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Use of feedback resistor & damping resistor in XOSC

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hung_wai_ming@hotmail.com

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Hi

For feedback resistor, as far as i know, is to help biasing to VDD/2 during start-up.
Any more use? How we choose the value?
Also, there's damping resistor at XOUT sometimes and claims not to overdrive the crystal, but I don't understand. Anyone can explain ?
 

Hi,

Regarding the value of the feedback resistor, it should be there from a DC but should not be there from an AC point of view.
For DC, the smallest value depends on how much dissipation we wish to make in the active device, and for AC the smallest value depends on causing the least negative feedback possible at the oscillation frequency.
For DC, the highest value depends on the active device input and output resistance, and when you bridge the two with the feedback resistor, it should be able to transfer any DC voltage changes between the two (i.e. keep the operating point at the half VDD value). In case of a TTL gate/inverter this value is a few kiloOhms but for a CMOS or HCMOS gate/inverter it can be in the MegaOhm range.
(Note: when a few kOhm is needed the feedback resistor must be made from two equal value resistors in series and use a capacitor from their middle point to the gnd to prevent (short-circuit) any AC feedback via the low value resistor. In case of a CMOS gate this is usually not problem with the MegaOhm (5.6-10MOhm) value feedback resistor, though the same capacitor shunt still can be used in need of using two series resistors.

Regarding the role of the 'damping' resistor: One role is indeed to reduce crystal dissipation (it is actually in series with the crystal), another role is to help create additional phase shift in the oscillator loop, workin together in this respect with the capacitor which is connected to one of ends and the gnd.

See additional info at two links on these:

**broken link removed**

http://www.fairchildsemi.com/an/AN/AN-340.pdf

Regards
unkarc
 
Thanks, unkarc.
I have more understanding.
I have one more question.
What do u think at start-up should the PMOS and NMOS inverter be ALWAYS biased at linear region for startup?
Since in reality, due to process variation. this may not be the case and unfornately one of the transisors may marginally biased at satuaration and one at linear region. What do u think?
 

Hi,

Well, at the very moment of switching on VDD, I think the inverter or gate is either in a logical high or low state either at its input or its output. In fact I do not think it all matters which is at what state because if one of them is ,say, in high the other one must in low state by default, right? Now comes the role of the feedback resistor being present between a logical high and low state i.e. voltage difference: because the value of this resistor is much lower with respect to that of the inverter or gate (that is how or why it is chosen that way) it will be able to pass current from the high state voltage side towards the low state side and thus bring (and keep) the operating point near in the middle of the DC transfer curve, hence forcing the inverter or gate to function as a near linear amplifier.
I do not think this operation would really depend on PMOS or NMOS devices, of course manufacturing tolarences may manifest as some shift on the linear transfer characteristic curve because these inverters was NOT designed for a linear operation. Some tinkering with the value of the feedback resistor may correct assymetry and may help startup condition (you may want to reduce the 10MOhm feedback or use another 5-10MOhm resistor from either the input or the output of an 'ill-behaving' inverter to the gnd to restore DC balance at that side).

unkarc
 

As I know,the value of the fb resistor help xosc to startup. If xosc wants to oscillate, the transconductance of the inverter fulfil some relationship. there is limitation ,uplimit and donwlimit. if gm between them , the xosc can work.
the larger value of the fb resistor, the more easy and faster the xosc can oscillate.
you can read the following papers for the detail.
 

Here is the direct download of your file:

**broken link removed**
(replace ** with tt of course)

unkarc
 

Hi unkarc

i have one more question.
why adding larger feedback resistor can reduce negative feedback? i don't quite understand this so-called negative feedback
 

hung_wai_ming(at)hotmail.com said:
Hi unkarc

i have one more question.
why adding larger feedback resistor can reduce negative feedback? i don't quite understand this so-called negative feedback

Hi,

For the shake of simplicity, let's talk DC and AC feedback.
If an inverter input is at a DC low logic level its output must be at a high logic level or vica versa. If you do not use a (feedback) resistor between input and output the voltage difference remains for ever. If you apply a 1000 GigaOhm fedback resistor its hugh value cannot disturb much the voltage difference because the current through it will be very very small. But if you reduce this to the MegaOhm range, then you have come well under the GigaOhm range which is characteristics for the CMOS gates or inverters and the current through the MegaOhm resistor will dominate and force the DC voltage difference to reduce/diminish towards a midpoint value between logic low and logic high voltage values. In case of TTL gates/inverters the resistor value is in the kOhm range because of the bipolar transistor process for TTL ICs.

For the AC feedback case, the more you reduce the feedback resistor's value the less AC amplification gain you experience. Why? Because there is an opposite phase signal at the output with respect to the input and the two signals "see" each other through the feedback resistor (signals of opposing phase tend to cancel each other depending on their amplitude). This is why we do not want too much AC feedback through the feedback resistor in case of these oscillators, we want only the correct DC feedback which is able to establish the linear operation point. And if you have to reduce the resistor's value to reach the correct DC point, you may wish to prevent the bigger AC feedback this reducement brings in. How can you do this? Once you have the correct resistor value, you may use two resistors in series with half values each and place a capacitor between the resistors midpoint and the ground to shunt any AC signal that may go through the series resistors. (This may bring in some additional phase shift at the output/ or input but usually these are small.)

Maybe these answer your question.

unkarc
 

Hi unkarc

One more question, what about 3-stage inverter oscillator normally seen in VCO ? Why they don't need feedback resistor then ?
 

Because as you said those are 3 in series - so they bias themself. In Sims you have to use some puls to start this kind of VCO. In real world noise is enough.
 

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