Dbx Vca Construction

[damnyankee]

Thank you, John.

That's the rub with my 119 200 VCA: R1 is not installed and R10 is only 27k (the 160VU's R1 is 13k & R10 is 100k).  My negative pull down resistor (R8) is 4.7k - the same value as in the 160 VU's VCA.  The JFET on both 119 & 160VU VCAs is the U2749. 

On paper, will I harm anything if I install the R1 (13k), change my R10 from 27k to 100k, and change my R2 from 5.6k to 17k which, in effective, turns the top circuit on?  Or will this negatively impact my Q4 & Q5 (BC479...which is BC263C on the 160VU VCA)?

I don't want to install/swap out resistors and take a chance on popping those matched transistors before I get a chance to check some voltages!

Thank you again, John!

 

[Biasrocks]

I don't want to install/swap out resistors and take a chance on popping those matched transistors before I get a chance to check some voltages!

I think what John is essentially saying is that the 119 and 160 VCA's are both equivalent even thought they use different transistors and resistor values. My guess is that the 27k resistor is all they needed to get the desired result from that particular VCA. Messing with those values will surely compromise the VCA.

My suggestion is to build it and see how it sounds and performs before you go messing with anything. I suspect you will find that your VCA will perform like the original 160.

Mark

 

[damnyankee]

I think what John is essentially saying is that the 119 and 160 VCA's are both equivalent even thought they use different transistors and resistor values. My guess is that the 27k resistor is all they needed to get the desired result from that particular VCA. Messing with those values will surely compromise the VCA.

My suggestion is to build it and see how it sounds and performs before you go messing with anything. I suspect you will find that your VCA will perform like the original 160.

Mark

Thanks, Mark.  I'm sure it will function fine.  But please understand I've been a casual geetar pedal builder, so most of my questions are more of an inquisitive nature.  Those BC479's in my VCA have half the gain as the BC263C.  Do the lower resistance components in my VCA push those BC479's?  Do the higher resistors in the 160VU VCA effectively lower the gain in the BC263C?  Can I swap resistors to "overclock" the JFET / opamps without creating more distortion or popping transistors?  I find electronics very fascinating and unfortunately, I only have limited schooling in Chemistry/Physics (E=IxR, stuff like that).

DY

 

[JohnRoberts]

Thank you, John.

That's the rub with my 119 200 VCA: R1 is not installed and R10 is only 27k (the 160VU's R1 is 13k & R10 is 100k).  My negative pull down resistor (R8) is 4.7k - the same value as in the 160 VU's VCA.  The JFET on both 119 & 160VU VCAs is the U2749. 

I don't know why this is so hard to follow...  all JFETs Idss or operating current when the voltage between gate and source is 0V will vary all over the place, even from the same production batch ... that's why they have to individually select the resistor values. One JFET draws .56mA (6.5V/11.5k) and the other .24 mA (6.5V/27k). They need the different "selected" resistors to hold the base of Q1 to the same voltage. If you have a VCA where these parts are already selected DON"T CHANGE THEM. If you use a third different JFET it will need a completely different new selected resistor value..

On paper, will I harm anything if I install the R1 (13k), change my R10 from 27k to 100k, and change my R2 from 5.6k to 17k which, in effective, turns the top circuit on?  Or will this negatively impact my Q4 & Q5 (BC479...which is BC263C on the 160VU VCA)?

I don't want to install/swap out resistors and take a chance on popping those matched transistors before I get a chance to check some voltages!

Thank you again, John!

What are you talking about... ?? Are you building a new VCA from scratch, or messing with one that already works..?  If it was working, and you change the selected resistors in it to resistors selected for another VCA it will not work properly.

R1 and R10 (in any order) are selected for the variation in JFET (u2749) Idss, used in the gain stage. R13, R11, and R2 are selected for the match of the VCA transistors


JR

 

[W DeMarco]

I have to say my DIY builds are far less noisey than any of the old school DBX comps we have.  To the point where I hate the bastards...

