Textbook Active Hi-Pass Filter Not working... what am I missing?

Thread Starter

dr.evil

Joined Aug 18, 2010
80
Gents,

Please have a look at my awesome schematics:
Screenshot from 2017-11-20 19-29-16.png
And this is the signal entering the non-inv input: y:2v/div x: 500ns/div
Screenshot from 2017-11-20 19-22-35.png
I know that since I have single supply, the negative spikes will be lost, and this is desired. But this is the output:

.... absolutely NOTHING, a complete flatliner 0V with some negligible crud on top.

Now I know the op-amp works, because if remove R3, I have a square signal of 4V p-p and it gets amplified to rail of 5.5V p-p

So what am I doing wrong here?

Evilly yours
Dr. Evil
 

MrChips

Joined Oct 2, 2009
19,912
If it is a high-pass filter, what is the expected corner frequency?
Set your generator to a low-frequency sine wave and monitor the output amplitude as the frequency is increased.
 

OBW0549

Joined Mar 2, 2015
3,135
I don't see anything inherently wrong with the circuit, other than the fact that the C1-R3 is only 5 nanoseconds and that may be too short a pulse for even a fast op amp like the AD8066 to process. But I'd expect to see something, at least, on its output anyway. If you're not seeing anything at all, I'd say check your connections.
 

Thread Starter

dr.evil

Joined Aug 18, 2010
80
If it is a high-pass filter, what is the expected corner frequency?
Set your generator to a low-frequency sine wave and monitor the output amplitude as the frequency is increased.
MrChips,

Is "corner frequency" the same as "cut-off frequency" ?

In any case I don't know, I just messed a bit with the R3 value until it produced the desired output signal. I have a measurable signal, so how would an expected corner frequency value benefit? (asking out of pure ignorance)
 

Thread Starter

dr.evil

Joined Aug 18, 2010
80
Mr. OBW0549

C1-R3 is only 5 nanoseconds and that may be too short a pulse for even a fast op amp like the AD8066 to process.
Yes, but if I change these values I won't get the desired spikes. But I do see that 5nS translates to 400V/uS and the datasheet says typical slew rate is 160V/uS, could that be it? If yes, what does a man have to do to get a 5nS pulse?

If you're not seeing anything at all, I'd say check your connections
If I remove R3, I have a square signal of 4V p-p and it gets amplified to rail of 5.5V p-p, so everything is a-ok.
 

crutschow

Joined Mar 14, 2008
24,056
Here's the LTspice simulation of your circuit.
It shows an output, so I don't understand why you are seeing nothing. :confused:
Interestingly, it shows a positive output for both edges of the input pulse, which I'm not sure is real.
(If this is being done on a plug-in breadboard, the high stray capacitances of that will seriously change the response).

upload_2017-11-20_12-40-34.png

Note that R1 and R2 are a little high for a high frequency signal due to stray capacitance at the (-) input node.
Better to change them to about 2kΩ or so.
The simulation below shows an improved response with that change.

upload_2017-11-20_12-44-36.png
 

MrChips

Joined Oct 2, 2009
19,912
MrChips,

Is "corner frequency" the same as "cut-off frequency" ?

In any case I don't know, I just messed a bit with the R3 value until it produced the desired output signal. I have a measurable signal, so how would an expected corner frequency value benefit? (asking out of pure ignorance)
Yes, corner frequency is the same as cut-off frequency.
This the frequency below which a high-pass filter will be attenuating sine waves.
 

Thread Starter

dr.evil

Joined Aug 18, 2010
80
@crutschow, thank you very much for your response.

Here's the LTspice simulation of your circuit.
It shows an output, so I don't understand why you are seeing nothing. :confused:
Interestingly, it shows a positive output for both edges of the input pulse, which I'm not sure is real.
(If this is being done on a plug-in breadboard, the high stray capacitances of that will seriously change the response).
Well, a simulation is a simulation i.e. theory. And you are right, the pulse on the falling edge is false. Anyway, it's not done on a breadboard, it's old-school tight soldering ;-)

Note that R1 and R2 are a little high for a high frequency signal due to stray capacitance at the (-) input node.
Better to change them to about 2kΩ or so.
I followed your advice and changed to 2K2, and you were right it is a capacitance issue, now at least something is getting through the op-amp, please see below, RED being the input pulse:Screenshot from 2017-11-21 08-28-45.png
 

crutschow

Joined Mar 14, 2008
24,056
Why are the two positive input pulses different amplitudes?
And why does the lower input amplitude have a higher output amplitude? :confused:
 

Thread Starter

dr.evil

Joined Aug 18, 2010
80
Why are the two positive input pulses different amplitudes?
And why does the lower input amplitude have a higher output amplitude? :confused:
I'm asking myself the same thing, I suspect it's some overtones (if I can call it that), probably need some more decoupling and grounding somewhere. I will be investigating the issue today and let you know.

Here's the full picture:

Screenshot from 2017-11-21 10-02-23.png
 

Thread Starter

dr.evil

Joined Aug 18, 2010
80
Why are the two positive input pulses different amplitudes?
And why does the lower input amplitude have a higher output amplitude? :confused:
I did some debuggin': The physical distance from square gen output to op-amp non-inv input is about 20mm (3/4 inch), so it's quite tight for a prototype. I then disconnected the op-amp non-inv input, and measured before and after C1 (see below scope dump). The square is rock steady, but after C1 all the hubbub begins and I have no idea why. I even switched to battery power to eliminate any potential noise from the power supply, but the problem persists.

Any ideas?

RED is after C1, and 0.4V/div
Screenshot from 2017-11-21 16-17-30.png
 

MrChips

Joined Oct 2, 2009
19,912
With
C = 22pF
R = 2200Ω
the cut-off frequency = 1/(2 x pi x R x C) = 3.3MHz

Your input square wave at 1MHz is below the cut-off frequency. Since this is a high-pass filter, 1MHz is in the stop-band of the filter.
What you are seeing is the derivative of the square wave, i.e. spikes on the positive and negative transitions.

Use a sine wave input and measure the attenuation Vout/Vin as a function of frequency.
 

Thread Starter

dr.evil

Joined Aug 18, 2010
80
With
C = 22pF
R = 2200Ω
the cut-off frequency = 1/(2 x pi x R x C) = 3.3MHz

Your input square wave at 1MHz is below the cut-off frequency. Since this is a high-pass filter, 1MHz is in the stop-band of the filter.
What you are seeing is the derivative of the square wave, i.e. spikes on the positive and negative transitions.
Thank you for your suggestion, but the resistor is only 220Ω, so the cut-off freq is about 330KHz which is way below the 1MHz I would say.
 

Thread Starter

dr.evil

Joined Aug 18, 2010
80
Your calculator needs batteries.
The cutoff frequency is 32.9MHz.
Let me try again, I never claimed meth err i mean math, was my core competency:

1/(2 x pi x R x C)

1/(2 * π * 220Ω * 2,2×10⁻¹¹F) = 32.899.931,56 Hz

OK, you are right, the result is 32.9MHz, but it wasn't a battery problem ;-)
 

Thread Starter

dr.evil

Joined Aug 18, 2010
80
OK, now C3 is 1nF which is 723KHz, and the scope looks like this:

RED: op-amp +IN BLUE: op-amp output
Screenshot from 2017-11-21 19-16-07.png

So how do I make it more spiky/pointy like it was before? Increase R3?
 
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