Is 'the changes in Vbe make a large difference' very intuitive? Could you please say more about it because it is not very obvious to me?If Vcc is 100 volts then the voltage across Rb is 100-(0.7) where 0.7 changes with temperature. In a low voltage, one battery, application the voltage on Rb is 1.5-Vbe. Now the changes in Vbe make a large difference.
Is the purpose here to determine a value of resistance Rb by a certain Vcc1 so that the transistor is operated in its forward active region, and use that constant Rb in the following work?We can call the lower or higher Vcc voltage Vcc1, and this allows us to set
the base resistor. For the either voltage we have:
Rb=(Vcc1-Vbe1)/Ib1
If Rb is a constant whose value is (Vcc1-Vbe1)/Ib1, where Vcc1, Vbe1 and Ib1 are all constants, I don't understand why dIb/dVbe=-Ib1/(Vcc1-Vbe1) is identical to dIb/dVbe=-Ib1/(Vcc-Vbe1), in which Vcc now is acting as a variable.The initial current Ib1 and initial base voltage Vbe1 will always be the same, the only thing that changes is Vcc1.
So now we substitute this (and Vcc1) into the first equation and we get:
Ib=(Vcc1-Vbe)/((Vcc1-Vbe1)/Ib1) ...[note Vbe a variable, Vbe1 a constant]
Now the change in collector current is proportional to the change in base current,
so all we have to do is evaluate the change in base current as the Vbe changes.
Taking the first derivative with respect to Vbe we get:
dIb/dVbe=-Ib1/(Vcc1-Vbe1)
or simply:
dIb/dVbe=-Ib1/(Vcc-Vbe1)
I guess Ib should be expressed as Ib=(Vcc-Vbe)/((Vcc1-Vbe1)/Ib1), not only Vbe but also Vcc is a variable?Ib=(Vcc1-Vbe)/((Vcc1-Vbe1)/Ib1) ...[note Vbe a variable, Vbe1 a constant]
Hello MrAl,
Is the purpose here to determine a value of resistance Rb by a certain Vcc1 so that the transistor is operated in its forward active region, and use that constant Rb in the following work?
If Rb is a constant whose value is (Vcc1-Vbe1)/Ib1, where Vcc1, Vbe1 and Ib1 are all constants, I don't understand why dIb/dVbe=-Ib1/(Vcc1-Vbe1) is identical to dIb/dVbe=-Ib1/(Vcc-Vbe1), in which Vcc now is acting as a variable.
I guess Ib should be expressed as Ib=(Vcc-Vbe)/((Vcc1-Vbe1)/Ib1), not only Vbe but also Vcc is a variable?
Wow, the experiments seem to be a little complicated to me. But first, I don't understand why we need to change the value of Rb to Rb2 in experiment 2 even when we use another Vcc?Hi,
Yes, Rb is not a true constant that is why we try to eliminate it. Rb is a constant within one experiment, but is not constant when we look for a formula that varies with Vcc. Another way of stating this is that Rb is a function of Vcc:
Rb(Vcc1) which then later becomes Rb(Vcc).
Let's look at the setup of a single experiment one time:
1. We start with some collector resistor Rc1, and say we want some output voltage. This requires a certain current in the collector.
2. Due to the Beta, that collector current requires a certain base current Ib1.
3. Due to Ib1 and Vbe1 and Vcc1, we choose a certain Rb that gives us Ib1, call it Rb1.
4. Rb1 gives the right base current for the desired operating point, and then finally we change Vbe to Vbe2 to find the change in base current and that tells us how much the base current has changed with that particular Vcc which we called Vcc1.
So in the above Vcc is a constant, Vbe is not.
Now later, we want to set up another experiment with a different Vcc, call it Vcc2. Now we have to perform steps 1 through 4 all over again because the old base resistor Rb1 is no longer acceptable, so we need Rb2. Now we want to know what happens when we change Vcc without having to go through all the trouble of setting up experiments.
Another way to look at this would be:
Exp1:
Ib1a=(Vcc1-Vbe1a)/Rb1
Ib1b=(Vcc1-Vbe1b)/Rb1
Exp2:
Ib2a=(Vcc2-Vbe1b)/Rb2
Ib2b=(Vcc2-Vbe2b)/Rb2
Note in either experiment only Ib and Vbe change, and in either of these two experiments we want to find:
(Ib1-Ib2)/(Vbe1-Vbe2)
and in BOTH experiments we end up with the same formula except that one contains Vcc1 and the other Vcc2, So we end up with two formulas in a form similar to this:
y1=f(a1)
y2=f(a2)
but in each of these a1 and a2 are always the same physical thing, namely the power supply voltage Vcc, so we can substitute:
y1=f(Vcc)
y2=f(Vcc)
I hope this helps but if not i'll get back here with a full example.
Hello, MrAl, thank you very much for your precious time to help me with my weird questions.Alternately we find how Ib changes with Vbe with a given Vcc, then find how Ib changes with Vbe wth a different Vcc, then compare to see which Vcc caused a bigger change in Ib. This is the more experimental approach and can actually be performed in the lab or in a simulator.
Hi, Ratch, thank you very much for mentioning that. I had noticed it a few days ago, and I had changed the collector resistor from 1k to 600 in my later circuit, in post #14, to drive the B-C junction reverse biased.I will say something about this circuit and the two others in post #3. The calculations are all suspect, because the transistors are in saturation. In post #3 you can tell that because the voltages at the base and collector are the same. The base-collector voltage should be reverse biased.
Yes, thanks again, I remember. This made me have another question: for the circuit below, how do we calculate the base current?Remember me telling you that the diode curve is highly nonlinear? The current is approximately Is*exp(Vbe/Vt), where Is is the saturation current and Vt=0.026 volts. If the Vbe goes from 0.76620 to 0.79466, the base current will increase by exp(0.79466/0.76620) = 2.821, an increase of almost 3 times.
Hello, MrAl, thank you very much for your precious time to help me with my weird questions.
I think I need to put the math part off until a later time because it seems to be a little complicated for me right now. Instead, I have done a simulation, but it seems to be showing a contradiction.
Here's what I did:
I use 3 different Vccs: Vcc1=1.5V, Vcc2=3V and Vcc3=6V.
For each Vcc, I plot the Ib-Vbe characteristics by altering Rb from 80k to 100k while keeping Rc not changed.
The result is as shown in the 2nd figure. The horizontal axis is Vbe and the vertical axis is Ib. The left straight line is for Vcc=1.5V, the middle for Vcc=3 and the right for Vcc=6.
View attachment 86621
View attachment 86622
For each Vcc, dIb/dVbe is a constant, and is bigger when Vcc is bigger which seems to be showing that at larger Vcc, the change in base current per unit change in Vbe is larger. Doesn't that mean at larger Vcc, the bsae (or collector) current is more sensitive to Vbe?
In PSpice, the emitter current reference direction is into the emitter terminal, so there's a negative sign there.Why is the emitter current negative? It should be almost the same as the collector current.
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