 

[damnyankee]

I don't know why this is so hard to follow...  all JFETs Idss or operating current when the voltage between gate and source is 0V will vary all over the place, even from the same production batch ... that's why they have to individually select the resistor values. One JFET draws .56mA (6.5V/11.5k) and the other .24 mA (6.5V/27k). They need the different "selected" resistors to hold the base of Q1 to the same voltage. If you have a VCA where these parts are already selected DON"T CHANGE THEM. If you use a third different JFET it will need a completely different new selected resistor value..


JR

You do this for a living, I do Medicine for a living - which is why it's a little difficult for me to follow.  Believe me, I get patients who ask me endless questions and I sometimes it takes me exploring several different angles until it finally bells in to them.  I know it can be frustrating - and I apologize for frustrating you.

If the JFET is drawing .56ma at 9V, then the selected resistor must be 16k, correct?

And, altering the resistor value affects the current going to the JFET, correct?

Higher valued resistors lowers current to the JFET and it won't operate correctly.  On the other hand, Lower valued resistors increases current to the JFET and it won't operate correctly...and if current is too high, *Pop! goes the JFET*.

In the simplest of terms, is this basically correct?

DY

 

[JohnRoberts]

You do this for a living, I do Medicine for a living - which is why it's a little difficult for me to follow.  Believe me, I get patients who ask me endless questions and I sometimes it takes me exploring several different angles until it finally bells in to them.  I know it can be frustrating - and I apologize for frustrating you.

Think of electricity as blood flow...

If the JFET is drawing .56ma at 9V, then the selected resistor must be 16k, correct?

That particular JFET draws .56 mA at any rail voltage when the gate to source voltage is 0V. This output current is like a current source so it doesn't care what the rail voltage is. In fact that drain is clamped by the base of Q1 so it can't go above 1V.  Since we want to bias the base of Q1 at roughly one volt (the Vbe of Q1, plus the drop of CR1), we use a resistor between the drain of the JFET and Q1 base  and +7.5V supply that will supply roughly the same current as the JFET is pulling down. If you know the beta of that transistor you can calculate that base input current too..

With a 9V supply we still get .56mA from the JFET and still want to end up with 1V to drive the transistor, so the resistor must now drop 9-1=8V. 8V/.56mA= 14.3k

And, altering the resistor value affects the current going to the JFET, correct?

No.. The current is a function of the FET gate voltage relative to source. The selected resistor defines the total current coming down from the plus supply into the base, which minus the current the JFET is sucking back out controls Q1.
 

Higher valued resistors lowers current to the JFET and it won't operate correctly.  On the other hand, Lower valued resistors increases current to the JFET and it won't operate correctly...and if current is too high, *Pop! goes the JFET*.

In the simplest of terms, is this basically correct?

DY

NO.. The current in the JFET is a function of the gate voltage relative to the source. If the bias resistor (R1,R10) is too large a value, it will not deliver enough current to overcome the current pulled down by the JFET Q1 will not bias up and turn on. If the resistor is too small a value it will drive the base of Q1 too hard for the JFET to overcome and control.

Now this static explanation is a little simplified, since the JFET will only have 0Vgs on it at equilibrium. If the VCA is getting too little or too much pull up current from Q2 it will pull the gate up or down one diode drop when it saturates the VCA transistors. But the JFET will draw less than Idss for negative gate drive, and slightly more than IDSS for a positive gate drive so this is additional negative feedback and stabilizes the current in the VCA. 

The important concept to grasp, is that the output of the JFET is a pull down current, and the selected resistor supplies a similar amount  of current pulling up on the base of Q1. Equilibrium is when these two currents net out close to zero, give or take a little base current to turn Q1 on.

JR

 

[abechap024]

JR Thanks for explaining all of this!
I know all this is answering a lot of my own questions too.

