Measuring/estimating a quartz crystal power dissipation (drive level) - Page 1
Forum section: electronics / beginners

author: shapirus
author_rank: Super Contributor
timestamp: February 18, 2024, 11:43:56 pm
content: I am experimenting with an oscillator and frequency divider circuit based on74HC4060with the oscillation frequency set by a quartz crystal.I started with the circuit shown on page 12 of the datasheet:It worked, but there was significant jittering on the outputs (horizontal position of the pulses before and after the one that the scope triggered on was unstable within +/- a few nanoseconds). I experimented with the component values, keeping the crystal the same, and found a combination that resulted in no jitter that I could observe or measure: there's now a nice clean time-wise stable oscillation at the expected frequency. There is voltage ripple on the outputs of the IC, both when they are high and low, which has the frequency of the clock -- I believe it's because there is no sufficient IC power supply bypassing, as the whole thing is built on a solderless breadboard and, furthermore, the IC is an SO-16 installed in a spring-loaded adapter.The component values currently are:Rbias= 510k (doesn't seem to affect anything really, I tried values from 91k to 700k)R2 = 680 OhmC2 = C3 = 22 pFThe crystal is 16.384 MHz.Crystal's datasheet values:CL= 16 pFC0max = 7 pFDrive level = 0.1 mWSeries resonance RRmax = 50 OhmNow, I want to make sure that I don't exceed the allowed drive level (aka power dissipation in the crystal, if I understand it correctly). What are the possible ways of measuring or estimating it?The RenesasAN830application note provides a formula to calculate it, but it requires to know the drive current flowing through the crystal.Here's what I tried. I don't have a current probe, which is used in the above app note's example, so need a different approach.1. Measure RMS voltage across the crystal (math A-B between two channels in the scope). This yields 1.7 V, and if I calculate power as (1.7 V)2/ 50 Ohm ~= 58 mW, then it does not make sense, because the entire circuit's power consumption is only 17 mW measured. Clearly it's not the right way of measuring it -- probably because the crystal's voltage and current are not in phase, and, I guess, not only that.2. Measure voltage across R2 to calculate current flowing through it and have an upper estimation of what can be flowing through the crystal. I say "upper" because I lack understanding of what portion of current flowing via R2 will flow into C2 and not into the crystal.Here we have ~980 mV RMS across R2, which yields 980 mV / 680 Ohm ~= 1.44 mA, and then, knowing C0and CL(assuming my values of the caps plus the parasitic capacitance result in CL~= 16 pF) and using the formula from the app note, we can calculate the value for power P = I2* RR* (1 + C0/CL) ~= 0.149 mW.3. A somewhat dubious method, because of the noise and required CMRR considerations: add a 10 Ohm resistor in series with one of the crystal's pins and measure the voltage drop across it, again, using two scope channels.(One more possible problem here is that I have no idea what this resistor's actual resistance will be at 16 MHz, but I'm assuming that it's still close enough to 10 Ohm -- but the assumption may be wrong, considering that it's a 6.3x2.3mm THT resistor with uncut leads, so it may have considerable parasitics.)Since the voltage difference between the channels relative to absolute values is too low (about 1.5%), the scope fails to perform meaningful math, and I had to do it manually: zoom in (a lot!) and do a series of single-shot waveform captures to measure a typical difference between the peaks of the waveforms of both channels. The differential noise/fluctuation was not too bad, and the Vpp difference was quite consistently around 30 mV, which, measured across a 10 Ohm resistor, makes it 3 mA peak to peak, or, assuming a sine wave (which it should be -- the crystal's oscillation appears to be pretty close to a sine wave), 1.06 mA RMS, which is on the same order of magnitude as the current via R2, measured by method 2, and it is also lower than the latter, which agrees with my assumption that a portion of the current flowing via R2 is diverted into C2.Update: there's a mistake (I think). The difference between the channels is +30 mV at their upper peaks and -30 mV at the lower peaks, which makes the differential voltage 60 mV p-p, not 30 mV, as I thought, so 30 mV is now the amplitude, and then the current, respectively, will be ~2.1 mA RMS. Still on the same order of magnitude as method 2, but actually higher current than measured by voltage drop across R2 (and the latter is more precise of course, because it didn't require chasing millivolts on top of ~2.2 Vpp signals).Do the methods 2 and 3 make sense? Assuming no current probe is available, are there any other methods of measuring the drive power of the crystal that would be viable at say 10-20 MHz?Is there a realistic way to calculate it, knowing the component values? If so, it would be interesting to compare the calculated value with what I measured. I suspect that I could answer the latter myself, only my brain is cowardly refusing to think and reach for some of the knowledge obtained over 20 years ago in school and university and never used since, so I need some help.

