Larger voltage swing from crystal oscillator after attenuation

Question

I use a MEMs crystal oscillator SiT8008BI-13-33E-24.000000E to drive two ICs.I power the oscillator from 3.3V for the first IC. For the second IC, a USB HUB (USB2640) I need to step the voltage down so the swing goes from ~0.3V to ~1.6V for the USB HUB to properly register the pulse, based on datasheet:

(It looks weird to me that it says "-10uA of Input leakage" on the datasheet, like the HUB itself biases the XTAL input with some current?)

My schematic and layout are as follows:

Simply using voltage dividers with resistors around 1k or 2k combinations on R3, R4 -> I get these waveforms at the input of the USB HUB CLKIN (GREEN is the output directly out of the oscillator, the GRAY waveform with a 1k/1k resistor divider, ORANGE is 1k/2k voltage divider and RED is a 1k resistor (R3) in series :

All these waveforms do not give me the range I want, as the output after the R3, R4 is either too high or too low for my desired ~0.3V to ~1.6V range.

I decided to play around on LTspice, with the 1k resistor that I placed in series and based on the oscilloscope's waveform and the 10uA as input leakage the datasheet indicates, the input capacitance of the pin, along with stray capacitances should be ~15pF:

Question: How do I get further, getting a larger output swing from the oscillator?

You're using 10X attenuator probes for those RIGOL waveforms?? Probes may add ~12pf capacitance. So expect those wave shapes to square-up somewhat when you remove the probe. I'd suggest the orange (looks brown) waveform is close to meeting JESD76-2 spec (in data sheet paragraph 10.3 External Clock) — glen_geek
– glen_geek, Commented Apr 9 at 17:39
@Justme ADUM3166. It does work as expected with 1.8V as clock but datasheet says: Input Logic High Threshold minimum 2.0V (page 3 at bottom): eu.mouser.com/datasheet/2/609/adum3165_3166-2946864.pdf So I prefer not to use it with less than 2V. — Christianidis Vasileios
– Christianidis Vasileios, Commented Apr 9 at 18:41
@ChristianidisVasilis Can I ask why do you do it this way? Why not use a level shifter from 3.3V to 1.8V to have both 3.3V and 1.8V clocks? If price is the problem, then for the price of one MEMS oscillator, you get two quartz crystals, and likely the load caps and bias resistors too. — Justme
– Justme, Commented Apr 9 at 19:37
The USB hub expects a 1.8V clock. Go lower impedance with the resistive divider, something like a 470 Ohm for the series and 560 Ohm for the shunt resistor. The clock input capacitance is around 2pF, 3pF if you consider additional external parasitic capacitance. If you really need to compensate for that, put a small cap (e.g. 3.3pF) across the 470 Ohm series resistor. I would go without the 3.3pF, considering the short trace up to the clock input pin being a parasitic inductor. — Raonoke
– Raonoke, Commented Apr 9 at 20:32

Justme · Accepted Answer · 2025-04-09 21:22:17Z

What you basically want is to feed one 3.3V square wave oscillator output to two chips, one wanting a 3.3V square wave and another wanting 1.8V square wave.

The oscillator output is rated for 4mA current to stay within 3.3V logic levels, so it means you can't load it with impedance lower than 825 ohms.

For example a DC coupled voltage divider with 470 and 560 ohms would divide the 3.3V squarw wave to 1.8V square wave but also adds a DC load.

Indeed in some cases you could capacitively divide the AC signal and rebias the 1.8V input to 0.9V weakly.

But it would be much simpler to use a logic buffer to take in 3.3V square wave and convert it down to 1.8V square wave. Or get a oscillator that outputs 1.8V and a logic buffer that converts it up to 3.3V square wave.

The MEMS oscillator plus any added components is expensive. Price of one MEMS oscillator is approximately equal to two quartz crystals which you can directly connect to XIN/XOUT pins of the chips, with accompanying load caps and bias/series resistors, so they would be asynchronous and independent of any other circuitry.

You are right. Initially, I did (and still) use crystals to drive each IC independently. I didnt have any issues, but always had in the back of my mind the disadvantages of the crystals. Maybe the largest disadvantage being the startup of the crystal not being defined (see a previous question of mine electronics.stackexchange.com/questions/696479/… ). I wanted a more-defined startup behavior, to rule out some of the issues we face with rare USB disconnections and very rare startup failures. In short, I try to improve the design — Christianidis Vasileios
– Christianidis Vasileios, Commented Apr 10 at 8:59
@ChristianidisVasilis I understand. Let me also tell about downsides of using oscillators - you are already finding out many of them, like different chips needing different levels, some chips requiring specific AC coupling networks if not natively made to support square wave inputs, then the problems of routing a 24 MHz square wave to various parts of the board with good signal integrity, with 24 MHz signal having at least 120 MHz bandwidth requirement to look like square wave, having light enough load for fast enough slew, the oscillator polluting the supply voltage, filtering, it's a mess. — Justme
– Justme, Commented Apr 10 at 9:06

Christianidis Vasileios · Accepted Answer · 2025-07-25 08:36:30Z

To answer my question, on a simulation-stage, I managed to combine a resistor and capacitor to get a waveform from ~0 to ~2.0V!

The way this works out is, a capacitor in series lets the rise and falling edges to pass through, which also contols the upper voltage of the output (the larger the capacitor the more charge it holds and the larger the output voltage). But with the capacitor only, over time the voltage becomes AC and gets below 0:

(340k resistor draws the 10uA leakge current).

So a series resistor is needed to keep the pulse above 0V and prohibit the capacitor on onverting the voltage below 0V: And then the resistor needs to get sized properly to keep the DC level.

I think I am happy with this waveform. Final values for my case are a 10pF for a 2.2Vpkpk pulse output in parallel 4MΩ to offset the pulse above 0V:

I assume without the probe's parasitic capacitance the pulse's low will be ~(-0.1V) and peak ~2.5V.

Stack Exchange Network

Larger voltage swing from crystal oscillator after attenuation

2 Answers 2

Linked

Hot Network Questions

Larger voltage swing from crystal oscillator after attenuation

2 Answers 2

Linked

Related

Hot Network Questions