Maker Pro
Maker Pro

74HCU04 LC Osc

A

Andrew Holme

Jan 1, 1970
0
I built this 5 MHz oscillator using a Toko KANK4174 inductor, which is
specified as having a Q of 100:



74HCU04
|\
.------| >O-----o----
| |/ |
| .-.
| | |
| | | 3k3
| 4.8uH '-'
| ___ |
o------UUU------o
| |
--- ---
--- 390p --- 390p
| |
| |
=== ===
GND GND
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

Peak-to-peak voltage at the output is about 3V. Peak-to-peak voltage across
either capacitor is about 0.5V

The 3k3 forms a potential divider with the impedance across the capacitor:

0.5 / 3 = Z / (3k3 + Z)
Z = 660 ohms

Dynamic impedance across the capacitively-tapped tuned circuit = Rp = 4*Z =
2640 ohms

Q = Rp/wL = 2640 / (2*pi*5e6*4.8e-6) = 17.5

Why is my calculated Q so low?

TIA
Andrew
 
J

John Popelish

Jan 1, 1970
0
Andrew said:
I built this 5 MHz oscillator using a Toko KANK4174 inductor, which is
specified as having a Q of 100:



74HCU04
|\
.------| >O-----o----
| |/ |
| .-.
| | |
| | | 3k3
| 4.8uH '-'
| ___ |
o------UUU------o
| |
--- ---
--- 390p --- 390p
| |
| |
=== ===
GND GND
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

Peak-to-peak voltage at the output is about 3V. Peak-to-peak voltage across
either capacitor is about 0.5V

The 3k3 forms a potential divider with the impedance across the capacitor:

0.5 / 3 = Z / (3k3 + Z)
Z = 660 ohms

Doesn't this formula assume that the 3 represents a zero
impedance 3 volt source? I think you need to replace this
voltage with the no load voltage from the inverter, and add
the inverter's effective (Thevenin equivalent) output
impedance to the 3k3 series resistor to approximate the
effective resistance in series with the resonator.
 
M

Mike Monett

Jan 1, 1970
0
Andrew Holme said:
I built this 5 MHz oscillator using a Toko KANK4174 inductor, which is
specified as having a Q of 100:



74HCU04
|\
.------| >O-----o----
| |/ |
| .-.
| | |
| | | 3k3
| 4.8uH '-'
| ___ |
o------UUU------o
| |
--- ---
--- 390p --- 390p
| |
| |
=== ===
GND GND
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

The calculated resonant frequency is 5.202142 MHz, so XL = 156.8929 ohms.

With an inductor Q of 100, you need to add a resistor in series or parallel
with the inductor. The series value RS = 1.568929 ohms, and the parallel
value RP = 15689.2908 ohms.
Peak-to-peak voltage at the output is about 3V. Peak-to-peak voltage
across either capacitor is about 0.5V

The voltage across the capacitors is 3V p-p. The inductor provides a 180
degree phase reversal, so the capacitor voltages are 180 degrees apart. The
voltage across the inductor is 6V p-p.
The 3k3 forms a potential divider with the impedance across the
capacitor:

0.5 / 3 = Z / (3k3 + Z)
Z = 660 ohms

The capacitive reactance XC = 78.446 ohms. I don't know where you got 660
ohms, but it is incorrect.

You are forgetting about resonance. Many people make this mistake. You
cannot use the capacitive reactance as part of a voltage divider. You have
to view the tank as part of a pi network. The inductor Q, plus the
capacitor ratio, determines the input impedance.

The inverter adds some delay, so the actual frequency of oscillation is
slightly lower than the calculated tank resonance frequency. This changes
the phase angle slightly, and makes the actual tank impedance very
difficult to calculate. It is better to do this in SPICE and let it take
care of these details.

You can use the technique described in my article below on Crystal
Oscillators to start the oscillator in steady-state. This speeds the
analysis, and is still quite useful for tank Q's as low as 100.

I use Microcap 8 since it has the peak search feature needed to locate the
steady-state oscillation amplitude. PSpice may have a similar feature, but
I have not figured how to do this in LTspice.
Dynamic impedance across the capacitively-tapped tuned circuit = Rp =
4*Z = 2640 ohms

Q = Rp/wL = 2640 / (2*pi*5e6*4.8e-6) = 17.5

Why is my calculated Q so low?

