low(est) frequency amplifier

[jazzalex]

Hello all,

I am looking for a circuit which allows amplification for lowest possible frequencies. I have been experimenting with the attached OP-amp based circuit which so far has the characteristics as also attached. You can see that it works well starting a 0.09 Hz, however, I would like to get rid of the stop band below as well - ideally starting at DC if (possible at all).

I am a newbie in this domain and would appreciate any hint.

Thanks a lot in advance,
best

Alex

[Kleinstein]

One can do amplification from DC on, just build the circuit without capacitors (except for supply decoupling). The OPs usually need a feedback network to set the gain - so it takes more than just the OP. At high gain the Offset of the amplifier can become a problem and may need adjustment.

The TL071 is not a good choice for DC or very low frequencies, as it has quite a lot of temperature drift and 1/f noise. If the signal source allows (not to high in impedance) I would prefer a BJT based OP like the OP07.

[pwlps]

Your circuit is strange : opamps are intended to be used with a feedback path controlling the gain.

You have two basic configurations :
inverting amplifier : https://www.electronics-tutorials.ws/opamp/opamp_2.html
non-inverting amplifier: https://www.electronics-tutorials.ws/opamp/opamp_3.html

both working down to DC.

If you want a high gain and good precision then choose a "precision" opamp with low offset voltage (for very low frequencies you can use zero-drift opamps featuring an automatic offset correction).

https://www.ti.com/amplifier-circuit/op-amps/precision/products.html
https://www.analog.com/en/products/amplifiers/operational-amplifiers/precision-op-amps-vos-1mv-tcvos-2uv-c.html

[jazzalex]

Allright, thanks a lot for the replies ! I will modify the circuit asap but one question so far remains: Can I use the TL071 as a proof of concept and then later change it to something else or is it completely unusable ?

Thanks,
best

Alex

[james_s]

Yes it will work fine as proof of concept, all op-amps do roughly the same thing.

[TimFox]

As pointed out above, there are two reasons why the circuit as drawn will not work.
1.  Without negative feedback from output to “-“ input, the finite input offset voltage will force the output to either extreme value (roughly the positive or negative supply voltage.
2.  There is no DC path for the “-“ input bias current.  Even a FET input (e.g. TL071) has a finite bias current.  This current will force that input pin to an extreme voltage after it charges the capacitor.
When choosing a part for the correct circuit, the choice between FET or BJT device depends on the source impedance driving the op-amp.  High source impedance requires the low input current of the JFET, but low source impedance will give better noise with the lower voltage noise of the BJT.

[jazzalex]

According to

https://www.electronics-tutorials.ws/opamp/opamp_2.html

I have now implemented the feedback loop and obviously can change the gain by changing the involved resistors - fine !

However, one more question arises:

My intention is to generate a steep slope at the moment when my gaussian signal *starts* rising. Right now the amplification starts when the input has reached - 0.5 V and now I wonder if it's possible to shift this threshold towards lower values (basically as early as possible).

Thanks,
best

Alex

[Kleinstein]

The curve showed a relative high amplification and with the relatively large input signal the amplifier runs into saturation on both ends. One can add an DC offset and this way shift the region where the amplifier is in the linear region.

[TimFox]

You did not post your circuit, with the feedback added, but it is not working.  A proper DC amplifier has a constant gain through zero input voltage, on both sides.  I suspect that you are using a single supply in the incorrect fashion.  Try split power supplies with the input connected at one end to the center (common) of the power supply.  Look at the standard references (or Google) about common-mode input and output range.

[jazzalex]

Just in case I reimplemented it with a non-inverting amp. Please see thee attachments for better clarification:

1.) actual circuit 
2.) simulated amplification via qucs
3.) current result

The behavior is the same as in the non-inverting case. Indeed - this is heavily saturated but this is fine for me. My current goal is to achieve such a steep slope at the beginning of the gaussian puls.

