Opamp biasing troubleshoot.

[TheWinterSnow]

I have designed a little circuit for direct recording without a guitar cabinet for tube amplifiers.  I have an op-amp buffer that can be switched from a Voltage Follower to a Non-Inverting Amp.  Because I am using a single 9V supply I use on part of a RC4558C to give a virtual ground.  I am using the virtual ground (4.9V) to bias the input of the non inverting pin to (3.5V).  The 3.5V on the input pin (A) is due to the 100K resistor to properly load the input of the previous stage.  Under the normal voltage follower condition, the output (C) is giving ~4.7V which is weird to me considering it is a voltage follower with an input of 3.5V; input (B) is also 4.7V which is expected considering it is directly connected to output (C).  When the switch is applied, the Op-Amp is now configured in a Non-Inverting configuration with a gain of ~40dB (a 20K Rf and a 1K Ri), this is where things get really weird.  The Non-Inverting Input (A) stays at its 3.5V which is expected.  The Inverting Input (B) rises from 4.7v to 4.9v as it is now connected to the virtual ground via voltage divider.  Because the pins (A) and (B) have a difference of ~1.5v, output (C) is now 1.5v and clips all the negative peaks of the signal regardless of the input signal.  I know this is because of the voltage difference of inputs (A) and (B).  My question is, how would this be remedied in design without going to a dual rail design?  Or is it even possible?  This would mean that the 40dB boost would have to be removed.  I am out of ideas as how to achieve virtually the same voltage of inputs (A) and (B).

To sum the DC operation in a chart:

VCC = 9v
Vr = 4.9v

Voltage Follower:
A (+) = 3.5v
B (-) = 4.7v
C (Out) = 4.7v

Non-Inverting Amp:
A (+) = 3.5v
B (-) = 4.9v
C (Out) = 1.5v

And the schematic


Hopefully someone can help.

[Gurner]

There's clearly a voltage drop going on across that 20k resistor in voltage follower mode .....IMHO that 20k resistor should be shorted when you want the opamp to be in true voltage follower mode (because only then can the -ve pin 'B' exactly follow the op pin 'C')

Re having 3.5V DC at the +ve pin 'A' in voltage follower mode....that doesn't look right to me - it should be pretty near 4.9V (on account there's cack all current flowing into that +ve pin, therefore very little voltage drop across the 100K resistor.....try lifting the feed into point A (+ve pin) & see what voltage you have at the top of the 100k resistor then.

Unless you get it functioning as expected as a voltage follower, then it's pointless even trying to troubleshoot in non inverting opamp mode.

[TheWinterSnow]

The Voltage follower is operating correctly from what I am gathering.  The output voltage does equal -ve.  That just goes out the window in Non-Inverting mode.  The 3.5v bias on +ve is a little odd considering the input impedance of the RC4558P is 5M, but when an AC signal is added, the output operates as if it was at 4.5v, meaning the output clips equally on both the positive and negative portion of the signal.  Everything only goes haywire when it does to the Non-Inverting amp.

You are right about +ve being near the expected 4.9v, as the voltage drop of the 100K resistor considering the Ri=5M of the chip should only be ~4.4mV.  The fact that there is substantial current flowing through the 100K resistor would lead me to believe that Ri of +ve is around double that of the 100K considering the voltage drop.

[merlinb]

The Inverting Input (B) rises from 4.7v to 4.9v as it is now connected to the virtual ground via voltage divider.  Because the pins (A) and (B) have a difference of ~1.5v, output (C) is now 1.5v and clips all the negative peaks of the signal regardless of the input signal.  My question is, how would this be remedied in design without going to a dual rail design?  

You need to add a large capacitor in series with the 1k resistor so that you are only amplifying AC, not the DC offset at the input.  10uF or so, should do it.

You are right about +ve being near the expected 4.9v, as the voltage drop of the 100K resistor considering the Ri=5M of the chip should only be ~4.4mV.  The fact that there is substantial current flowing through the 100K resistor would lead me to believe that Ri of +ve is around double that of the 100K considering the voltage drop.

Could it be your voltmeter loading the +ve input, thereby giving a false reading?

[Gurner]

The Voltage follower is operating correctly from what I am gathering.  The output voltage does equal -ve.  That just goes out the window in Non-Inverting mode.  

Actually I missed that hadn't had my morning coffee!

The out of whack readings (eg 3.5V on +ve pin A ....& the output pin C being different to the input +ve pin) suggests the opamp might be suspect here  ...have you tried another in its place?

[TheWinterSnow]

You need to add a large capacitor in series with the 1k resistor so that you are only amplifying AC, not the DC offset at the input.  10uF or so, should do it.

So you are saying that you should not have a reference voltage at inverting input?  I think I understand, by adding a cap in series, we are only applying the voltage divider to the AC signal, as the AC signal coming out of the output of the OP amp I assume already has a DC offset from the +ve input, therefore there is no need for the reference voltage at -ve, the output will provide that and will match that of +ve.

Could it be your voltmeter loading the +ve input, thereby giving a false reading?

I thought that.  I am using an old oscilloscope to measure my DC values and I have no clue what its input impedance is or if it has drifted over the years.  When I went into dual channel mode to see the difference between ve+ and ve-, in v-follower mode the inputs go down and up respectively to match each other at around 4.7v.  In the Non-Inverting mode, however, they stay at their respective levels.

OT:  Merlin, I love your books, I have learned more from the preamp book when it comes to design than my two year AS Degree in Electrical Engineering.

The out of whack readings (eg 3.5V on +ve pin A ....& the output pin C being different to the input +ve pin) suggests the opamp might be suspect here  ...have you tried another in its place?

When I was soldering I didn't notice but one of the chips got screaming hot, I forgot which one it was.  That may have been the one and it may have gone when I was soldering it in.

[merlinb]

You need to add a large capacitor in series with the 1k resistor so that you are only amplifying AC, not the DC offset at the input.  10uF or so, should do it.

So you are saying that you should not have a reference voltage at inverting input?

In theory it will work the way you have it, and it is often done that way, but adding the cap at least eliminates one possible cause of the problem by reducing the gain to unity at DC, so input offsets are not amplified.

[PRR]

> A (+) = 3.5v

What kind of meter? Digital or needle-meter? Cheap/rugged needle meter?

A DMM or VTVM will show 4.85V.... the drop is due to 10Meg meter loading on 100K resistor.

A 10K/V passive meter on the 25V range will be a 250K load on the 100K resistor and will give 3.5V when connected.

[teemuk]

So you are saying that you should not have a reference voltage at inverting input?

You already have one at non-inverting input. The output and inverting input will automatically (at least try to) bias to that.

Without capacitive coupling in the negative feedback the stage will amplify any DC offset, meaning DC voltage difference between input and reference. Equally as it is amplifying any difference between AC voltage and reference.

That is not a problem if offset is very low but when you have a circuit that performs well as a unity gain follower and goes off balance in gain mode then it very much hints about considerable amplification of DC offsets being a problem.

By the way, the inputs should see a reference of 4.9V all times. If they don't something is wrong in the circuit. You don't make a 4.9V reference to bias the inputs to 3.5V. The 4.9V reference SHOULD bias the input to 4.9V. If it doesn't you are doing something wrong.


...oh, and do pull off that 1uF cap at the buffer output. Or at least put some resistance at the buffer output. Opamps do not want to see a direct capacitive load, the charge currents of that cap might trip the current limiters within the opamp, and it's really doing no good there in the first place. The OpAmp buffer is already doing its best pushing output to equal the voltage at non-inverting input. The extra filter cap will likely just fight against that.