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What would be AC load line if AC is blocked? (biasing resistors not loaded with AC)

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AC equivalent circuit of BJT/FET amplifier is usually given like that:
AC-equivalent-circuit-of-CE-amplifier.jpg
(image from http://ecetutorials.com/analog-elec...tabilization-biasing-circuitsthermal-runaway/ )

But at high frequencies, biasing is done through thin lines with quarterwave stubs,
so biasing resistors not loaded by AC signal, and they disappear from AC equivalent circuit:
Biasing_GaAs-FET.gif
(image from http://radioaficionsdr.es.tl/Circuitos-Microondas-y-Microstrip.htm )
The only resistance is left is Rload and Rsource (i think).
In all examples AC load line calculated for Rcollector||Rload. So only Rload left for AC circuit analysis?

I know how to calculate DC load line.
Now i am trying to calculate AC load line, but do not know what AC equivalent circuit would be. Should i consider only one resistance Rload=50 Ohm?
For VCC of 8v and Q point
Vce=4v
Ic=15ma
my DC load line goes through Vcc=8v, IcQ=30mA

Then assuming Rl=50 Ohm (guess it is wrong, because matching network is not assumed)
Vc=Vce+Ic*50 = 4.75 v
Ic= Ic+Vce/50= 175 mA
 
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Terminator3 said:
So only Rload left for AC circuit analysis?
....................
Should i consider only one resistance Rload=50 Ohm?
Click to expand...
Yes, as far as if you ignore incompleteness of quarterwave stubs
 
Thank for your reply! How about matching network?
For example, i decided to use Zload= (70 +j*4) Ohm. With 50 Ohm load.
So i transform 50 Ohm to Zload using stubs/lines/inductance/etc.
Although load is 50 Ohm, it is transformed to (70 +j*4) Ohm
As i understand, it is enough to use real part of final impedance (70 Ohm), that looks into transistor's collector. Not sure if +j*4 complex part can be used somehow.
 

Terminator3 said:
For example, i decided to use Zload= (70 +j*4) Ohm. With 50 Ohm load.
Click to expand...
AC load impedance is (70 +j*4) Ohm.

Terminator3 said:
As i understand, it is enough to use real part of final impedance (70 Ohm), that looks into transistor's collector.
Click to expand...
Wrong.
Transistor's collector looks AC load impedance is (70 +j*4) Ohm.

Terminator3 said:
Not sure if +j*4 complex part can be used somehow.
Click to expand...
If Zload is real number, locus of (Vload, Iload) forms line which passes DC Operation Point, (Vload_DC, Iload_DC).

If Zload is complex number, locus of (Vload, Iload) forms elliptic which never passes DC Operation Point.

Try to plot dynamic locus of (Vload, Iload), using following two methods.

(1) Transient Analysis.
This locus starts from DC Operation Point and forms elliptic after long time.

(2) Large Signal Periodic Steady State Analysis such as Harmonic-Balance or Shooting-Newton.
This locus is elliptic and never passes DC Operation Point.
 
Thank you! I see it now.
I have additional question about oscillator:
when i draw IV curve for oscillator, it looks almost like circle, also pretty distorted.
Can i make some conclusions by looking at IV dynamic locus shape?
I found something here:
https://www2.ece.ohio-state.edu/~roblin/papers/NVNA_MwMag_05725592.pdf
by feeding transistor with sine signal i obtained elliptic curve (very slim, almost like line).
but oscillator have very "distorted huge circle" IV curve...

If someone interested in drawing AC load line, i tried it in SpeqMath and superimposed on Transient analysis curve (biasing as in S53MV's bfp420 VCO, but AC is blocked for biasing resistors):
Code: 
QpointVce=Vce
	QpointVce = 4

QpointIc=Ic
	QpointIc = 0.015

QpointIc=0.014
	QpointIc = 0.014


dcLoadLineVce=Vcc
	dcLoadLineVce = 8

dcLoadLineIc=Vcc/(Rc+R_e) 'A
	dcLoadLineIc = 0.03

acLoadLineVce=QpointVce+QpointIc*50
	acLoadLineVce = 4.7

acLoadLineIc= QpointIc+Vce/50
	acLoadLineIc = 0.094


x1=dcLoadLineVce;
y1=0;

x2=0;
y2=dcLoadLineIc;

slope=(y2-y1)/(x2-x1)
	slope = -3.75e-3

dcLoadLineIc
	Ans = 0.03
dcLoadLineVce
	Ans = 8

m=-dcLoadLineIc/dcLoadLineVce
	m = -3.75e-3

dcLoadLineVce
	Ans = 8


' y-y1=m(x-x1)
' y = m*(x-dcLoadLineVce)

f(x)=m*(x-dcLoadLineVce)
	Function f(x) is defined

Plot(f(x))
	Plot done

acLoadLineIc
	Ans = 0.094
acLoadLineVce
	Ans = 4.7

m2=-acLoadLineIc/acLoadLineVce
	m2 = -0.02
f2(x)=m2*(x-acLoadLineVce)
	Function f2(x) is defined

Plot(f2(x))
	Plot done

f2(x) is AC load line. x is vce. Curve is almost exact match to transient simulation!
I think it is correct, but not sure.
 
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Terminator3 said:
Can i make some conclusions by looking at IV dynamic locus shape?
Click to expand...
I can not understand what you want to mean at all.

Terminator3 said:
but oscillator have very "distorted huge circle" IV curve...
Click to expand...
What analysis do you use ?
If you use Transient Analysis, (Vload, Iload) include higher harmonics. Here you can not use constant value reactance such as Zload=70+j*4.
See https://www.edaboard.com/threads/202948/#4

On the other hand, you can plot (Vload, Iload) locus regarding single frequency, if you use Large Signal Steady State Analsis.
Here you can use constant value reactance such as Zload=70+j*4.

Show me your (Vload, Iload) locus.

Terminator3 said:
If someone interested in drawing AC load line, i tried it in SpeqMath and superimposed on Transient analysis curve (biasing as in S53MV's bfp420 VCO, but AC is blocked for biasing resistors):
........................................................
f2(x) is AC load line. x is vce. Curve is almost exact match to transient simulation!
I think it is correct, but not sure.
Click to expand...
What simulator do you use ?
You can plot (Vload, Iload) locus very easily, if you use Keysight ADS, Keysight GoldenGate, Synopsys HspiceRF, Mentor eldoRF, AWR MicrowaveOffice, Ansoft Designer, Nexxim, Cadence Spectre, etc.
 
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