@Kleinstein: So, you think a DC offset is needed to achieve this or does my approach have a conceptual problem ? 
@TimFox: I am already using two power supplies - one +16V, the other -16V.

[jazzalex]

Hello again,

watching my own post I realized that the slope occurs once the pulse has reached 0 V and now I understand what Kleinstein wanted to say: a DC offset of the input signal. After having applied this the pulse starts at 0 V the the amp follows respectively (see attachment).

Thanks a lot for your help - I might probably start another thread based on this little experiment.

Best

Alex

[TimFox]

Your posted circuit has a gain of +21 V/V, but your input pulse goes from -4V to +4V, which would imply -84V to +84V output.  Since you have +/- 16V supplies, the amplifier is in heavy saturation.  If I read your drawing correctly, you are loading the output with 60 ohms, so the actual voltage is determined by the output current limiting.  The TL071 limits at approximately 10mA, and the feedback is defeated when the current limit is active.

[jazzalex]

Hej Tim,

I applied the DC offset and now the pulse goes from 0 V to 2 V and the OP amp ranges between +16 V and -16 V (just added an attachment to the last post). For now this is fine for me – it's just a little experiment and most likely this was not my last question related to it :-)

Thanks again and best

Alex

[TimFox]

If you don't want linear amplification, you can use an op amp with non-linear feedback (e.g., Zener diodes) and overdrive the amplifier.  Are you trying to obtain linear amplification?  Is there really a 60 ohm load on the device?

[jazzalex]

The circuit I sent was just a drawing I did within the qucs software simulator and for the simulation it requires a *theoretical* 60 Ohm Input. In real life my input is the output of a Picoscope AWG – so basically a conventional signal generator I don't know the impedance of.

In fact I don't care for linear amplification at all. The main purpose of my experiment is the generation of a steep slop at the very first beginning of a comparably long (7 seconds) Gauss puls. Don't ask me for details yet - it is just the brainstorming process of a computer science project (I am a media computer scientist with limited EE knowledge).

That being said the purpose of this thread is fulfilled - thanks to anyone who helped !

Regarding further upcoming questions shall I create a new thread or keep this one ?

These are:

1.) Right now I am consciously saturating the amp but what is your favorite alternative recommendation to solve the issue mentioned above more effectively ?

2.) More generally: Which OP amps provide lowest possible noise while still being affordable ?

Best

Alex

[TimFox]

Your circuit drawing shows a 60 ohm generator connected to the output—that must be wrong.  Your picoscope input is not a generator, and probably has a higher impedance (normal oscilloscope input impedance is 1 megohm).
I would suggest using diodes in parallel with your feedback resistor (start with 1N4148s back-to-back across the resistor).  It’s not a good idea to force the op amp into saturation, since the output stage may take a surprisingly long time to recover.
If your source impedance is 50 or 60 ohms, then a BJT input device will have lower noise, but the 5000 ohm resistor will dominate the noise and drift calculation (until the diodes conduct).  When the output voltage is close to zero, the voltage across the diodes will not result in diode current, and the circuit will have high voltage gain.  With “high” output voltage, one diode conducts and the “incremental” gain goes to unity, until the input reaches approximately the maximum allowable output voltage minus the diode drop, above which the amplifier will saturate.
If your source impedance is really close to 50 ohms, the circuit would have lower noise if you reduce both resistors in the network, reducing the total resistance seen by the amplifier (before diode conduction).

[jazzalex]

I just double checked:

>>Your circuit drawing shows a 60 ohm generator connected to the output—that must be wrong.

This is just a way to implement complete circuit analysis via the software emulator qucs. It must be attached to in- and output and compares in- vs. output over frequency.

>> Your picoscope input is not a generator, and probably has a higher impedance (normal oscilloscope input impedance is 1 megohm).

The picoscope also has an AWG with 600 ohm impedance but I also have another AWG with 50 ohm impedance. 

