I was just looking at the following preamp with tone controls on eBay.
It uses two AD827 op amps and has supposedly a gain of 10.
Because of the AD827's high input bias current, wouldn't this design have a large amount of DC offset at its outputs or do the green NP elctrolytics filter out the DC offset?
Also, wouldn't this circuit have the ole' "swishing" sound when rotating the volume pot because of DC on the wiper?
I personally like the AD827, but I've yet to find a design where someone has placed a volume pot in front of one.
It uses two AD827 op amps and has supposedly a gain of 10.
Because of the AD827's high input bias current, wouldn't this design have a large amount of DC offset at its outputs or do the green NP elctrolytics filter out the DC offset?
Also, wouldn't this circuit have the ole' "swishing" sound when rotating the volume pot because of DC on the wiper?
I personally like the AD827, but I've yet to find a design where someone has placed a volume pot in front of one.
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If the DC resistance is balanced on the +in an the -in of an opamp, then the DC offset is determined by the input offset current. That is the difference between what the +in is drawing and what the -in is drawing in current. (Opamps are designed to amplify differences between the inputs.) As long as you block DC from passing through the pot with a capacitor, there should be no rustling.
It's hard to guess what's what by that picture alone.
It's hard to guess what's what by that picture alone.
How would you balance the inverting input offset current if the noninverting input current changes because of a volume pot in front of it?
I'm not questioning your reply in any way. I'm trying to understand how you can use a higher input bias current with a volume pot in front of the noninverting input.
I'm not questioning your reply in any way. I'm trying to understand how you can use a higher input bias current with a volume pot in front of the noninverting input.
Oops I forgot the half of the point I was trying to make. 
If a capacitor is between the pot and the opamp, the opamp can't draw DC through the cap, so for the purpose of offset, it's like the pot isn't there! Win-win: it fixes rustle and offset issues. (If you plug something with DC problems directly into the pot, you would still get rustle. It's a compromise.)
If a capacitor is between the pot and the opamp, the opamp can't draw DC through the cap, so for the purpose of offset, it's like the pot isn't there! Win-win: it fixes rustle and offset issues. (If you plug something with DC problems directly into the pot, you would still get rustle. It's a compromise.)Input offset current is the difference between the two input currents. A volume pot is unlikely to change the input currents and hence the input offset current, but it may change the effective input offset voltage due to Ohm's law.ammel68 said:
A circuit would be nice. We are good, but we are not that good!
Yeah...it would be nice but it isn't posted, so I think we can forget a schematic for that circuit.
Getting back to what I'm trying to learn here...how would you modify the AD827 circuit below in order to minimize DC offset at the outputs? I put a capacitor in right after the volume pot, as barefootwhistler suggested, but not sure what value to use. I typically see smaller values like 2.2uF used.
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DF96, I have read my own thread several times and I'm not asking anyone to "redesign" the eBay circuit originally posted.
Once again...I'm asking how you can use a higher input bias current op amp as simple preamp with a gain of 2, have a volume control hung on the front of the noninverting input and have a reasonable level of DC offset at the outputs. If it isn't possible, please just say so.
Okay...I need to equalize the DC resistance seen by the 2 inputs and I need to add a capacitor to the feedback network.
Being there are many resistor and capacitor values, this doesn't tell me much.
How would I choose, or calculate, what resistors would be needed for a gain of only 2 using the AD827? I typically use values like 2K for both resistors, but I'm assuming these aren't going to be the correct values so that the -input resistance is the same as the + input.
Same question for the capacitor going to ground...no clue as to what value to try.
I even drew a circuit hoping someone would help me determine what value components to try.
Once again...I'm asking how you can use a higher input bias current op amp as simple preamp with a gain of 2, have a volume control hung on the front of the noninverting input and have a reasonable level of DC offset at the outputs. If it isn't possible, please just say so.
Okay...I need to equalize the DC resistance seen by the 2 inputs and I need to add a capacitor to the feedback network.
Being there are many resistor and capacitor values, this doesn't tell me much.
How would I choose, or calculate, what resistors would be needed for a gain of only 2 using the AD827? I typically use values like 2K for both resistors, but I'm assuming these aren't going to be the correct values so that the -input resistance is the same as the + input.
Same question for the capacitor going to ground...no clue as to what value to try.
I even drew a circuit hoping someone would help me determine what value components to try.
What exactly does a RC circuit & time constants have to do with calculating the value of the RF and RG in the circuit I drew above?
Not trying to be sarcastic, just wondering how it factors in.
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Could'a' fooled me.
The -3dB rolloff is where R=Xc. This is sometimes, but more rarely expressed in terms of the time constant, which is more commonly thought of when considering the step response.
Really?? I don't know about you, but for me it gets frustrating when I get replies that almost seem to be in some foreign language.
