Low noise differentiator

[jewdai]

What is the best way to create a LOW NOISE differentiator? I am using it for a controller.

One Idea I came up with was to use a negative feedback integrator to differentiate, but the transfer function still remains the same as a regular differentiator. I was wondering what were the effects of this?


The System transfer function is

SRC/SRC+1 Where R and C correspond to the integrators components.

for those of you who dont know, the top amplifier is a summing amplifier and the bottom one is the integrator.

 

[crutschow]

Don't know that the circuit you show has any particular advantage over a standard differentiator.

A pure differentiator tends to be noisy because it's gain continues to increase with frequency (within the op amps gain-bandwidth), so you get a lot of high frequency noise. You can minimize this by rolling off the frequency response above the frequencies of interest for your application. In a standard op amp differentiator circuit this can be a resistor in series with the input capacitor and a capacitor across the feedback resistor (if each has the same RC time-constant, then the gain will roll off 6dB/octave above the rolloff point).

 

[jewdai]

well I know its going to amplify high frequency noise... but is there any benefit of using a different configuration?


an alternate design was to use a two poll system (it would essentially look like a band pass filter)

 

[Hero999]

Yes if you want to reject, low frequencies and very high frequencies then you need a band pass filter but you need to decide both the upper and lower cut-off points.

 

[BrownOut]

I don't see how your circuit makes a differentiator. A differentiator should give just SRC.

 

[crutschow]

A band-pass filter is not a differentiator.

 

[BrownOut]

In addition, U1 isn't really a summing amplifier.

 

[jewdai]

actually it is:


S/(S+a)(S+b) is a differentiator from frequencies Wc = 0-a and an integrator from frequencies Wc = b-infinity the flat band point is between Wc = a-b.


draw a bode plot and see. you need to introduce a little bit of gain to deal with A and B but thats manageable.

 

[jewdai]

well if you actually spent the time to derive the equation for the voltage out of the top op amp:
if you assume all the resistors to the V+ terminal are the same
the voltage at the V+ node is
(Vin - V+ )/ R + (Vf - V+)/R = 0

multiply by R

Vin - V+ + Vf - V+ = 0

(Vin +Vf)/2 = V+

so thats the voltage at the V+ terminal

the Op amp provides a gain of 2 (1 + Rf/Rg) to eliminate the divide by two and then your system equation is:

Vin + Vf = Vo


i think it is a summer, my PSPICE models agree with me.

 

[jewdai]

I think you are too familiar with working with inverting summers. this is a NON-INVERTING summer.

 

[crutschow]

According to my simulation, your circuit is a differentiator up to a corner frequency of about 49kHz. Above that the gain is constant at 1 (0dB). Below the corner frequency the gain drops at 6dB/octave. For example at 1Hz the gain is about -94dB (likely the noise level of the circuit). For more gain you can increase the feedback resistor of the top op amp, but that also changes the corner frequency.

I see no advantage to your circuit over that of a single op amp differentiator circuit with a resistor in series with the input capacitor to give a high frequency rolloff at the same corner frequency (which is standard practice with a differentiator).

 

[BrownOut]

Please show your bode plots. I'm still not getting it.

 

[BrownOut]

BTW, for your circuit:

V = vin + vf

Vf = -(v/SRC)

V = vin –v/SRC

V(1 + SRC) = vin

v/vin = 1/(SRC + 1)

I know I'm probably making a mistake somewhere, and destroying what little rep I have left.

 

[jewdai]

**broken link removed**



in the lower frequencies the two polls may be ignored, therefore you look at it like: S which is the initial riseing part of the graph which is a differentiator

then when the first pole turns on at A you get S/(S/a +1) which is the flat part of the graph

when the secondpole is active you get S/(s/a +1)(s/B+1) which coresponds to the decreasing slope

if you are still lost please review your S domain analysis of BODE plots.

 

[BrownOut]

Your bode plot looks like just a bandpass filter to me. A differentiator would have zero output a DC and a constant -90 degree phase angle, until parasitics kicked in.

 

[jewdai]

you might want to ask your self, what does the bode plot of a differentiator look like.... you are really frustrating. look at any standard differentiator circuit or even a high pass filter circuit, there is a cut off frequency at some point.

https://upload.wikimedia.org/wikipedia/commons/6/63/Bode_High-Pass.PNG

it acts as an differentiator in lower frequencies... the cut off frequency is when it starts turning into a flat band.

 

[jewdai]

BTW, for your circuit:

V = vin + vf

Vf = -(v/SRC)

V = vin –v/SRC

V(1 + SRC) = vin

v/vin = 1/(SRC + 1)

I know I'm probably making a mistake somewhere, and destroying what little rep I have left.

Wrong:

Vf =- Vout*(1/RSC)

Vout = Vin - Vout(1/RSC)

Vout (1 + 1/RSC) = Vin

Vout = Vin /(1+1/RSC)

multiply numerator and denominator by RSC

Vout = Vin RSC/(RSC+1)

one pole one zero

review your control system theory: the integrator applies negative feed back to the system

so you can modle the plant as K and the feed back as 1/RSC

the block diagram reduction would reduce to

plant/(1+plant*controler)

Let K equal 1

1/(1+1/RSC) multiply by RSC

RSC/(RSC+1)

 

[crutschow]

Attached is the Bode plot of the Op's circuit. It has a rising gain and a 90 phase shift until the corner frequency, thus showing it has a standard differentiator output until the corner.

 

[jewdai]

yup, I knew that and said that by the S diagram. Is there a way to make a cleaner differenciator. I understand that it boosts high frequency noise, but would that be only from the input? Is there any benefit of using the integrator configuration of an op amp instead of differenciator.

 

[crutschow]

Yes, for most applications the noise is primarily high frequency noise from the input, not the op amp. So using the integrator configuration would have no effect on that. The integrator configuration actually has two op amps in the loop instead of one, so the op amp contributed noise would likely be higher.