i did some reading on wikipedia and apparently an inverting feedback amplifier layout has lower distortion compared to a non inverting feedback loop, so assume i make something like the design below, will it work? (picture only shows 1 channel)
the first op amp will invert the signal and handle the gain where gain = 1 + R2/R1
the second op amp will have no gain and just invert the signal again
assume that the thing above actually works, do i really need R3?
following the above question, is the reduced distortion actually worth doing? will using 2 inverting feedback loops actually have more distortion compared to just 1 non inverting feedback loop?
im kind of lost here as i am quite new to diy audio, please help, any help will be appreciated
the first op amp will invert the signal and handle the gain where gain = 1 + R2/R1
the second op amp will have no gain and just invert the signal again
assume that the thing above actually works, do i really need R3?
following the above question, is the reduced distortion actually worth doing? will using 2 inverting feedback loops actually have more distortion compared to just 1 non inverting feedback loop?
im kind of lost here as i am quite new to diy audio, please help, any help will be appreciated
That equation is incorrect. The gain is -R2/R1 or if you prefer R2/R1 *-1
thanks, i just realised what's wrong with my schematics, so basically the second op amp inverts the inverted signal but Vout would be 0 since 0/R3 = 0
so what if i do something like the schematic below?
where:
R2 = R4
R1 = R3 = R2/2
gain would be 4 if i didnt mess things up right?
sidenote, what happens if i use resistors with larger resistances but remain proportional for all the numbers, like
R2 = 1000, R1 = 500
compared to
R2 = 10, R1 = 5
so what if i do something like the schematic below?
where:
R2 = R4
R1 = R3 = R2/2
gain would be 4 if i didnt mess things up right?
i dont know, i used search around the forums as well as on google and there are no conclusive answers, people did say that a lot of preamps and CD players are inverting however
sidenote, what happens if i use resistors with larger resistances but remain proportional for all the numbers, like
R2 = 1000, R1 = 500
compared to
R2 = 10, R1 = 5
You calculate the gain for each stage. Lets say R1 is 68k, R2 is 390k, R3 is 1.5k and R4 is 4.7k.
With R1 at 68k and R2 at 390k then the gain of that stage is -5.7
With R3 at 1.5k and R4 at 4.7k then the gain of the second stage is -3.1
Multiply the two together and you get the overall gain which is +17.67
Yes
With R1 at 68k and R2 at 390k then the gain of that stage is -5.7
With R3 at 1.5k and R4 at 4.7k then the gain of the second stage is -3.1
Multiply the two together and you get the overall gain which is +17.67
Yes
sorry for being away for quite some time (was busy dealing with midterm)
i see that using 10k ohms is popular (as various cmoy designs use 10k ohms) for R1 and R3
i remember reading somewhere that an ideal op amp should have infinite input impedance, so since higher is better, when is it too high (say i go crazy and get something like 2200k ohms resistors)
i see from the datasheet for a OPA2227 that it has an input voltage range of + or - 2V, is this the limiting factor preventing ultra high input impedance in op amps?
i see that using 10k ohms is popular (as various cmoy designs use 10k ohms) for R1 and R3
i remember reading somewhere that an ideal op amp should have infinite input impedance, so since higher is better, when is it too high (say i go crazy and get something like 2200k ohms resistors)
i see from the datasheet for a OPA2227 that it has an input voltage range of + or - 2V, is this the limiting factor preventing ultra high input impedance in op amps?
Do not confuse op amp input impedance with the input resistors or with the stage input impedance. They are 3 different quantities.
- op amp input impedance depends on the device you choose
- input resistor is a value you must choose
- stage input impedance is the parallel combination of input and feedback resistances but of course is dominated by the input resistor which is usually the smaller.
Op Amps are not perfect, they have a small but significant input current (that is they do not have infinite input impedance).
So this current flowing through your input resistor creates an error voltage that is also amplified along with the signal and appears at the output.
Therefore it is wise to keep the input resistance a low as practical, that is somewhere between about 2k and 10k. If you required a stage input impedance to be a certain value, to suit the signal source for example, you will need to calculate it.
Just as importantly and for the same reasons, the non-inverting input to the op amp should also have an input resistor, even though it goes to ground. The value of this resistor should equal to the parallel combination of the inverting input and the feedback resistors. This value allows the input current generated voltages at the inputs to be equal. However, at the output the 2 offsets are self cancelling (the inverted error cancels the non-inverted error).
