# Postive input ==> Positive and Negative output

I was watching some tutorials on thesignalpath.com and came across something that I didn't quite understand (even after it was explained).

toddrharrison

August 21, 2011 at 6:29 PM

Answer to first op-amp: You are using it as a high impedance voltage follower to split your single power supply into a plus and minus dual source.

Details: Your single power supply source will be feed into a 50/50 voltage divider circuit with relatively high resistance to keep the voltage divider’s current to a minim. Then this divided voltage is feed into the first op-amp’s non-inverting terminal. The out of this op-amp is feed back into the inverting terminal with a short creating wait is called a voltage follower. The output of this op-amp then becomes your new common for the other three op-amps.

Results: This makes the single source power supply’s ground look like a minus voltage supply and the positive looks like 1/2 the original positive supply to the part of the circuit using the voltage follower’s output as common. From this you can power your other op-amps with equal + and – power rails. What do I win?

http://thesignalpath.com/blogs/2011/08/21/op-amps-pwm-and-ignition-coils-tutorial/#comment-71

Is this a matter of connecting Vout of the voltage divider to the Vin of the voltage follower?
So the voltage follower Vout is -1/2 voltage divider Vout, yes? Would I have a 2nd output on the voltage divider between R2 and GND for my +V rail?

I hope this is not confusing.

• This isn't an exact duplicate by any means, but since everyone is talking about what negative voltage is, you might want to check out this question: electronics.stackexchange.com/questions/10322/… Feb 28 '12 at 19:56

Yes, you connect the output of the divider to the input of the opamp. If you would use the output of the divider directly the voltage would vary depending on what load you connect to it. With the opamp you have a high impedance load, meaning that there doesn't flow any current into the opamp and the voltage on its input will remain at Vin/2.
The opamp will set its output so that both inputs are equal, so for the voltage follower this means that the output voltage is equal to the input voltage. The opamp can supply some current (not much really) and a small load won't change the Vin/2 at the output.
I'm not sure I understand your last question.

• So is the output of the opamp -Vin/2 or Vin/2? Feb 28 '12 at 16:16
• @Hair_of_the_Dog - It's +Vin/2. Everything is between 0V and Vin. There's no way to get a negative voltage there. Like Uiy says, voltage is relative. If you decide to make the opamp's output your reference, so 0V, then what was your 0V becomes -Vin/2 and what was your Vin becomes +Vin/2. It all depends where you place your reference. Feb 28 '12 at 16:23
• 10V-5V Feb 28 '12 at 17:23
• I think I have the -Vin/2 part using the opamp output as ground. (see circuit sim link above). The negative part just seems to indicate the direction of current flow. i.e. a larger voltage will cause current to flow in the direction of a smaller voltage, yes? Feb 28 '12 at 17:29
• @Hair_of_the_Dog - Yes, The 10V is +5V above your virtual ground, so it makes sense that the lower diode conducts. If you would connect the resistor to 0V (-5V with respect to your virtual ground) the other diode would conduct. Feb 28 '12 at 18:01

Ok, this is very simple:

1. Voltage is relative. You can call anything ground and use it as ground(well, almost).

2. Your voltage divider splits the supply into two equal parts(you could do any parts or even use caps, inductors(not a good idea), tubes, or anything that will give you Vcc/2.

This is called "rail splitting". We need to have a + and - voltage for op amp/analog stuff when we want to be able have negative swings. If we don't our signal's end up only working on the positive half(and generally nothing really works like it should).

Assuming you understand how the resistive voltage divider works and that the middle should be Vcc/2 if the two resistors are the same. The second thing to notice is that we could use the middle of the voltage divider, Vout in your pic, to act as ground(and the gnd in your pic would then become -Vcc/2).

BUT!! There are some problems with this. The resistors limit the total amount of current we can use AND if we use things that pull current from the resistors it will effect the values. So either we have to use very low value resistors, wasting a lot of power, use very efficient devices, or fine some way to provide more current.

This is where you 2nd picture comes in. It is an op amp configured as a buffer. You are missing one key part of the drawing. The op amp has power pins supplied to it. I'm not going to explain how an op amp works but there a plenty of things online.

