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.

Need help from pros (emergency)(desperate)

Status
Not open for further replies.

ricopt

Member level 5
Joined
Jun 8, 2010
Messages
80
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,286
Activity points
1,807
Any pro out there, pls pls pls help me with my design
I need to wrap up my design tomorrow.Need to settle with this problem
I am currently designing shunt feedback transimpedance with cascode gain stage and source follower.

I am able to get the gain at 525.6 for Rf=1k

but when I simulating Ac analysis with magnitude 1A , my Vout went to large negative value .and my biasing is correct.I can't find any solution to it.I also duno why it happen.Can someone guide me ?







These two are the info regarding the transistors



Transimpedance gain



Ac analysis circuit



bandwidth
 

Break your feedback loop!

If you want to run an AC analysis in order to measure gain & phase, you must break the feedback loop, which you didn't do in your simulation. Hence your (negative) transimpedance of -526Ω is superimposed by a positive one of +1kΩ (your feedback resistor), which probably causes this weird gain magnitude result.

You might want to study this thread, and especially my explanation attempt - which was wrong for that thread -, but might give you some enlightenment for your problem.

So break your feedback loop - either by the LC method shown, or - even much easier - by using the STB (stability) analysis, breaking the feedback loop by an iprobe.
 

don't quite understand with your explanation....sorry with my noob skill
I really need more detail explanation....like how do i simulate bandwidth for shunt-feedback transimpedance amplifier ?
 

You should break the loop at 'in2' node. As what erikl suggested, using LC method, you can put a big inductance (1kH) and big capacitor (100uF) and injected your ac test signal.
 

Sorry to trouble you, but can you do the LC method based on my circuit ?Like draw for me ?
I still dun get your meaning
how exactly break the in2 node ?
 

Re: Break your feedback loop!

If you want to run an AC analysis in order to measure gain & phase, you must break the feedback loop, which you didn't do in your simulation. Hence your (negative) transimpedance of -526Ω is superimposed by a positive one of +1kΩ (your feedback resistor), which probably causes this weird gain magnitude result.

You might want to study this thread, and especially my explanation attempt - which was wrong for that thread -, but might give you some enlightenment for your problem.

So break your feedback loop - either by the LC method shown, or - even much easier - by using the STB (stability) analysis, breaking the feedback loop by an iprobe.

bro, i have seen your reply , but i dun understand...how i implement the LC method into my circuit ?
can draw for me ?or sketch for me ? pls ?
 

Re: Break your feedback loop!

i dun understand...how i implement the LC method into my circuit ?
can draw for me ?or sketch for me ? pls ?
I've already given you a very concrete hint in and use this LC connection (bottom left corner; the inductance L0 ≥ 1H in series with your 1kΩ feedback resistor, capacitor C2 ≥ 1F, instead of the voltage source V23 your unit (1A) current source.

HTH! erikl
 

What current source should i replace with ?
I understand with the LC component, but what about the source ?
DC voltage source ?or AC voltage source ?Ac current source ?DC current source ?
How much value should i set for my source ?
 

Anyone who can read performs better

DC voltage source ?or AC voltage source ?Ac current source ?DC current source ?
How much value should i set for my source ?
So (re)read the last sentence of my former posting!! AC of course for an ac analysis; the DC bias will be provided by your feedback.
 

instead of the voltage source V23 your unit (1A) current source ??????????

what you mean is it replace the V23 with my current 1A source ?
 

transimpedance circuit!

Hey! Is it a transimpedance circuit - or not?
A transimpedance amp converts input current into output voltage, doesn't it?
 

oh ya, sorry
thx glad to have you all to help
 

Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top