# Crystal current limiting resistor

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I need help to calculate the current limiting resistor needed to interface a crystal to X1(in) and X2(out) for a IC .. the IC specifies a minimum drive level of 100uW and a maximum as 500uW .. i want to use a crystal with a specified drivelevel of 50uW .. how can i calculate the current limiting resistor which is needed between the X2 of the IC and the crystal?

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well thats the question .. i dosent specify the drive voltage :/ ... the load capacitance is 18pf

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Find out the motational inductance and/or capacitance of the crystal, the holder capacitance, and the Q at frequency (or the motational resistance)

-- manufacturers usually publish this information somewhere on their site; if one doesn't then you know that one isn't supporting you well and should be left off of your approved vendor list.

Now you know enough to make a model of the crystal -- do so. If you don't know how, search the web -- it's out there. If I remember correctly, International Crystal shows you how to do this on their site.

Then slap the model of the circuit, with the crystal model, into your favorite version of SPICE, check the magnitude of the current in the crystal's internal 'resistor', and calculate power. Now play with the current limiting resistor until everything works out right.

Have fun. Remember that when you actually build the board everything will be different, and touching a scope probe to an oscillator pin usually makes it stop oscillating.

```--
Tim Wescott
Control systems and communications consulting```
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Using a simulation will be lots of fun but perhaps not much help when there isn't a model for the chip containing the oscillator circuit. Of course, the real answer in that case is to beat up the chip vendor for not providing the appropriate data in their data sheet.

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this is what the chip producer support answered :

"calculating the current limiting resistor is typically depending on the crystal manufacture's recommendations!

R1 is the drive limiting resistor, the primary function of this resistor is to limit the output of the inverter so that the crystal is not over driven. R1 and C1 form a voltage dividing circuit, the values of these components are chosen in such a way that the output of the inverter goes close to rail-to-rail and the input to the crystal is 60% of rail-to-rail, usual practice is to make resistance of R1 and reactance of C1 equal at the operating frequency, i.e. R1 >> XC1. This makes the input to the crystal half that of the inverter output. Always make sure that the power dissipated by the crystal is with-in the crystal manufacturer's specifications. Over-driving the crystal may damage the crystal.

Please refer to the crystal manufacturer's recommendations."

So as i understand: If i need to have rail-to-rail voltage (Vcc) at the X2 output, then i need to have a impedance of R1+XC1 = (VCC^2)/P .. then the voltage at the X2 chip output will be Vcc, and if i choose R1=XC1 then i will have Vcc/2 at the crystal. but how can i calculate the current throug the crystal ? then i need to know the inductance, capacitance and ESR of the crystal which is not always available.

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Wow ! How complicated !

Xstal is parallel resonant ckt ( series would be

lower Q .)

Overdriving it with a Chip is not a problem as

chips input is hi-Z , just like Xstal .

Xstal is parallel Res' , hi-Z , thus microamps flow

out of the chips "output" to drive xstal .

The xstals internal parallel cap is very temp

dependent thats why you can add another

component to reduce tempco .

20 pf at both sides of xstal to ground

will make it start up slow , but better stability ,

up to about 50 pf ...will start fast but be maybe

unstable ( xstals need stability ).

xstals above about 12 mhz are too thin ,

so they oscilate in a overtone mode that has

enough loss of Q each side of that higher

freq to be stable .

Notice in the ARRL hand book , the cap's

Xc calcs out to about 400 ohms , at the design

freq .

KC7CC ....

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It's probably safe to assume that the oscillator output goes close to the rails -- were I doing this for serious I'd measure it with a scope, probably at the bottom end of a capacitive voltage divider to keep the scope probe from loading down the oscillator.

```--
Tim Wescott
Control systems and communications consulting```

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