OpAmp Wonders

[cowboybob]

As a proud grad of "OP Amps 101, Real World Addition" from KISS University, I thought I was now qualified to do OpAmpy stuff with a genaral understanding of the basics.

Wrong. I need "OP Amps 102".

I SIMed the below TL081 virtual ground Split Rail power supply circuit from a similar 1/2 TL082 circuit in the most recent issue of the ARRL's QST April 2014 magazine (page 41) by David L. Chute, KG4BZW. It is used in a very nice, 2 chip TL082 home brew ESR meter:

I built this circuit this morning using 1/2 of a TL082 and the real thing confirms the SIM, except that the rail values are slightly (20mV or so) lower.

No load circuit draws about 18mA. Rail loads of 20mA or so reduce rail voltage levels 15% to 20%.

I am assuming there is some cap charging going on, but I have seen some similar circuits from google searches without the caps. Didn't build one of those.

My questions are:

1. how does this particular circuit work? (Simple answers, please, suitable for a Level 102 student...)
2. can the load carrying capacity of the circuit be increased?

(Wish I'd known about this before my radiator fan fault monitoring post(s) where I bought some MAX1044s).

 

[ronsimpson]

>You have a 9 V battery.
>The two 10k resistors make a voltage at 1/2 the battery voltage. It is 'soft' in that you can not pull power from this point with out effecting the 1/2 voltage.
>The two capacitors are much liked the two resistors and want to make a 1/2 v point.
>Often I see the capacitors across the resistors but what you have will work.
>The op-amp has (-) and (out) connect together so it has a gain of one. The op-amp will try to make the voltage on (out) be the voltage on (+).
>The voltage on op-amp-out will be at 1/2 battery. (you can pull power from this point up to what the amplifier can deliver and it will not drop in voltage)
>You understand the op-amp is not perfect. The input to put put voltage may be off by 10mV. If you pull power hard the output may sag a little.
>This circuit is trying to make your 9V battery look like two 4.5V batteries.

>You can get a more power full op-amp or add a two transistor buffer on the output. (more power)

 

[Ratchit]

cowboybob,

"1. how does this particular circuit work? (Simple answers, please, suitable for a Level 102 student...)"

It looks straight forward enough. The opamp is simply a noninverting unity gain amplifier. The two caps connected to virtual ground are simply filters across the positive and negative voltages. They have no effect on how the opamp regulates the of the voltage of the virtual ground. The junction of the resistors, the + and - terminals, and the output terminal of the opamp are all at the same voltage, specifically 1/2 the battery voltage, which is designated a virtual ground. Any load you apply to either the +4.5 or - 4.5 will be supplied and regulated by the opamp. Any 9 volt load will be supplied directly by the battery with no interaction with the opamp.

"2. can the load carrying capacity of the circuit be increased?"

Yes, but any booster will have to have a plus/minus voltage capability.

Ratch

 

[KeepItSimpleStupid]

You can BUY rail splitters too: https://www.ti.com/lit/ds/symlink/tle2426.pdf

So why is the TL081 a poor choice? Datasheet here: https://www.ti.com/lit/gpn/tl081

PS: I was going to mention a rail splitter for that project.

 

[KeepItSimpleStupid]

CBB:

I wanted you to think about my question, but since you didn't look at the equivalent resistance in the output. It's nearly 200 ohms. That's why the TL081 s not a very good rail splitter.

 

[cowboybob]

CBB:

I wanted you to think about my question, but since you didn't look at the equivalent resistance in the output. It's nearly 200 ohms. That's why the TL081 s not a very good rail splitter.

KISS, are we talking about the comparator circuit or the dual rail supply circuit?

I'll be honest, I don't think I know how to determine equivalent resistance for an OpAmp output. I can determine equivalent resistance, of course, of a resistance network and I assume what you're talking about is similar.

From the SIM specs, the output resistance of the TL082 s 75 ohms. With a resistive load I'm pretty sure I could figure that total resistance for the comparator. But with the power supply I don't have a clue.

Or are you talking about something else. Don't mean to be dense...

 

[Ratchit]

KISS,

Would you and could you explain your phrase "equivalent resistance in the output", and how you obtained the 200 ohm value?

Ratch

 

[KeepItSimpleStupid]

Sure. Look at the datasheet of the TL081. The equivalent internal circuit shows a 200 ohm resistor DIRECT to the output, so the output has to pass DIRECTLY through this resistor. There is also a 100 ohm resistor to each of the rails. So, on the order of 300 ohms. So, maybe I wrote the 200 wrong. The point is, is that it's high.

No, i did not try to compute the effects of feedback, but even so it's unity gain.

If you were designing a rail splitter, the output Z of the rail splitter would be an important selection criteria.

 

[Ratchit]

KISS,

Are you looking at the schemat of the op amp? You did not specify. I see a 128 ohm resistor connected to 2 64 ohm resistors, which appear to be in an internal feedback loop of the op amp, so who knows what their real output resistance is. The manufacture does not say, but I will give you 300 ohms, because the output resistance of an op amp is usually a few decades of resistance. So, a noninverting amp has feedback which reduces the output resistance by a factor of 1/Vol, in other words divide the output resistance of the op amp by its open loop gain to get its output resistance. At room temperature, that particular op amp has a typical open loop gain of 200. Therefore, the output resistance of the op amp is about 300/200 or 1.5 ohms due to its feedback. That should be low enough to be a fair enough voltage source for the power supply.

Ratch

 

[KeepItSimpleStupid]

PDF page #2: https://www.google.com/url?sa=t&rct...=1Eck5fJR5DRZzWirTw5upA&bvm=bv.62922401,d.dmQ

See the 200 and the 100 ohm resistor?

And noted, feedback reduces the output Z,

Point really is, there are better OP amps to use as a rail splitter.

 

[Ratchit]

PDF page #2: https://www.google.com/url?sa=t&rct...=1Eck5fJR5DRZzWirTw5upA&bvm=bv.62922401,d.dmQ

See the 200 and the 100 ohm resistor?

And noted, feedback reduces the output Z,

Point really is, there are better OP amps to use as a rail splitter.

OK, that is a different datasheet than your original link. Anyway, the output impedance will be around 1 ohm or less no matter what op amp you use. So how would a "better" op amp improve the function for which the OP wants to use it?

Ratch

 

[MrAl]

Hi,

Where are you guys getting an open loop gain of 200 from? That's got to be way too low an estimate. I'd say 100000 is more like it, with maybe 20000 as a min. That makes the output impedance very low, as with most op amps with feedback. The internal series source resistance is another story.

Those resistors could present a problem depending on supply voltages and output current demand, because the circuit would eventually degenerate into (as an approximation) two 300 ohm resistors in series with the output (of the op amp) taken from the junction of the two resistors, but with impedance of 300 ohms total, and that's even with great feedback. With supply voltages of plus and minus 10 volts that means max current to ground would have to be limited to 33ma, but since the op amp rating is less then it should work.
With lower voltages the resistors would become a problem because the max output to ground would be less.
Low story short, the circuit should only be used at low currents like 10ma or so. 10ma causes a 3v drop in the 300 ohm 'resistor', which isnt bad.