Need help with opamp bias

[CPmkI]

Hello!

I'm trying to learn a bit more about power supplies and op amps by both simulating and building a simple buffer circuit and measuring the voltages. I've drawn this up in circuitlab, and all the voltages are exactly where I want them within insignificant error.



Then I breadboarded it, and the voltages are way off.

I measured 8.98v before the diode, and 8.75v after. Then 8.58 after R2. I get 4.3v after the voltage divider, and then 3.85 at the non-inverting input. Isn't this too far off from the 4.5v I'm aiming for?

I searched this forum before posting this question, and found this article by RG Keen: http://www.geofex.com/circuits/Biasnet.htm but I wasn't able to understand the role played by input bias current. I can't find the input bias current on the TL071 datasheet anyway, nor would I know what to do with it if I had it.

The first big drop seems to happen after the diode (though I thought I read somewhere that 5817s were already on the low side for voltage drops). I've read an article that made a case for keeping the protection diode in series with the power supply (rather than in parallel to ground as I usually see). Is this really necessary?   

Then, from some article on Electrosmash, I know that R2 forms a LPF with the C4 and C5 to filter out power supply ripple, but I don't know if I would compromise this functionality by lowering its value to avoid as big a drop across it. Also on Electrosmash, I found the suggestion to keep R1 at least 10x bigger than the resistors forming the divider, which I've done with R1 at 1M. That 1M seems pretty standard across many schematics I've come across, but it's also causing a decrease in the bias voltage.

Does anyone have any suggestions for where I can look next so that I can better understand what's happening here?

[Vivek]

The time constant of the voltage divider seems to be very high

100k ohms and 47 uF.

Maybe reduce the 100k to 10k

If you have large time constant, you need to measure voltages 30 seconds after power up


Voltage are + pin will be Vref minus the drop on the 1M resistor.

If you found the bias current, use Ohms law V=RI

R is 1M, I is the bias current

V will then be the voltage drop on the 1M.


That 1M also helps determine the input impedance of your circuit. You guitar purposes, it needs to be high, let's say at least 500k for a clean buffer (maybe one can get away with even 30k for fuzz applications that create a lot of high frequencies)

[antonis]

For TL071, consider actual voltage drop across R1 1M bias resistor negligible..
3.85 instead of 4.3 is due to DMM loading..
(meter's impedance set in parallel with R1)

230mV drop for 1N5817 seems fine..

Although R6 & R7 could be as low as 10k or so (as Vivek said..) I can't see any obvious DC issue..

AS for AC, C1 could be ten times lower with no input HPF issue..
(16Hz instead of present 1.6Hz..)

edit: Welcome, of course..

[ElectricDruid]

Welcome CP!

+1 agree what Antonis said - the difference you're seeing between the sim and reality is the multimeter loading the circuit and changing the voltage you're reading. Obviously that doesn't happen in the sim, so you can read the exact voltage.

[PRR]

....3.85 at the non-inverting input. Isn't this too far off from the 4.5v I'm aiming for? ...

You want "centered", yes, IF you need maximum signal capability.

In fact, for e-guitar at unity gain, you probably don't.

But go on. I have a 9 foot garage and I want to divide it "in half" for two storage units. 4.5 feet exactly? Or would 3.85 and 5.15 be acceptable? Only a 33% difference. 16% each side. Many many audio problems have no "exact right" answer. If you are really hurting for 16% more signal swing, get a bigger battery.

And as said: never neglect meter loading. When you have MEG resistors, expect meter loading! Look up the specs for your particular meter. On your simulation, put a resistor that size where you will be putting the meter.


I guessed 10Meg, 5Meg, and came to about 6Meg loading in your meter. Is that what your specs say?

Ah, and I neglected your power diode drop, but I'll leave the small details to you.

(The 100Meg across the inputs at "Ideal" covers a quirk of this simulator model. It thinks I should have more wires. For extra credit, show how little difference this resistor makes.)

