Inverting Amplifier Bias Resistor

Started by aziltz, November 05, 2009, 06:25:50 PM

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aziltz

With an Inverting Amplifier, with the V+ input tied to Vcc/2, is there any reason why I can't/shouldn't use a resistor to tie the V+ to Vcc/2?

I've always seen it like this, http://en.wikipedia.org/wiki/Operational_amplifier_applications#Inverting_amplifier

V+ is a high-impedence input, and draws no current so inserting series resistance won't change anything that I can see.  It would just make board routing a bit easier if I had resistors to jump over traces to get there.

What do you think?

Cliff Schecht

The voltage drop with a 1 meg resistor into a JFET op amp + terminal is still a fraction of a volt. Some people place a resistor in series with the + input to balance the currents into the differential pair but this usually isn't necessary, it's just done to minimize input offset currents appearing at the output.

You know they make 0 Ohm resistors right? They're used as jumpers wherever a bare piece of metal can't be used or when a product is being mass produced and wire jumpers aren't an option (resistor is easier to automate than a wire jumper!).

JKowalski

Quote from: Cliff Schecht on November 05, 2009, 07:00:13 PM
The voltage drop with a 1 meg resistor into a JFET op amp + terminal is still a fraction of a volt. Some people place a resistor in series with the + input to balance the currents into the differential pair but this usually isn't necessary, it's just done to minimize input offset currents appearing at the output.

I thought I read that this was only a technique used on bipolar op amps?

aziltz

Thanks Cliff.

I've seen the 0 ohms before.  I'm not sure its worth it to me to try to get a hold of them.  I was trying to avoid jumpers in general, but i think they just might work best.

Now I just need to figure out how to add something to the eagle board that's not on the schematic.

Cliff Schecht

Quote from: JKowalski on November 05, 2009, 07:35:06 PM
Quote from: Cliff Schecht on November 05, 2009, 07:00:13 PM
The voltage drop with a 1 meg resistor into a JFET op amp + terminal is still a fraction of a volt. Some people place a resistor in series with the + input to balance the currents into the differential pair but this usually isn't necessary, it's just done to minimize input offset currents appearing at the output.

I thought I read that this was only a technique used on bipolar op amps?

Really it just depends on your application. Offset currents aren't a problem in JFET amplifiers because the FETs have such little current draw, down in the picoAmps typically, but I'm sure there are cases where even an imbalance this small could present a problem (like in the circuits where they test for bias current in pA). BJT's draw much more base current and so even in less critical applications, a resistance is sometimes used to minimize offsets.

My point wasn't that JFET's don't need this resistor though, I was just explaining what it is used for typically and why he can use it without harm. All I was trying to point out was that even with a large resistance, it doesn't usually drop enough voltage to cause any concern.

R.G.

Quote from: aziltz on November 05, 2009, 07:43:30 PM
Now I just need to figure out how to add something to the eagle board that's not on the schematic.
I'd suggest making component that's a jumper. Works on several board CAD systems I've used. Haven't tried it on Eagle, though.

Perhaps an even simpler way is to lay out the board as though it's double sided, and put ONLY the jumpers on the top side. Then leave out the top layer when you have boards made. Put wires in the holes when you get the physical board.
R.G.

In response to the questions in the forum - PCB Layout for Musical Effects is available from The Book Patch. Search "PCB Layout" and it ought to appear.

aziltz

Thanks RG.  I had come to that same conclusion as well.  Use Via's to locate the holes of the jumpers, pretending they are traces on the Top when I'm only going to print the bottom.

R.G.

Hit "send" too soon.

The resistor from the + input sees the input bias current. There is a number for this specified on every opamp datasheet.

The voltage drop in this resistor is, as Cliff says, V = Ibias*R. For FET amps with input biases of picoamperes (1*10^-12 amps), you'd need to have a resistors of 1000 Megohms to drop a millivolt per picoampere. Input in nanoamperes would cause drops of  one millivolt per megohm per nanoamp. Ignore it.

In  fact, the practice is to ignore it for all AC amplifiers. For AC amplifiers, it's generally better to tie the + input to a low impedance point to get lower thermal noise than you'd get from a high impedance resistor.

The only reasons to use input biasing resistors are (a) you're injecting a signal into the + input, so the resistor forms most of the input impedance, at least for FET input amps, and (b) if both the + and - inputs have the same equivalent resistance to bias, then the DC drops caused by the bias current in the bias resistors balance and do not add an offset voltage to upset DC accuracy. There are very, very few effects applications where this degree of DC accuracy is needed.
R.G.

In response to the questions in the forum - PCB Layout for Musical Effects is available from The Book Patch. Search "PCB Layout" and it ought to appear.