[TimFox]

I am not familiar with qucs.  What does the 60 ohm resistor shown at the output represent?
Going back to my diode feedback suggestion (to keep the amplifier itself from saturation), if you start with, say, 4V maximum input signal, you can reduce the resistance from the inverting input to ground to keep the feedback current below the amplifier’s maximum output current, change the feedback resistance to obtain your desired gain near zero input, and connect the feedback diodes you need to limit the output voltage, remembering that the diode forward voltage adds to the input voltage for the non-inverting amplifier.
The 50 ohm and 600 ohm AWG values you quote are presumably the generator output impedances, but not the ‘scope input impedance.

[pwlps]

In fact I don't care for linear amplification at all. The main purpose of my experiment is the generation of a steep slop at the very first beginning of a comparably long (7 seconds) Gauss puls.

It's not to be pedantic about the title of your post, but it might be interesting for you to note :

1) if you have a pulse it is not "low frequency" anymore. The pulse contains a broad spectrum of frequencies, the steeper it is the more high frequencies it contains. For an opamp to handle it the first relevant parameter is the gain-bandwidth product. The TL071 has a GBP of 3MHz which means that with a gain of 20 the bandwidth will be 3MHz/20.
2) the slope will be limited by another opamp parameter, the slew rate. For TL071 it is 13V/microsecond, whatever the configuration the output voltage cannot change any faster. 
3)  it seems you are using the opamp as a comparator.  There are specialized ic's for this, comparators are similar to opamps but designed for very fast switching.  You don't use any feedback with comparators as they are designed to work saturated.

[jazzalex]

>>1) if you have a pulse it is not "low frequency" anymore.

I know what you mean but in my case with a Gaussian signal of 7 s the slope is very flat and in turn in involved frequencies accordingly low. However, it's not a pure low-frequency sine for sure.


>>2) the slope will be limited by another opamp parameter, the slew rate. For TL071 it is 13V/microsecond, whatever the configuration the
>>output voltage cannot change any faster.

Very valuable - thanks !


>>3)  it seems you are using the opamp as a comparator.  There are specialized ic's for this, comparators are similar to opamps but
>>designed for very fast switching.  You don't use any feedback with comparators as they are designed to work saturated.

Completely right - this is how I am using it and indeed in the meantime I have already removed the feedback resistor because it lead to less amplification.


Puh - this is all very useful in my current brainstorming ! Thanks again to all - I might come up with another therad at some point.

Best

Alex

[jazzalex]

>>I am not familiar with qucs.  What does the 60 ohm resistor shown at the output represent?

It's a tool for analysis – basically a software network analyzer which eventually gives info in terms of amplitude, phase etc. for the observed frequencies.

>>The 50 ohm and 600 ohm AWG values you quote are presumably the generator output impedances, but not the ‘scope input impedance.

Right, these are the outputs of the AWG. The scope itself has an input impedance of 1 MegOhm.

[TimFox]

Does the analysis present a 60 ohm load resistance applied to the amplifier?  This would be normal for a radio-frequency network analyzer, but makes a big difference to the performance of an op amp.

[Vovk_Z]

Completely right - this is how I am using it and indeed in the meantime I have already removed the feedback resistor because it lead to less amplification.

But we don't use opamp without feedback. It will be completely saturated and won't amplify a signal at all.

[pwlps]

Completely right - this is how I am using it and indeed in the meantime I have already removed the feedback resistor because it lead to less amplification.

But we don't use opamp without feedback. It will be completely saturated and won't amplify a signal at all.

I think he already got the point.  He uses an opamp as a comparator which is a sort of sacrilege but it's ok for testing the ideas.

[jazzalex]

Does the analysis present a 60 ohm load resistance applied to the amplifier? 

Yes, I think this is the case. At least it works for now and analyses the actual circuit well.

This would be normal for a radio-frequency network analyzer, but makes a big difference to the performance of an op amp.

I have been using it for any kind of simulation from DC up to GHz but I am not familiar with the technical details of the implementation. They might be documented here: http://qucs.sourceforge.net/

Best

Alex