Ring Oscillators are very commonly used circuits in SoCs, where they find their use in Voltage Controlled Oscillators used inside the PLLs, and also used as Silicon odometer circuits- which are used to track various parameters of the device like variation of timing with ageing etc.

It is also a well-known fact that ring oscillators have odd number of inverters connected in form of a chain, as shown below:

And the frequency of oscillations is given by the expression:

*f = 1/(2NT)*

*N= Number of inverters (odd)*

*T= Propagation delay of a single inverter*
Can you comment on the below circuit? Maybe an expression for it's oscillation frequency? Or perhaps any analytical expression (without bothering about the intricacies of various transistor parameters) for the voltage? Assume the operating voltage to be 1V.

Please post your answers here. I'd be happy to share the solution and my thought process in a couple of days.

vout=vin=vm

ReplyDeletebias generator

Yes! That is right!

DeleteCan you elaborate?

DeleteI surely will. However, I'd like to wait for a day or two to post it here. If you want, I can mail it to you.

Deleteplz send to remesh.manu0@gmail.com

Deletefunctionally both the circuits are equal ... bt the later 1 ... ie ., the oscialtor with single invertor has highest frequency compared to any other oscilators using invertors with the given propagation delay.

DeleteI understood the above theory ... bt did not get vout=vin=vm ... pls elaborate in mail gautamireddy1990@gmail.com

Plz comment or mail to gautamireddy1990@gmail.com

vout=vin=vm ???... pls elaborate this and bias generator in mail prashantguptaengg@gmail.com

DeleteVout = Vin = Vdd/2..?? half the bias voltage..

ReplyDeleteYes, that is correct. No oscillator like function.

Deleteno, that's wrong.. read it again...

Deleteand not an oscillator like performance in achieved... I suppose

ReplyDeleteHi Naman,

ReplyDeleteCan you please elaborate how Vout equals to Vin when an inverter is established in between input and output

And how the propagation delay of an inverter effects the above circuit

ReplyDeletejkam273@gmail.com

ReplyDeleteplease mail the explaination

for oscillation atleast three stages req. to satisfy Barkhausen criteria of oscilation. in which phase shift should above 180 degree. to get above one degree we need to connect three stages one inverter will give only 90. if we use two inverter 180 but even bcz of that no change in output and even formation of latchup.

ReplyDeletesingle inverter wont provide any oscillation

Please mail me the explanation @ montu.bhavsar75@gmail.com

ReplyDeleteHi,

ReplyDeletePlease email me the solution on jaiswalbhaskar@gmail.com

hey,

ReplyDeleteplease emial me the answer to rathod.hari31@gmail.com

hi

ReplyDeletePlease email me the solution peeyushtapadiya@gmail.com

If I go by control theory the second circuit is only having a single pole (not a pair of complex/imaginary poles) as required for marginally stable systems. Hence the system will always settle immediately to a steady state value (likely tobe closure to the 50% threshold between high and low. thus frequency is 0, i.e a DC.

ReplyDeleteHi,

ReplyDeletePlease mail me the solution.

shethketan30692@gmail.com

Thanks in advance..

i think it is about bias generator, but it requires atleast 3 inverters.

ReplyDeleteplease mail me

contactsanjai483@gmail.com

Explanation can be - you have two different function ( 1st function VTC Char of inverter and 2nd function Vout= vin) applied on same inverter.so you have "one single solution".It means two different curve meet at one single point that is Vdd/2 if the inverter sizing correctly.

ReplyDeleteExplanation can be - you have two different function ( 1st function VTC Char of inverter and 2nd function Vout= vin) applied on same inverter.so you have "one single solution".It means two different curve meet at one single point that is Vdd/2 if the inverter sizing correctly.

ReplyDelete