Large Signal Bandwidth

[samiran_dam]

Could somebody please explain to me the large signal bandwidth of opamp, its relation with slewrate and any other detail that would help me understand this intuitively.

Cheers!!!
Samiran

 

[dick_freebird]

May also see it as "full power bandwidth" or a closed loop -1dB
or -3dB bandwidth. You apply an input signal varying frequency
and looking at output amplitude rolloff.

Slew rate is one fundamental limiter, you can calculate the max
dV/dt that the full power output sine wave ought to exhibit as
it crosses zero (d(sin(wt)/dt) and when that approaches the
slew rate of the part, you're done (or at the least, distorted).

 

[LvW]

Quote: May also see it as "full power bandwidth" or a closed loop -1dB
or -3dB bandwidth. You apply an input signal varying frequency
and looking at output amplitude rolloff.

I think, it's not quite correct. You must not observe the "amplitude roll-off" but you have to watch the sinusoidal form of the output signal. As soon as you observe some distortions similar to a triangle wave you have to decrease the input amplitude until the output becomes again sinusoidal.
The following formula connects the cause of the full-power bandwidth BW (which is the slew rate SR) with the maximum possible output amplitude Vmax (given by the supply voltages):

BW=SR/(2*Pi*Vmax) .

 

[samiran_dam]

That means, it is applicable only to sinusoidal signal. And to measure the FPBW, do I need to connect the opamp in unity-gain mode?

Added after 1 minutes:

And how FPBW can be measured? I mean from gain-freq plot do I need to measure the -1dB bw?

 

[LvW]

That means, it is applicable only to sinusoidal signal. And to measure the FPBW, do I need to connect the opamp in unity-gain mode?

Yes, of course, since all bandwidth and distortion related parameters are referenced to sinusoidal signals.
In theory, negative feedback has not much influence on the full-power BW (FPBW). Because of non-linear effects, calculations are not easy to show how and if feedback influences FPBW. You are on the safe side when you measure/simulate this function Vmax,sinusoidal=f(frequency) in the actual opamp configuration. On the other hand, it is not false to use the unity gain mode.

And how FPBW can be measured? I mean from gain-freq plot do I need to measure the -1dB bw?

No, don't measure/simulate the gain. The parameter FPBW is related only to the output amplitude - regardless on gain! As mentioned in my first posting, increase the output amplitude at lower frequencies until it reaches the maximum Vout,max and, then, follow this procedure:
- increase the frequency and watch the output,
- if the amplitude decreases due to lower gain (but is still sinusoidal) increase the input signal again until Vout,max,
- Further increase the frequency,
- as soon as you observe distortions (like triangle), reduce the input amplitude somewhat until the output is again sinusoidal (and has lower amplitude),
- repeat these steps for increasing frequencies,
- Thus, you get the curve which enables you to find the -1dB or -3dB point (which ever you prever). That's the FPBW.

 

[samiran_dam]

In one of my earlier post you suggested that, for high frequency filter I need to take care of slew-rate limitation. So, I guess, by calculating the FPBW, I can set the min slew-rate required for a given input amplitude. Am I correct?

 

[LvW]

In one of my earlier post you suggested that, for high frequency filter I need to take care of slew-rate limitation. So, I guess, by calculating the FPBW, I can set the min slew-rate required for a given input amplitude. Am I correct?

...for a given OUTPUT amplitude (which leads to the input amplitude if you know the gain).

 

[samiran_dam]

Hi LvW,

I have a question. I have done a transient simulation of an opamp connected in unity gain mode with a sinusoidal small-signal at input. The opamp is the model opamp written in cadence ahdlLib. The Opamp specs are:

1. gain = 100
2. ugf = 30 MHz
3. SR = 8 v/µS

Input signal is: 2 mV p-p, freq (fin) = 5 MHz.

The output I am getting is as follows:

Input: Red-color curve
Output: Blue-color curve

Why the output is skewed? Has already slewing started? But SR >> 2π * fin * Vm = 0.0314 V/µS.

Added after 7 minutes:

I observed that if signal frequency increases the skewness increases. And If I increase the UGF of the opamp to a sufficiently large value (e.g. 10 GHz), then rate of change at output is aligning with that of input.