Anyways, the buffer takes a voltage in and produces the exact same voltage at the output BUT it allows for larger loads. A buffer is also called a UNITY GAIN Voltage amplifier. That is, it doesn't really amplify the voltage but that's exactly what we want. It does this by pulling current from the power pins "adds" it to the output.

So, we have our crappy voltage divider that doesn't work well for much of anything, but we use a device, in this case the op amp, to provide us with more current drive. This way we can drive things like maybe an LED, a motor, or other op amps or whatever with Vcc/2(or whatever ratio we wanted.

The Op amp will be happy to provide a lot of current to those devices if they need it. The resistor voltage divider can't do it and it will screw things up big time.

Realize, all we are doing is BUFFERING the voltage Vout on the voltage divider. You can sort of forget the op amp... it's just a tool to do that. We could have used a bjt, npn, jfet, TL431 or many other things to do it. The point is now we have created a sort of minature power supply with voltages A, (A - B)/2, and B.

Hopefully you got that.

Suppose we had Vcc = 10V, Gnd = 0V then a 1:1 divider gives us Vout = Vcc/2 = 5V!

BUT since voltage is relative(you can basically add any value to all the voltages)... we can also view it as

Vcc = 0V, Gnd = -10V, Vout = -5V.

or better yet

Vcc = 5V, Gnd = -5V, Vout = 0V.

Then to make it even easier we will relabel things to get

Vcc = 5V, Vss = -5V, Gnd = 0V.

Now we have a +- power supply. We got this from a + power supply. But we can now forget we even have a + power supply if we want and just work with the +- supply.

This makes live much easier since lets us shift the signals to be centered at 0V instead of Vcc/2... which is really not that difficult to think about but it is unnecessary.

There are some caveats to this method but there are some very good things on the net that talk about them and I'll leave you to find them when you are ready. Just remember once you get the circuit down, even if you don't understand it fully, you can forget how it works and just think of it as a +- power supply... as if that is what you started with in the first place.... cause really that is what you have created with the voltage divider and you've improved it's power supply ability by using the op amp. you can even make it better by adding more junk to fix things if you want.

AGAIN!!!!

Voltage is relative: This means you can add any fixed constant to ALL the voltages without changing any of the physics.

Labels can be relabeled: You can relabel things all you want as long as they are a permutation(no duplicating them or dropping them). If you want to call rename GND to VCC and VCC to GND you can. Of course this is confusing because it will flip the currents and they won't go in the direction we are used to.. But if you order the labels in terms of voltage you can sort of shift them any direction but not rearrange and it will make sense:

So

..., ] Vcc, Gnd, [Vss, ...

can be shifted

. ..., ] Vcc, Gnd, [Vss, ...

..., Vcc ] Gnd, Vss [, ...

So Vcc became Gnd and Gnd become Vss. (Note, the bracket stuff are labels I've added. The point is, they are ordered in terms of voltage).

For our +- supply

Vcc, Vcc/2, Gnd, -Vcc/2 (Vcc/2 doesn't exist in our original labeling but it's what we want to go to)

can become

Vcc , Vcc/2, Gnd
Vcc/2, Gnd , -Vcc/2

So Vcc -> Vcc/2 Vcc/2 -> Gnd Gnd -> -Vcc/2

THIS IS EXACTLY THE SAME as when I subtracted 5V from the 10V supply we had. So the methods are the same. The labeling stuff might be more confusing but maybe not.

• It's not true that Op-Amps can provide "a lot of current" (In engineering "a lot" is meaningless), especially for driving motors; some amps can drive some motors, but you can't generalize. And labeling is arbitrary, but there are conventions, saying that Vcc (Vdd) is the upper rail (from collector or drain), and Vee (Vss) is the lower (from emitter or source). And you have mixed CMOS-like labels with TTL-like labels. The same way, GND is 0 by convention, and chosen that you fix the other levels. Feb 28 '12 at 9:54
• Feb 28 '12 at 16:18
• @clabacchio - Thanks for your help in removing the useless comments. Discussions like these are not productive. Your first comments says everything that needs to be said here. Feb 28 '12 at 16:35