Hope this helps
- op amp input impedance depends on the device you choose
- input resistor is a value you must choose
- stage input impedance is the parallel combination of input and feedback resistances but of course is dominated by the input resistor which is usually the smaller.
Op Amps are not perfect, they have a small but significant input current (that is they do not have infinite input impedance).
So this current flowing through your input resistor creates an error voltage that is also amplified along with the signal and appears at the output.
Therefore it is wise to keep the input resistance a low as practical, that is somewhere between about 2k and 10k. If you required a stage input impedance to be a certain value, to suit the signal source for example, you will need to calculate it.
Just as importantly and for the same reasons, the non-inverting input to the op amp should also have an input resistor, even though it goes to ground. The value of this resistor should equal to the parallel combination of the inverting input and the feedback resistors. This value allows the input current generated voltages at the inputs to be equal. However, at the output the 2 offsets are self cancelling (the inverted error cancels the non-inverted error).
Hope this helps
If you go to high on the resistors for your inverting amp drawn above then you will run into problems with loss of bandwidth and increased noise.
Not sure what you are looking at to interpret the "input voltage range" of the OPA2227. I think the -/+2 volts you mean is how near to each rail you can take the input voltage, and that's measured "open loop" with no feedback.
i see, but since resistors come in certain values only (and i only have certain values only), when i place a resistor connected to the positive input and ground and assuming i do not have the exact value for the parallel combination of R1 and R2, then should i overshoot and use a bigger value? or should i actually use the closest value possible to get the best noise canceling effect?
the example below should illustrate what i meant nicely
assume:
R1 = R3 = 5.7k ohm
R2 = R4 = 10k ohm
gain = approximately 3.1
if i use the layout on the top of the image, then
ideally the resistors to positive input should be
R5 = R6 = 57/15.7 = 3.63k ohms
but i only have 3k and 3.9k resistors, i also have 680 ohms resistors so should i use the 3.9k resistor or should i put two resistors in series for R5 and R6 so that
R5 = R6 = 3.68k ohm which is really close to 3.63k ohms?
or alternatively, could i follow the layout on the bottom of the image where
R7 = R6/2
which ideally should be 1.315k
then should i use a single 1.5k resistor or combine a 1k resistor to a 330 ohm resistor?
sidenote: is the noise generated by the resistors really that significant? and would trimpots work better since they are trimmable?
thanks in advance
EDIT: just realised i forgot about the resistors having an error rating of + or - 1%, so i guess i could actually manually pair all the resistors with a multimeter and get to the ideal value as close as possible
the example below should illustrate what i meant nicely
assume:
R1 = R3 = 5.7k ohm
R2 = R4 = 10k ohm
gain = approximately 3.1
if i use the layout on the top of the image, then
ideally the resistors to positive input should be
R5 = R6 = 57/15.7 = 3.63k ohms
but i only have 3k and 3.9k resistors, i also have 680 ohms resistors so should i use the 3.9k resistor or should i put two resistors in series for R5 and R6 so that
R5 = R6 = 3.68k ohm which is really close to 3.63k ohms?
or alternatively, could i follow the layout on the bottom of the image where
R7 = R6/2
which ideally should be 1.315k
then should i use a single 1.5k resistor or combine a 1k resistor to a 330 ohm resistor?
sidenote: is the noise generated by the resistors really that significant? and would trimpots work better since they are trimmable?
thanks in advance
EDIT: just realised i forgot about the resistors having an error rating of + or - 1%, so i guess i could actually manually pair all the resistors with a multimeter and get to the ideal value as close as possible
Last edited:
Top circuit- R5 = 3k9 would be fine, or you might consider simply using 2 resistors in parallel (same values as R1 and R2 of course) which will be very close to 3k63 and therefore give the lowest error.
Bottom image/circuit is not something I would recommend.
Recommend you avoid using trimpots - they can be unreliable in very low current signal paths like this.
Resistor noise is unlikely to be measurable with these value resistors and op amp gains. Obviously there are occasions with very high gain circuits that all noise sources need to be considered with usually high complexity solutions.
Bottom image/circuit is not something I would recommend.
Recommend you avoid using trimpots - they can be unreliable in very low current signal paths like this.
