a. Given that αdc=0.980, determine the corresponding value of βdc b. Given βdc=120, determine th

A. Given that αdc=0.980, determine the corresponding value...

Key Concepts

Current Relationships in BJTs

The currents in a BJT are interconnected through the fundamental equation I_E = I_C + I_B, where I_E is the emitter current, I_C is the collector current, and I_B is the base current. This relationship is used to determine unknown current values when at least two of these currents are known, and it forms the basis for analyzing the performance and biasing of transistor circuits.

Transistor Current Gain (? and ? Relationship)

In bipolar junction transistors (BJTs), ? (alpha) is the common-base current gain which represents the fraction of the emitter current that becomes the collector current. ? (beta) is the common-emitter current gain, defined as the ratio of the collector current to the base current. They are mathematically related by the formulas: ? = ? / (1 - ?) and, conversely, ? = ? / (? + 1). This relationship is critical for converting between the two parameters for analysis and design of transistor circuits.

Transcript

00:01 In this problem, we are asked to find i .b or the current in the base of the transistor, and vc, or the voltage in the collector, given this figure, and given the values of beta equals 100, and the voltage in the base emitter is 0 .7 volts.

00:23 Now, we first start with the equation of omslough as iv is equal to voltage in the resistor over the resistance.

00:47 Let's change this.

00:50 Now, the voltage across the base can be associated with the voltage in the 10 -kilome resistor, and that is equal to the voltage at node c minus the voltage in the base emitter node over 10 kohoms.

01:14 So we have an unknown here, which is the voltage at node c.

01:22 He could get this from the equation of voltage division, or that is vc, c, is equal to, vcc or our voltage source minus the voltage in the 5 kilo -on resistor.

01:42 So 12 volts minus the voltage in the 5 -kilome resistor will give you the voltage at the node c.

01:51 But what is the voltage across the 5 -kilome resistor? now, going back in the circuit, if this is ib and this is ic, we know that the current passing through the 5 kilo on this is just the sum of these two currents.

02:09 So that is ib plus ic.

02:14 So the equation now becomes bc or the voltage at node c is equal to 12 volts minus ib plus ic times 5 kilowatoms.

02:28 But we'll have to put an assumption here to simplify the problem, and that assumption would be ic is much, much greater than i -b, which normally happens in a transistor.

02:45 If that's the case, we say that the voltage in node c is approximately equal to 12 volts minus just ic times 5 kilo -onns, because ib is so small that we could remove it from the addition term here now this is our third equation let's put in another this is our second equation now let's put in a third equation and that is the relationship between the current in the base and the current in the collector which is given by ib is equal to ic over beta now substituting the second and then the third equations in our first equation, we'll get the current in the base as equal to 12 minus 5 kilo -oom ic minus vbe over 10 kilooms.

03:48 Now ib from the third equation is equal to i .c over beta.

03:58 Isolating the ic will get a term on the right side here as times 10 kilomes plus 5 kilomes ic is equal to 12 minus de...

Please give Ace some feedback