Non-inverting amp design, input resistor, input bias current etc. - Page 1
Forum section: electronics / projects

author: Plecto
author_rank: Regular Contributor
timestamp: September 27, 2013, 02:08:09 pm
content: Hi. I've been designing different types of small amplifiers (mostly headphone amps), but there is one element of the inverting amplifier design that I really don't understand. Take a look at this design:http://tangentsoft.net/audio/cmoy-tutorial/misc/cmoy-tangent-sch.pdfWhat I'm curious about is R2 and it's purpose. I've been using NE5532's as op-amps (bought a reel of 100pcs) and I know that having R2=100kOhm just doesn't seem to work, but 10kOhm works fine. It's been a while since I've tried using 100kOhm so I don't know exactly what's happening or why it doesn't work. I've also added filters in the input path of these op-amps with limited success (tried to implement a bass-boost function at some point). I'm planning on making a input control board (with a digi-pot, different equalizers etc.) and it would be really nice to implement all of this in the input path rather than pulling the feedback path through this board. In order to do this though, I really need to know what's going on when different component's are added to the input path, could someone give me some help?

author: kizzap
author_rank: Unknown
timestamp: September 27, 2013, 03:25:22 pm
content: R2 paired with C2 form a Highpass filter. Roll off frequency is roughly around 16Hz.-kizzap

author: Rufus
author_rank: Super Contributor
timestamp: September 27, 2013, 03:32:10 pm
content: Opamp inputs require bias current and so require a dc path for that current. Without R2 the bias current would just charge C2 and the opamp output ramp towards one voltage rail or the other.The resistance of the dc path and bias current produce an offset voltage which may be amplified along with the signal.The amount of bias current varies over a huge range with the type of opamp - look at datasheets.

author: Plecto
author_rank: Regular Contributor
timestamp: September 27, 2013, 03:52:12 pm
content: Thanks for the reply. Are you saying that there's a current flowing out of the input of the op-amp? When looking at the schematic for the NE5532, how is that possible?http://www.ti.com/lit/ds/symlink/ne5532.pdfThat would mean that current flows from the base of the bjt's, from the cathode to the anode of the PN-junction? Anyway, what would be the effects of having different values of R2? Am I right to say that the higher the resistance value, the higher the offset voltage (thinking of Iib*R2)? I'm not actually sure of how to properly handle the bias currentQuoteYes, this is why I want to understand the purpose of R2. I want to start using film caps instead of ceramic caps, but film caps are so much bigger in size. It would be nice to use something like a 0.1uF film cap, but that will require R2 to be quite big.

author: AlfBaz
author_rank: Super Contributor
timestamp: September 27, 2013, 04:07:00 pm
content: This link may helphttp://www.analog.com/library/analogdialogue/archives/41-08/amplifier_circuits.html

author: Rufus
author_rank: Super Contributor
timestamp: September 27, 2013, 04:30:01 pm
content: Quote from: Plecto on September 27, 2013, 03:52:12 pmCurrent flows in. If it didn't that first transistor would be completely turned off and be unable to have any effect on the rest of the circuit.The datasheet says 800nA max at 25C bias current. Through 100k that would be 80mV and 800mV at the output with a gain of 10.

author: Plecto
author_rank: Regular Contributor
timestamp: September 27, 2013, 04:57:25 pm
content: That sounds more reasonable, but what about a single sided supply and an inverting amp design with a resistor divider connected to the non-inverting input? The current will in that case flow from +V to the input so that if high value bias resistor's are chosen, the bias point will shift? I'm also wondering if this current is constant and if not, what variables will change it (other than temperature)?Is this the only thing take into account here? Find the bias current in the datasheet and assume that this current will flow into the input regardless and then do some simple calculation to find out what voltage changes the bias current leads to? Sounds pretty simple to me

author: electros6
author_rank: Regular Contributor
timestamp: September 27, 2013, 04:58:00 pm
content: The op amp you are using accept only 0.6V differential signal due to diode clamp in the input section as mentioned in the data sheet. The R2 is used for the DC return path or called biasing. When you using 100K the bias current produce the voltage drop high enough so that it also amplified along with the signal with the gain of 10 times . That's why your sircuit is not working with 100K and work with 10K.

author: Plecto
author_rank: Regular Contributor
timestamp: September 27, 2013, 07:40:34 pm
content: I'm aware of the diodes at the input, but I can't see how it matters. The op-amp will respond to the voltage created by the bias current so that the inverting input gets the same voltage as the non-inverting input. I do see how a 800mV constant DC output would have ruined some of my previous amps thoughHow about input offset current? How does that fit into the mix?

author: Rufus
author_rank: Super Contributor
timestamp: September 27, 2013, 08:13:49 pm
content: Quote from: Plecto on September 27, 2013, 04:57:25 pmIt will likely change with supply voltage and voltage on the input (usually called common mode voltage). The general idea is to design the circuit so the worst case bias current has negligible effect. FET and MOSFET input opamps can have bias currents so small they are difficult to measure.Quote from: Plecto on September 27, 2013, 07:40:34 pmInput offset current specifies the maximum difference between the bias current of each input. If you match the resistance feeding each input bias current effects tend to cancel out.

author: Plecto
author_rank: Regular Contributor
timestamp: September 27, 2013, 11:07:49 pm
content: Another question though. How can there be a bias current and an input impedance? If the bias current is more or less constant, how can there also be a resistance?

author: Plecto
author_rank: Regular Contributor
timestamp: September 27, 2013, 11:09:58 pm
content: Quote from: Plecto on September 27, 2013, 11:07:49 pm

author: dannyf
author_rank: Super Contributor
timestamp: September 28, 2013, 12:28:27 am
content: QuoteR2 is to provide the pin with DC bias.The design you linked to is actually wrong, in that it failed to pick a R2 that minimizes the imbalance caused by the bias current flowing through R2 and R3//R4: to minimize that, you should pick R2 = R3//R4.There are quite a few errors on that site and I would take anything there with a big grain of salt.

author: dannyf
author_rank: Super Contributor
timestamp: September 28, 2013, 12:31:54 am
content: QuoteWhether it flows in or out (or not at all) depends on the construction of the opamp's input stage.Quoteapples and oranges.Quote