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Help needed! op amp current and voltage offset

Discussion in 'DIY' started by from lotus, Jan 19, 2018.

  1. from lotus

    from lotus New Member

    Messages:
    19
    Location:
    No. Ca
    Ok I'm new to all this. i figured out how to bias a transistor, doing ok with R/C/ circuits and filtering.
    doing ok building transistor pre amps for guitar, clean and distorted.
    PROBLEM, OP AMP
    i cant figure out the offset voltage/current
    the best i can tell is. no resistor connecting the + or - power supplys to the op amp
    i understand the neg feed back and the gain ratio. just not sure if i shouid use 5K and 50K or 15K and 150 K for a gain 10
    and explain the voltage divider for single power supply. i know how to make a divider, just not sure where to put it in the circuit.
     

     

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  2. TerryS

    TerryS AK Subscriber Subscriber

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    The input offset voltage / current is simply accounting for some non-ideal characteristics of real world op amps. If the op amp was ideal, then no current would flow into or out of the input pins (the +/- inputs), and the output would be zero volts for zero volt inputs. In this case, the 5k and 50k would act like the 15k and 150k. But in the real world, there is always some (usually very small) current that flows either into or out of the input pins. This is the input bias current. It is usually approximately equal on the two pins. Since the op amp amplifies the voltage difference between the two pins, you want to keep the voltage at the two pins identical (except of course for the signal you wish to amplify. So you want to try to equalize the resistance at the two input pins so that the voltage developed across the input resistors (caused by the input bias current) is equal. To take an example, if you use the 5k and 50k resistors connected to the - input to set the gain, then the - input 'ses' a resistance of 4.545k (the two resistors are effectively in parallel to the bias current). So instead of just grounding the + input, you would add a 4.545k resistor to ground so the the input voltages due to the bias current would be equal.
    That's the theory. In practice, it might not matter. That is because the input bias current is usually pretty small to begin with, so if your circuit is not particularly sensitive to D.C. offsets at the output, it might not matter and grounding the + input would be just fine. Also, even if you make the two input voltages the same by ensuring the two resistances are the same, the output still will not be at zero volts D.C. That is because again the real world interferes and even with the two inputs exactly the same, there will still be a D.C. output due to non-ideal op amp behavior. It should be pretty small, and hopefully your circuit can tolerate a small D.C. offset at the output, otherwise you will need extra circuitry to null the D.C. offset.
    Another reason to care about the 5k 50k or 15k 150k decision is the input impedance of the circuit. When the op amp is working correctly, the - input will be at the same voltage as the + input (the feedback ensures this). This means that if there is zero volts at the + input, there is zero volts at the -input. So the input impedance (resistance in this case) is either 5k Ohms or 15k Ohms depending on which you chose. That might be important to you depending on what the circuit driving the - input looks like.
    The last reason is noise. Smaller resistors contribute less noise. For noise critical applications (phone pre-amp) use the smallest resistance values you can tolerate in your circuit. Of course smaller resistances mean more current, so there is always a trade-off.

    As far as the divider for a single supply, there are a couple of ways it can be used. One is to 'create' a dual supply from a single supply voltage by making a virtual ground. So a single 9 volt batter can become two (+4.5V and -4.5V) supplies. The other way is to use the divider to set the voltage to the middle of a single supply to center the signals between the supply voltage and ground (Vcc/2). Op amps are happiest when the input signals stay near the center of the two supply rails as opposed to near one rail or the other.

    Here is a pretty good discussion of the first approach (making a split supply from a single supply).
    https://tangentsoft.net/elec/vgrounds.html

    This shows the second approach
    http://www.swarthmore.edu/NatSci/echeeve1/Ref/SingleSupply/SingleSupply.html

    Which way you go is going to depend on the rest of the circuit (what is driving the op amp and what it is driving). It seems like a pain, but it is often easier to just use two supplies (+ and -) and avoid the issue altogether.
     
  3. from lotus

    from lotus New Member

    Messages:
    19
    Location:
    No. Ca
    So if I use a 10k and a 180k I would put a 9.5k from noninverting terminal to common.
    Looking at the img. Is this correct
    C1- blocks DC and allows higher Hz’s to pass
    R1 and R2 adjust gain
    R3 = parallel of R1 and R2. (R3 is to offset current / voltage).
    C2 and C3 allows higher Hz’s to pass so they are not amplified.
    C5 blocks DC and lower Hz’s
    C4 bypasses higher Hz’s to ground R4 determines how much of that get by passed to ground.
    Using the LM 1875
    Offset voltage; +-1 typical +- 15 tested mV
    Input bias current; +-.2 typ +-2 tested uA
    Input offset current; +-.05 tested uA
     

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  4. from lotus

    from lotus New Member

    Messages:
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    do i add voltage to the non-inverting pin. if i just equalize the resistance why all the fuss with the data info on offset's ect.
     
