integrator macro model of cadence

[najd]

op amp integrator cadence

hi every body

please i need some help

i'm trying to modeling a switched capacitor integrator with cadence

in fact i use an opamp macro in the library functional i have many problems with this macromodel beacause the out put swings are always very big even i adjusted the parametre at +1 v et -1v.
i dont know what is the problem everything is alraight u will find the pictures of the macro and his parametre joined here

please anyone who worked with it before

thanks a lot

 

[LvW]

One reason for your problems is certainly the fact that your opamp has a POSITIVE capacitive feedback. This causes instability. Try to interchange the opamp inputs.

 

[najd]

i had interchanged the inputs but the outputs always big and this time negative -50 v

Added after 1 hours 4 minutes:

i had a stable amplop coz after ac simulation i have a 64 as phase margin

i dont understand why the ampli op give heig outputs in a closed loop raelly it make me crazy
please some one help me

 

[LvW]

Do you use an offset-free opamp model ?
In this case (and if the cap in the feedback loop is connected to the inv. terminal) your circuit should work.
However, if you use a real model with offset, the output goes into saturation cause there is no dc feedback.
Forget the ac analysis and the phase margin, because you can get "good" results even for unstable cicuits. At first, bias point and tran analysis must be OK, then you can trust the ac analysis.

Added after 18 minutes:

There is another point I just have discovered: Has your opamp single supply ??? That´s another problem for the bias point.

 

[najd]

My opamp use double supply as shown in the pictures ( pvref=1.2 and nvref=-1.2).
and for the offset i putted the output voltage offset parameter at 0.

how can i make the dc feedback?

 

[JoannesPaulus]

In this case, shouldn't your ideal amplifier be a transconductor (v input, i output)? I had a similar problem some time ago.

 

[LvW]

My opamp use double supply as shown in the pictures ( pvref=1.2 and nvref=-1.2).
and for the offset i putted the output voltage offset parameter at 0.
how can i make the dc feedback?

Do you joke ? DC feedback is made by a part which can conduct a dc current - a resistor.
(Sorry, but if you really do not know what dc feedback means, I doubt that you can design an s/c circuit succesfully)

I don´t know how it works: to put output voltage offset parameter at 0 .
The safe method is to measure the dc transfer curve and to see what the input offset is. Then you can make an offset correction.

 

[najd]

im a beginner in electronic so thank you to help me even i ask bad questions i want to understand .

for the input offset every thing is alright cause the output dc is as i like 600mv

the problem is if i put the aopmaco in a circuit of analog integrator (resistor and capacitance) and i use q square input i obtain a square output amplified .

normally i should obtain a triangle signal ?
any idea?

 

[LvW]

You can expect a sawtooth or triangle only if
a) the time constant RC is appropriate with respect to the duration of the square,
b) the amplitude does not drive the opamp into saturation.

Perhaps it helps when you show us a circuit diagram.

 

[najd]

ok the first figure show the diagrame of the opamp macro and an ideal aop in an integrator circuit.

the second figure show the results of dc et trans simulations .
net27 is the voltage juste after the resistor
dc response are in fuction of the openloopgain of the opmacro

Added after 59 seconds:

this is the figures

 

[LvW]

Questions:

1.) Do you feed the inv. opamp input with 600 mV ? Why ?
2.) Power supply only 1.2 volts ? Does your model work with these small voltages ?
3.) Try a smaller R instead of 1Megohm
4.) Why you are interested in "net24" at the opamp input ?
5.) Which voltage do you tune for dc analysis ?
6.) Don´t you have any operating point feature in your program ?

 

[najd]

1) no i feed it with 0v and i puted the amplitudes [10mv and -10mv]for a square signal.
2)yes the model work with this small supply.
3)ok same problem
5)0v
6)yes i have used the openloopgain as a operating point


the dc response of the integrator is always on -40v even i change the input bias current

the trans response as you see of the integrator is a constant signal on -40 v

Added after 6 minutes:

5) input voltage and the dc response is cetred at -40 v

Added after 53 seconds:

6) ihad used now the input voltage as an operating point

 

[LvW]

1) no i feed it with 0v and i puted the amplitudes [10mv and -10mv]for a square signal.

You feed the NON-INVERTING input. That´s wrong.
What potential has the OTHER input ?

2)yes the model work with this small supply.
3)ok same problem
5)0v
I don´t understand. For dc analysis you must tune a voltage source or any part.

6)yes i have used the openloopgain as a operating point
Again I don´t understand. The operating point is a VOLTAGE and not any gain.

the dc response of the integrator is always on -40v even i change the input bias current

How do you manipulate the bias current ? Confusing !

6) ihad used now the input voltage as an operating point
Again confusing. I really don´t know what you mean.

Here are some basics to the operating point of an opamp:
When it has dual supply and both inputs are grounded the output voltage (that is the operating point) is at 0 volts if the opamp is ideal and has no input offset error.
Since your opamp is ideal the output must be zero if both inputs are at ground potential. Check this !

 

[najd]

ok i cheked this and i have the output voltage=15nv it mean equal to 0 beacause this is a real mode ok
and then?

 

[LvW]

As a next step I propose to check a simple inverting amplifier - for example with 10 kOhms (negative !) feedback and 1kohms between input and inv. terminal. Pos. terminal at ground.
Input voltage 10 mV must give an output of -100 mV.

If this works, replace the 10k resistor by a capacitor of 1mikroF. Now you have an inverting integrator with an 1msec time constant.

 

[najd]

really it makes me crazy

if i used it as an inverting amplifier he become an integrator
with a square input it gives me a triangle ouput

 

[LvW]

really it makes me crazy

if i used it as an inverting amplifier he become an integrator
with a square input it gives me a triangle ouput

Show me the inverting amp diagram.

 

[najd]

ok if i use a square signal input amplitude 1mv with a period of 1u s

and a resistor of 10kohm and a capacitor 1 uF

how can i be sure that the output is right ?

Added after 1 minutes:

i mean what should i mesure in the output triangle signal ? and how much i should find as a value?

 

[LvW]

ok if i use a square signal input amplitude 1mv with a period of 1u s
and a resistor of 10kohm and a capacitor 1 uF
how can i be sure that the output is right ?
Added after 1 minutes:
i mean what should i mesure in the output triangle signal ? and how much i should find as a value?

1.) The form resp. the mean value of the sqare is important: between 0 and 1mV ?
or between -1mV and +1mv ?

2.) Don´t you know the mathematical solution of an integral ?
If your time constant is RC=10millisec, the output reaches the input amplitude after 10msec. The output goes back when the input goes negative and it remains constant if the input disappears towards zero.
Thats basic math !!