Resistor noise is unlikely to be measurable with these value resistors and op amp gains. Obviously there are occasions with very high gain circuits that all noise sources need to be considered with usually high complexity solutions.
i see
one last question just out of curiosity, assume that i am actually building the amplifier and decided to use 2 dual op amp chips (such as say opa2134/2227 instead of 4 opa134/227s)
lets call the 2 dual op amps X and Y, X1 meaning the first channel of X and X2 meaning the second channel of X and etc
for my left and right input channels, should i have it such that:
each channel for the op amp is used for the same stage for both input channel as shown
or should it be that each op amp is used for a separated input channel with each channel for the op amp being used for two different stages of a single channel as shown?
it seems to me the method on top would allow me to use different op amps for X and Y for a unique mixed sound (such as using AD8620 for X and OPA2227 for Y) whilst the method on the bottom would reduce crossfeed issues sometimes related to dual op amp chips
one last question just out of curiosity, assume that i am actually building the amplifier and decided to use 2 dual op amp chips (such as say opa2134/2227 instead of 4 opa134/227s)
lets call the 2 dual op amps X and Y, X1 meaning the first channel of X and X2 meaning the second channel of X and etc
for my left and right input channels, should i have it such that:
each channel for the op amp is used for the same stage for both input channel as shown
or should it be that each op amp is used for a separated input channel with each channel for the op amp being used for two different stages of a single channel as shown?
it seems to me the method on top would allow me to use different op amps for X and Y for a unique mixed sound (such as using AD8620 for X and OPA2227 for Y) whilst the method on the bottom would reduce crossfeed issues sometimes related to dual op amp chips
Last edited:
We tend to design so that the op amps are not mixed across channels. Also makes physical separation of the layout easier if it was a large amp.
That said, I am not sure that I could pick the difference, either by measurement or listening, between mixed and not-mixed. Intuitively I guess cross talk noise and distortions reduce.
That said, I am not sure that I could pick the difference, either by measurement or listening, between mixed and not-mixed. Intuitively I guess cross talk noise and distortions reduce.
I'm happy to use one IC for both channels. You will get far more interaction from wiring and PCB print issues than from the opamps. You don't need the resistors connected to the non inverting inputs, certainly not with FET devices. All they do is raise the impedance and create a possible point for noise pickup. Link them out.
i thought the resistors connected to the non inverting inputs are supposed to cancel out potential noise generated by the input resistors and feedback resistors at the inverting inputs
now im confused
Here's how it works... simplified
for bjt (transistor input) opamps there is a small but significant current that flows into or out of the two input pins. Its the opamp input bias current. That current develops a small volt drop across any resistors that connect to those pins. If the resistance seen by the two inputs is unequal then the input pins then have an unequal voltage between them. So in order for the complete opamp (with its feedback network) to work the output of the opamp must assume a voltage that will push or pull current through the feedback network to cancel that "input offset" out. The result of that is that the opamp out has a small DC offset voltage. For these kind of opamps we can minimise that by ensuring each input sees a similar resistance. Similar resistance... the error is the same for both inputs and cancels out.
FET opamps (such as the OPAxxxx) have zero bias current and so there is no possibility of DC offset occurring due to this input offset current that affects bjts. So there is no reason to have that extra resistance. All the resistor does is raise the impedance (how it looks at AC) of the pin and makes it more prone to noise pickup.
for bjt (transistor input) opamps there is a small but significant current that flows into or out of the two input pins. Its the opamp input bias current. That current develops a small volt drop across any resistors that connect to those pins. If the resistance seen by the two inputs is unequal then the input pins then have an unequal voltage between them. So in order for the complete opamp (with its feedback network) to work the output of the opamp must assume a voltage that will push or pull current through the feedback network to cancel that "input offset" out. The result of that is that the opamp out has a small DC offset voltage. For these kind of opamps we can minimise that by ensuring each input sees a similar resistance. Similar resistance... the error is the same for both inputs and cancels out.
FET opamps (such as the OPAxxxx) have zero bias current and so there is no possibility of DC offset occurring due to this input offset current that affects bjts. So there is no reason to have that extra resistance. All the resistor does is raise the impedance (how it looks at AC) of the pin and makes it more prone to noise pickup.
inverting op amp connection only "fixes" one of the least important distortion mechanisms
simply using low, matching Z for both input and feedback already reduces most op amp common mode input nonlinearities in noninverting circuits
and sometimes you really want the high input Z of noninverting circuits
there is still i_b with fets - just very low, even CMOS op amps need DC path to both inputs to make up leakage currents
simply using low, matching Z for both input and feedback already reduces most op amp common mode input nonlinearities in noninverting circuits
and sometimes you really want the high input Z of noninverting circuits
there is still i_b with fets - just very low, even CMOS op amps need DC path to both inputs to make up leakage currents
Last edited:
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.