  5. TerryS

    TerryS AK Subscriber Subscriber

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    Sorry, I should have made it clear that it is the D.C. impedance that matters to the offset. So in your drawing, R1 is not in the D.C. path since C3 blocks the D.C. current. So the - input 'sees' 180k (R2).
    In this case, the D.C. gain of the op amp is unity, so the input offset is not much of a concern. The input offset is multiplied by the D.C. gain of the amp and that is the value that appears at the output. But since C3 is there, the D.C. gain is 1. So the output offset will be the same as the input offset, which is usually a few milliVolts at most.

    Most of your assumptions are correct.
    C1 blocks DC and allows D.C. to pass.
    R1 and R2 adjust the A.C. gain.
    R3 should be 180k (same as R2) and should be connected to ground (C3 is not required).
    C5 needs to move as you show. If it stays where you show it now, the D.C. gain of the op amp would be near infinite (same as the open loop gain of the op amp at D.C.). This means the op amp output would sit on one rail or the other. Not good.

    I didn't realize we were talking about a LM1875. That is a bit of a special case. Most op amps could not drive a speaker. They just don't have enough output current capability. The LM1875 is different. It has a beefy output drive capability.
    There is a very good circuit shown in the typical applications drawing of the T.I. data sheet.
    http://www.ti.com/lit/ds/symlink/lm1875.pdf
    You see that C5 isn't used. This is because the output offset voltage would be very nearly zero as I state above. Other than that, it mirrors your circuit pretty well. You notice that their R2 is about equal to R4. I don't know why they are not exactly equal (20k vs 22k), but the small difference isn't significant.

    The R5 and C5 on the output of their circuit just improves high frequency stability which can be an issue when driving speakers, which look inductive at high frequencies. The goal is to prevent oscillations.
    Note that if you did include C5 in your circuit, (moved to the proper location of course), it would work. This is commonly done as a D.C. block to keep the output D.C. offset from driving the speaker. As mentioned, you don't need this since your circuit has low D.C. gain so not enough offset to worry about. But if you did include it, it affects the low frequency response. If the speaker is 8 Ohms, the output to the speaker will start to be attenuated at frequencies below where the impedance of the cap starts to equal the speaker impedance (8 Ohms). So if you wanted the low frequency response to be at 20Hz or below, the capacitance would have to be very large. At least 1000uF, and would need to be non-polar since you don't know which polarity the output would be at D.C. So better off leaving it out.
    Pay attention to the power supply decoupling like they suggest in the data sheet. Since it is supplying lots of current, the decoupling is important.
    And the Virtual ground circuit I mentioned in the earlier post is not usually a good option for a higher current application like this. It can be done, but it requires some care to make sure it is capable of maintaining the virual ground under high current conditions.
     
  6. TerryS

    TerryS AK Subscriber Subscriber

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    You would only have to worry about it if your circuit had very high gain at D.C. (yours is a gain of 1, or no gain), or if your load was very sensitive to D.C. voltage so that a few millivolts would matter. The speaker won't care. If you need to, there are ways to 'null' the D.C. offset. Adding a voltage at the + input is one way that would work, but there are other ways that are probably easier.
     

     

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  7. TerryS

    TerryS AK Subscriber Subscriber

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    Sometimes the D.C. offset spec matters a lot. Lets say you wanted the response of your amp to extend all the way down to D.C. Then you would leave out C3 and the D.C. gain woul be be the same as the A.C. gain (18 times). Then the D.C. offset would be multiplied by the gain. It could be as high as 15mV times 18 = 270mV. That might start to be a concern (still probably OK driving a speaker), so that's why they spec it. Sometimes it matters. In a high gain circuit like a phono pre amp, it gets to be an issue pretty fast.
     
  8. from lotus

    from lotus New Member

    Messages:
    19
    Location:
    No. Ca
    thanks for the help. i read slow and type even slower. give me some time to sort this out
    I've built 15-20 guitar pre amp transistor circuits got to the point where most of the ones i built worked.
    some with 2 and 3 transistors. haven't used inductors yet. got to go my brain is going onto fri mood.
     