Using SPICE, the bandwidth at the -3dB points is 26.330 kHz. The center
frequency is 5.20188 MHz, so the Q = 5.20188e6/(2*26.33e3) = 98.782

This shows the 3.3K resistor has little effect on the loaded tank Q.
TIA
Andrew

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
 
J

Jim Thompson

Jan 1, 1970
0
The calculated resonant frequency is 5.202142 MHz, so XL = 156.8929 ohms.

OK so far.
With an inductor Q of 100, you need to add a resistor in series or parallel
with the inductor. The series value RS = 1.568929 ohms, and the parallel
value RP = 15689.2908 ohms.

Neeerp! What's the Q before any added resistor?

[snip]
The inverter adds some delay, so the actual frequency of oscillation is
slightly lower than the calculated tank resonance frequency. This changes
the phase angle slightly, and makes the actual tank impedance very
difficult to calculate. It is better to do this in SPICE and let it take
care of these details.

Not discernibly lower at 5MHz.

[snip]

...Jim Thompson
 
K

Ken Smith

Jan 1, 1970
0
Andrew Holme said:
I built this 5 MHz oscillator using a Toko KANK4174 inductor, which is
specified as having a Q of 100:



74HCU04
|\
.------| >O-----o----
| |/ |
| .-.
| | |
| | | 3k3
| 4.8uH '-'
| ___ |
o------UUU------o
| |
--- ---
--- 390p --- 390p
| |
| |
=== ===
GND GND
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)
[...]

You are forgetting about resonance. Many people make this mistake. You
cannot use the capacitive reactance as part of a voltage divider.

Actually you can use capacitors and inductors and combinations there of as
"voltage dividers". You just have to be prepared to have higher voltages
coming out of the divider than went in etc because the impedances are not
in phase. You have to include the "j"s in the math to get the right
answers.
 
J

John - KD5YI

Jan 1, 1970
0
Andrew Holme said:
I built this 5 MHz oscillator using a Toko KANK4174 inductor, which is
specified as having a Q of 100:



74HCU04
|\
.------| >O-----o----
| |/ |
| .-.
| | |
| | | 3k3
| 4.8uH '-'
| ___ |
o------UUU------o
| |
--- ---
--- 390p --- 390p
| |
| |
=== ===
GND GND
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

Peak-to-peak voltage at the output is about 3V. Peak-to-peak voltage
across either capacitor is about 0.5V

The 3k3 forms a potential divider with the impedance across the capacitor:

0.5 / 3 = Z / (3k3 + Z)
Z = 660 ohms

Dynamic impedance across the capacitively-tapped tuned circuit = Rp = 4*Z
= 2640 ohms

Q = Rp/wL = 2640 / (2*pi*5e6*4.8e-6) = 17.5

Why is my calculated Q so low?

TIA
Andrew

Hi, Andrew -

Is the inductor's Q specified at 5 MHz? Be sure to account for the different
intrinsic skin effect, if not.

I have played with your circuit in LTSpice but I cannot reproduce your
voltages. Inductor Q at 5 MHz may have something to do with it, but also
Trise, Tfall, Tdelay of the inverter has a large effect. I used the Philips
data sheet to specify those items and then adjusted them to get your values.
I assumed Vcc of 3V since the inverter in LTSpice is a behavorial one and
your output was 3V. What instrument did you use to measure the voltage? What
was the probe impedance?

If you want the LTSpice netlist, let me know. Good luck.

John
 
J

Jim Thompson

Jan 1, 1970
0
Hi, Andrew -

Is the inductor's Q specified at 5 MHz? Be sure to account for the different
intrinsic skin effect, if not.

I have played with your circuit in LTSpice but I cannot reproduce your
voltages. Inductor Q at 5 MHz may have something to do with it, but also
Trise, Tfall, Tdelay of the inverter has a large effect. I used the Philips
data sheet to specify those items and then adjusted them to get your values.
I assumed Vcc of 3V since the inverter in LTSpice is a behavorial one and
your output was 3V. What instrument did you use to measure the voltage? What
was the probe impedance?

