replacing a split supply op amp with a single supply. - Page 1
Forum section: electronics / beginners

author: Adhith
author_rank: Frequent Contributor
timestamp: July 06, 2017, 07:22:27 am
content: hello everyone..I'm working with an active audio crossover using a dual split supply OP amp LM833. since I only have a 12 dc adapter to power this thing up I can only have a single supply rail. So is there something that I can do like replacing it with a ICs like MC33172 which is a low noise single dual op amp single supply IC. The schematics is attached within this.

author: Zero999
author_rank: Super Contributor
timestamp: July 06, 2017, 08:57:02 am
content: Quote from: Adhith on July 06, 2017, 07:22:27 amI don't know why you'd want to replace the LM833 with the MC33172, unless it's to reduce cost. It's inferior in every single way.The only thing the MC33172 can do, which the LM833 can't is work with its inputs at the negative rail and output voltages near there too (hence the name single supply op-amp) but you don't need to do that.You can use a rail splitter to give +/-6V from your 12V supply. A good IC for this is the TLE2426 but you can make your own with an op-amp.https://tangentsoft.net/elec/vgrounds.html

author: capt bullshot
author_rank: Super Contributor
timestamp: July 06, 2017, 09:16:16 am
content: There's no need to replace the LM833 with an other OP Amp.This circuit is simple to convert to single supply, keeping the LM833:All the + inputs of the amps and no other circuit nodes are connected to ground.Since the + inputs are high impedance, the only thing you need is a voltage divider to the middle point of your power supply.Connect the +15V node to your positive power supply.Connect the -15V node to your supply ground.Use a resistive divider made of two identical resistors (e.g. 10k each) to create a half-supply potential. Connect this potential to the "+" inputs of the amps (instead of GND). Placea decoupling cap from the half-supply node to your supply ground (e.g. 100n).Place suitable dc blocking caps into the circuits inputs and outputs (calculate the capacity from your circuits impedance and required low frequency response - about 10Hz for audio)Refer your input and output signals to power suppy ground.Check maximum peak-to-peak input and output voltage swing (will be significantly lower than the original circuit, with a 12V supply you should assume a 7.5Vpp max. for input and ouput and internal circuit nodes, should work with typical 1Vrms audio signals)

author: Zero999
author_rank: Super Contributor
timestamp: July 06, 2017, 12:50:44 pm
content: AC coupling is a good idea but you need to know what you're doing more than just adding a rail splitter. It's important to know what the load impedance is, in order to select the values of the output capacitors.See attached.

author: Adhith
author_rank: Frequent Contributor
timestamp: July 06, 2017, 06:03:23 pm
content: Quote from: Hero999 on July 06, 2017, 08:57:02 amThank you very much for the help. The reason why I thought of MC33172 was that it uses a single supply and I was under the false assumption that it could output a negative audio signal without a negative power supply. Now I would definitely go with the rail splitter. Also do I need a regulated supply for the rail splitter circuit??

author: Adhith
author_rank: Frequent Contributor
timestamp: July 06, 2017, 06:06:44 pm
content: Quote from: Hero999 on July 06, 2017, 12:50:44 pmSo if I use a rail splitter with the TLE2426 then there is no need to calculate the load impedence right??

author: w2aew
author_rank: Super Contributor
timestamp: July 06, 2017, 06:55:39 pm
content: Have you watched my video on the subject of op amp power supplies?

author: Zero999
author_rank: Super Contributor
timestamp: July 06, 2017, 10:00:04 pm
content: Quote from: Adhith on July 06, 2017, 06:06:44 pmYou don't calculate the load impedance. It should be known.What is it driving? Presumably another amplifier? If so, it's probably quite high, at least 10k.

author: Adhith
author_rank: Frequent Contributor
timestamp: July 07, 2017, 05:55:05 am
content: Quote from: w2aew on July 06, 2017, 06:55:39 pmThank you w2aew. The video was very helpful to understand the basics of op amps.

author: Adhith
author_rank: Frequent Contributor
timestamp: July 07, 2017, 06:02:31 am
content: Quote from: Hero999 on July 06, 2017, 10:00:04 pmThank you for your valuable reply.I'm using two of these circuit for two channel..one for driving a LA4440 amplifier in stereo mode whereas the other low frequency left and right signals is passed through a resistor mixer and then into LA4440 in bridged mode to drive a woofer.

