Need Help with Integrator

Hi,

I am trying to implement the integrator shown in the file integrator.jpg. But I don't know why the output of the integrator is always constant and doesn't vary with time. I have attached the schematic of the OTA used in this integrator. Below is my spice code. Could anyone please take a look and tell me what the problem is?



*
.protect
.lib 'mm0355v.l' TT

.unprotect

.option ACCURATE METHOD=gear


***Current Mirror***

Rs vdd q2s 25k
MQ2 q0g q2g q2s q2s pch W= 3.6um L= 1um
MQ3 q2g q2g vdd vdd pch W= 3um L= 1um
MQ0 q2g q0g gnd gnd nch W= 1um L= 1um
MQ1 q0g q0g gnd gnd nch W= 1um L= 1um


*** First stage ***

M1 o1 vin cm cm pch W= 6um L= 1um, M=4
M2 o2 vip cm cm pch W= 6um L= 1um, M=4

M3 o1 o1 gnd gnd nch W= 2um L= 4um
M4 o2 o2 gnd gnd nch W= 2um L= 4um

M10 cm q2g vdd vdd pch W= 3um L= 1um, M=4

*** Second stage ***

M7 m7d m7d vdd vdd pch W= 1um L= 5um
M8 out m7d vdd vdd pch W= 1um L= 5um

M5 m7d o1 gnd gnd nch W= 2um L= 4um
M6 out o2 gnd gnd nch W= 2um L= 4um

Cout out gnd 1p

*************************************************************
C vin out 1.28p
R ref vin 100k

Vref ref gnd -1V $ac =1V


****** SOURCE ******
vdd vdd gnd dc=3.3v

$Vin vin gnd dc= 1.65V AC =0.5V, 180
Vip vip gnd dc= 1.65V $AC =0.5V, 0 $



***SIMULATED COMMAND

.op

.TRAN 0.1U 250U


.probe tran v(out)

.end

At the bottom of your figure a classical MILLER integrator is shown.
However, this circuit requires an OPAMP with a low resistance voltage output.
The OTA integrator requires only a load capacitor (without feedback). That´s all.

The capacitor is connected between the negative input of the op and the op output so that the output of the op will continue to rise with time.

However, i don't know why the output of the op doesn't vary with time.

chichan said:

The capacitor is connected between the negative input of the op and the op output so that the output of the op will continue to rise with time.
However, i don't know why the output of the op doesn't vary with time.

Click to expand...

In your first posting you spoke abot an OTA.
Now it´s an opamp.
Do you know about the difference between OTA and opamp ?

An OTA can be regarded as an OP with high output impedande as well. I don't know, if the circuit is stable and has sufficient loop gain, but it should work as an integrator somehow. Typically however, an OTA integrator is build in an open loop style, as LvW mentioned.

There's a different problem, not related to OP versus OTA. Your circuit will already achieve saturated output during operation point calculation and stay saturated, according to the shown simulation setup. An integrator without DC feedback or reset circuit needs an .IC statement to specify a suitable initial condidtion, e.g.

FvM said:

........................................
it should work as an integrator somehow.

Click to expand...

The transfer function of an ideal OTA with C-R feedback is

H(s)=(1-gm/sC)/(1+gmR) with gm=transconductance.

As in each first order circuit with an general/overall low pass characteristic there may be a limited region with a phase shift around 90 deg. , but I doubt
a) that such a circuit could be called "integrator" and
b) that it can be used as a kind of approximation to an integrating circuit.

I don't agree with the circuit classification. It has basically a PI characteristic, with negative P parameter. Below the corner frequency it's acting as an integrator.
The presented H(s) form isn't suitable to show the behaviour. The general form is
H(s) = a - b/sRC, b ≈ 1 for gmR >> 1
A similar error term is also present with an OP integrator due to the nonzero output impedance.

I didn't want to suggest the circuit as an integrator, however. I only wanted to clarify, that the simulation problem hasn't to do with the circuit's bad integrator characteristic.

Thank you so much for your suggestions. Special thanks to FvM. After I added a switch to the integrator, the integrator output can vary with time. You are so smart!