If you want the LTSpice netlist, let me know. Good luck.

John

In PSpice, assuming QL=100, the effective Q is ~45.

A real 74HCU04 (I have the device-level models) is quite good, and has
a barely noticeable effect at 5.2MHz.

...Jim Thompson
 
M

Mark

Jan 1, 1970
0
100 is the UNLOADED Q of the inductor.

In the circuit the Q is the LOADED Q.

Mark
 
M

Mike Monett

Jan 1, 1970
0
John - KD5YI said:
Hi, Andrew -
Is the inductor's Q specified at 5 MHz? Be sure to account for the
different intrinsic skin effect, if not.
I have played with your circuit in LTSpice but I cannot reproduce your
voltages. Inductor Q at 5 MHz may have something to do with it, but
also Trise, Tfall, Tdelay of the inverter has a large effect. I used
the Philips data sheet to specify those items and then adjusted them
to get your values. I assumed Vcc of 3V since the inverter in LTSpice
is a behavorial one and your output was 3V. What instrument did you
use to measure the voltage? What was the probe impedance?
If you want the LTSpice netlist, let me know. Good luck.

John,

I tried to find the inductor on the toyo site but it was not listed.

Usually the Q for that value is measured at 7.9MHz. This is no too far from
5.2MHz, so it should be fairly close. But I find a Q of 100 a bit high.
Usually it's around 30 to 50.

I did my analysis at 5V, so it will be a bit different from yours. Could
you post your ASC file so we can compare the results from MicroCap 8?

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
 
D

Didi

Jan 1, 1970
0
This triggered a 15+ year old memory... Back then I needed a 110-120
MHz
clock generator (TTL levels); I did it using a 74AS00 (or was it
04?...), an inductor
I had routed on the PCB, a crystal - I had to locate something suitable
to work
between 115 and 120 MHz at 5-th, and a tiny trimcap... Obviously each
unit
had to be trimmed so the LC frequency would match the crystal, the
"lock"
was easily identifiable on a scope. The footprint was that of a typical
metal
crystal oscillator (DIP-14 corner pins only style), the height somehat
more.
It worked and managed to stay "locked" to the crystal over a pretty
wide
temperature range (that AS or F chip was quite an oven, especially
at about 120 MHz... :).
Just a related memory, hopefully on something mad enough to be worth
remembering :).

Dimiter
 
J

John - KD5YI

Jan 1, 1970
0
Mike Monett said:
John,

I tried to find the inductor on the toyo site but it was not listed.

Yes, I tried to find it as well. No luck.
Usually the Q for that value is measured at 7.9MHz. This is no too far
from
5.2MHz, so it should be fairly close. But I find a Q of 100 a bit high.
Usually it's around 30 to 50.


I agree that the Q seems to be specified unusually high.

I did my analysis at 5V, so it will be a bit different from yours. Could
you post your ASC file so we can compare the results from MicroCap 8?

Yes, of course. See below.
Regards,

Mike Monett

I also added some series resistance to each capacitor to make their Q about
1000. The inductor is specified to be 1.5 ohms.

The Trise (equals Tfall) is a bit longer thant the Philips data sheet says.
Also, the Tdelay is longer. I don't remember any more gotchas at this time.
If you have questions, please ask. My email is groups5 at verizon dot com.

I have been interested in the difficulties of measuring Q for some time now.
I will be following this thread as closely as my situation permits.

John

***********************************************

Version 4
SHEET 1 880 680
WIRE 96 128 -96 128
WIRE 208 128 160 128
WIRE 96 160 96 144
WIRE 208 160 208 128
WIRE -96 272 -96 128
WIRE 0 272 -96 272
WIRE 128 272 80 272
WIRE 208 272 208 240
WIRE 208 272 128 272
WIRE -96 288 -96 272
WIRE 128 288 128 272
WIRE -96 368 -96 352
WIRE 128 368 128 352
FLAG 96 160 0
FLAG 128 368 0
FLAG -96 368 0
SYMBOL Digital\\inv 96 64 R0
WINDOW 3 -199 2 Left 0
SYMATTR InstName A1
SYMATTR Value Trise=20n Td=30n Vhigh=3
SYMBOL res 192 144 R0
SYMATTR InstName R1
SYMATTR Value 3k3
SYMBOL cap 112 288 R0
SYMATTR InstName C1
SYMATTR Value 390p
SYMATTR SpiceLine Rser=.08
SYMBOL ind 96 256 R90
WINDOW 0 5 56 VBottom 0
WINDOW 3 32 56 VTop 0
SYMATTR InstName L1
SYMATTR Value 4µ8
SYMATTR SpiceLine Rser=1.5
SYMBOL cap -112 288 R0
SYMATTR InstName C2
SYMATTR Value 390p
SYMATTR SpiceLine Rser=.08
TEXT -316 160 Left 0 !.tran 0 20u 0 1n
 
