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.

How to amplify a sin wave signal having 0.8 - 1V amplitude?

Status
Not open for further replies.
Joined
Sep 3, 2007
Messages
848
Helped
66
Reputation
132
Reaction score
16
Trophy points
1,298
Activity points
0
Amplifier

Hi,
I need to amplify a sin wave signal having 0.8V amplitude to 1 V. Who can help me ?
 

Amplifier

What is your application? Accurency, bandwidth etc are key parameters that you should consider first.
 

Amplifier

Thanks my friend for your response.
It is about a pll that deliver a weak signal and i need to amplify the signal without affecting it's frequency.
 

Re: Amplifier

A.Anand Srinivasan said:
what is the frequency of operation of the pll... i mean the range...

I don't understand what is the relation between the freq and Amplificaaation.
Could you explain my friend ?

The operating frequency is 1 Ghz.
 

Re: Amplifier

Hi,
I don't know where you are going to put this amp. I mean at the input or output of the PLL.

You can use a simple common source amp with current source load.
Lets say your supply voltage=1.2v
Design your amp such that the current source takes 0.1 Vdsat and the input device takes 0.1 Vdsat. So the output can swing is 1.2-0.1-0.1=1v(i.e is it can swing from 0.1V to 1.1V). Your output common mode voltage has to be 0.6V(may be you need to use a CMFB to set your Vocm to 0.6v). Now coming to your input, if your Vicm=0.6V , then you can go from 0.2v to 1v. But I am telling you, your output will have harmonic components along with your sine output(Because of large input signal). You can use linearization techniques like resistive degeneration but this creates headroom. So Try different circuits.

I hope this helps you.

Thanks
 
Amplifier

dear master,
you spoke about pll so i wanted to know the frequency of operation because if the frequency of operation is very high as in your case of 1GHz then you have to go for different amplifier circuit owing to the bandwidth of the amplifier....

i mean the gain has to be stable in your frequency of operation....
 
Re: Amplifier

swarnavenu said:
Hi,
I don't know where you are going to put this amp. I mean at the input or output of the PLL.

You can use a simple common source amp with current source load.
Lets say your supply voltage=1.2v
Design your amp such that the current source takes 0.1 Vdsat and the input device takes 0.1 Vdsat. So the output can swing is 1.2-0.1-0.1=1v(i.e is it can swing from 0.1V to 1.1V). Your output common mode voltage has to be 0.6V(may be you need to use a CMFB to set your Vocm to 0.6v). Now coming to your input, if your Vicm=0.6V , then you can go from 0.2v to 1v. But I am telling you, your output will have harmonic components along with your sine output(Because of large input signal). You can use linearization techniques like resistive degeneration but this creates headroom. So Try different circuits.

I hope this helps you.

Thanks

Thanks My friend for your elaborated reply. A particular thank also for A.Anand Srinivasan.
I want to amplify the output of the pll. Could you please swarnavenu send me the shematic of the amplification circuit at transistor level ?
Thanks in advance.
 

Re: Amplifier

Hi,

Is your PLL's VCO ring oscillator based?
if yes, then you can use a pair of inverters whose supply is 1v. This gives you the required output.


If your PLL'S VCO is a LCVCO then use a common source amplifier. You can find common source amplifier with current source load circuit in "Design of Analog CMOS Integrated Circuits" - by Razavi(page 58). If you want more linearity and your PLL's output is differential then use a simple differential pair with current source load. This circuit is given in page 124(figure 4.32(b)) of Razavi book.
The advantage of differential circuit is each leg of your differential output only need to swing 0.5v(so the total output swing will be 1v). If you want further more linearity use resistive degenaration for your differential pair. Differential pair will give good PSRR and CMRR but device generated noise.

So the design depends on the power supply voltage(for headroom), required linearity(i.e whether you want an exact amplified replica of your input or not) and availability of differential inputs and finally most impotantly noise(power supply noise,device noise etc..).

Read Razavi chapter 3 and 4. You get good insight.


Thanks
 

Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top