author: Benta
author_rank: Super Contributor
timestamp: February 19, 2024, 12:41:32 am
content: It's not too hard to calculate once you have the crystal specs.First, the bias resistor is just there to set a DC equilibrium, for CMOS 500k...1M is ballpark.Second, the load capacitance is computed from:[C3 in parallel with IC input C] in series with C2 (you'll get input C from the IC data sheet).The joker is the power limiting resistor R2. But the value is not important.It's a Pierce oscillator, and as such the crystal is in parallel resonant mode (= very high impedance). The resistor only plays a role during start up, and AT-cut crystals are very robust. 2.2k is a good value based on experience.No worries.

author: David Hess
author_rank: Super Contributor
timestamp: February 19, 2024, 10:18:29 am
content: Quote from: shapirus on February 18, 2024, 11:43:56 pmJim Williams had to measure the crystal drive current for a micropower 32kHz oscillator which is even more demanding. He did it using a Tektronix current probe and custom amplifier to get enough sensitivity.What I might do is the same thing but using a ferrite toroid to make a current transformer.

author: shapirus
author_rank: Super Contributor
timestamp: February 19, 2024, 10:41:59 am
content: Quote from: Benta on February 19, 2024, 12:41:32 amIt is. Anything above ~820 Ohm, and the oscillation does not start, when the load capacitors' values are as I specified above. With the values from the 4060 datasheet it does start, but then there is that jitter, which I mentioned, and that's why I started decreasing the capacitors (which aso forced me to reduce R2) until jittering stopped.Quote from: Benta on February 19, 2024, 12:41:32 amNot really. From further info that I found (e.g.http://stades.co.uk/XTAL%20Oscillator/XTAL%20Oscillator.html), it oscillates above series resonance and below parallel resonance. This makes me wonder whether R1specified in the crystal's datasheet (50 Ohm) is still applicable to calculate the power, provided that we know the current.Speaking of the current measurement, I found one more article on this topic that provides more formulas:https://www.allaboutcircuits.com/technical-articles/measuring-drive-level-of-a-quartz-crystal/The two methods -- via the amplifier input voltage and via Vrms across the crystal -- yield similar results of ~2..4 mA RMS, which results in the estimated power of ~0.5-0.6 mW, assuming the value of Rm=50 Ohm.The method of measuring current via the voltage drop across a small resistor in series with the oscillator does not unfortunately yield consistent results when the resistance is changed (same +/- 30 mV peak with either 10 or 20 Ohm).The question remains which value of R should be used in the ESR calculation, however, but even with the lowest value of 50 Ohm I'm getting power levels exceeding the allowed one, if the current value of 2 to 4 mA is correct.Quote from: Benta on February 19, 2024, 12:41:32 amSo far the estimation methods I used have been giving, worst case, at least 5x the rated power (> 0.5 mW), which doesn't look good. I can of course implement the oscillator separately using an inverting gate with a bigger power limiting resistor (if that combination works well, which remains to be seen) and feed its output to the 4060, but it kind of sucks, because the 4060 is supposed to be able to work with the crystal directly.

author: shapirus
author_rank: Super Contributor
timestamp: February 19, 2024, 10:46:53 am
content: Quote from: David Hess on February 19, 2024, 10:18:29 amThat's a good idea. However creating a suitable amplifier for it may in itself become a complicated project (this is essentially designing an active current probe), and then characterizing it at the target frequency will probably also be a challenge.