M

Mike Monett

Jan 1, 1970
0
Yes, of course. See below.

OK, Thanks. I'll take a look and see what the differences are.
I also added some series resistance to each capacitor to make
their Q about 1000.

That's probably realistic Q at 5MHz, but the inductor Q will
dominate.
The inductor is specified to be 1.5 ohms.

That's probably the max DC resistance. That won't have much effect in
this circuit where the input bias is probably in the attoamp region:)

I redid the analysis using a Q of 30 and VCC of 3V.

The inductor series resistance went to 5.229 ohms, and the cap
voltage dropped to 585.192 mV which is close to Andrew's value. So I
think the problem is an inductor Q of 100 was way too high.

The AC analysis showed the center frequency dropped slightly from
5.202142MHz to 5.199250 MHz. I'm using NMOS and PMOS devices to
model the 74U04, so probably the prop delay increased slightly. The
rise and fall time doesn't seem to have much effect since the
inverter output is aligned with the sinusoidal waveform across the
tank input cap.

The bandwidth at the -3.015dB points is 174.5454 KHz, so the loaded
Q is 5.199250e6 / 174.5454e3 = 29.787, which is very close to the
original Q of 30. This shows as the inductor Q decreases, the 3.3k
resistor has even less effect on bandwidth. Which is what you expect.
The Trise (equals Tfall) is a bit longer than the Philips data
sheet says. Also, the Tdelay is longer. I don't remember any more
gotchas at this time. If you have questions, please ask. My email
is groups5 at verizon dot com.

Thanks, I'll look at the LTspice version and see how it compares to
MicroCap 8.
I have been interested in the difficulties of measuring Q for some
time now. I will be following this thread as closely as my
situation permits.

Thanks, John. Don't work too hard:)

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
 
M

Mike Monett

Jan 1, 1970
0
Mike Monett said:
The bandwidth at the -3.015dB points is 174.5454 KHz, so the
loaded Q is 5.199250e6 / 174.5454e3 = 29.787, which is very close
to the original Q of 30. This shows as the inductor Q decreases,
the 3.3k resistor has even less effect on bandwidth. Which is what
you expect.

Something didn't seem right with these high Q values, so I went back
and checked. It turns out I was measuring the bandwidth wrong. To do
the AC analysis, I simply broke the loop and injected the AC source
to the input of the inverter. However, the output of the inverter
changes with load when operating in the linear region, so the
apparent Q was too high.

Since the output is normally close to saturation when the oscillator
is running at normal amplitude, we can consider the top of the 3.3k
resistor is effectively at RF ground.

Eliminating the inverter and connecting the AC source to the top of
the 3.3k resistor gives a bandwidth of 236.158 KHz, and a resulting
Q of 5.197476e6 / 236.158e3 = 22.0084.

So the 3.3k has more effect on the tank Q than previous calculations
showed. This makes sense.

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
 
A

Andrew Holme

Jan 1, 1970
0
John - KD5YI wrote:
[snip]
Hi, Andrew -

Is the inductor's Q specified at 5 MHz? Be sure to account for the different
intrinsic skin effect, if not.

I have played with your circuit in LTSpice but I cannot reproduce your
voltages. Inductor Q at 5 MHz may have something to do with it, but also
Trise, Tfall, Tdelay of the inverter has a large effect. I used the Philips
data sheet to specify those items and then adjusted them to get your values.
I assumed Vcc of 3V since the inverter in LTSpice is a behavorial one and
your output was 3V. What instrument did you use to measure the voltage? What
was the probe impedance?

