virtual ground circuit - Page 1
Forum section: electronics / beginners

author: vivi-d
author_rank: Regular Contributor
timestamp: October 02, 2024, 03:23:49 pm
content: Just a quick question... I was reading this article:https://sound-au.com/project43.htmWhat is the purpose of the 10 ohm resistors in the 2nd circuit (attached below). I know that they help the op-amps share the load, but how?

author: Andy Chee
author_rank: Super Contributor
timestamp: October 02, 2024, 03:49:10 pm
content: It is explained in the article text:Quote

author: vivi-d
author_rank: Regular Contributor
timestamp: October 02, 2024, 04:37:24 pm
content: Yes, but how does this happen? Why would one do all the work?

author: TimFox
author_rank: Super Contributor
timestamp: October 02, 2024, 04:39:50 pm
content: The two amplifiers, with the 10 ohm resistors disconnected, would have slightly different output voltages due to their different input offset voltages.If supplying positive current to the load, then the output stage of the higher-positive-output amp would conduct, but the output stage of the other would not conduct very much.The resistors allow both amplifiers to contribute slightly different values to the output current.Also, each amplifier would be "unhappy" directly connected (no 10 ohms) to the large 100 uF capacitor, since its open-loop phase shift would be affected and probably cause oscillation with the 100% feedback.

author: Andy Chee
author_rank: Super Contributor
timestamp: October 02, 2024, 05:18:06 pm
content: Quote from: vivi-d on October 02, 2024, 04:37:24 pmManufacturing tolerances means that it's practically impossible to produce two opamps that perform identically, even when they're on the same piece of silicon die wafer.

author: Zero999
author_rank: Super Contributor
timestamp: October 02, 2024, 05:40:13 pm
content: If the voltage drop across the resistors is an issue, then it can be reduced to near-zero using negative feedback. Unfortunately the capacitors on the output increase the risk of instability, although 2×100µF might be large enough to damp them.

author: David Hess
author_rank: Super Contributor
timestamp: October 02, 2024, 08:36:22 pm
content: Here is the napkin analysis that I would do.The two amplifiers each have a typical input offset voltage of 1 millivolt, so 2 millivolts total of offset between their outputs is reasonable. The output offsets could be as high as 12 millivolts total.The output resistance of each amplifier is about 25 ohms, so 2 millivolts across 50 ohms is 40 microamps. That does not seem so bad, however the output resistance of each amplifier is within its feedback loop, so gets divided by the excess loop gain, or about 200k for a voltage follower. So it is really 2 millivolts across 250 microhms, or 8 milliamps, which is almost the short circuit current limit of the amplifiers, and it could be several times higher, and that current is not available to service the load.

author: vivi-d
author_rank: Regular Contributor
timestamp: October 04, 2024, 01:28:19 pm
content: Much to learn I still have...1st question, why are the 25 ohms output resistance considered in series?2nd question, what is excess open loop gain?

author: Zero999
author_rank: Super Contributor
timestamp: October 04, 2024, 01:38:32 pm
content: Quote from: vivi-d on October 04, 2024, 01:28:19 pmThe internal resistance of the op-amp. Basically a couple of resistors in the output stage of the op-amp, which are used as part of the over-current protection circuitary and biasing.QuoteThe factor of the closed loop gain, due to negative feedback, which is close to 1 in this case and the open loop gain i.e. the difference between the inverting and non-inverting inputs.

author: vivi-d
author_rank: Regular Contributor
timestamp: October 04, 2024, 01:45:04 pm
content: If each op-amp has a 25 ohm output resistance, how are they connected in series in this circuit?Thanks for the insight!

author: MrAl
author_rank: Super Contributor
timestamp: October 04, 2024, 03:17:51 pm
content: Quote from: vivi-d on October 02, 2024, 03:23:49 pmHello,First, as another post mentioned, the culprit is the input offset voltage of each op amp. They may be very close when on the same chip, but normally we don't design that way we more or less go with the worst case, which is an output offset difference. That's very common in power supplies when they have to be paralleled.The short answer as to why is the dynamic output impedance is very low for both op amps, but the output voltages could be different. They should not be different by much however, possibly 4mv as a max. Because the outputs are different, the output with the highest voltage will TRY to supply the entire load current. If it fails to do that, the second op amp will start outputting some current also, so it would probably work without any resistors. The difference though is that that one op amp might be stressed too much, you'd have to check the data sheet.In the meantime, using resistors is not a bad idea, but 10 Ohms on each output may be more than needed because the output voltage difference will not be very much.The drawback to using resistors on the output is the output can change due the load, and how much it changes depends on the value of the resistors. Note this is not due to the op amps, it is due to an unbalanced load. With 10ma from each op amp, each resistor will drop around 0.1 volt, which may or may not be too much depending on your application. The ground lead would then appear to change if the load changes. That's why the idea is to use as low a value as possible for the two resistors. For example, 5 Ohms instead of 10 Ohms should work, and even 2 Ohms for each resistor might work. You could test it and see.

author: David Hess
author_rank: Super Contributor
timestamp: October 04, 2024, 03:21:24 pm
content: Quote from: vivi-d on October 04, 2024, 01:45:04 pmThey are connected in series in the sense that they are in series between the outputs of the operational amplifiers.

author: vivi-d
author_rank: Regular Contributor
timestamp: October 04, 2024, 06:21:51 pm
content: Another question...Are there two op-amps in this circuit because there's two in a package? Could you just buffer the virtual ground with one op-amp?

author: Andy Chee
author_rank: Super Contributor
timestamp: October 04, 2024, 07:09:02 pm
content: Quote from: vivi-d on October 04, 2024, 06:21:51 pmYes, however the load capacity would obviously be half as much. You could go crazy and substitute one bigger opamp e.g. LM1875

author: Zero999
author_rank: Super Contributor
timestamp: October 04, 2024, 07:25:52 pm
content: The voltage offset doesn't have to be much to cause a high enough current flow to cause the op-amp to current limit.I had a play with LTSpice.In all the simulations. U1a has an input voltage offset of 0.1mV, compared to U1b which is zero. Their open-loop gain is 200k and output resistance (Rout) is 25R and current limit 25mASimply connecting both op-amps in parallel results in the 25mA current limit.Virtual ground 2 op-amp bad.asc(2.07 kB - downloaded 32 times.)Connecting a 10R to the output of each op-amp reduces the currents to 5µA, but now the output resistance will increase ot 5R.Virtual ground 2 op-amp 5R out.asc(2.39 kB - downloaded 25 times.)Connecting one of the op-amps with the feedback loop around the 10R resistor, reduces the current further and will also virtually eliminate the output resistance.Virtual ground 2 op-amp 0R out.asc(2.43 kB - downloaded 24 times.)