Apparently most of the people on this site expect that the OP must have some sort of degree in E.E. or advanced knowledge in electronics. Well...I don't. I do know some of the basics about op amps and that's about as far it goes.
Other than receiving 2 replies telling me that I need a cap in series with the volume control and a cap from a resistor to ground, which I DO understand, I haven't received a reply yet suggesting how to calculate any of the values for the components in the simple circuit I posted earlier. I'm also not looking for someone to do the math for me since I wouldn't learn anything from that.
Did a quick search for R=Xc and came up with nothing. I did learn that Xc is capacitive reactance and found a formula for calculating it.
Is the formula below what you're referring to?
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Your circuit in post #10
There are no hard and fast rules... you need to decide on the midand gain you want. Lets say you want a voltage gain of 15.
The gain is (Rf+Rg)/Rg and so you need resistors in a 14:1 ratio. i.e. if Rf were 14 ohms then Rg would be 1 ohm.
So what are suitable values. Typically we would set Rf at the 10k to 30k region. Any higher and noise contribution might start to become significant. Go lower and the opamp has to work harder to drive the feedback network and distortion rises.
So its a compromise that depends much on the opamp used.
Lets say we use a 10k. That means Rg has to be 714 ohms. We would use 680 as a nearest preferred value.
What about the cap in series with Rg. You decide the -3db point you want the amp to roll off at. Lets say 2Hz. With a 714 ohm resistor that means you need a 110uf cap. That would ensure the response was 3db down at 2Hz.
The resistor you have shown as 100k on the input needs to match Rf for minimum DC offset. This is because for bipolar opamps a steady DC bias current either flows out of or into the opamp input pin from ground. That current develops a volt drop across the 100k which causes the DC offset when the voltage at the other input pin is unequal. So to equalise them, Rf and the input bias resistor need to be the same value.
Use a FET opamp with essentially zero DC input bias current and the offset problem vanishes.
The input cap is chosen in the same way as the feedback cap. You then have two filter networks in the chain and the actual -3db point would shift from 2Hz to nearer 3Hz with the combined effect of both networks acting on the amp.
Does that help ?
There are no hard and fast rules... you need to decide on the midand gain you want. Lets say you want a voltage gain of 15.
The gain is (Rf+Rg)/Rg and so you need resistors in a 14:1 ratio. i.e. if Rf were 14 ohms then Rg would be 1 ohm.
So what are suitable values. Typically we would set Rf at the 10k to 30k region. Any higher and noise contribution might start to become significant. Go lower and the opamp has to work harder to drive the feedback network and distortion rises.
So its a compromise that depends much on the opamp used.
Lets say we use a 10k. That means Rg has to be 714 ohms. We would use 680 as a nearest preferred value.
What about the cap in series with Rg. You decide the -3db point you want the amp to roll off at. Lets say 2Hz. With a 714 ohm resistor that means you need a 110uf cap. That would ensure the response was 3db down at 2Hz.
The resistor you have shown as 100k on the input needs to match Rf for minimum DC offset. This is because for bipolar opamps a steady DC bias current either flows out of or into the opamp input pin from ground. That current develops a volt drop across the 100k which causes the DC offset when the voltage at the other input pin is unequal. So to equalise them, Rf and the input bias resistor need to be the same value.
Use a FET opamp with essentially zero DC input bias current and the offset problem vanishes.
The input cap is chosen in the same way as the feedback cap. You then have two filter networks in the chain and the actual -3db point would shift from 2Hz to nearer 3Hz with the combined effect of both networks acting on the amp.
Does that help ?
Barefoot gave the right answer in the 2nd post.
Nigel is almost right. The capacitor and the impedance that follows it form a high pass filter. The corner frequency of the filter is dependent on both values. On a preamp output you have to deal with the reality that the load might be as little as 600 ohms, so fairly large capacitor values are often used, such as 10uf.
On the input, the resistor between the input and signal ground sets the impedance, so you just do the math or use a calculator for a corner frequency below 20hz or so.
I'm not sure what sreten is on about.
As for balancing the currents, Tangent's calculator page has a good calculator for that. As the input currents go up, the values of the feedback resistors have to be more carefully chosen to balance the currents, and you will probably need some fairly specific values to limit offset. you will need the opamp datasheet.
Electronics Calculators
As the resistor values in the feedback network approach 1M, the amount of RF noise in the circuit will increase as well, so bypassing feedback resistors with double-digits of picofarads becomes helpful. Ferrite beads on inputs and outputs can be helpful. learning the difference between a faraday cage and a big freakin' antenna is helpful too.
And yeah, i am very far from an EE. I just studied hard under the likes of tangent, chu moy, prr, runeight, amb, etc, years ago.
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