author: Zero999
author_rank: Super Contributor
timestamp: July 07, 2017, 08:28:39 am
content: Quote from: Adhith on July 07, 2017, 06:02:31 amThe input resistance is on the datasheet for the LA4440. The output capacitors have to be big enough to avoid attenuating the signal. Make them around twice the minimum value required.C = 1/(2pi*F*R)http://www.onsemi.com/pub/Collateral/EN750-D.PDF

author: Adhith
author_rank: Frequent Contributor
timestamp: July 07, 2017, 09:33:24 am
content: Quote from: Hero999 on July 07, 2017, 08:28:39 am

author: Zero999
author_rank: Super Contributor
timestamp: July 09, 2017, 09:09:02 am
content: Quote from: Adhith on July 07, 2017, 09:33:24 amI think you've confused this active filter with a passive crossover.The R & C values used in the circuit make virtually no difference to the output impedance of the circuit, which should be very low, perhaps a couple of Ohms at most. Keep all the resistor values in the circuit the same. Use the input impedance of the LA4440 IC to calculate the value of the output AC coupling capacitor.The input of the circuit has two 24k resistors, each connected to the op-amp's input, which will be held at the same voltage by negative feedback, so the input impedance can assumed to be 12k and should be used as the R in the calculation for the input AC coupling capacitor.

author: Adhith
author_rank: Frequent Contributor
timestamp: July 27, 2017, 06:10:57 pm
content: I couldn't find a 24k and 6.2k from the electronic shop. They have only 22k and 27k in stock. Also instead of the 6.2k they have only 6.8k. So could I proceed with these values for building the circuit??

author: Adhith
author_rank: Frequent Contributor
timestamp: July 27, 2017, 06:17:42 pm
content: Quote from: Hero999 on July 06, 2017, 12:50:44 pmcorrect me if I'm wrong, So the sound signals extends over both positive and negative ranges which needs to be handled by an op amp which works on both positive & negative ranges. So could an ac coupling provide both the ranges??

author: capt bullshot
author_rank: Super Contributor
timestamp: July 28, 2017, 05:28:56 am
content: The circuit relies on the near 1:4 ratio of the resistor values (24k : 6k2). So if you use different values, try to get as close as possible to the 1:4 ratio.This particular circuit is AC coupled, shifting the audio signal "zero" to half of the supply line. The audio range is positive and negative in reference to the 1/2 VSS node. In reference to GND, the audio signal within the circuit is alway positive.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: July 28, 2017, 02:40:33 pm
content: I think you might be a bit confused when it comes to negative voltages.Voltage is all relative. You can think of it like elevation. A positive voltage would be when something is at an elevation higher than what you call "home", while a negative voltage would be when something is at an elevation lower than what you call "home". It's all relative to your "home" though, call somewhere else "home" and the relative elevation of everything else shifts up or down accordingly.Voltage works the same way, it's all relative. "Zero volts" is whatever node in the circuit you point at and say "that's zero volts". Anything above that is therefore positive, anything below it is negative. Point to a different part of the circuit and say "nope, THAT'S zero volts", and the relative voltage of everything else changes accordingly.Op-amps only have two power inputs, a "more positive" and a "more negative" input. It has no "ground" reference, it doesn't care. As long as the "more positive" terminal is more positive than the "more negative" terminal, by at least the minimum and no more than the maximum values specified in the datasheet, it's happy. There is no such thing as a "single supply" op-amp or a "dual supply" op-amp, there are just op-amps. Every op-amp is both "single supply" and "dual supply", because whether the voltages you feed it are "ground and V+" or "V- and V+" is completely up to your own naming convention, the op-amp doesn't care. You could call the supplies "+/- 5V", or "0V/10V", or "20V/30V", doesn't matter, as long as you don't exceed the maximum voltage difference for the op-amp, and whatever signals you're feeding into the + and - terminals are between the rails and follow the recommended input/output ranges specified in the datasheet (namely, how close you can get to the rails on the input and output).If you're feeding an op-amp what you call "0V" and "10V" on its supply lines, and you want it to buffer an AC signal, then your best bet is to center the AC signal at "5V", so it swings between say "3V" and "7V". As far as the op-amp is concerned, this isexactly the sameas if you had fed it "-5V" and "+5V" and centered your AC signal at "0V", so it swings between "-2V" and "+2V". The only difference is which part of the circuit you chose to call "0V".

author: Zero999
author_rank: Super Contributor
timestamp: July 28, 2017, 03:30:30 pm
content: Just use 27k and 6k8, which is closer to the 4:1 ratio, than the original component values.As mentioned above, all voltages are relative. The potential divider and coupling capacitors, just shift the signal up to half the power supply voltage.A single supply op-amp is just a an op-amp which can work with its inputs down to its negative rail. They're generally used in DC applications, such as monitoring the voltage of a battery. You don't need a single supply op-amp, because in this circuit, both inputs are held at around half the supply voltage and never reach the op-amp's negative rail.

author: rstofer
author_rank: Super Contributor
timestamp: July 28, 2017, 10:50:37 pm
content: Since you create a 'virtual ground' at 1/2 the supply voltage, you need to block that DC component when you insert a signal because the signal source is probably centered around 0V. So you use a capacitor on the input.At the output, the signal is still centered around 1/2 the supply voltage and you probably don't want that DC component heading off to wherever. So you use a capacitor on the output.You have to remove the DC components at the input and output.