DOD 250 vs distortion plus power supply voltage divider

Started by Dboy78, May 08, 2021, 06:49:10 AM

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Dboy78

Hi all

I've been playing around with the DOD 250 and distortion plus circuits on a breadboard.  In the DOD 250, R2, R3 of 22k and R4 of 470k sets the voltage at pin 3 of the Lm741 at about 4.5V. In the distortion plus these three resistors are all 1M and sets the voltage at pin 3 at about 2.5V.

Is this correct?  I understand R2 and R3 act as a voltage divider.  Is the ratio between R2/R3 and R4 acting as a second voltage divider or is something else causing the difference?

I've really enjoyed building umpteen pedals in the last year, and I'm trying to improve my understanding by digging into a bit more detail.

Thanks



teemuk

Are you using FET input (high impedance) or BJT input (low impedance) opamp?

R4 should make no significant difference as long as opamp's internal impedance is high enough. The bias point should be about 4.5V in both circuits.

Dboy78

I'm fast approaching the limits of my knowledge here!  It's a LM741CN I've used and I've tried a few different ones.

Rob Strand

#3
It's your multimeter loading down the divider.  The multimeter probably has 1M ohm input impedance.

If you measure the voltage on the opamp output pin 6, assuming the opamp isn't faulty and the opamp is working as an amplifier, it is usually a good indication of the voltage on pin 3.    If the resistances on pin 2 are low then you can measure the pin 2 voltage and the multimeter will load it down less and that will also be an indication of pin 3; that's not so good in this case.

A better way is to estimate how much your meter is loading the circuit then undo that effect to estimate the true voltage.

The way that's done is through a Thevenin equivalent circuit.    The two 1M divider resistors + 9V will look like 4.5V in series with 500k.  Then you add another 1M resistor for the resistor to pin 3 giving 1M + 500k = 1.5M ohm in series with a 4.5V source.   Now if your meter is 1M impedance it will form a divider with the 1.5M resistor.    So you would predict Vmeasured(with meter loading) = 4.5V * 1M /(1M + 1.5M) = 4.5 * (1 / 2.5) = 1.8V.    So that's a little low.   If your multimeter was 2M ohm input impedance then you would get 2.5V.   2M input impedance isn't common so it's not quite adding up.  Sometimes your can check you multimeter manual.


The opamp input bias current can affect the measurement but  normally it contributes somewhat less than 1V error.   

A simple test for the multimeter impedance is to measure the 9V battery then put a 1M in series with the meter leads a measure it again.  If you get 4.5V then the meter is 1M input impedance if it you get 8.2V it's a 10M input impedance meter.  In between you have to do some calculations.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

Mark Hammer

One can provide a Vref with any pair of equal-value resistances you want, from 470R/470R to 3M3/3M3 and beyond.  The resistance between their midpoint and V+ limits the current available to feed those points requiring a ground/Vref connection.  This limitation of current is further reduced/limited by the resistance between the Vref point and the input.

In the case of the DOD250 those two resistances are 22k+470k.  In the case of the Dist+, those two resistances are 1M+1M.  So the bias voltage (4.5VDC) may be the same in both instances, but the bias current is MUCH lower in the Dist+.  I don't know nearly enough about such matters but suspect that this lesser bias current plays a role in the character of the Dist+.  Many people who have built it note that it is nigh impossible to get them to sound "clean", even when there are no clipping diodes in circuit.  I suspect the chip is "crippled" by the low bias current.  That does not mean one can simply deploy this strategy with other op-amps.  It may well be unique to earlier op-amps like the 741.

mozz

About 250 micro amps (ua) i think, just toyed around with these a few weeks ago. I don't know if the current changes with signal but if it does the higher ohm resistors will sag somewhat compared to the circuit with the 22k's. Always found the voltage to be somewhat slightly less than half, don't know if that makes it clip earlier or not.
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iainpunk

its probably a combination of the opamp's input current and probe loading.
what DC voltage do you read on the output pin of the opamp? the DC operating point is on both sides of the opamp the same, so to negate the loading caused by the multi meter, you can measure the output voltage in DC mode.

a main reason for the ''never getting quite clean'' of the DIST+ is that they use the LM741 opamp in stead of the uA741 used in the DOD250. the LM has generally more crossover distortion, where the uA has quite a bit less.
i know this is controversial to state, since in theory they have the same internal schematic, but having tested them on the scope* for crossover distortion1 , the batch of LM741's were almost as bad as the LM358**, and the uA741 i tested had  crossover at the falling edge only just like the only JRC741 in my batch

cheers, Iain

* using a 1V peak triangle wave in to a voltage follower configuration and a capacitive load of 100nF in series with a 100ohm resistor.
** the LM358 does have a different ''kind'' of crossover distortion than a LM741, a kind of ''zig zag'' shape, instead of a wobbly flattening
1 i'm a bit of a crossover connoisseur
friendly reminder: all holes are positive and have negative weight, despite not being there.

cheers

mozz

Don't ever look at EL84's on a scope then! I have UA741 and LM741 here (and a few others), going to see what you're talking about.
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Dboy78

Thanks all.