If you want the LTSpice netlist, let me know. Good luck.

John

Hi John,

Vdd is 5V. The voltages were measured with 1Mohm x10 'scope probes.

The coil is from the old 10K series, and Q is 100 according to my old
Cirkit catalogue.

I reproduced the voltages in LTSpice (see below). I think my original
(back of envelope) maths was wrong; and my circuit is functioning
correctly.

Thanks,
Andrew.



Version 4
SHEET 1 880 680
WIRE -112 144 -336 144
WIRE 16 144 -32 144
WIRE 96 144 16 144
WIRE 208 144 176 144
WIRE 320 144 288 144
WIRE -336 192 -336 144
WIRE 16 208 16 144
WIRE 320 208 320 144
WIRE -336 320 -336 272
WIRE 16 320 16 272
WIRE 320 320 320 272
FLAG -336 320 0
FLAG 16 320 0
FLAG 320 320 0
SYMBOL cap 304 208 R0
SYMATTR InstName C1
SYMATTR Value 390p
SYMBOL cap 0 208 R0
SYMATTR InstName C2
SYMATTR Value 390p
SYMBOL ind 80 160 R270
WINDOW 0 32 56 VTop 0
WINDOW 3 5 56 VBottom 0
SYMATTR InstName L1
SYMATTR Value 4.8µH
SYMBOL res 304 128 R90
WINDOW 0 0 56 VBottom 0
WINDOW 3 32 56 VTop 0
SYMATTR InstName R1
SYMATTR Value 1.5
SYMBOL res -16 128 R90
WINDOW 0 0 56 VBottom 0
WINDOW 3 32 56 VTop 0
SYMATTR InstName R2
SYMATTR Value 3k3
SYMBOL voltage -336 176 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value SINE(0 1.5 5000000)
TEXT -370 506 Left 0 !.tran 20us
 
A

Andrew Holme

Jan 1, 1970
0
Andrew said:
John - KD5YI wrote:
[snip]
Hi, Andrew -

Is the inductor's Q specified at 5 MHz? Be sure to account for the different
intrinsic skin effect, if not.

I have played with your circuit in LTSpice but I cannot reproduce your
voltages. Inductor Q at 5 MHz may have something to do with it, but also
Trise, Tfall, Tdelay of the inverter has a large effect. I used the Philips
data sheet to specify those items and then adjusted them to get your values.
I assumed Vcc of 3V since the inverter in LTSpice is a behavorial one and
your output was 3V. What instrument did you use to measure the voltage? What
was the probe impedance?

If you want the LTSpice netlist, let me know. Good luck.

John

Hi John,

Vdd is 5V. The voltages were measured with 1Mohm x10 'scope probes.

The coil is from the old 10K series, and Q is 100 according to my old
Cirkit catalogue.

I reproduced the voltages in LTSpice (see below). I think my original
(back of envelope) maths was wrong; and my circuit is functioning
correctly.

Thanks,
Andrew.

[snip]

I've done some more simulation in LTSpice and plotted the Bode response
of the LC network, and I now lean back towards thinking that my
original math *was* correct, and something is wrong with my circuit.
The key thing is that the voltages at either end of the 3k3 resistor
are in phase. The only way I can get this in simulation is by placing
a large value resistor in series with the inductor. Could the Toko
coil be saturating?
 
M

Mike Monett

Jan 1, 1970
0
Andrew Holme said:
Vdd is 5V. The voltages were measured with 1Mohm x10 'scope probes.

The coil is from the old 10K series, and Q is 100 according to my old
Cirkit catalogue.

I reproduced the voltages in LTSpice (see below). I think my
original (back of envelope) maths was wrong; and my circuit is
functioning correctly.

Thanks,
Andrew.

[snip]

I've done some more simulation in LTSpice and plotted the Bode
response of the LC network, and I now lean back towards thinking that
my original math *was* correct, and something is wrong with my
circuit. The key thing is that the voltages at either end of the 3k3
resistor are in phase. The only way I can get this in simulation is
by placing a large value resistor in series with the inductor. Could
the Toko coil be saturating?

