Hi All!

I have a question that I can't seem to answer on my own...

What are the practical differences between a ...

1st order non-inverting active high pass filter (aka passive RC circuit followed by a non-inverting amplifier)

and a ...

1st order inverting active high pass filter?

( I of course mean besides the obvious inverting vs non-inverting nature. )

Are there differences in their non-ideal design considerations (e.g. sensitivity to resister noise, non-ideal op-amp performance, ...)?

Cheers!

 

The non-inverting circuit has a minimum gain of 2, while the inverting circuit can have gains greater than and less than 1.

The output impedance of the signal source affects the gain of each circuit differently.

Non-inverting - input offset voltage
Inverting - input bias current

Separate from that, the first drawing is incorrect. R3 is not a shunt path for the low frequencies to keep them from reaching the amplifier input. It is C1's increase in impedance with increasing frequency, changing the attenuation ratio of the C1-R3 voltage divider, that attenuates the high frequencies relative to the low ones. This is more apparent in the 2nd schematic, where R1 is in series with C for all frequencies.

ak

 

The non-inverting circuit has a minimum gain of 2, while the inverting circuit can have gains greater than and less than 1.

How do you figure a minimum gain of 2? If R1 = ∞ and R2 = 0, the gain is 1 (i.e., the op amp is operating as a simple unity-gain buffer).

 

The output impedance of the signal source affects the gain of each circuit differently.

Non-inverting - input offset voltage
Inverting - input bias current

Can you explain this in more details? I'm not sure I follow you.

 

Min G of 1 occurs when R2 = 0, independent of R1, for the ideal OpAmp -

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https://www.electronics-tutorials.ws/opamp/opamp_3.html

Imperfect OpAmp, finite Aol, results in G < 1. Also imperfect OpAmp
would draw input current, there fore there would be a V drop across
R1, again affecting G.

Regards, Dana.

 

Hello,

The transfer function for circuit 1 is:
Vout/Vin=(s*A*C*R3)/(s*C*R3+1)
(where A=1+R2/R1)

and for circuit 2 (the lower one) is:
Vout/Vin=-(s*C*R2)/(s*C*R1+1)

We can see the AC similarities here as they are both really the same.

One of the differences though is that the first circuit takes 1 more resistor.

Another difference is that the second circuit has less output offset DC voltage. That is because the gain for DC is 1 while for the top circuit it is 1+R3/R2 so the input offset gets amplified less in the lower circuit. That could make a big difference when the circuit has to have AC high gain.

If i remember right, the second circuit has less bandwidth for low gains but it may only be a little bit different for gains even as low as 2. I'd have to look this up again.

 

If i remember right, the second circuit has less bandwidth for low gains but it may only be a little bit different for gains even as low as 2.

The difference is greater for low gains.
The bandwidth of a non-inverting opamp circuit is the GBW / Gain.
The bandwidth of an inverting opamp circuit is the GBW/ (1+Gain).

 

The difference is greater for low gains.
The bandwidth of a non-inverting opamp circuit is the GBW / Gain.
The bandwidth of an inverting opamp circuit is the GBW/ (1+Gain).

Hi,

Thanks for reminding me. So there's a small difference like i remembered, for low gains.

The DC offset is sometimes more important.

 

So there's a small difference like i remembered, for low gains.

Depends upon you you look at it.
For a fixed GBW op amp, there's a 50% difference in bandwidth between an inverting and non-inverting amplifier with a gain of 1, but there's only a 1% difference in bandwidth between an inverting and n0n-inverting amplifier with a gain of 100.

 

Depends upon you you look at it.
For a fixed GBW op amp, there's a 50% difference in bandwidth between an inverting and non-inverting amplifier with a gain of 1, but there's only a 1% difference in bandwidth between an inverting and n0n-inverting amplifier with a gain of 100.

Hi,

Yes i meant to say that there is a difference at low gains, and at high gains there is s small difference.

Case in point also assuming fixed internal gain:
Gain of 2 externally set: about 50 percent for non invert amp config.
Gain of 2 externally set, about 33 percent for invert amp config.

So sometimes it matters, sometimes not.

The DC offset will be affected in a similar manner. With low gains the DC offset wont be too bad in the non invert case, but the higher the gain the worse it gets. The invert amp should stay about the same for all gains except the settling time will probably vary, which is another interesting point to look into because of today's extensive use of ADC's.

 

The intrinsic op amp noise is also multiplied by a similar factor being [Vn * Gain] for a non-inverting amp and [Vn * (Gain +1)] for an inverting amp.

 

Ref material -

http://www.ecircuitcenter.com/Circuits/op_bandwidth1/op_bandwidth1.htm

There are lots of approximations that can get people into trouble. Take for example
a voltage follower OpAmp. At DC its commonly considered to have a G = 1. That's
an approximation, its actually less than one by a factor of 1/ [1 +1/Aol]. At AC it gets
worse because the Gain error is rising due to loss of gain.