 

[LvW]

Why the output is skewed?

I don't think the output is "skewed". For my opinion, it is a rather good sinus, which is somewhat shifted in time as a result of the phase behaviour of your circuit. That's all. But why is your opamp gain (open loop) only 40 db?

 

[samiran_dam]

But why is your opamp gain (open loop) only 40 db?

I just set the gain as 40 dB. Even if I increase the gain to say 60 or80 dB, I will only have the output amplitude more closer to that of input but the skewness will be there.

..shifted in time as a result of the phase behavior of your circuit.

I don't think there is any phase margin problem, because there is only one dominant pole so phase margin should be well above 90° at the frequency of interest.


I just want to confirm one thing, if signal amplitude is very small, and if I increase the frequency to an extent so that ωVm exceeds the opamp slewrate, then according to the FPBW concept o/p will slew but before that it will be affected by the opamp's small-signal bandwidth, and o/p will be attenuated, isn't it?

 

[LvW]

I just set the gain as 40 dB. Even if I increase the gain to say 60 or80 dB, I will only have the output amplitude more closer to that of input but the skewness will be there.

Where do you see "skewness"? It is simply a time shift. If you increase the open loop gain, the time shift should decrease (because less phase influence).

I don't think there is any phase margin problem, because there is only one dominant pole so phase margin should be well above 90° at the frequency of interest.


Yes, that`s true. However, I did not speak about phase margin. I was referring to a small phase shift introduced by the opamp.

I just want to confirm one thing, if signal amplitude is very small, and if I increase the frequency to an extent so that ωVm exceeds the opamp slewrate, then according to the FPBW concept o/p will slew but before that it will be affected by the opamp's small-signal bandwidth, and o/p will be attenuated, isn't it?

No, not in general. It depends on the gain (with feedback) as well as on the output amplitude. In many, many cases the slew effect starts before the gain goes down due to the small signal bandwidth.

 

[samiran_dam]

1. What might be the reason due to which the opamp introduces the small phase shift? Ididn't observe change in the shift if I increase the open-loop gain, rather as I mentioned the shift is decreasing with increasing UGF.

2. What may be the reason that slewing starts before small-signal 3 dB roll-off starts? I guess, it can only happen iff SR < ωVm.

Added after 48 minutes:

Yes even if the SR >> ωVm, slew starts for relatively large amplitude signal as I the large signal drives one of the i/p transistor into cutoff. The waveform I got for i/p 2 Vp-p, 2.5MHz signal with opamp's ugf=50MHz, SR = 10V/µS:

If I increase the ugf then I got the following waveform:

What I have understood from my intuition that, when the output gets closer to the input the differential i/p to the opamp becomes smaller and thus it follows the small-signal behavior; and as ugf increases in the latter case, the speed of response of the opamp is faster and the output tends to settle according to the exponential law with time constant = 1/(2π*bw). Am I correct?

 

[LvW]

1. What might be the reason due to which the opamp introduces the small phase shift? Ididn't observe change in the shift if I increase the open-loop gain, rather as I mentioned the shift is decreasing with increasing UGF.


Your opamp gain function is identical to a first order low pass. Don't you expect a phase shift from such a low pass - particularly when the UGF is 30 MHz and the signal is 5 MHz?

What I have understood from my intuition that, when the output gets closer to the input the differential i/p to the opamp becomes smaller and thus it follows the small-signal behavior; and as ugf increases in the latter case, the speed of response of the opamp is faster and the output tends to settle according to the exponential law with time constant = 1/(2π*bw). Am I correct?


I don't know how the slew rate is incorporated into your model and, therefore, I cannot know if and how it is related to other parameters (like UGF).

 

[samiran_dam]

1. Yes, I checked the phase response and found that @ 2.5 MHz there is a phase shift and that is exactly what is reflected in time-domain plot. I should have checked this before posting my query.