Hi Andrew,

It's unlikely the inductor is saturating. However, there are a number of
other problems that should be addressed. Here are some of them:

1. It is unlikely the inductor is 4.8uH. The nearest standard value is
4.7uH. This means the center frequency should be 5.257 MHz, not 5.00 MHz as
your LTspice file assumes.

2. A Q of 100 is certainly possible at these frequencies, but probably not
in a standard commercial inductor. The expected Q is probably between 30
and 50. You should be able to measure the actual Q and inductance of your
coil with a simple sig generator, inexpensive frequency counter, and scope.

3. The 1.5 ohm resistor in your LTspice file is probably the maximum dc
resistance. This gives a much larger Q than expected for standard
inductors. You can easily calculate the proper value to use in your LTspice
analysis using more realistic Q values.

4. Your LTspice file drives the tank with a 5.00 MHz sine wave. This is
below the resonant frequency of the tank, so the resulting signal
amplitude you measure in SPICE has little relation to your actual circuit.

5. If you did a Bode plot, you should have noticed the resonant frequency
was well above 5.00 MHz. This should have alerted you that something was
wrong with your analysis.

And so on...

With the large number of basic issues shown above, there are probably even
more involved with your bench measurements. These would likely take a long
time to resolve.

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
 
J

Joop

Jan 1, 1970
0
Andrew said:
John - KD5YI wrote:
[snip]
Hi, Andrew -

Is the inductor's Q specified at 5 MHz? Be sure to account for the different
intrinsic skin effect, if not.

I have played with your circuit in LTSpice but I cannot reproduce your
voltages. Inductor Q at 5 MHz may have something to do with it, but also
Trise, Tfall, Tdelay of the inverter has a large effect. I used the Philips
data sheet to specify those items and then adjusted them to get your values.
I assumed Vcc of 3V since the inverter in LTSpice is a behavorial one and
your output was 3V. What instrument did you use to measure the voltage? What
was the probe impedance?

If you want the LTSpice netlist, let me know. Good luck.

John

Hi John,

Vdd is 5V. The voltages were measured with 1Mohm x10 'scope probes.

The coil is from the old 10K series, and Q is 100 according to my old
Cirkit catalogue.

I reproduced the voltages in LTSpice (see below). I think my original
(back of envelope) maths was wrong; and my circuit is functioning
correctly.

Thanks,
Andrew.

[snip]

I've done some more simulation in LTSpice and plotted the Bode response
of the LC network, and I now lean back towards thinking that my
original math *was* correct, and something is wrong with my circuit.
The key thing is that the voltages at either end of the 3k3 resistor
are in phase. The only way I can get this in simulation is by placing
a large value resistor in series with the inductor. Could the Toko
coil be saturating?
Your math was probably correct.

I used Netcalc to translate the components to parallel form. This
gives (with RL=1.52 ohm and 74HCU04 Rout=0):

4.8uH // 195pF // 13200 ohm

Q=Rp*w*C => 16.8

Which is pretty close to yours.

Cheers,

Joop
 
J

Joop

Jan 1, 1970
0
I've done some more simulation in LTSpice and plotted the Bode response
Your math was probably correct.

I used Netcalc to translate the components to parallel form. This
gives (with RL=1.52 ohm and 74HCU04 Rout=0):

4.8uH // 195pF // 13200 ohm

Q=Rp*w*C => 16.8

Which is pretty close to yours.

Cheers,

Joop
Sorry I made some wrong calculations there.
It should have been:

4.8uH // 195pF // 7277 ohm

And Q = 7277*2*pi*195pico = 46.3

After the messy calculations I though I'd better check it in LT-spice.
The -3dBV bandwidth = 112K

Which gives Q = f/delta-f = 5200/112 = 46.4
That should be better.
 
T

Tony Williams

Jan 1, 1970
0
Andrew Holme said:
I've done some more simulation in LTSpice and plotted the Bode
response of the LC network, and I now lean back towards thinking
that my original math *was* correct, and something is wrong with
my circuit. The key thing is that the voltages at either end of
the 3k3 resistor are in phase. The only way I can get this in
simulation is by placing a large value resistor in series with
the inductor. Could the Toko coil be saturating?


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.
 
M

Mike Monett

Jan 1, 1970
0
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
 
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