You multiply two resistances two get Vi ???20 = Rin*300Ω
Use instead:
Rin (ohms)=Vi*300 (ohms)
You multiply two resistances two get Vi ???20 = Rin*300Ω
This is how you (or anybody) should test it.Ok, I'll switch them around and give it another test tomorrow but I'm still confused as to how we test it.
Max,The only bit that confuses me is the Rin, so input resistance. On the equipment it says there will be a resistance of 50 Ohms. Is that what you're referring to?
To my understanding this is not fully correct (see "Equivalent circuit INPUT side.jpg").Vi=Iout/Iin.
Using Iin=Vin/Rin and Iout=Vout/Rload
we get
Vi=Vout*Rin/Vin*Rload=Rin/Rload if Vout/Vin=1.
(Rload=R,emitter||RL,extern).
Isn`t it a matter of definition? Because Rin is the dynamical input resistance (small signal ac) I think it is only logical to use also the differential load resistance to calculate the small signal current gain of the complete circuit.To my understanding this is not fully correct (see "Equivalent circuit INPUT side.jpg").
Iout = Vout / RL and not Vout / Rload.
Max, the lower frequency limit of the circuit is app. at 300 Hz.I attempted to simulate it. I got these results:
But for the frequency instead of using 1Khz I used 10Hz with an amplitude of 1 otherwise nothing showed.
Tweaked RE a bit and got:
Stick to your DESIGN values until you can demonstrate that the DESIGN is not correct.I attempted to simulate it. I got these results:
But for the frequency instead of using 1Khz I used 10Hz with an amplitude of 1 otherwise nothing showed.
Tweaked RE a bit and got:
See my note in post #27.Yeah I was wondering why I couldn't get the same results as Efron.
I guess you will be using an oscilloscope. It will measure the voltage difference between two points. One of these two points will be the reference (oscilloscope ground) and the other will be the measured signal. It will be positive if voltage in the measured signal is bigger than oscilloscope reference/ground and negative otherwise.if I was doing this on the real instruments. What would I do then? In reference to what you just stated LvW
Let's first focus on the output (signal 2 in the oscilloscope I guess). As I told you before you cannot directly measure currents in an oscilloscope, but only voltages.I tested it this morning and got these results...don't really know what I'm looking at:
GOOD point The Electrician,In the simulations that have been posted so far, it appears that the built-in model for a 2N3904 is being used. I suspect the β for that model is not 100. But, your problem requires the β to be 100, so using the LT spice simulation is not going to lead to a design that meets the problem requirements.
I'm not an LT spice user, and perhaps the simulations shown have included a change to make the 2N3904 β equal to 100, but I don't see it.
For an emitter follower like this, the current gain is very dependent on β.
Let's do the math. Using the component values from post #27, and letting re be 1.7 ohms:GOOD point The Electrician,
You're right if you focus only at the transistor. But thanks to the bias, the current amplification is not so dependent of BETA.
See my post #4
Ai = Zi / RL (assumed Av = 1) and Zi = Rb // ( rπ + (1 + beta)*RL' ).
So, mathematically speaking, yes, current gain depends on BETA. But do the maths and you'll see that the implication of BETA is very very small.
At the end, the value of Rb (parallel between R1 and R2) is small compared to (rπ + (1 + beta)*RL') and this is even more true with bigger BETA.
So Rb is predominant for Zi.
We can say this is true for BETA >= 100. Special care should be taken when using power transistors in this configuration (power transistors have lower BETA values , e.g. 20).
β Zi Ai
100 16951 16.951
200 22586 22.586
300 25421 25.421
400 27128 27.128
The Electrician, you're right. I made a mistake in my calculations, yours are correct.It doesn't look to me like "...the implication of BETA is very very small."
Of course, that depends on what you mean by "very, very small".