Load Resistor across Crystal - though that's for a passive crystal and not an integrated XO.


I'm considering using an Epson TG-5035C XO and am figuring out the circuit necessary to couple it to an Atmel AT89LP428.

Oscillator Specs

The XO shows these specs for the passives to connect:

Output load components

I assume that the "DC cut capacitor" is a coupling, or DC block, capacitor to be inserted inline between the output of the oscillator and the XTAL1 crystal amp input of the MCU.

Microcontroller Specs

The MCU shows this connection diagram:

MCU connections

which is only partially applicable in my case. As determined in a previous question, XTAL2 will be left open, and the output of the XO will go through a 10nF cap to XTAL1.

What I don't know how to handle is the load resistance. The XO shows it as a load "condition" of 10k, and I don't know whether that's an equality requirement, a minimum, a maximum, or simply what they tested at - and if it's what they tested at, it's not clear which other values on the specsheet rely on that 10k.

Most other literature on crystal oscillators that I've read suggest a load resistor in the 1-5Mohm range, and the MCU spec agrees. So which makes more sense in this scenario - a 10k or a 4.7M load resistor?


2 Answers 2


Go with the higher value. The "output load condition" is the maximum load allowed to guarantee the minimum p-p voltage, specified at 0.8 volts. The resistor is probably not needed at all but possibly provides a path to bleed off charge.

  • \$\begingroup\$ Does that also mean that the cited Ccut=10nF is a maximum value, and that decreasing it would unload the XO? \$\endgroup\$
    – Reinderien
    Commented Mar 5, 2019 at 19:27
  • \$\begingroup\$ I would say yes; at least if you go with a higher capacitance, the output is not guaranteed. I have used oscillators with the output load condition resistor spec'd as low as 100 ohms. The measurement is made with the specified parts as load and driven through the "DC cut capacitor" so that the oscillator bias is not pulled to ground. \$\endgroup\$ Commented Mar 5, 2019 at 19:40

You are confusing "crystal" with "crystal[base] oscillator".

The MCU circuit shows how to connect a "crystal" (which is a passive piece of quartz mineral with metal pating) to the MCU to make an oscillator. By itself a crystal doesn't produce any oscillations and requires active amplifier circuitry to do so. The 4-5 MOhms to ground is a bit unusual requirement, which is likely here to compensate for some deficiency of inputs on this MCU (due to universality of I/O ports I guess). Usually any resistor on XTAL1 (XIN, input) is a killer for the oscillator.

The device TG-5035 is a self-contained OSCILLATOR. Upon applied voltage it will produce digital meander that can be used as external oscillator to the MCU (which has a different mode for this). As a self-contained circuit, the TG-5035 can drive loads as strong as 10k and above (to infinity). Therefore you don't need any resistor at all when connecting the TG-5035 to your MCU.

CORRECTION: it appears that the TG-5035 is not suitable for direct interface to digital electronics, since it seems to output voltage as low as +- 0.4 V relative to ground (0.8V p-p). The MCU claims to have VL = 0.2 Vcc and VH = 0.7 Vcc, or 0.5 Vcc p-p. Which is 1.8V p-p. Yet the TG-5035 has only 0.8 V p-p. It is likely that the "crystal mode" of the MCU doesn't have these 0.2-0.7 minimum swing requirement, and just have one typical CMOS threshold at Vcc/2. But again these all are guesses, which shouldn't be acceptable in engineering. If you really need to use TG-5035, you do need a DC-decoupling, level shifting, buffer, etc. Or you can select a different oscillator with output levels that are compatible with CMOS inputs.

Instead of selecting VXCOs with "clipped sine wave" you should pick oscillators with "CMOS" or "LVCMOS" or "HCMOS" outputs like ASTX-H11.

  • \$\begingroup\$ That's fair. That being said, this answer seems to contravene this advice - electronics.stackexchange.com/a/425648/10008 - which suggests that oscillator output can be fed to the crystal amp. Is that not the case? \$\endgroup\$
    – Reinderien
    Commented Mar 5, 2019 at 19:31
  • \$\begingroup\$ @Reinderien, the cited suggestion doesn't contradict to anything, but you don't need any decoupling (DC-cut) caps, this is wrong idea and unnecessary . For more details of oscillator technology see this, electronics.stackexchange.com/a/371101/117785 \$\endgroup\$ Commented Mar 5, 2019 at 19:46
  • \$\begingroup\$ That suggestion is to use crystal mode. This suggestion says don't use crystal mode... maybe I'm missing something. Also, @JohnBirckhead says above that connecting the output without a sufficiently small cut cap can prevent it from working. Is that not true? \$\endgroup\$
    – Reinderien
    Commented Mar 5, 2019 at 19:49
  • \$\begingroup\$ @Reinderien, nothing is true. The only true is that the MCU is a huge 5-V device with huge signal thresholds, while you have selected a tiny 3-V oscillator-generator, which might have not enough output amplitude to drive your MCU input. Get a 5-V crystal oscillator chip or use the MCU at Vcc=3.3V \$\endgroup\$ Commented Mar 5, 2019 at 19:53
  • \$\begingroup\$ Even though the MCU can be driven by 5V, it isn't. The actual Vcc used makes a big difference, and in this case it's 3.6V. \$\endgroup\$
    – Reinderien
    Commented Mar 5, 2019 at 19:54

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