20 = Rin*300Ω

You multiply two resistances two get Vi ???

Use instead:

Rin (ohms)=Vi*300 (ohms)

 

Ok, I'll switch them around and give it another test tomorrow but I'm still confused as to how we test it.

This is how you (or anybody) should test it.

1/ Simulate your circuit in a computer (e.g. you can use LTspice for that)

2/ Verify all real components are of correct value and are placed and connected correctly in your prototype board. Usually it is better to use breadboard for this so you don't have to worry about soldering (changes are easier in breadboard).

3/ Check the Power supply output voltage without connecting it to the circuit.

4/ Connect the Power supply to your circuit and verify that your circuit is correctly powered. DO NOT connect Vin for the moment.

5/ Check all voltages at all strategic pins with a multimeter (e.g. VE, VB, VC).

6/ If all previous checks are OK, it is time to connect Vin into the circuit (Vin should also be checked in standalone previous to connecting it to the circuit).

Only after this you can get valuable conclusions of your design.

 

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?

Max,

When you get confused, use and abuse of schematic drawings.

This is the equivalent circuit you have, as seen from the Vin source ("Equivalent circuit OUTPUT 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).

To my understanding this is not fully correct (see "Equivalent circuit INPUT side.jpg").

Iout = Vout / RL and not Vout / Rload.

So the correct current amplification is

Ai = Zi / RL = 20K / 1K = 20, as per design

 

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:

 

To my understanding this is not fully correct (see "Equivalent circuit INPUT side.jpg").
Iout = Vout / RL and not Vout / Rload.

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.
But - as mentioned above: Matter of definition.

 

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:

Max, the lower frequency limit of the circuit is app. at 300 Hz.
Thus, no gain around unity can be expected at this low frequency signal of 10 Hz.

In case you cannot see anything for 1 kHz ("nothing showed" - what does this mean??) there must be an error in your simulation setup.
Did you ever try an ac analysis?
For transient analysis with 1 kHz use a time period of 10 millisec and a time step limit of 10 µsec (or less).

Did you check the DC operating point ?

Edit: I think it is OK (dc quiescent current). Why didn`t you show the input and output voltages? They are more relevant than the currents through Re and RL.

 

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 simulated as well and everything is upon design.

You measured the output current value at emitter side, probably at RE. Remember that Iout is the current flowing through RL (your load).

In my simulation, I've got a Iout = 1mA current amplitud over 40uA amplitud for I(Vin). If you do the maths, that's make a current gain of 25 !!!! Spot on !!!

Note: By default, LTspice asigns a current sens to Vin. Therefore, I've used -I(Vin) for input current.

 

I didn't really know how to show two separate readings, but when RL is graphed alone it has an amplitude of 40uA. That makes a lot more sense now actually seeing it in practice, when I tested it on the Marconi instruments though all it showed was voltages. I couldn't figure out how to make it display current. Which is what intially confused me on how am I actually supposed to demonstrate if it works. If all its showing is two readings that were the same. I have lab access tomorrow so I'll take some pictures of the results now that the caps are in the correct polarity. Thank you again guys. I got a perfect gain of 20 by switching R1/R2 to 48K..but don't think I can find that value in store.

 

Max, one last comment:
I don`t know if you have realized the phase shift (180 deg) between both currents.
In reality both currents must be in-phase.
The reason is as follows:
The current I(V2) is NOT the current into the transistor circuit but the current that flows internally within the source from "+" to "-".
Thus, to see the current into the circuit use -I(V2).

 

Yeah I was wondering why I couldn't get the same results as Efron...to be honest I've never used this program before so don't really know how to set that -I(V2).

Thanks for pointing that out though!

Quick question though..if I was doing this on the real instruments. What would I do then? In reference to what you just stated LvW

 

Yeah I was wondering why I couldn't get the same results as Efron.

See my note in post #27.

if I was doing this on the real instruments. What would I do then? In reference to what you just stated LvW

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.

You need to know what to measure, how and evaluate the results. Before measuring something, try to determine what the output should be; it will help you a lot.

 

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.

 

I tested it this morning and got these results...don't really know what I'm looking at:

 

I tested it this morning and got these results...don't really know what I'm looking at:

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.

For each signal you were measuring, between what and what?

Assuming you were measuring at the output RL, you have a Vpp = 920mV (let's assume it is almost 1V). This means a current flowing through RL of about 1mA pp (because RL = 1K).

If you remember our LTspice simulation, we had a 2mA pp at RL when using a 1V amplitud at Vin. If all these assumptions are correct, you should be using an input voltage source of 0.5V amplitude - or 1Vpp. Is this correct?

If this is correct, everything seems fine, but tell us, for each channel, between what and what you measured.

I guess you signal 1 corresponds to the input voltage. As you can see there is no much difference between signal 1 and 2. This is normal as the common-collector amplifier has a voltage gain near 1.

Something strange for me, what is the frequency of the input signal? Is it 1KHz or 1000KHz?

 

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.

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).

 

Yes that's correct I used 1Vpp and I'm pretty sure its 1Khz..it is a '.' not a ','

Also on that Beta note..1mA /40uA is a gain of 25. I needed 20. Could that be because of the different Beta value?

And what do you mean by "For each signal you were measuring, between what and what?"

This is how I hooked it all up.

 

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).

Let's do the math. Using the component values from post #27, and letting re be 1.7 ohms:

rπ~(β+1)*re
RL'=333.33 ohms
R1=68k
R2=68k
RB=34k

So for several values of β:

 β       Zi      Ai
100    16951   16.951
200    22586   22.586
300    25421   25.421
400    27128   27.128

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".

 

When I did the calculations along with efron using 100 as beta..we got a gain of 20 spot on. So how did it change?

 

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".

The Electrician, you're right. I made a mistake in my calculations, yours are correct.

Good point again.

 

Does that mean all previous calculations are wrong?