Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

voltage controlled oscillator calculations

Status
Not open for further replies.

Techman_7

Full Member level 2
Full Member level 2
Joined
Dec 21, 2013
Messages
138
Helped
0
Reputation
0
Reaction score
1
Trophy points
1,298
Activity points
2,420
Hello,

I wonder how the attached VCO circuit works. Its output frequency is 300KHz. Can anyone pls tell me the calculations? Its an integrator but component values are not matching with the output. VCO.PNG
 

The integrator produces a ramp, the comparator, with hysteresis, trips
at its edge set by the + fdbk R divider. So you would calc ramp from
RC and input V, and determine when it hits comparator trip point. That
gives you one half the period, do the same for the other side, then you have
the period.

Comparator calculator -



Regards, Dana.
 
Thanks for your reply.
I understand the theory, but similar to comparator can you please tell integrator calculations.
Because I need to change the frequency from 300KHz to 100KHz , so that calculations will help me to find component values.
 

Hi,

Without schematic and integrator calculations:
What happens if you increase the capacitor by a factor of 3?
Since 300kHz/100kHz = 3

Just as simple approach.

Klaus
 
The relationship is fairly simple, a step input to an integrator results in
a ramp .


So we have Q = C x V, I = C x dV/dT or dT = (C x dV)/ I

Because of virtual ground at OpAmp I = Vcomparator / R. So you have two
dT's, the ramp up and the ramp down to compute, Vcomparator is the trip
point of the comparator.


Regards, Dana.
 

Thanks for your reply.
I understand the theory, but similar to comparator can you please tell integrator calculations.
Because I need to change the frequency from 300KHz to 100KHz , so that calculations will help me to find component values.
I have referred calculations at below link but the calculations & frequencies do not match as per the above schematic.

Link - https://www.eeeguide.com/op-amp-integrator-circuit/
--- Updated ---

But
The relationship is fairly simple, a step input to an integrator results in
a ramp .


So we have Q = C x V, I = C x dV/dT or dT = (C x dV)/ I

Because of virtual ground at OpAmp I = Vcomparator / R. So you have two
dT's, the ramp up and the ramp down to compute, Vcomparator is the trip
point of the comparator.


RegardsT

But, these calculations not justify with the component values & output frequency.
 

Have you measured your R & C values ? And what type of caps are you using ?
Ceramic or Teflon or Mica or Polymer or.......

Some caps have a strong V dependency, like ceramic caps. So if you measure them
you have to do that in the conditions they operate in to get their "effective" value.


You could back into the problem by taking a known R & C, measure freq, and use that in a
linear equation to predict freq from other values.

1657877882188.png


Freq = a *C + K where C equal C value, K is slope offset.

Note this assumes the F to V relationship in circuit operation stays linear over F and V.....

Lastly this ignores the slew rate of the comparator solution which could add significant delay
at higher freqs, hence alter the ideal mathematical relationships, describing equations.

When you measure freg, how are you doing that ?


Regards, Dana.
 
Last edited:

HI,

I don´t see your calculations. Thus I can´t verify it. Show them.

Where is the schematic of post#1 from? give a link.

The schematic of post #1 misses some important informations for the frequency calculations.
I miss the supply voltage and the exact OPAMP type. Both have impact on resulting frequency.
Thus we can´t do the math.

Klaus
 

Hi,

Without schematic and integrator calculations:
What happens if you increase the capacitor by a factor of 3?
Since 300kHz/100kHz = 3

Just as simple approach.

Klaus
Hello KlausST,

Actually, I am working on CLLC DC-DC converter & want to vary output as per input voltage to opamp. This VCO is in the feedback path. So, if I get calculations then effectively feedback circuit can be designed. The input to the circuit is constantly varying as per load & also to minimize ripple.
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top