Ring Osc Example

8/2/2019 Ring Osc Example

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Sheet1 of 6

RING OSCILLATOR EXAMPLE

Determine the frequency of operation of a 7-stage free running ring oscillator using thefollowing MOSFET device data and a 3V supply rail. Also assume that a 1pF capacitoris connected to each output to ground.

Cmosn

KPn=6.32664E-5; Vto=0.622490; W=8um; L=5um

Cmosp

KPp=2.63544E-5; Vto=-0.63025; W=12um; L=5um

Calculate Rn & Rp

( ) ( )

( )

( )Ω=

−=

Ω=−

⎥⎥

⎥

⎥

⎦

⎤

⎢⎢

⎢

⎢

⎣

⎡

−⇒

−=

16.892K 63025.03

512.

22.63544E

3

10.486K 622490.03

58

.2

6.32664E3

VTHNVDD.2

KPnVDD

WL

VTHNVDD

LW

.2

KPnVDD

25-

25-

22

Rp

Rn

We now need to calculate the parasitic capacitance of each cell:

( )

( ) 204fF.5E12E6.5E8E1.36E23

1.36EE225

E85.8*46.3TOXEox

Cox

Cox.W.L'

WpLpWnLn'23

6663-

3-10

21

=+−=

===

=

+=

−−−

−

−

Cin

Cox

Cox Cin



Sheet2 of 6

( ) 136fF.5E12E6.5E8E1.36E

)WpLpWnLn('

6663- =+−=

+=

−−−

Cox Cout

Ctotal = 204fF + 136fF = 340fF

Calculate frequency of operation assuming 7 stages:

( ) ( )

( )( ) 3.8MHz E3401E.16.892E10.486E7.

1

N.1

tdN.1

T1

f

τ pτ n(s)stageeachof delaytimetdstagesof number theN

151233=

++=

++===

+===

−−

Ctotal Cload Rp Rn

We can now verify this result by running an ADS transient simulation shown in Figure 1 andFigure 2.

The resulting output waveform is shown in Figure 3 showing that the frequency will be3.805MHz as predicted from the hand calculations.

The final plot of Figure 4 shows the phase difference through each cell with each stagecontributing a phase shift of 51.4 degrees, which is equal to 360 degrees (or zero degreesphase shift) divided by 7 cells.

Conclusion

The hand calculation of the ring oscillator using the spice model parameters gives a prettyaccurate prediction of ring oscillator oscillating frequency. However, care must be taken inensuring the parasitic capacitances of the MOSFET devices are included in the frequencycalculations to avoid large errors.The ring oscillator example given here is only capable of operating at a single frequency butcan be made variable by adjusting the current (and hence resistance) through the devices byusing a current mirror of each inverter – This will be the subject of another tutorial on ringoscillator VCO (variable controlled oscillator).



Sheet3 of 6

CC1C=1 pF

V_DCSRC1Vdc=3 V

LEVEL2_Modecmosn

Tnom=27Delta=0Neff=4.68830Vmax=17726Uexp=4.612355e-

Ucrit=174667Uo=1215.74Ld=0Xj=0.9 umTpg=1Nfs=4.55168e1Nss=3e10

Nsub=1.066e+1Tox=225e-Mjsw=0.195 Cjsw=1.74e-10 Mj=0.1067 Cj=3.27e-04 Rsh=60Cgdo=2.89e-10 Cgso=2.89e-10 Lambda=0 Phi=0.31Gamma=0.63924Kp=6.32664e-05 Vto=0.622490 NMOS=yes

LEVEL2_Modecmosp

Tnom=27Delta=0Vmax=63253.Uexp=8.886957e-Ucrit=637449Uo=361.941Ld=0.03e-6Xj=0.112799Tpg=-1Nfs=1.668437e+1Nss=3e10

Nsub=6.575441e1Tox=225e-Mjsw=0.307 Cjsw=2.23e-10Mj=0.341Cj=4.75e-04 Rsh=150Cgdo=3.35e-10 Cgso=3.35e-10 Lambda=0 Phi=0.541111 Gamma=0.61810Kp=2.63544e-05 Vto=-0.63025 PMOS=yes

MOSFET_NMOMOSFET7

Temp=27 Width=8 um Length=5 um Model=cmos

MOSFET_PMOMOSFET2

Temp=27 Width=12 um Length=5 um Model=cmos

PortP2Num=2

Port P1 Num=1

I_ProbeI_Probe1

Figure 1 Single cell design showing inverter circuit with Spice models .



Sheet4 of 6

a1

DCDC1

DC

Tran Tran1

MaxTimeStep=0.05 nsecStopTime=800 nsec

TRANSIENT

Single_CellX7

Single_CellX2

Single_CellX4

Single_Cell X6

Single_CellX

Single_CellX3

Single_Cell X1

InitCondInitCond1V=2 V

InitCond

Figure 2 ADS transient setup for analysing the 7-stage ring oscillator. Each cell(Single_cell) is the same as the circuit shown in Figure 1. The measurement node hasbeen set to ‘a1’ and an initial startup condition of 2V has been placed on the oscillatoroutput.



Sheet5 of 6

m1 time=76.65nsec TRAN.a7=3.000 V

m2time=339.5nsecTRAN.a7=3.000 V

50 100 150 200 250 300 350 400 450 500 550 600 650 700 7500 800

0.0

0.5

1.0

1.5

2.0

2.5

-0.5

3.0

time, nsec

TRAN.a7,V

m1 m2

Eqn Freq=1/(indep(m2)-indep(m1)) Freq

3.805MHz

Figure 3 Simulation result from the ADS simulation setup of Figure 2. The two markershave been placed at identical points on the signal points and the frequency calculatedfrom ‘Freq’ equation. The simulated frequency being 3.8MHz, which agrees well withthe hand calculations of our example.



Sheet6 of 6

m1 time=39.37nsec TRAN.a5=3.000 V

m2time=76.95nsecTRAN.a7=3.000 V

10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210 220 230 2400 250

0.0

0.5

1.0

1.5

2.0

2.5

-0.5

3.0

time, nsec

TRAN

.a5, V

m1

TRAN

.a6, V

TRAN

.a7, V

m2

TRAN

.a1, V

TRAN

.a2, V

TRAN

.a3, V

RN

.a4, V

Eqn Delay=indep(m2)-indep(m1) Delay 3.758E-8 Cell_Phase

51.477Eqn Cell_Phase=(Delay/(1/3.805E6))*360

Figure 4 Showing the waveform on each cell output. The phase difference betweeneach cell is 51.4 degrees. This phase is equal to 360 degrees divided by 7 cells.

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Ring Osc Example