Virtual ground resistor choice - Page 1
Forum section: electronics / beginners

author: TimNJ
author_rank: Super Contributor
timestamp: May 20, 2013, 11:44:25 pm
content: A voltage divider is a simple way to create a split supply for a circuit.My question is, what do you have to consider when choosing the two resistors? What are the pros and cons of using high/medium/low value resistors?

author: Paul Price
author_rank: Super Contributor
timestamp: May 20, 2013, 11:52:57 pm
content: Using low value resistors increases power drain, something you don't want to do if you are running on a battery. Use the highest value resistors as possible and they both should be identical in value. Too high a resistor and the offset currents of the opamp or the inputs of the virtual ground generating device could cause a small offset in virtual ground or cause your circuit to be more likely to be upset by electrical interference.

author: c4757p
author_rank: Super Contributor
timestamp: May 21, 2013, 12:07:06 am
content: Quote from: Paul Price on May 20, 2013, 11:52:57 pmAn interesting, semi-related note on this - one of the option boards for the Tek 24x5B series of scopes has a number of very low impedance voltage dividers, for level shifting around an ECL chip. Seeing 70 ohms across the 5V rail on a digital board was enough to make me think it was faulty...The board as a whole takes 1A just to sit and twiddle its silicon thumbs. I am guessing there was a speed-related reason for this, though (it's the counter/timer board). Not familiar with ECL.

author: Paul Price
author_rank: Super Contributor
timestamp: May 21, 2013, 12:12:49 am
content: C4757P: ECL operates at GHz frequencies and the lower the impedance, the closer all wiring approaches a transmission line low impedance, the better the rise and fall times and the fewer the reflections from point to point. ECL is power hungry and performance poor and so it is no more.

author: c4757p
author_rank: Super Contributor
timestamp: May 21, 2013, 12:18:18 am
content: The really fun part is the level shifting back from ECL levels to TTL. In the high speed locations, they've accomplished that with discrete long-tailed pairs made out of 2N3906 (not sure if they're supposed to be matched or anything). The slow signals get a 339 comparator. Were there really no affordable ECL-TTL level shift chips in the late 80s when ECL was common?

author: c4757p
author_rank: Super Contributor
timestamp: May 21, 2013, 12:23:53 am
content: Sorry for the little hijack, TimNJ. I'll try to contribute something. Generally, I try to make voltage dividers as high impedance as possible, keeping in mind the increased thermal noise with high impedance, the effect of the op amp bias current, and for higher frequencies, the increased effects of parasitics in poor layouts (and elsewhere - input capacitance too!) at high impedance nodes. I'd just calculate the highest impedance you can use for the bias current you have and precision you want, and then cut that off around 5Meg or so to keep noise down. (You can obviously reduce it further if it ends up being a problem.) Alternatively, if the impedance at the other input is already determined, try to match it.

author: Paul Price
author_rank: Super Contributor
timestamp: May 21, 2013, 12:30:08 am
content: C4757p: Yes, there were a few ECL to TTL level shifters/interface chips, but the direction of speed might be from fast to slow, so slower will do, and these now hard to find chips might not be as cheap, have the small footprints developed years later, not have the package needed to not to take up too much space or else fit the available power supply voltages and other conditions of PCB layout as well as a discrete transistor/resistor interface approach.

author: c4757p
author_rank: Super Contributor
timestamp: May 21, 2013, 12:33:51 am
content: Quote from: Paul Price on May 21, 2013, 12:30:08 amI think you've hit it. The board is completely packed to the gills and these poor little transistors are just tossed off in a corner. Didn't even occur to meI guess I'm just used to SMD now where you could practically fit the whole scope on that little board.

author: TimNJ
author_rank: Super Contributor
timestamp: May 21, 2013, 12:38:49 am
content: I'm talking about the V+ and V- power supply inputs. So how does that relate to the input bias current?I know there's a certain minimum operating current. I think an LM358 needs around 0.5mA to work with a +5/-5 split supply. (Quick side note, is that current figure the combined current, so 0.25mA on the +5V and 0.25 on the -5V, or is it 0.5mA on each?).Is that the bias current you're talking about. So assuming 0.25mA on each rail, I would need 2x 20k resistors to maintain those currents.Am I wrong?

