Tony Williams said:
V1 V2
3k3 | R L |
---/\/\----(1)---+---/\/\---))))---+
| |
Z12--> C1=== ===C2
| |
-----------(2)---+-----------------+--0v
If I've done the algebra right, Z12 is not resistive
if calculated at w.L = 1/(w.C1) + 1/(w.C2), which
is the resonant frequency of the pi-network.
Z12 = 1/R*(w.C1)^2 - j/(w.C1).
That looks like a resistor in series with a capacitor.
If the Q is high then R is low and the resistive part
is very high compared to the capacitive part.
To get your 0.5V for V1 and V2 then R is about 9.4 ohms.
Hi Tony,
A series resistance of 9.4 ohms would produce an inductor Q of 15.7. This
is lower than typical commercial inductors, which may have typical Q's of
30 to 50. This indicates there may be some error in the measurement. For
example, the nearest standard inductor is 4.7uH, so the tank resonance will
occur at 5.257MHz, not at 5.00MHz as Andrew assumes.
Below is an LTspice ASC file of a pi network at 5.00 MHz using a 4.7uH
inductor and a typical Q of 30. The second file is the plt file with the
phase of the current through the 3.3k resistor shown on a separate graph.
As shown in the AC analysis, the phase angle of the current through the
resistor around resonance is very close to zero degrees. This indicates the
input impedance of the tank near resonance is resistive, and not capacitive
as many people seem to think.
The reason is the inductor provides a 180 degree phase reversal, so the
capacitive currents cancel at the input to the pi network. This places the
resistive portion of the tank impedance in series with the 3.3k input
resistor.
Since the phase angle changes rapidly around resonance, an attempt to
calculate the input impedance quickly runs into trouble as you don't know
the exact frequency the circuit will oscillate at. This means SPICE is the
best and easiest way to analyze these circuits.
Here's the ASC file:
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Version 4
SHEET 1 880 680
WIRE -192 144 -256 144
WIRE -112 144 -192 144
WIRE 16 144 -32 144
WIRE 96 144 16 144
WIRE 208 144 176 144
WIRE 320 144 288 144
WIRE -256 192 -256 144
WIRE 16 208 16 144
WIRE 320 208 320 144
WIRE -256 320 -256 272
WIRE 16 320 16 272
WIRE 320 320 320 272
FLAG -256 320 0
FLAG 16 320 0
FLAG 320 320 0
FLAG -192 144 Vin
FLAG 320 144 Vout
FLAG 16 144 VCap
SYMBOL cap 304 208 R0
SYMATTR InstName C1
SYMATTR Value 431.1pf
SYMBOL cap 0 208 R0
SYMATTR InstName C2
SYMATTR Value 431.1pf
SYMBOL ind 80 160 R270
WINDOW 0 32 56 VTop 0
WINDOW 3 5 56 VBottom 0
SYMATTR InstName L1
SYMATTR Value 4.7µh
SYMBOL res 304 128 R90
WINDOW 0 0 56 VBottom 0
WINDOW 3 32 56 VTop 0
SYMATTR InstName R2
SYMATTR Value 4.92
SYMBOL res -128 128 M90
WINDOW 0 0 56 VBottom 0
WINDOW 3 32 56 VTop 0
SYMATTR InstName R1
SYMATTR Value 3k3
SYMBOL voltage -256 176 R0
WINDOW 123 24 134 Left 0
WINDOW 39 0 0 Left 0
SYMATTR Value2 AC 1
SYMATTR InstName V1
SYMATTR Value SINE(0 1.5 5000000)
TEXT -80 72 Left 0 ;'5.0 MHz pi Network
TEXT -216 360 Left 0 !.ac lin 10k 4.8e6 5.2e6
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Here's the PLT file
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
[AC Analysis]
{
Npanes: 2
{
traces: 1 {34603011,0,"I(R1)"}
X: ('M',2,4.8e+006,40000,5.2e+006)
Y[0]: (' ',1,0.000223872113856834,0.2,0.000288403150312661)
Y[1]: (' ',0,-8,2,10)
Log: 0 2 0
GridStyle: 1
PltPhi: 1
},
{
traces: 2 {524290,0,"V(vout)"} {524293,0,"V(vcap)"}
X: ('M',2,4.8e+006,40000,5.2e+006)
Y[0]: (' ',1,0.10471285480509,0.7,0.25409727055493)
Y[1]: (' ',0,-240,30,60)
Log: 0 2 0
GridStyle: 1
PltMag: 1
}
}
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Regards,
Mike Monett
Antiviral, Antibacterial Silver Solution:
http://silversol.freewebpage.org/index.htm
SPICE Analysis of Crystal Oscillators:
http://silversol.freewebpage.org/spice/xtal/clapp.htm
Noise-Rejecting Wideband Sampler:
http://www3.sympatico.ca/add.automation/sampler/intro.htm