2. Following is the veriloga code I am using for the model. It's same as the code written in cadence ahdlLib (except the soft output limiting part.):

`include "constants.vams"
`include "disciplines.vams"

`define PI 3.141592653

module myopamp(vin_p,vin_n,vout,vref);

    // small-signal parameters for single-poll roll-off model
    //*********************************************************
    parameter real gain=10000;
    parameter real ugf=10G;
    parameter real rin=100M;
    parameter real rout=500;

    // Large-signal parameters required for slew-rate model
    //**********************************************************
    parameter i_max = 20u;
    parameter real slewrate = 5e6;
    parameter real vsoft = 0.05;

    input vin_p, vin_n, vref;
    output vout;
    electrical vin_n, vin_p, vout, inode, vref;

    real c_pole, r_pole, gm, vin_max, vin_val;

    analog begin    
    
        @ ( initial_step or initial_step("dc") ) begin

            c_pole = i_max/slewrate;
            gm = 2*`PI*ugf*c_pole;
            r_pole = gain/gm;
            vin_max = i_max/gm;

        end

        // Input Stage:
        //*************
        I(vin_p,vin_n) <+ V(vin_p,vin_n)/rin;

        // Gain Stage:
        //*************
        vin_val = V(vin_p,vin_n); 
        if (vin_val > vin_max)
                 I(vref,inode) <+ i_max;
        else if (vin_val < -vin_max)
                 I(vref,inode) <+ -i_max;
        else
                 I(vref,inode) <+ gm*vin_val;

        // Dominant Pole:
        //*****************
        I(inode,vref) <+ ddt(c_pole*V(inode,vref));
        I(inode,vref) <+ V(inode,vref)/r_pole;

        // Output Stage:
        //***************
        I(vref,vout) <+ V(inode,vref)/rout;
        I(vout,vref) <+ V(vout,vref)/rout;
    end                                                                                                                                                  
endmodule

The code is based on the following pictorial description:

Please have a look at the code and let me know whether it is fine to take care of slew-rate factor.

 

[LvW]

In one of your earlier postings you have mentioned a value SR=8 V/us.
However, in your schematic of the opamp model I cannot see any part which models the slew rate. I didn't check the verilog code as I am not familiar with verilog.

 

[samiran_dam]

actually in the following piece of code the philosophy behind the slew-implementation is achieved:

// Gain Stage:
//*************
vin_val = V(vin_p,vin_n);
if (vin_val > vin_max)
I(vref,inode) <+ i_max;
else if (vin_val < -vin_max)
I(vref,inode) <+ -i_max;
else
I(vref,inode) <+ gm*vin_val;

Here, I ensured that whenever absolute input differential voltage value (vin_val) exceeds a certain level (decided by the maximum saturated output current i_max; i.e. vin_max = i_max/gm), output current does not follow the small signal current gain formula, rather it gets fixed to a constant value, i_max. That takes care of the slewing. The same philosophy is explained in 1978 paper by Phil Allen. [ref: "A Model for Slew-Induced Distortion in Single-Amplifier Active Filters", IEEE TRANSACTIONS ON CIRCUITS AND SYSTEMS, VOL. CAS-25, NO. 8, AUGUST 1978].

 

[LvW]

For my opinion, this sounds reasonable. And it is well confirmed by your simulated transient response with app. 2 volts per 0.2 µs (10 V/µs).

 

[samiran_dam]

Does the output amplitude limitation come from the output-swing of the opamp? And hence in-turn the input signal handling capacity will put a limit on applied input signal at filter input...right???

Added after 12 minutes:

P.S. : Here I assumed that opamp is working linearly and input is sinusoidal with small amplitude and I am concerned about high-frequency slewing.

 

[LvW]

Does the output amplitude limitation come from the output-swing of the opamp? And hence in-turn the input signal handling capacity will put a limit on applied input signal at filter input...right???

In general or in the case as shown in your figure May 14th ? (In the latter case the cause of limitation is slew rate).

 

[samiran_dam]

I think I get two things mixed - 1). due to limitation output swing of opamp, filter o/p will be saturated - that is one kind of distortion (is harmonic distortion analysis required for this?), the other is 2) Due to slew limitation sinus can be triangular at output which also causes distortion. Am I correct?

Actually my doubt was - as you mentioned on 13th may: "...for a given OUTPUT amplitude (which leads to the input amplitude if you know the gain), what did you mean by a "given OUPUT amplitued"???

One more question: When I apply an i/p to the filter should there be certain common-model level maintained within the ICMR of the opamp?