  9. from lotus

    from lotus New Member

    Messages:
    19
    Location:
    No. Ca
    To take an example, if you use the 5k and 50k resistors connected to the - input to set the gain, then the - input 'ses' a resistance of 4.545k (the two resistors are effectively in parallel to the bias current). So instead of just grounding the + input, you would add a 4.545k resistor to ground so the the input voltages due to the bias current would be equal.

    Sorry, I should have made it clear that it is the D.C. impedance that matters to the offset. So in your drawing, R1 is not in the D.C. path since C3 blocks the D.C. current. So the - input 'sees' 180k (R2).

    is this the difference of not having C3 and having C3
    -------------

    R3 should be 180k (same as R2) and should be connected to ground (C3 is not required).
    did you mean C2 not required?
     
  10. TerryS

    TerryS AK Subscriber Subscriber

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    C3 is there to reduce the gain of the circuit at D.C. Having C3 there means the gain at very low frequencies is reduced, and at D.C. the gain is 1. This eliminates concerns with the D.C. offset due to the op amp input offsets. The gain remains 1+(R2/R1) at higher frequencies, where C3 acts like a short.
    Since C3 blocks the D.C. current, it means R3 should equal R2. No D.C. current can flow through R1, so it is not 'in the circuit' at D.C. where the effects of D.C. offset matter.
    Yes, I should have said C2 is not required.
     
  11. from lotus

    from lotus New Member

    Messages:
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    Location:
    No. Ca
    wouldn't C2 let high pass to ground and not be amplified? do long guitar cords pick high Hz's. or maybe a pre amp could have noise?
    I will bring in the batteries in (car batteries) and try to build a circuit.
    with all the amps available from the batteries, should i use some resistors to limit the current available to the circuit.

    whats the diff with a quote and a reply

    Thank for the help
     

     

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  12. TerryS

    TerryS AK Subscriber Subscriber

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    Yes C2 would tend to have a low pass effect on the circuit, although with R3 in place and no resistance in series with C1, the effect is pretty small and unpredictable. If you want a low pass, you need a series resistance (in series with the input, like C1 is) and a shunt C (like C2, but without R3). But we were talking about balancing the op amp input bias current. For R3 to have any effect on the D.C. bias voltage, C2 can't be there. It blocks the D.C. current.

    I would be pretty nervous using car batteries to do experimentation. They have the ability to put out dangerous amounts of current. You could limit the current with resistors, but if you do, you need to add large amounts of filter capacitance after the resistors to give the supply low impedance to A.C. Much better to use a couple of 9 volt batteries instead. Better yet, get a small power supply for experimentation.
     
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  13. fred soop

    fred soop Super Member

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    I am in the process of doing some preamp redesign and building using LM4562 op amps. There was some thought of eliminating most of the coupling capacitors by balancing the inputs to eliminate the offset. Unfortunately, component tolerances make that essentially impossible and any DC on intermediate pots or switches will produce noise, so the intermediate capacitors remain. However, modern op amps have an extremely low input current. Today, I set up a simple experiment with the LM4562 using a 47K input resistor with one end grounded. Feedback was via another 47K resistor (gain of 1). Then the feedback resistor was replaced with a direct wire connection. The difference in offset voltage at the output was 0.1 mV. For any audio purposes, this can be ignored and the balancing of input resistors is totally unnecessary.
     
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  14. from lotus

    from lotus New Member

    Messages:
    19
    Location:
    No. Ca
    Thanks for the help

    i made 2) 12 volt battery pack from AA batteries.

    got the Lm1875 working with an MPSA13 transistor pre stage. with both a single and a duel power supply.:beerchug:

    Getting ready to order parts., help me if you can this is what i'm thinking.
    Transistors; MPSA-14,18,63,64,65... BC-108,327,337,547,557... 2N-2219,2222,2907,3904,3906,4401,4403,5457... PN2369,... TIP-41,42
    op amps; TDA-2009,2050... NE5532... lm1875... opa2604... 1458... 4558

    more interested in the transistors than the op amps.

    a little more work and late night studying and maybe *probably* some more burned up parts
    than i can move on to the tubes.
    maybe i should post this in a new post.
     
  15. TerryS

    TerryS AK Subscriber Subscriber

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    Sounds like you've made some good progress. Without knowing more about what you are doing, I can't be any help with transistor selection. To tell the truth, I usually just grap what I've got handy. Most audio applications aren't that finicky about transistor types unless you get into low noise applications (like phono preamps or microphone amps). The MPSA-14 and 3904/39066 cover most applications at low voltages.
    Probably best to start a new thread and describe the project. SOmeone with experience in that will likely jump in.
     

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