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I want to build SIM Card Reader by following a schematic from here as follow:

here http://www.ladyada.net/make/simreader/simreaderv1_0.png

I realized that the clock generator (upper center) form a Pierce oscillator but with an additional resistor (R2 symbol in the schematic), which different than Pierce Oscillator circuit at wikipedia:

sim reader schematic.

I've also found other variant of pierce oscillator as follow: pierce oscillator

Which tapping output from the inverter instead of the capacitor like in SIM card reader schematic.

Could you please explain the use of R2? What if I removed the R2 from the schematic? And what the differences between the first and the last Pierce oscillator circuit?

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R2 in your main circuit forms an ~8 MHz low pass filter with the 20 pF capacitor it connects with. It has nothing to do with the oscillator circuit and my guess is that it's acting (with the 20pF cap) as a small delay to the inverter (IC1B) it feeds.

In this circuit: -

enter image description here

R1 is needed to bias the inverter into its linear region - now it acts like a linear amplifier and the smallish semi-sinusoidal voltage that the xtal produces at the input can be turned into a square wave at the output. Without R1, there is nothing to align the input dc voltage with the inverters linear region and it wouldn't work.

In your final circuit (or any xtal oscillator circuit like this), Rs is always present but usually the output resistance of the gate itself forms this resistor. Some while ago I did an analysis of a crystal fed like this and looked at the phase shift: -

enter image description here

If you CTRL-right_click on the picture you can open it in another browser page. For the equivalent circuit I chose values like so: -

  • L = 8.45mH
  • C1 = 0.03pF
  • C2 = 6pF

It was designed to produce 10MHz at a phase shift of 180 degrees.

Regarding Rs (R1 in my circuit sim) it is mainly thought of as limiting the power going into the xtal so as not to overstress is. This is of course very true however, if you look closely at the phase shift in the simulation, 180 degrees phase shift isn't quite reached and, by adding Rs (and Cb) you get a few more degrees and bingo, it oscillates.

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  • \$\begingroup\$ I've actually only used that resistor as a way to reduce power levels to avoid damaging a crystal, didn't think about its other effects -- thanks! \$\endgroup\$ – Krunal Desai Dec 25 '15 at 23:19
  • \$\begingroup\$ @KrunalDesai well happy christmas as well! \$\endgroup\$ – Andy aka Dec 25 '15 at 23:23
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Each oscillator with an inverter as active element needs a feedback circuit that allows a phase shift of -180deg at the desired oscillation frequency. In case of the PIERCE oscillator, this is established by a third order lowpass in form of R-C-L-C ladder network (both capacitors grounded). The first resistor is not necessary in case of a sufficiently large output resistance of the inverter circuit (example: BJT inverting amplifier).

However, in case of an opamp this resistor is always required. Otherwise, the following capacitor would act simply as a load without contributing to the frequency properties of the feedback network. Please note that in these circuits the crystal is used as a high-quality inductor only.

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