Printed inductor: is it short circuited or open-circuited stub on the base of BJT?

Status
Not open for further replies.
T

Terminator3

Advanced Member level 3
Joined
Feb 25, 2012
Messages
802
Helped
71
Reputation
142
Reaction score
63
Trophy points
1,308
Visit site
Activity points
9,027
Hello!

I can't understand what is inductance of microstrip W=1mm,P=3mm.
I assume that resonant frequency is determined by 0.2pF capacitor and microstrip inductance, and for 6GHz we need inductance:
Code: 
L=3.5*10^-9 ' nanohenrys
	L = 3.5e-9
C=0.2*10^-12 ' picofarad
	C = 2e-13

F=1/(2*Pi*Sqrt(L*C))
	F = 6.015491419e9
of 3.5 nanohenrys, that can be made using open or short microstrip:
Code: 
short=ATan(omega*L/Z0)
	short = 6.924628078e1

open=ATan(-Z0/(omega*L))
	open = -2.075371922e1

open+180
	Ans = 1.592462808e2

lambda=7.44*4
	lambda = 2.976e1

lambda*(open+180)/360
	Ans = 1.316435921e1
lambda*short/360
	Ans = 5.724359211e0
so it is open 13mm stub or 5.7 short stub for Er eff = 2.82.
Where i am wrong?
 

First of all the circuit presented will never oscillate without feedback capacitors, whatever the series inductor is made by microstrip or lumped. Just put the circuit in a simple simulator for confirmation. Even the circuit is considered a Negative Resistance VCO, would need some feedback components.
Second, the DC current bias of BFP420 (30mA) is set too high, very close to maximum burn out value of 35mA.
For best overall performances (stability, phase noise, harmonics, thermal runaway, etc), the BFP series transistors working as oscillators needs a DC current set to about 5 to 10mA.
 
Status
Not open for further replies.

Similar threads

Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…