Calculate collector currents for Q1 and Q2 in Fig. 146 . |...

Name: Problem 50, Chapter 4: Electronic Devices and Circuit Theory
Uploaded: 2023-08-09T08:34:29-08:00
Duration: 14 min 54 s
Channel: Brandy Heflin
Description: Calculate collector currents for Q1 and Q2 in Fig. 146 .

Key Concepts

Transistor Biasing

Biasing in transistor circuits sets the proper DC operating point by providing the necessary base-emitter voltage. Correct biasing ensures that the transistor remains in the desired region—usually the active region—so that its current and voltage parameters can be accurately controlled and predicted.

Bipolar Junction Transistor (BJT) Operation

BJT operation is based on the control of a large current between the collector and emitter terminals using a smaller current at the base terminal. Understanding its active, saturation, and cutoff regions is crucial for ensuring the transistor functions as intended in amplification and switching roles.

Collector Current Calculation

Calculating the collector current typically involves using the relation I_C = ?I_B in the active region, where ? (current gain) represents the amplification factor of the base current. This fundamental relationship allows for determining the operating point of a transistor in various circuit configurations.

Current Mirror Concept

In circuits where matching current levels across transistors is needed, a current mirror configuration is often utilized. This concept involves replicating a reference current from one branch of the circuit to another, leveraging the similar characteristics of matched transistors to achieve consistent current levels.

Transcript

00:02 In this question, we are asked to find three currents pictured in the circuit diagram that's given.

00:14 We're going to do that by applying kirchhoff's rules, both the loop rule and the junction rule.

00:22 And we are going to need to establish some reference points here.

00:33 So we're going to need to note the junction right here at the left edge.

00:39 I'm going to call that point a.

00:42 And i'm going to call the junction at the left edge point b.

00:46 I'm going to call the junction at the top point c.

00:50 And then i'm going to call the junction right there in the middle point d.

00:53 And the purpose of this is because i'm going to need to write junction rules for a handful of these to solve for these currents.

01:09 And the other thing we're going to need to do is establish some currents that are not presently listed.

01:19 So when i2 flows over to point a, clearly i3 is going to split off and flow downward.

01:27 We're going to need one that flows upward.

01:31 Similarly, at point b, when i1 flows across, we're going to have i3 coming in, and there's going to be an i4 here that goes up the side.

01:44 And then i called i5 the one that goes up the left side exiting junction a.

01:52 And then at point c, 4 and 5 are going to come together to form one more that i'm going to call i6, heading down between c and d.

02:10 And then 6 will split into one end.

02:24 Let's go ahead and take our loop rules.

02:30 I do tend to like to color code these.

02:37 So let's see how we can do that.

02:39 Let me grab the orange because that's not a common color on here.

02:45 So the first loop i'm going to take is going to be just this top left corner where we're going to have a 12 -volt potential rise minus current i5 flowing through the 5 -ohm resistor as a potential drop minus i2 flowing through a 1 -ohm resistor as a potential drop, and then we're back to the start of our loop, so that all equals zero.

03:33 Similarly, we're going to do our upper right loop where we're going to have a 9 -volt potential rise minus current i4 traveling through the 8 -ohm resistor minus current i1 traveling through a 1 -ohm resistor equals zero.

04:06 And now we need a third loop that includes current i3 because we haven't done that yet.

04:14 And what i did that looks a little bit different, looking at how 1 and 3 are going to come together to make 4, and in order to avoid having to go backward through any batteries or anything like that, i took, maybe gray will work, i came across i3 up the outside of the circuit through the 8 -ohms, down through the middle, through the 1 -ohm resistor on the left, and then through the 12 -volt battery, although i actually started at the battery coming up with this loop.

04:57 So here's my third loop.

05:01 And what i have for this, starting at the battery, we have a 12 -volt potential rise minus i3 times 10 -ohms minus i4 times 8 -ohms minus i2 times 1 -ohm, and that brings us back to the battery, and that equals zero.

05:31 Now let's do a handful of our junctions.

05:36 We can take at point a, we have i2 coming in minus i3 minus i5 equals zero.

06:00 At junction b, we have i1 plus i3 coming in minus i4 leaving equals zero.

06:12 At junction c, we're going to have i4 plus i5 coming in minus i6 leaving, and then i6 will split into i1 and i2.

06:34 And so, one of the things i noticed is that we actually could have eliminated i6.

06:39 I realized that c and d, they're going to be at the same potential, and so we don't even have to worry about physically whether there's a wire there or not, since there's no resistance there.

06:54 We really have i4 and i5 coming into junction c, and then i1 and i2 leaving.

07:05 So, we could go ahead and combine these last two.

07:10 We could have i4 plus i5 coming in minus i1 and minus i2 coming out.

07:22 And so, this now gives us six equations that we can use, although since we eliminated i6, we would only need five of them.

07:47 I'm just looking through, just double checking really quickly everything before i move on to my next step.

07:54 So, my preferred way to solve a system of equations like this is to put all of the coefficients in one matrix and then put all of the constants in the other...

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