author: shapirus
author_rank: Super Contributor
timestamp: February 19, 2024, 02:39:33 pm
content: Quote from: David Hess on February 19, 2024, 10:18:29 amTo follow up on this, I did a quick experiment: took a 150 uH toroid core inductor, attached an oscilloscope probe to it and passed a single wire through its center, then connected the wire to the output of a signal generator, outputting a sine wave, via a DMM (BM869s) in AC ammeter mode, which is spec'd up to 100 kHz (and seems to work fine to at least 500 kHz).Well, the result surprised me and made it clear that I don't understand something crucial. Yes, the magnitude of the output voltage seen as a waveform on the scope screen is, as expected, proportional to the current flowing in the primary (the wire going through the toroid center), but it is also proportional to the frequency! If Vrms on the secondary is, say, 10 mV at 10 kHz, then it becomes 50 mV at 50 kHz and 100 mV at 100 kHz, and the RMS current in the primary, as measured by the DMM, does not change. Adding a 100 Ohm burden resistor across the secondary does not affect this behavior.This continues to some 5-6 MHz, where it peaks (at 5-6 Vrms, continuing the trend), and then it begins to fall, but I have no way of measuring current at those frequencies.So either I am doing it wrong, or the voltage-frequency response of a simple inductor used as a CT cannot be automatically expected to be anywhere near flat. In fact, that it is proportional to frequency, suggests something, but I don't know what. But there has to be a way to work around it, because, well, wideband CTs are a thing.

author: gf
author_rank: Super Contributor
timestamp: February 19, 2024, 02:49:07 pm
content: A CT always needs a (low enough) load resistor connected to the secondary. Some off-the-shelf CTs have a built-in load resistor, or at least built-in clamping diodes to prevent over-voltage if you forget to connect an external load resistor.

author: shapirus
author_rank: Super Contributor
timestamp: February 19, 2024, 02:53:12 pm
content: Quote from: gf on February 19, 2024, 02:49:07 pmWell I did connect a resistor, as I mentioned -- it didn't change anything. Maybe 100 Ohm was too much, I'll retry with a lower value later. The proportional relation between frequency and output voltage is puzzling me.

author: shapirus
author_rank: Super Contributor
timestamp: February 19, 2024, 04:39:35 pm
content: Quote from: shapirus on February 19, 2024, 02:53:12 pmThis!With a 10 Ohm resistor frequency response becomes quite flat starting at about 50 kHz.If the signal generator's output voltage is adjusted so that at 100 kHz the CT's output is 1 mV RMS (while the measured current in the primary is ~3.8 mA), then the frequency vs output voltage relation becomes this:50 kHz: 967 uV100 kHz: 1 mV500 kHz: 1 mV1 MHz: 1 mV2 MHz: 1 mV4 MHz: 1 mV8 MHz: 980 uV12 MHz: 970 uV16 MHz: 950 uV24 MHz: 810 uVOf course this required the probe to be set to 1x and turning on averaging of the waveform to cancel out the noise (which seems to work very well).Current to voltage transfer ratio also seems to be quite flat, but I haven't yet looked into it in more details.So the idea seems to work pretty well, and the frequency response can probably be further improved (and it remains to be determined how much of the drop at the higher frequencies is caused not by the CT itself, but by the cable with which the primary is connected, and by the signal generator's output stage). Definitely good enough to try this at home. With a bit of signal amplification this may make a decent DIY AC current probe. Sounds like the crystal circuit will have to wait...

author: David Hess
author_rank: Super Contributor
timestamp: February 19, 2024, 05:15:52 pm
content: The current transformer must operate into a low impedance load. Jim Williams used a Tektronix CT-1 which is internally terminated into 50 ohms and would normally be connected to a 50 ohm input:https://www.analog.com/en/resources/technical-articles/sub-ua-rms-current-measurement-for-quarts-crystals.htmlI would do something a little different and terminate the current transformer directly into the inverting input of a transimpedance amplifier, which has an input resistance of essentially zero ohms. Calibration can be done with a leveled sine source driving a resistor to make a current, but the output of the current transformer driving a low or zero impedance should be flat with frequency.When the output of the current transformer is properly loaded, it operates like a transformer with a shorted turn and looked like a short to the current being measured, which is exactly what is desired when measuring current. If the current transformer has a ratio of say 1:100, and a 100 ohm resistor is used as the load, then on the primary side it will look like 1 ohm of resistance.