Rob - I've got two multimeters, and testing as you suggested shows one has 1M and the other 10M impedance. When I've gone back to measure input and output of the op amp with each multimeter I get this:

10M DMM
Pin 3 3.8V
Pin 6 4.4V

1M DMM
Pin 3 1.8V
Pin 6 4.5V

So just as you said. I think knowledge has improve slightly!

If I was going to build one of these with either sockets or switching between Ge and Si diodes, would you go with the DOD 250 component values or the distortion plus? Or does it make no odds?

Thanks

iainpunk

Quote from: mozz on May 08, 2021, 01:01:41 PM
Don't ever look at EL84's on a scope then! I have UA741 and LM741 here (and a few others), going to see what you're talking about.
a single EL84 can't do crossover distortion if you don't go out of your way to actually design it in to the circuit.
for crossover distortion in opamps; the bigger the capacitance is compared to the resistance, the worse it distorts.
the point and shape of cross over also shifts at larger capacitance.
even the Falstad simulation shows this crossover distortion:

the green one is what comes out, red is the input signal.

now i think about it, i don't think any 741 should have crossover distortion at such a high resistance circuit, maybe its worth testing out in this specific circuit as well.
i used to be obsessed with crossover distortion, in all its variants, so i was experimenting with how to create it, not in this specific circuit, so i'm not sure about this circuits crossover.

sorry for the Off Topic above
i suggest you go with the DOD250 values, since they seem to be a bit better behaved when it comes to noise and such.

cheers.
friendly reminder: all holes are positive and have negative weight, despite not being there.

cheers

antonis

The lower the value of R2/R3 the more stiff the voltage divider..

I never got the reason for using 1M/1M divider with 1M bias resistor..
It looks to me as a very bad design, despite its origin..
(in the mean of, any 9V pedal which can't afford 450μA current consumption (10k/10k) for bias purpose isn't an "upright" design one..)
"I'm getting older while being taught all the time" Solon the Athenian..
"I don't mind  being taught all the time but I do mind a lot getting old" Antonis the Thessalonian..

Rob Strand

Quote10M DMM
Pin 3 3.8V
Pin 6 4.4V

1M DMM
Pin 3 1.8V
Pin 6 4.5V
Looks like the answer.   (FYI, low voltages due to meter loading must have have come up on this forum a few hundred times now.)

QuoteIf I was going to build one of these with either sockets or switching between Ge and Si diodes, would you go with the DOD 250 component values or the distortion plus? Or does it make no odds?
The 2x22k vs 2x1M won't be noticeable.   Perhaps very small improvement in noise immunity with 22k and the larger cap.
The 470k to 1M  might be noticeable.  It affects the low and highs by very small amounts.


QuoteThe lower the value of R2/R3 the more stiff the voltage divider..

I never got the reason for using 1M/1M divider with 1M bias resistor..
It looks to me as a very bad design, despite its origin..
(in the mean of, any 9V pedal which can't afford 450μA current consumption (10k/10k) for bias purpose isn't an "upright" design one..)

I don't think it's a "bad" design.  For a single opamp circuit with nothing else on Vref there's no obvious problem using the 1M's.
Also, for a single opamp circuit the 450uA with 10k's would contribute a reasonable percentage of the power without offering much in return.   I agree the 1M's are probably going too far and perhaps 100k's would be a better trade off.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

mozz

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iainpunk

crossover distortion is a lot of times not actually at the 0v point, that point shifts when the load isn't perfectly on the real axis. it shifts ''back'' when the load is more capacitive, and it shifts ''further'' when induction takes over.

cheers
friendly reminder: all holes are positive and have negative weight, despite not being there.

cheers

Rob Strand

Quotecrossover distortion is a lot of times not actually at the 0v point, that point shifts when the load isn't perfectly on the real axis. it shifts ''back'' when the load is more capacitive, and it shifts ''further'' when induction takes over.
Zero crossing for current would be closer.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.