Hi
I need some help with my assignment please.
My task is to investigate the operation of the amplifier using sin. frequency of 1KHz and amplitude of 4V.(the circuit diagram is attached). I can see that is a class B, but what is confusing me that there is no voltage gain. Is it a current amplifier? (output current is higher, i can see it from multi-meter) And why the bode plotter show negative gain?
Thanks!

 

Yes, it's a current amplifier.
The imitation oscilloscope shows no inversion. The bode plotter is either frankly wrong or both of us don't know how to read the labels on it. (A real possibility)

 

I am not sure what does the plotter shows, the voltage gain or current. If its shows voltage gain then its make sense, because unity of gain is 0 db, but i am really not sure...

 

Well, maybe you should change your question to: "How to read this bode plotter?"
You will get better answers than I can give you.

 

Well i can read the plotter , its shows a negative gain. The question still the same . Why does it shows negative gain?

 

OK. The question is, "Why does this plotter show the wrong answer?"

I don't know. Somebody else will have to answer that.

 

The Bode plot is correct. It's not negative gain, minus dB is just gain less than 1. The transistors are biased at zero current so their small signal gain is basically 0 (a measured value of -101.585dB or 8.3e-6 which may be due to capacitive feedthrough).

You have large signal gain of near 1 showing on the virtual oscilloscope since that signal is large enough to move through the zero bias point (which gives the observed crossover distortion). If you use a small signal from the generator (say less than 100mV) then you will observe a very low gain from the oscilloscope, similar to the Bode plot.

 

Well the gain had flattened out (presumably to almost unity) at about 1GHz.

 

Well spotted crutschow.

Perhaps a useful exercise for the OP:

Convert to class AB, with the addition of a couple of series diodes replacing the short / link between the two bases and you will see an entirely different picture.

 

May I add a short comment to Chrutschow`s answer?
Yes - as mentioned, the BODE plot is correct, however useless.
The reason is as follows: The BODE plot is the result of a LNEAR SMALL-SIGNAL analysis. However, the circuit under discussion is strongly NON-LINEAR.
The amplifying properties of such a circuit can be analysed in the time domain only.

 

1Ghz for a 2N3904?

 

1Ghz for a 2N3904?

One of the reasons I don't simulate.

 

1Ghz for a 2N3904?

Again, I believe that's just capacitive coupling between input and output through the base-emitter junction. The transistor is cutoff as noted in my previous post.

 

One of the reasons I don't simulate.

More often then not the simulation is correct. You just have to determine why it appears not to be.

 

I suppose that if I took the time to learn the details I would get something besides garbage out of a simulator.

 

I suppose that if I took the time to learn the details I would get something besides garbage out of a simulator.

It's well worth the effort if you do any analog design. A lot faster and more efficient then trying to troubleshoot the circuit after you built it and it doesn't work as you anticipated (my experience has been that many circuits don't but perhaps that's just a result of my less than exacting design ability) . It's especially useful to determine frequency response and optimize negative feedback parameters to get proper transient response and gain/phase margins, also to try out different circuit options to help determine which is best.

 

It's well worth the effort if you do any analog design. ......... It's especially useful to determine frequency response and optimize negative feedback parameters to get proper transient response and gain/phase margins, also to try out different circuit options to help determine which is best.

Yes - I fully support this opinion.

 

Thank you for all of your replies, very much appreciated. I did different simulations with the input signal 0.9v and the voltage gain got a lot smaller. So in assignment i will go for the version that the bode plotter is correct and i use large signal. As crustchow said.
And thanks again.

 

Thank you for all of your replies, very much appreciated. I did different simulations with the input signal 0.9v and the voltage gain got a lot smaller. So in assignment i will go for the version that the bode plotter is correct and i use large signal. As crustchow said.
And thanks again.

What is voltage in dB:
Voltage in dB=20*log10(voltage in volts)

Example 1:
20*log10(1 volt)= 0 dB

Example 2:
20*log10(2 volts)= 6 dB

Example 3:
20*log10(0.5 volts)= -6 dB

Like someone said above, the fact that you have negative dB values means that you have less than 1 volt.

 

What is voltage in dB:
Voltage in dB=20*log10(voltage in volts)

Example 1:
20*log10(1 volt)= 0 dB

Example 2:
20*log10(2 volts)= 6 dB

Example 3:
20*log10(0.5 volts)= -6 dB

Like someone said above, the fact that you have negative dB values means that you have less than 1 volt.

This is nonsense.