Voltage follower bias current (opamp)

[bour1992]

So in the input of this buffer circuit I have a small AC signal (μV) that is capacitively coupled to an op amp.
What is the role of R?
I read that it is to provide a bias current path and it has to be large in general but I can't understand it.

Any help much appreciated

 

[Baluncore]

The CR pair is a high pass filter.
Low input frequencies appear across the capacitor and so do not reach the follower.
High input frequencies appear across the resistor and drive the voltage input to the follower.
Over long periods the follower input will average zero volts.
The value of R should be low to minimise voltage offset due to op-amp input bias current.
The value of R should be high so a small capacitor can be used to get the C * R product for the cut-off frequency needed.

 

[bour1992]

Thank you for the answer.
I know that the CR pair in the input forms a high pass filter and the consequences of that.
The thing that I still can't understand is that we put the R as a path for bias current.
Somewhere I read that without R, the op amp would be in saturation constantly.
Why is that?

 

[Baluncore]

without R, the op amp would be in saturation

Because the capacitor would be charged by the input bias current until the op-amp input voltage exceeded the input voltage compliance range.

 

[Formagella]

the op amp needs a bias current and if there is no R, where does this constant current come from/go to? Remember that the capacitor behaves like an open circuit at DC frequency.

In reality it will charge up, and thus the output of the buffer will grow too until it saturates, then stuff will happen but you'd have to look at the op amp internals and see what happens to the transistors to know exactly.
You don't care though, the point is that the op amp always saturates after a transient and so it's useless.

 

[bour1992]

I begin to get it now.
But I am still a bit confused because if we wanted the bias current to go to the op amp we would connect R to +Vdd.

So the bias current flows out of the op amp and to the ground through R?

 

[Jony130]

Watch this video

 

[Baluncore]

So the bias current flows out of the op amp and to the ground through R?

Yes. R is used to prevent the continued integration of bias current in C.
The steady state is reached when the average voltage across R is stable, when V/R = I_bias for the op-amp input.

The voltage offset resulting from the product R * I_bias, can be canceled by replacing the op-amp negative feedback link with a parallel RC pair having the same values as the high pass filter.

 

[jim hardy]

see if this helps - section on 'input bias current'

http://www.physics.unlv.edu/~bill/PHYS483/op_amp_datasheet.pdf

Opamps being man-made thus imperfect have a tiny current flowing into or out of their input pins, not the zero we were taught in lectures on ideal opamps..
That resistor gives it someplace to go without causing much voltage drop.
"Input Offset current" is the difference between bias currents for the two input pins.

Bias current ranges from microamps to femtoamps depending on the opamp family.
R must be low enough that the bias current won't make significant voltage across it.
In a good design you'd add an equal R in series with the -input pin and so that bias currents make equal voltage drop in the paths to both input pins..

It's worth spending some hours looking at datasheets, figuring out every line.

See "OpAmps for Everyone", by Texas Instruments Staff.

old jim

 

[NascentOxygen]

What is the role of R?
I read that it is to provide a bias current path and it has to be large in general but I can't understand it.

All amplifier inputs fundamentally require a tiny amount of DC. It either flows into the input, or it flows out. It's usually tiny, but you still have to allow for it, otherwise the amplifier won't work. In comparison with your signal currents to that input, the bias current is usually tiny (and easily overlooked).

Just don't forget it.

 

[sophiecentaur]

I seem to remember that OpAmps should have the same DC bias current into both inputs - to help with common mode rejection. Not always relevant - as in this case, perhaps.

 

[jim hardy]

Ideally the bias currents would be zero, failing that exactly equal.
In practice they are nearly equal, look at parameter"Input Offset current".
which is the difference between the two Input Bias currents.
for good opamps it's typically an order of magnitude smaller than the bias currents .
http://www.ti.com/lit/ds/symlink/lm741.pdf
for 741A typical bias current = 30na, offset = 3na.
Of course, the A suffix is the cream of the crop...

 

[Averagesupernova]

When I worked for a test equipment manufacturer we used some op-amps that had FET inputs. Now admittedly nothing is a 100% open circuit but those op-amps did the job of an ideal op-amp better than anything I have ever seen.

 

[psparky]

All that above is correct and neat stuff.

Here's some things right out of the college handbook...
I always like to consider the transfer function and bode plot.

A voltage divider tells us this:
R/(R+ 1/jwc)

Simplifies to your basic high pass filter of:
JWRC/(JWRC+1)

When W is zero, your output is zero. Makes sense because in DC that capacitor is just going to charge up and there will be no input.
Put W to infinity and you get a gain of 1. Makes sense as well since capacitors essentially short at super high frequencies.

You asked what is the point of R in this circuit.

Well, the break frequency of the circut occurs when the frequency =1/(2*pie*R*C)...or when W=1/RC.

So you can see that you can adjust when you want our frequency to no longer attenuate.

 

[Baluncore]

Without R present, R will have an infinite value, so the C*R high pass circuit will have an infinite time constant. When turned on, the capacitor charge will be zero. Without R there will be no way of charging C to the voltage needed for AC coupling or DC level conversion. A high pass filter requires both C and R be present. Without R, you may as well replace C with a wire link.

Op-amps are imperfect and have an input bias current. That requires bias current cancellation, or some way to prevent integration of bias charge in C. I_bias*R decides the maximum acceptable value of R and hence the physical size of C needed for a particular signal frequency.

Very high impedance AC coupled amplifiers used in the biological sciences often require a “de-block” state in which the input capacitors are charged to bring their voltages close to that needed for the common mode input range of the amplifier. Once “de-blocked” the very high impedance inputs place minimal loading on the biological system. The shields on the input cables are “bootstrapped” from buffer amplifier outputs to neutralise the cable capacitance.

 

[psparky]

It's probably obvious to most, but since no current can go thru an ideal amp, not having a place for the current to flow there isn't much going on. In other words, Vin would equal Vout.

So at steady state, all the current flows thru the "C" then thru the "R".

Do we even need the voltage follower in this circuit? Without a gain what's it's purpose in this circuit?
I know you mention the trickle current thru amp at transient points...does this do anything to "help" the circuit?

Sorry if its mentioned above or in the video (haven't had chance to see video yet). But still not clear what the voltage follower is for in this circuit.

 

[jim hardy]

Without a gain what's it's purpose in this circuit?

Voltage follower is often used as a "buffer" .
Imagine if you tried to read the voltage across R with an analog voltmeter. You'd change the circuit by touching the meter lead to top of R,, R would become R in parallel with whatever is resistance of the meter. A good meter is 20kohms per volt, so on 10 volt scale it'd be 200K in parallel with R. A good 'scope is 1 meg...

So the follower "buffers" the signal , assuring that your measuring device doesn't eaffect what you're using it to measure.
http://www.diffen.com/difference/Affect_vs_Effect

I know you mention the trickle current thru amp at transient points

That trickle isn't just at transients, it's continuous. That's one of the differences between ideal and real opamps. But it's typically nano to picoamperes so it gets ignored until you try to make a long time constant integrator or sample&hold from a general purpose opamp..
See "Input Bias Current" on any opamp datasheet

 

[psparky]

Ahhhh... Gotcha... Gotcha.

 

[Baluncore]

Do we even need the voltage follower in this circuit? Without a gain what's it's purpose in this circuit?

Another way to look at this is that a voltage follower is an impedance converter. It converts the high input impedance to the lower output impedance without changing the voltage. The output power is then greater than the input power because, while the voltages remain the same, the lower output impedance has a greater current flowing than flows in the input impedance. That is why an active amplifier device with an external power supply is needed to implement a voltage follower.