Coupling Capacitor in microphone preamplifier causing voltage drop - Page 1
Forum section: electronics / projects

author: tapasxplore
author_rank: Contributor
timestamp: October 27, 2017, 07:39:57 pm
content: Hello everyone,I am using OPA1664 a op-amp IC from TI. I have simulated low pass sallen key topology (dual power supply). Whenever microphone capacitor is connected with the op-amp input I am getting negaliable output in uV, but if I am removing the capacitor then voltage is 100 mV which is required.Please view the attachement for more details, I am unable to insert images directly in the post.Attachment details:1) without cap2) without cap DSO (both the images shows schematic and DSO output while capacitor is not placed.3) with cap4) cap DSO (both the images shows schematic and DSO output while capacitor (C2) in placed, & hence voltage drop significantly)5) How microphone will be attached in the circuit, showing coupling capacitor also C3Please suggest any changes,Regards,

author: wraper
author_rank: Unknown
timestamp: October 27, 2017, 07:50:24 pm
content: If you use electret microphone, it needs power. Both circuits are incorrect if you connect microphone exactly as signal source on the circuit.

author: tapasxplore
author_rank: Contributor
timestamp: October 27, 2017, 07:57:27 pm
content: Hi Wraper,Yes, I will be using microphone as show by you. But in my post, I am using circuit simulating to produce output waveform using AC voltage source.

author: wraper
author_rank: Unknown
timestamp: October 27, 2017, 08:01:42 pm
content: I see that you attached mic connection pic. The issue is that you leave non inverting input completely floating. Opamp cannot work normally like this. Use some high resistance resistor (like 100k) and tie input after capacitor to ground.

author: Zero999
author_rank: Super Contributor
timestamp: October 27, 2017, 10:28:34 pm
content: As mentioned above, both inputs of the op-amp need to have a DC path to 0V, otherwise the bias currents will upset the DC operating point, causing it to behave unpredictably. It will be worse in real life than the simulator shows. According to the data sheed for the OPA1664, it has a bias current of 600nA. With the inverting input connected to 0V and the non-inverting input floating, at a higher undetermined voltage, the output will saturate at the positive rail.http://www.ti.com/lit/ds/symlink/opa1662.pdfEven with a 100k bias resistor, the output will still be biased at around 88mV, because the impedance seen at the two inputs will differ and have different voltages because of the bias currents.To keep the output as close to 0V as possible, the impedance seen at both inputs should match as closely as possible. If the biasing resistor were 133k, then the input impedance seen by the +input would be 180k1. The values of R1 and R2 could be changed to 360k, making the impedance seen by the -input 180k, which is close enough.

author: duak
author_rank: Super Contributor
timestamp: October 27, 2017, 11:02:39 pm
content: It also seems to me that on the schematic, V2 is reversed and is positive relative to circuit common.

author: wraper
author_rank: Unknown
timestamp: October 27, 2017, 11:06:18 pm
content: Quote from: duak on October 27, 2017, 11:02:39 pmIt is a negative voltage source, so fine for simulation.QuoteI did not check the datasheet. Yep, this opamp is not bias current compensated unlike most of modern opamps. So resistors should be selected accordingly.

author: vk6zgo
author_rank: Super Contributor
timestamp: October 28, 2017, 12:55:15 am
content: Quote from: wraper on October 27, 2017, 11:06:18 pmThis misuse of symbols is what I sorely hate with simulations.If the designers are either too ignorant or too lazy to put a battery symbol the right way round, it casts doubt on the credibility of the whole thing.Another classic stuff up which I see a lot is using a DC generator symbol for an ac source.It's OK to say "fine for simulation", but if beginners see this sort of thing, they may use the incorrect symbols in real schematics, causing needless confusion.

author: Zero999
author_rank: Super Contributor
timestamp: October 28, 2017, 08:44:28 am
content: Quote from: vk6zgo on October 28, 2017, 12:55:15 amI don't see the issue with inverted battery symbol and negative value. To some beginners it might make it clearer it's a negative voltage source. In real life it might not even be a battery, but the output of a mains powered supply.I wouldn't recommended powering the OPA1664 from two 1.5V batteries because it's not specified for operation below +/-1.5V and it doesn't give any headroom for the voltage dropping, as the batteries discharge. Use four AA cells or three and a rail splitter to give +/-3V or +/-2.25V, respectively.QuoteI did that with LTSpice, until fairly recently, when I discovered the AC generator symbol. I don't see the problem because it's blindingly obvious it's AC because of the text.QuoteDepends on how bad it is. I've seen J-FET and MOSFET symbols being confused, which I agree is bad. Another one of my pet hates is to just use a box symbol for something like an op-amp, especially for a dual or quad op-amp package, because it makes the circuit harder to understand.

author: tapasxplore
author_rank: Contributor
timestamp: October 28, 2017, 04:27:50 pm
content: Hi all (especially wraper, Hero999, vk6zgo),I did the simulation again with the multiple feedback topology and I am getting the output even with the microphone coupling capacitor (without placing additional resistor). Below I have attached pictures of both schematic and output waveform.Regards,Tapas

author: Zero999
author_rank: Super Contributor
timestamp: October 28, 2017, 04:53:00 pm
content: Quote from: tapasxplore on October 28, 2017, 04:27:50 pmThat looks good to me. The outputs now both have a DC path to 0V. If the remaining offset is an issue, add a resistor in series, equal to R6 + R2, with the on-inverting input.

author: tapasxplore
author_rank: Contributor
timestamp: October 28, 2017, 05:23:57 pm
content: Quote from: Hero999 on October 28, 2017, 04:53:00 pmI understand that non-inverting op-amp has DC path to 0, but how come inverting terminal has DC path to 0, as there is no ground connected directly or indirectly to inverting terminal.Regards,

author: Audioguru
author_rank: Super Contributor
timestamp: October 28, 2017, 05:29:35 pm
content: The inverting opamp amplifier in the inverting lowpass filter circuit (it is not a multiple feedback bandpass filter) has its (+) input at 0V which is its DC input reference voltage. The inverting input of the circuit has negative feedback from the output of the opamp so it does not need a bias resistor to 0V because its DC gain is 1 (its DC output is the same as its DC (+) input.

author: Zero999
author_rank: Super Contributor
timestamp: October 28, 2017, 06:06:21 pm
content: Quote from: tapasxplore on October 28, 2017, 05:23:57 pmDon't forget, for the purposes of biasing the op-amp, only the steady state, DC conditions are important. The output sits near 0V, so can be considered to be ground, at DC. The inverting input is connected to the output, via R6, in series with R2.In reality, both inputs will be within the op-amp's voltage offset rating (0.5 to 1.5mV), around 0V and the output voltage will be equal to, the bias current multiplied, by the difference in impedance seen by the inputs. The impedance seen at the non-inverting input is 0R and 18k44 is seen at the inverting input, a difference of -18k44, so given a typical bias current of 600nA, that's -18k44×600nA = -11mV.As mentioned above, the output can be taken closer to 0V by adding a resistor equal to R6 + R2 between the non-inverting input and 0V. 18k would be near enough. There's no point in getting it perfect, since the voltage and bias current offsets will dominate.

author: tapasxplore
author_rank: Contributor
timestamp: October 28, 2017, 08:01:47 pm
content: Hi,Do I have to care about input bias current for the case of OPA1664 as typical value is 600nA I think which is pretty high. I think I have to connect resistor of appropriate value in the non inverting terminal to offset offset voltage .Regards,