EDIT: re-running with 4.3V at the half-tap suggests a 9Meg meter.

[CPmkI]

Wow, thank you everyone for the help and the welcome!

I now not only have a more clear picture of what's happening in my physical circuit, but I also have new avenues to go studying down. Thanks for taking the time to explain this and to draw out diagrams, etc. Y'all're the best!

Here's to the least gatekept hobby on the internet!

[CPmkI]

EDIT: re-running with 4.3V at the half-tap suggests a 9Meg meter.

I will check this prediction when I get home.

[m4268588]

If you have any questions in bias, please do this.

However, do not do this from the beginning. You may miss a serious build error. (e.g. R1 is open)

[antonis]

I will check this prediction when I get home.

It's not a prediction rather than some simple maths..
You have 8.75V after diode so you should ideally (for exactly same divider resistors values) read 4.375V..
75mV offset (4.375 theoretical - 4.3 actual) might be due to both resistors values unequality and slight meter loading (100k is loaded about 1/10th of that of 1M)
As for 3.85 bias measurement, PRR explained you the deviation from theoretical value..
(who also told you not to worry about exact mid-supply bias both for unity gain buffer and not directy bias next stage, due to C3 DC blocking cap..)

P.S.
A DC measurement on TL071 output (pin 6) should be more accurate about pin 3  actuall voltage..

[CPmkI]

P.S.
A DC measurement on TL071 output (pin 6) should be more accurate about pin 3  actuall voltage..

So this makes total sense, thank you! However, after trying this, I got sucked down another troubleshooting rabbit hole... The voltage on pin 6 was hovering just over 5v. Not only that, but the 071 wasn't passing my guitar signal. With an audio probe, I found that my guitar was getting to pin 3, but wasn't getting through the op amp. I re-checked all the pins to make sure everything was getting the power it was supposed to. No problems with pins 4 or 7... but pin 8 - which should be left disconnected - read about 7.3v. The offset pins also had about a volt and a half on them. Now, I'm not an expert on this chip, but I figured something was wrong. I swapped it out for another 071, and it had exactly the same problem.

Instead, I re-routed a few things and plugged in through half of a 072 instead (different supplier) - and it worked perfectly. Could there be something off about these 071s I've got? I have been getting them from sketchy Chinese sellers on Taobao... but who would spoof such a common op amp?

The time constant of the voltage divider seems to be very high
100k ohms and 47 uF.
Maybe reduce the 100k to 10k
If you have large time constant, you need to measure voltages 30 seconds after power up

This is one of the paths I went to go study down after reading this thread. I'm still not sure I'm understanding how to calculate this. After the derivation of the formula, it should be as easy as R*C, right? When I do that, I get (100,000 Ohms)(0.000047 Farad) = 4.7 seconds. Yet, I tried measuring the voltages upon first powering up, and sure enough, it rose slowly over the course of about 30 seconds before stabilizing just like you said. Is this because 4.7 seconds only takes it to 63.2% charged? Am I correct to visualize a slow, logarithmic increase after that? You suggest lowering the resistor values; is that because 47uF is an appropriate value for this second filter cap? Other than choosing a cutoff frequency for noise in the power line, I'm not entirely sure how to go about choosing the value for the cap.

[antonis]

Other than choosing a cutoff frequency for noise in the power line, I'm not entirely sure how to go about choosing the value for the cap.

First of all, power consumption together with noise acceptable level set divider resistor values..
Considering input bias current to be reckoned with leads to resistor values inequality for effective mid-supply bias point AND appropriate cap value for acceptable bias voltage ripple, following the formulae: C = I_bias / (2 * f * V_ripple), where f = twice the mains frequency..
(Battery powered circuits clearly deduce the absence of respective cap..)

Another factor of cap value (and presence) is the possible use for AC ground point..
(in such a case, its value should be much bigger than that of individual hinged caps..)

Or just ignore all the above and plagiarize..