DC Biasing and blocking... I need re-education. - Page 1
Forum section: electronics / projects

author: Phu
author_rank: Newbie
timestamp: March 27, 2021, 04:04:45 pm
content: Hi,Long time subscriber to the EEVBlog YouTube channel, first time user of the forums, so please be gentleConsider the following circuit (drawn by my fair hand) that is part of a bigger circuit I'm working on. The principle here is to apply a digitally controlled pre-mix volume control to an audio signal (actually a pair of signals, so you can considered this duplicated for each channel). IC2 is a MAX5410EEE SPI-controlled 10K logarithmic digital potentiometer with zero-crossing detection. IC1 is a TLE2426 virtual ground - essentially a potential divider, but with buffering so the resulting output is (is as much as is relevant) is electrically isolated. The output is 0.5 x ( VCC - GND ), so in this case 2.5V. Initially, pretend R3 isn't in the circuit. I've drawn in the downstream circuit for completeness in case it's relevant - essentially this circuit is one of 16 channels, which are all summed using a basic op-amp circuit.The principal is as follows: The digitpot can't take an AGND lower than 0V, or more accurately, lower than DGND. So my idea was simple: biasing it up to 2.5V using a potential divider, then use the TLE2426 to supply the "ground" at the bottom of the pot, and then capacitively decouple the output to remove the DC bias.And here's where my knowledge seems to collapseFirst, C1 and C2 were 22nF. This apparently is fantastically too low, and blocked pretty much all audio frequencies - like literally the right hand end of C1 and C2 was always a solid 2.5V. A cursory glance of the web shows I need to be aiming for 10µF, but the only ones of those are electrolytic. Fine, they'll do for testing. Initially I had then oriented with - to the left and + to the right. Progress! Now signal goes into the pot, and comes out, but now there is no signal on the right side of C2.It seems the capacitors are acting a little like diodes - if I reverse the polarity of C2, I get a signal at the point marked with a ?. But, it's still centred around 2.5V! In order toactuallyremove the DC bias, it seems I need R3 in place to "bleed" the DC bias to ground! But then....The sound output sounded clippy... and indeed, on the oscilloscope I can see that anything below 0V is absent. More diodish behaviour from the capacitors!So, please tell me what I'm doing wrong here. My basic questions consist of:1) Am I right that the electrolytics are acting like diodes and clipping the bottom have of the waveform?2) Will it work if I simply replace C1 and C2 with unpolarised ceramic 10µF capacitors?3) Am I supposed to need R3? Is 270K a suitable value? (it was to hand, and seemed to do the job)4) Am I barking up the incorrect proverbial, and need to be slapped with Electronics for Beginners again?TIARebecca

author: magic
author_rank: Super Contributor
timestamp: March 27, 2021, 04:51:58 pm
content: Yes, electrolytics act like diodes if charged in reverse direction. They can also act like zeners (or geysers) if overvolted.Not sure what's the noise of that TLE thing and whether it could affect your signal, unless filtered heavily.I think I would consider the feasibility of regulating ±2.5V rails for the digipot and level shifting SPI instead.

author: bson
author_rank: Unknown
timestamp: March 27, 2021, 04:53:26 pm
content: C1 will see 0-5V on the left side and 2.5V on the right. This is less than ideal for an electrolytic capacitor since with anything less than 2.5V on the left side if that's + it will be negatively biased. And as you've noted that won't work with a polarized capacitor.C2 will see 0V in the right side and 0-5V on the left. It will work as an electrolytic with the positive terminal to the left.Electrolytics are very non-linear, meaning the capacitance changes with the charge held. That becomes audible very quickly. Polypropylene capacitors are an example of a far more linear non-polarized variety that are often used for audio signal coupling. They're not as cheap though, and are physically bigger for the same (printed) capacitance.C2-R3 is a low-pass RC filter. It won't change the biasing.When a capacitor charges there is a current on both terminals. It is this current that causes charge to collect; a positive charge current into the left terminal of C2 and out the negative terminal. This causes positive charge to collect on the left side and negative on the right (from the outflow of charge). This happens because the two sides aren't electrically connected, and so the charges collect until the capacitor is saturated and can't hold any more. With R3=270kΩ you significantly limit the current on the right side, that is the current out of the negative terminal to 0V. If there's 5V across C2+R3, C2 will start at 0V and R3 will have 5V, so the current is 5/270k = 18.5µA. Needless to say it will take forever to charge a 10µF capacitor to 5V at such a minuscule current so it will seem to just get stuck at whatever voltage it's at. Note also that whatever is downstream will supply a current which at these levels (µA) might be significant. As a starting point, try a much smaller R3, like maybe 1k and 1µF. That will give you a peak current of 5mA (5V/1kΩ).Since the current in both capacitor terminals is the same, the same current has to be supplied on the left side - by the digital pot. Look at the datasheet to see what it can handle and give yourself some margin. Use this to size R3, and then size C2.

author: srb1954
author_rank: Super Contributor
timestamp: March 28, 2021, 07:31:55 am
content: Apart from the other problems you will also need to ensure that the mixer op amp has a DC path for the input bias current for the +ve input. Even FET input op amps need a return path for their bias current, low as that might be.Since all inputs are shown as capacitively coupled there is no DC path for the bias current and, depending on the direction of the bias current, the +ve input will eventually float up or down to one of the supply rails and the op amp will go into saturation.

author: mikerj
author_rank: Super Contributor
timestamp: March 28, 2021, 08:17:21 am
content: Why do you need C2/R3 if the inputs to your mixer section are already AC coupled?The rail splitter IC is may add significant noise to your signal, and biasing both the upper and lower terminals of a divider to the same voltage is redundant. I'd ditch the TLE and replace it with a suitable bypassed divider and get rid of the 100k/100k divider at the front end.