author: c4757p
author_rank: Super Contributor
timestamp: May 21, 2013, 12:40:58 am
content: The bias current is the current taken by the inputs. If you put two 1Meg resistors on an op amp input making up a virtual ground (for a Thevenin impedance of 500k), and the op amp pulls 500nA to V-, the virtual ground is going to shift downward by 0.25V.The power supply inputs themselves don't need any dividers...As an additional point - don't think you can be clever by "compensating" for that shift. It varies widely from op amp to op amp (even of the same type), over temperature, over common mode input voltage, and over power supply voltage. It's also not always to the negative rail (though this shouldn't vary within an amplifier family...). All you can do is make sure that even at the maximum, it's not too much.

author: Paul Price
author_rank: Super Contributor
timestamp: May 21, 2013, 12:56:22 am
content: The LM358 VDD(LM358 + power pin and VSS LM358 ground pins) usually do not connect to the virtual ground, so do not source or sink power into the virtual ground. The virtual ground sets a ground level for the inputs and the "bias" in bias currents mentioned here is about the tiny leakage current mismatch of the transistors of the input pins of the virtual ground device which can possibly create a small offset, but this is often not a problem. The higher the virtual ground setting resistor values, the higher the injected noise appearing on the virtual ground relative to either to the VDD or VSS bus as a reference.

author: TimNJ
author_rank: Super Contributor
timestamp: May 21, 2013, 03:41:52 am
content: Quote from: c4757p on May 21, 2013, 12:40:58 amPerhaps I didn't make myself clear. I'm talking about a single supply powering a split supply op amp (or some split supply device). So a battery powering a LM358 is my example. You need to allow positive and negative swing if you plan on using an AC input, right? So you would need a way of creating a virtual ground split supply.

author: c4757p
author_rank: Super Contributor
timestamp: May 21, 2013, 04:32:44 am
content: Yes, that is what I am talking about too. Doesn't need to be a supply though, just a reference. The op amp doesn't care if you have a split supply or not.I'd try to help more but I am going to bed now!

author: w2aew
author_rank: Super Contributor
timestamp: May 21, 2013, 04:41:13 am
content: Quote from: c4757p on May 21, 2013, 12:07:06 amFYI - what you were likely seeing was not really a voltage divider, but the termination network for the open-emitter outputs of the ECL circuits. These outputs are designed to drive 50 ohms, but not 50ohms directly connected to ground. The outputs ideally want to "see" 50 ohms connected to VCC-2V (where VCC is the positive supply of the ECL chip). A "split termination" was often used, which was a pair of resistors in series across the supply (82 and 130 ohm as I recall) which created a thevenin equivalent of the 50ohm to -2V termination for the ECL outputs.

author: c4757p
author_rank: Super Contributor
timestamp: May 21, 2013, 04:57:16 am
content: Thank you! I never knew that.

author: TimNJ
author_rank: Super Contributor
timestamp: May 21, 2013, 07:31:28 pm
content: But say your signal is low level audio, if you don't have a split supply, aren't you going to lose the bottom half of the waveform?

author: free_electron
author_rank: Super Contributor
timestamp: May 21, 2013, 07:45:55 pm
content: use two 1k resistors. put 100nf // 1uf ceramic across one of them to keep that point stable. then connect that node to an opamp in unity gain. preferrably an opamp that can sink/source some current on its output.there you go. you now have a 'hard' midpoint reference

author: quantumvolt
author_rank: Frequent Contributor
timestamp: May 21, 2013, 09:02:24 pm
content: Neither the title of the tread nor your introductory answer to your own question are clear. May be as a result - most of the answers you get are imo somewhat foggy and off point.In general - if you are looking for designs - do not ask hobbyists. In this case your obvious first place to look for water is the datasheet:http://www.ti.com/lit/ds/symlink/lm158-n.pdfVirtual ground and battery single supply is mentioned on page 11. If you DC-couple your circuit then your virtual ground must be set as a zero-ref. This is shown on page 18.After you have studied this, you will know that what you want to learn about is something like " DC coupled single supply battery virtual ground". Searching this you find many info sources - a good advice is on page A-4 in this:http://www.ti.com/lit/ml/sloa091/sloa091.pdfYou write:"Perhaps I didn't make myself clear. I'm talking about a single supply powering a split supply op amp (or some split supply device). So a battery powering a LM358 is my example. You need to allow positive and negative swing if you plan on using an AC input, right? So you would need a way of creating a virtual ground split supply."No need for defined positive / negative swing or a split supply if the circuit is AC-coupled with capacitors on input and output. This will decouple your circuit from any ground references in input sources and output sinks. So your needs are based on whether you must use DC-coupling or not.In order to answer your question further, it should be better formulated. Or better - tell us what you want to do exactly, and present all information, specs and schematics that you have.