author: shapirus
author_rank: Super Contributor
timestamp: February 20, 2024, 10:08:13 am
content: So I have experimented further with this makeshift CT.First, I added a 51 Ohm resistor in series with the piece of wire that's going through the toroid core in order to measure voltage across it with a scope to make sure that the current sourced by the signal generator remains stable across the frequency range of interest -- up to 16 MHz -- and it does.Then I replaced the 10 Ohm burden resistor that was sitting across the leads of the inductor with an 1 Ohm one, and it had an interesting effect: the usable frequency range had shifted downwards. While with the 10 Ohm resistor the range of almost flat frequency response was from 50 kHz to about 12 MHz, with the 1 Ohm resistor it is now from ~8 kHz to about 4 MHz, after which the output voltage begins to rise significantly. So, at least with this particular inductor, shorting its secondary winding will not work -- at least for my frequency of interest of up to say 16 MHz (higher if possible), and starting from 100 kHz, to take over from the point up to which my DMM's AC ammeter is specified. It looks like the optimal value for the burden resistor in this case may be somewhere between 10 and 20 Ohms.Would a different inductor be better? What parameters do I need to look for? I have a few toroidal inductors of the same type, but different inductance values, will try them to see what it changes.

author: David Hess
author_rank: Super Contributor
timestamp: February 20, 2024, 01:26:31 pm
content: Quote from: shapirus on February 20, 2024, 10:08:13 amCurrent transformers are specified by a volt-sec product which can be calculated from the saturation current and inductance. Using a lower burden resistance extends the time before saturation, which extends the low frequency response.QuoteMy guess is that the core material is not suited for operation at 16 MHz because it has too much loss.

author: shapirus
author_rank: Super Contributor
timestamp: February 20, 2024, 05:53:01 pm
content: I have built a characteristic plot for this CT for two parameters versus frequency: a) mV/mA transfer ratio; b) phase shift between the current measured as voltage drop across a series resistor and the voltage measured on the output of the CT.There are actually two curves for mV/mA: one is as seen directly on the scope screen, taken with the probe switch set to 1x, another is with the values divided by the 1x probe attenuation ratio (which I measured at 0.71 at 21 MHz). The 1x position was required to make meaningful readings at all possible (low-mV values).This is all passive, without amplification. Though it'll be a bit on the edge of the scope's capabilities, it should be usable as it is for measuring low-mA currents. I will try this to measure the crystal drive current later.When combined with the 1x probe, the mV/mA curve is surprisingly good from 100 kHz to about 20 MHz -- good enough for what I'm doing at the moment.Not so good if it's to be amplified to be usable with a 10x probe without having to do waveform averaging -- it's flat only up to about 5-6 MHz.I will order more different inductors to play with them at a later time, and if I find a better combination (say, from 10 kHz to maybe 30-40 MHz?), then it'll be totally worth it to build an amplifier for it.Additionally, I will increase the signal amplitude by the factor of 10 and repeat the measurements with the probe switch set to 10x to see 2 things: a) whether the frequency response rise after 5-6 MHz remains; b) whether 10x the current will not screw everything up.

author: shapirus
author_rank: Super Contributor
timestamp: February 22, 2024, 10:51:08 am
content: All right, so current measured using a makeshift toroidal core CT ranges from 2 mA to 4 mA depending, which is a surprise to me, on the polarity of the probe connection to the CT's output. Either way this allows to estimate the order of magnitude of drive power: from ~0.3 mW to ~1.2 mW, which exceeds the allowed drive level -- and this result agrees with the previous calculations obtained via different methods.So apparently I'll have to create a dedicated Pierce oscillator using an inverting gate for this crystal with a lower drive level and feed its output into the 4060. I wonder if the jitter issue at lower drive levels will be the same though, but it'll be a different topic.

author: voltsandjolts
author_rank: Unknown
timestamp: February 22, 2024, 10:55:02 am
content: Can't you just place a suitable resistor in series with the crystal to reduce drive level, as per picture above in the OP?

author: shapirus
author_rank: Super Contributor
timestamp: February 22, 2024, 11:03:06 am
content: Quote from: voltsandjolts on February 22, 2024, 10:55:02 amOscillation doesn't start until the resistor becomes low enough -- and at that point drive level is already too high.With the increased capacitances the resistor can be made larger (the values from the 4060 datasheet with a 2.2k resistor work), but then there is jitter, which goes away as the capacitances are made smaller.

author: magic
author_rank: Super Contributor
timestamp: February 22, 2024, 06:14:56 pm
content: I recall reading about the following trick:Measure RMS voltage on the input side of the inverter gate, where current only flows between various capacitances and the crystal.Measure combined capacitance there, including gate input, scope probe and whatnot, use it to calculate RMS current.The same current flows through the crystal, so dissipation is I²·ESR.