:$$I_C = \frac{V_C + 5V}{R_C} = \frac{(V_B + 0.7V - V_{EC}) + 5V}{2.2k}$$Then ##I_E = I_C + I_B## yields:$$\frac{4.3V - V_B}{2.2 k} = \frac{(V_B + 0.7V - V_{EC}) + 5V}{2.2k} + \frac{V_B}{20k}$$Usually this would spit out ##V_B##, which I could solve all the equations with. The problem is I don't know ##V_{EC}## for this transistor. Consider where ##\beta## is in all this. You need to involve it somehow to solve your assignment. ##I_C = \beta I_B,I_C = I_E - I_B## is two equations in three unknowns. You should be able to find a third equation using KVL to solve for the device currents. So I somehow need a KVL equation involving ##I_E##? Would that mean I apply KVL to the top loop? Also is my prior method incorrect? Zondrina said: So I somehow need a KVL equation involving ##I_E##? Would that mean I apply KVL to the top loop? Consider the loop starting at the 5 V source and moving down through the base-emitter junction to ground. Zondrina said: Also is my prior method incorrect? It is if you need to involve ##\beta##. If you assume ##V_B = 0##, then ##\beta## has no effect on your circuit. milesyoung said: Consider the loop starting at the 5 V source and moving down through the base-emitter junction to ground. It is if you need to involve ##\beta##. If you assume ##V_B = 0##, then ##\beta## has no effect on your circuit. I thought the circuit depended on ##\beta## if ##V_B = 0##. If ##V_B = 0##, the whole base is a ground. Then ##V_{EB} = 0.7V## for active mode operation would imply ##V_E = 0.7V##, and you can find ##I_E## from there. Then since ##I_C = \alpha i_E = \frac{\beta}{1 + \beta} i_E##, the collector current does depend on ##\beta##. I get the feeling ##\beta = 50## is somewhat large, so I think that's why you're saying it has no effect. As for the loop:$$\sum V = 0 \Rightarrow 5V + I_E R_E + V_{EB} + I_B R_B = 0I_E R_E + I_B R_B = - 4.3V$$So the three equations would be:$$I_E R_E + I_B R_B = - 4.3VI_C = \beta I_BI_E = I_C + I_B$$Zondrina said: I get the feeling ##\beta = 50## is somewhat large, so I think that's why you're saying it has no effect. Sure, but I was thinking more in terms of your assignment. To put it another way: If you just short out RB, what actually changes in your circuit when you set RB = 100 kΩ? Zondrina said: As for the loop:$$\sum V = 0 \Rightarrow 5V + I_E R_E + V_{EB} + I_B R_B = 0$$There's a sign issue with the source, but otherwise it looks fine. milesyoung said: Sure, but I was thinking more in terms of your assignment. To put it another way: If you just short out RB, what actually changes in your circuit when you set RB = 100 kΩ?There's a sign issue with the source, but otherwise it looks fine. Oh derp on the sign there sorry. I forgot the source is oriented so the negative side is traversed. So it would be ##+4.3 V##. After solving the equations and obtaining the currents, simple application of KVL will be enough to find ##V_E, V_B, V_C##.$$I_E R_E + I_B R_B = 4.3VI_C = \beta I_BI_E = I_C + I_B$$Plugging the second into the third:$$I_E = \beta I_B + I_B = (\beta + 1)I_B$$Now plugging the above into the first:$$(\beta + 1)I_B R_E + I_B R_B = 4.3VI_B \left[(\beta + 1) R_E + R_B \right] = 4.3VI_B = \frac{4.3V}{\left[(\beta + 1) R_E + R_B \right]} = 32.53 \mu A$$Plugging back:$$I_E = (\beta + 1)I_B = 1.659 mA$$One more plug:$$I_C = I_E - I_B = 1.627 mA

Now simple application of KVL will find the voltages.

As for the change in ##R_B##, nothing would happen because ##I_B = 0## if ##V_B = 0##.

Zondrina said:
As for the change in ##R_B##, nothing would happen because ##I_B = 0## if ##V_B = 0##.
If ##V_B \equiv 0##, then you've effectively just replaced ##R_B## with an ideal wire, so there can still be base current.

The point of this assignment is to show you the effect of bad biasing when ##R_B## becomes large compared with the input impedance ##\beta R_E## of the BJT (it's a voltage divider).

## 1. What is a PNP BJT and how does it work?

A PNP BJT (Bipolar Junction Transistor) is a type of transistor that consists of three doped semiconductor layers. It works by controlling the flow of current between two of the layers (the collector and emitter) by varying the voltage applied to the third layer (the base).

## 2. What is active/saturation mode and how do you determine which mode the transistor is in?

Active mode refers to the state of a transistor when it is amplifying a signal, while saturation mode is when the transistor is fully conducting and cannot amplify any further. To determine which mode the PNP BJT is in, you can use the relationship between the base-emitter voltage and the collector-emitter voltage. If the base-emitter voltage is greater than the collector-emitter voltage, the transistor is in active mode. If the base-emitter voltage is equal to or less than the collector-emitter voltage, the transistor is in saturation mode.

## 3. What is V_EC and how is it calculated for a PNP BJT?

V_EC (Early Voltage) is a parameter used to describe the slope of the output characteristics of a transistor. It is a measure of how the output current changes with respect to the output voltage. It is calculated by finding the slope of the output characteristics curve for different values of collector current and plotting them on a graph.

## 4. Can you explain how to analyze a PNP BJT circuit in active/saturation mode?

To analyze a PNP BJT circuit in active/saturation mode, you must first determine the mode of the transistor (as described in question 2). Then, you can use Kirchhoff's laws to analyze the circuit and determine the voltages and currents at different points. It is important to also take into account the biasing of the transistor and any resistors or other components in the circuit.

## 5. What are some common applications of PNP BJTs?

PNP BJTs have a wide range of applications, including amplifiers, switches, and oscillators. They are commonly used in audio amplifiers, power supplies, and control circuits. They are also used in digital logic circuits, although they are not as commonly used as NPN BJTs for this purpose.

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