When does it matter, 16 bit and higher A/D signal path error for example, depending
on Aol of the part. Even << 16 bits at frequency.

So buyer be aware of approximations.

https://electronics.stackexchange.c...ce=google_rich_qa&utm_campaign=google_rich_qa


Regards, Dana.

 

Take for example a voltage follower OpAmp. At DC its commonly considered to have a G = 1. That's an approximation,

Here is something to consider that rarely is spelled out in class or books. An opamp's gain is fixed. For whatever the operating conditions of *the chip* are, its gain is fixed. It is the circuit gain that is varies with external components.

Example:
Opamp open loop gain is 120 dB (1,000,000). DC voltage follower, 1 V input, 1 V output. Seems simple enough. Here is what is happening. At the opamp input stage, the 1 V signal is reduced to 1 uV (1 microvolt). *That* signal is then amplified by the opamp's internal fixed gain of 1 million to produce the 1 V output. The 1 V input is not attenuated down to 1 uV. It is reduced by subtracting 0.999,999 V from it in the differential input stage. At its heart, a differential amplifier is a subtractor, not a divider.

Negative feedback sets the gain of the overall circuit by reducing (attenuating / dividing) the amount of output signal that is available to be subtracted from the input signal. If the output is attenuated by 10:1 in the feedback components, than the output has to make just over 10 V to produce just over 1 V at the input pin so that the input 1 V is reduced to 10 uV, that is amplified by 1 million to produce a 10 V output. To danadak's point, with a perfect 1.000,000,000 V input and theoretically perfect feedback resistors, the output will not be exactly 10.000,000,000 V.

In the real world, none of this matters much of the time. However, when amplifying AC signals that are approaching an intersection with the opamp's open loop gain plot, the fact that neither the gain nor the bandwidth of the opamp are infinite (two of the assumptions in standard opamp gain equations) has very real-world consequences.

ak

Disclaimer: I'm writing this on zero sleep, so my heart is pure but my English might be crap.

 

Ref material -

http://www.ecircuitcenter.com/Circuits/op_bandwidth1/op_bandwidth1.htm

There are lots of approximations that can get people into trouble. Take for example
a voltage follower OpAmp. At DC its commonly considered to have a G = 1. That's
an approximation, its actually less than one by a factor of 1/ [1 +1/Aol]. At AC it gets
worse because the Gain error is rising due to loss of gain.

When does it matter, 16 bit and higher A/D signal path error for example, depending
on Aol of the part. Even << 16 bits at frequency.

So buyer be aware of approximations.

https://electronics.stackexchange.c...ce=google_rich_qa&utm_campaign=google_rich_qa


Regards, Dana.

Hi,

Yes there are some finer points to all this, we are just mentioning some of the more important points.

Yes, the gain of the voltage follower is:
Vout=0.9999900001*Vin

with an internal DC gain of 100000. At DC that's probably good enough for 16 bit ADC operating with a reference voltage of 5vdc.

Some of the 1GHz amplifiers i have worked with in the past have more issues to worry about than DC gain error though

 

Here is something to consider that rarely is spelled out in class or books. An opamp's gain is fixed. For whatever the operating conditions of *the chip* are, its gain is fixed. It is the circuit gain that is varies with external components.

Example:
Opamp open loop gain is 120 dB (1,000,000). DC voltage follower, 1 V input, 1 V output. Seems simple enough. Here is what is happening. At the opamp input stage, the 1 V signal is reduced to 1 uV (1 microvolt). *That* signal is then amplified by the opamp's internal fixed gain of 1 million to produce the 1 V output. The 1 V input is not attenuated down to 1 uV. It is reduced by subtracting 0.999,999 V from it in the differential input stage. At its heart, a differential amplifier is a subtractor, not a divider.

Negative feedback sets the gain of the overall circuit by reducing (attenuating / dividing) the amount of output signal that is available to be subtracted from the input signal. If the output is attenuated by 10:1 in the feedback components, than the output has to make just over 10 V to produce just over 1 V at the input pin so that the input 1 V is reduced to 10 uV, that is amplified by 1 million to produce a 10 V output. To danadak's point, with a perfect 1.000,000,000 V input and theoretically perfect feedback resistors, the output will not be exactly 10.000,000,000 V.

In the real world, none of this matters much of the time. However, when amplifying AC signals that are approaching an intersection with the opamp's open loop gain plot, the fact that neither the gain nor the bandwidth of the opamp are infinite (two of the assumptions in standard opamp gain equations) has very real-world consequences.

ak

Disclaimer: I'm writing this on zero sleep, so my heart is pure but my English might be crap.

Hi,

Yes, the gain is not exactly perfect. I calculated one such situation in the post just before this one.