a. Referring to Fig. 57 , determine the collector current above which the switching time does no

A. Referring to Fig. 57 , determine the collector current...

Key Concepts

Transistor Switching Dynamics

This concept involves the understanding of how a transistor transitions between its various states — cutoff, active, and saturation — and the factors that determine the speed (switching time) of these transitions. It encompasses the analysis of internal capacitances and circuit resistances that affect how quickly a transistor can respond to changes in input signals.

Effect of Load Resistance on Switching

Load resistance plays a pivotal role in determining the switching speed of a transistor circuit. By influencing the RC time constant (the product of resistance and capacitance), different load resistor values alter the rate at which voltage levels change during the switching process, thereby affecting overall response times.

RC Time Constant

The RC time constant is a fundamental concept that defines how quickly a capacitor charges or discharges through a resistor. It is central to analyzing timing and transient responses in circuits, particularly in the context of transistor switching, where both intrinsic and load-related resistances work with capacitances to dictate switching speeds.

Collector Current Impact

Collector current affects the operating region of the transistor and can influence switching behavior. At certain current levels, particularly when approaching saturation, the switching time may become relatively insensitive to further increases in current due to the inherent characteristics of the device's transition between its operating states.

Transcript

00:01 So for the first part, the 12 oom resistor and 25 oom resistor are both in parallel.

00:13 So the net resistance or the equivalent resistance for these two, let's call it 1 -2.

00:27 So this will be equal to.

00:29 So we go with the reciprocal rule.

00:32 So we add the reciprocals of the resistance and then take an inverse.

00:40 And this comes out to be 8 .108 oms.

00:46 And then this combination, this parallel combination, so basically this equivalent resistance is in series with the 4 .5 oom resistor.

01:01 So for the equivalent resistance of this combination, will be let's call it let me just write the notation so let's call it r1 let's call this r2 and let's call this r3 so r1 and r2 are parallel and the equivalent resistance of r1 and r2 is r1 2 and r1 2 and r1 2 and r1 2 is in series with r3 so the equivalent resistance of this whole combination will be r1 2 less r3 because these two are in series and this comes up to be equal to 12 .608 oms.

01:52 Now this resistance is again in parallel with so r2 3 is in parallel with resistor r4 which let's say is equal to so that is given to be equal to 18 ooms and therefore our equivalent which is basically the equivalent resistance of the whole circuit will be equal to so these two are in parallel so we go by the parallel method again the reciprocal method again so we add the reciprocals of these two resistance that are in parallel and we take the inverse so doing that and plugging these two values over here we get the equivalent resistance to be equal to 7 .4 oms.

02:51 Now coming to the second part, so we need to find the current in the 18 oom resistor.

03:01 So let's use kchhoff's loop rule for the loop containing the battery and the 18 oom register.

03:09 So we have the battery emf minus current flowing through the 18 -oom resistor and resistance of the 18 -home resistor so that i denoted to be r4 and this is equal to 0.

03:33 So this means that i -18 will be equal to emf over r4 and the emf is 6 volt, r4 is 18 -ooms and current flowing through that through the 18 -oom register comes up to be 0 .33 ampers.

03:55 So there goes part three.

04:00 Then we have to find the current flowing through this combination, the parallel combination.

04:09 So in short, we have to find the current flowing through r1 -2.

04:14 So we use kitchchhoff's loop rule again for the outer loop containing the battery and the resistor, this parallel register combination.

04:27 So we have 1 -2r -1 -2.

04:34 And then the 4 .5 resistor is also included.

04:42 We are also solving for that.

04:43 So we are applying kitchf's loop rule for the outer loop containing the battery and the resistor r -2 and the 4 -2.

04:51 And the 4 .5 resistor r3.

04:58 So we have i3r3.

05:01 This is equal to 0.

05:05 So from here we can find now since these two are in series, therefore the current flowing through these two should be same.

05:21 Therefore i3 should be equal to i12.

05:27 This means that emf minus i12 times r12 minus 0.

05:37 Now we can solve this to find the current flowing through the parallel combination r1 and r2.

05:47 So let's plug the values now.

05:56 This is the emf r12.

05:58 What's the equivalent resistance that we found? 8 .108...

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