load capacitance value for 8mhz crystal in msp430f155

Hi, you have to take care about the correct load capacitance for the selected crystal. In the crystal datasheet you find an effective load capacitance. This parameter basically says that if the crystal sees such a capacitive load it will oscillate at the given frequency. Any deviation from the capacitive load will cause a frequency deviation.

For example, if you selected a 32.768kHz crystal with an effective load capacitance of 6pF you should choose two 12pF capacitors that are on one site connected to Xin and the second one to Xout and the other side of the caps are connected to ground. This basically means the crystal sees these two caps that are connected in series. So the two 12pF series capacitors causes the 6pF load that is seen by the crystal. Not that the parasitic capacitance of your printed-circuit-board as well as the parasitics of the MCU itself (package: leadframe, bond wires) has to be considered.

The same calculation and selection of the caps is done for the 8MHz crystal.

so then  if i  choose any  8 mhz crystal with an 8pf load capcitance means then my c1,c2 value will be  the as per the following calculation

c1=c2=2*cl-(cp+ci)

cl=load capacitance =8 pf

(cp+ci)~ 5pf

c1=c2=16-5= 11pf..

whether it is wright or wrong??

then one more thing whether i can choose 8mhz crystal with an different (other than 8pf) load capacitance  as well as the c1,c2 value as per the above calculation.

or else any condition is there to choose load capacitance????

Hi, your thoughts are not completely correct... Here is a figure that may help to understand how the load capacitance is calculated:

The load capacitance is the capacitive load that is seen by the crystal. So remove the crystal and look into the pins where the crystal was connected (right side in above figure). You see the capacitors C1 and C2 connected together in series (there is a connection between the ground pins of C1 and C2).

Usually the same value is used for C1 and C2 (Cx=C1=C2).

      Cload =(Cx * Cx) / (Cx + Cx)

      Cload =(Cx * Cx) / (2*Cx)

      Cload = Cx / 2

Let's assume the load capacitance of the crystal is 8pF:

      C1=C2=Cx= 2*8pF=16pF

In case you want to add the parasitic capacitance on your design you would calculate appropriate. So my assumption would be that your design you would see the following parasitics:

So if you would include the parasitic capacitance in your calculation you have to consider that C1 and the parasitic capacitance are parallel:

Cload = ( (C1+Cparasitic) * (C2+Cparasitic) ) / ( (C1+Cparasitic) + (C2+Cparasitic) )

Regards.

Sorry... it seems that there was an issue with pasting of the figures. I try it in this post with another format for the pictures.... hope this help.

Hi, your thoughts are not completely correct... Here is a figure that may help to understand how the load capacitance is calculated:

The load capacitance is the capacitive load that is seen by the crystal. So remove the crystal and look into the pins where the crystal was connected (right side in above figure). You see the capacitors C1 and C2 connected together in series (there is a connection between the ground pins of C1 and C2).

Usually the same value is used for C1 and C2 (Cx=C1=C2).

      Cload =(Cx * Cx) / (Cx + Cx)

      Cload =(Cx * Cx) / (2*Cx)

      Cload = Cx / 2

Let's assume the load capacitance of the crystal is 8pF:

      C1=C2=Cx= 2*8pF=16pF

In case you want to add the parasitic capacitance on your design you would calculate appropriate. So my assumption would be that your design you would see the following parasitics:

So if you would include the parasitic capacitance in your calculation you have to consider that C1 and the parasitic capacitance are parallel:

Cload = ( (C1+Cparasitic) * (C2+Cparasitic) ) / ( (C1+Cparasitic) + (C2+Cparasitic) )

Regards.