74HCU04 LC Osc

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.

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.

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 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.

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.

Regards,

Mike Monett

Antiviral, Antibacterial Silver Solution:

OK so far.

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

Not discernibly lower at 5MHz.

...Jim Thompson

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.

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

snip

In the circuit the Q is the LOADED Q.

Mark

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

Regards,

Mike Monett

Antiviral, Antibacterial Silver Solution:

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

Dimiter

------------------------------------------------------ Dimiter Popoff Transgalactic Instruments

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Jim Thomps> >

Yes, I tried to find it as well. No luck.

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

Yes, of course. See below.

I also added some series resistance to each capacitor to make their Q about

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

OK, Thanks. I'll take a look and see what the differences are.

That's probably realistic Q at 5MHz, but the inductor Q will dominate.

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.

Thanks, I'll look at the LTspice version and see how it compares to MicroCap 8.

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

Regards,

Mike Monett

Antiviral, Antibacterial Silver Solution:

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:

John - KD5YI wrote: [snip]

ent

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hat

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 4.8=B5H 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

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

1. 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.

2. 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.

1. 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.

2. 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:

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):

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:

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.

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

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: