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Learning how to do DC and AC operating point biasing for a single transistor amplifie

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promach

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I am trying to learn how to do proper DC and AC operating point biasing for a simple transistor amplifier circuitry. Could anyone guide me ?
https://www.circuitlab.com/editor/#?id=429hqwg8327t

Screenshot from 2017-10-02 20-19-21.png
 

You show a class-C radio frequency amplifier. A class-C transistor is not biased, it is turned off most of the time. A class-A transistor has its input (base) biased so that it never turns off and its output can swing equally up and down.
 

Could anyone explain why I am not getting the desired output conductance angle for class-C amplifier ?

Screenshot from 2017-10-03 10-50-52.png

Screenshot from 2017-10-03 10-31-55.png
 

By including a parallel tank resistance, I got the correct conduction angle in ngspice as below:

Screenshot from 2017-10-03 13-01-08.png

However, I am quite confused about the AC analysis result that I got though. Any comment ?

Screenshot from 2017-10-03 12-58-58.png
 

Not sure how you determined the conduction angle in the previous measurement that doesn't show a current waveform? It's however unlikely to get reasonable waveforms without any load connected to the class C amplifier. Similarly, your driving source should include a source impedance, otherwise the simulation isn't realistic.

AC analysis is small signal and useless for class C amplifier. It calculates the gain for an infinitesimal small input voltage at the actual bias point, giving about no gain at zero transistor current, just a bit Cbc feedforward. Advanced analysis programs have harmonic balance analysis to determine gain and harmonics of nonlinear circuits.
 

Please refer to the current probe V_IP1

I could not understand about your explanation on why small-signal analysis is not suitable for class-C amplifier.

According to **broken link removed** , we cannot use the small signal model of transistor due to non-linearity issue.

Screenshot from 2017-10-03 15-13-34.png
 

I didn't attempt to explain verbosely why you can't use AC analysis for class C. But apparently you have found the same statement in the paper.

My "explanation" was only one sentence
It calculates the gain for an infinitesimal small input voltage at the actual bias point, giving about no gain at zero transistor current.
Click to expand...
By looking at the circuit operation point, you can see that AC analysis is actually working at zero transistor current.


You should take the time to read a tutorial about SPICE analysis to understand the different nature of AC and transient analysis. A short answer in this thread can't compete with a full tutorial or text book.

I see that ngspice offers an "experimental" .PSS (periodic steady state) analysis which can be used for non-linear circuits.

Please refer to the current probe V_IP1
Click to expand...
I believe that you measured Ic somehow, just noticed that no current waveform is shown in the thread, so we can't know how the observed waveform is different from expectation.
 

AC analysis is actually working at zero transistor current.
Click to expand...

@FvM : I could not understand the above statement.

According to https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.22.7143&rep=rep1&type=pdf or the attached file View attachment Computer-Aided Circuit Analysis Tools for RFIC Simulation - Algorithms, Features, and Limitation.pdf ,

consider the simulation of high-Q oscillators. These also require very long transient simulations; a Q of 10 000 suggests that the turn-on transient time starting from a zero initial state will be of the order of 10 000 cycles of the oscillation period
Click to expand...

I am scratching my head now. I suppose I need some reference on this statement to read on this topic throughoutly.
 
Last edited:

I could not understand the above statement.
Click to expand...
I say that AC analysis is performed in your circuit with no transistor current and hence no gain.

This fact shows that AC analysis can't be used for this kind of circuits. Full stop.

- - - Updated - - -

You wrongly show "R Tank" parallel to the tank instead of in series with the inductor.
Click to expand...
Why series R? Coil resistance can be expected very small for a 100 MHz class C amplifier, parallel R is representing the actual amplifier load. The L and C values in the circuit are far off from suitable values, though.
 

The AC analysis calculates the DC bias conditions of the circuit and then, using linear models for the devices, does an AC analysis.
Since a Class C amp has no DC bias current in the transistor, its gain is considered to be zero for the purposes of the AC analysis.
So that's why an AC analysis doesn't give useable results for a Class C amp.
You can only do a large-signal, transient analysis.
 

FvM said:
Why series R? Coil resistance can be expected very small for a 100 MHz class C amplifier, parallel R is representing the actual amplifier load. The L and C values in the circuit are far off from suitable values, though.
Click to expand...
If "R tank" is the load then it should be called "R load". It should be capacitor-coupled and not shorting the DC of the transistor's collector to the supply voltage. A load of only 10 ohms is not suitable for a low power little transistor.
 

It should be capacitor-coupled and not shorting the DC of the transistor's collector to the supply voltage. A load of only 10 ohms is not suitable for a low power little transistor.
Click to expand...
As already mentioned, the actual L and C values are arbitrary and meaningless. So is the R load. Having this said, we can state that the circuit topology is basically correct. There's no (or very little) DC voltage drop across the inductor, respectively no AC coupling for the load needed. The collector voltage swings symmetrically above and below the supply voltage level.
 

Why does the voltage-across-Cb vary sinusoidally around 85mV ?

Screenshot from 2017-10-04 22-45-48.png

Screenshot from 2017-10-04 22-45-41.png

Note: i(v_ip1) is the collector current.
 

According to the diagram annotation, the blue waveform is showing Vcin, not Vcb. If so, it corresponds to an (almost) sinusoidal input current.
 

I think that if your circuit has a continuous input signal then the input capacitor will charge quickly by the base-emitter diode of the transistor then discharge slowly into the 1k resistor to ground. Then the base will develop a negative voltage that will keep it turned off for almost all the time.

Here is a class-C RF amplifier circuit with an additional diode to balance the charge and discharge times of the input capacitor:

OOPS. I cannot remember my password so I cannot post the schematic, what a nuisance. When I have more time then I will ask for my password to be reset.
 

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