I really liked the juicy details of how to set up the Jfet! That was really what I was hoping for when I started this post. It doesn't make 100% sense to me at the moment, But that is because I don't have the circuit in front of me and trimmers and a voltmeter. But when I do have all the stuff and am starting to trim, those posts are going to be invaluable!

 

[damnyankee]

Hi John,

Thank you for the blood analogy:

Current = blood flow
Voltage = blood pressure
Resistance = vascular system (vasoconstriction/vasodilation/plaque, etc)

I think it's finally starting to click.  It's not current necessarily, but VOLTAGE.  So suppose we have an 11V rail.  Our JFET is only going to draw .56 mA.  We want the base of Q1 at ~1V (the voltage we need to drive it).  So we need that resistor to drop 10 Volts, so that resistor value should be about 17.8k.

Now the big picture is, different transistors may have different current/voltage needs (in the case of my original question regarding BC263C vs BC479).  As such, this is why the resistor configurations differ between them.  I know there's alot more to it, but if I have the big picture, then I can dig down to the next layer and learn more of the details.  It's more complicated than Ohm's Law - and I know this is probably boring stuff to you, but I find this is extremely fascinating.  I understand now more than ever the importance of calculations and voltage measurements in circuits.

Thank you again,

Steve

 

[JohnRoberts]

Hi John,

Thank you for the blood analogy:

Current = blood flow
Voltage = blood pressure
Resistance = vascular system (vasoconstriction/vasodilation/plaque, etc)

Water is more commonly use in this analogy, but that works.

I think it's finally starting to click.  It's not current necessarily, but VOLTAGE.  So suppose we have an 11V rail.  Our JFET is only going to draw .56 mA.  We want the base of Q1 at ~1V (the voltage we need to drive it).  So we need that resistor to drop 10 Volts, so that resistor value should be about 17.8k.

Close enough,,,  FETs are mainly voltage input devices, while transistors are mainly current input, but not to confuse things, the VCA application is specifically using transistors with voltage inputs, to control how the pairs of devices divide the current between themselves. There is a logarithmic relationship between voltage and current in transistors Vbe that is used for multipliers and sundry computational circuits.

I know this sounds confusing the base of transistor Q1 is being driven by the difference in current between what the JFET is sucking out of the base and the resistor from 7.5V is feeding in. The base will end up at roughly two diodes drops above ground as a consequence of circuit topology. 

Now the big picture is, different transistors may have different current/voltage needs (in the case of my original question regarding BC263C vs BC479).  As such, this is why the resistor configurations differ between them. 

It appears they are seeing the same voltages and current. Max VCA current is defined by R8 and R4. Any differences in R2, R11, and R13 are due to quality of device matching, not the nature of the devices.

It seems that significantly different transistors if substituted might benefit from slightly different r6 and r7 that adjusts class A current transition from class A to class A/B VCA. Higher class A current trades more noise for less HF distortion, also scaling of the input resistor has a similar effect of trading noise for distortion. So when making noise comparisons between similar VCAs the input resistors need to be the same value to make an apples to apples comparison. 

I know there's alot more to it, but if I have the big picture, then I can dig down to the next layer and learn more of the details.  It's more complicated than Ohm's Law - and I know this is probably boring stuff to you, but I find this is extremely fascinating.  I understand now more than ever the importance of calculations and voltage measurements in circuits.

Thank you again,

Steve

It has been years since I thought about this stuff... I would never try to better what they have accomplished in the new generation of IC VCAs.  That said I am not a huge fan of that JFET gain stage but it might be part of the sound character (for better or worse) that people are trying to recreate. 

JR

 

[damnyankee]

Thanks again, John.  There are a number of us who have gained a great deal of knowledge based on your explanations on the inner workings of these old VCAs.  I know we greatly appreciate you sharing your expertise!

DY

 

[damnyankee]

John,

I have another question - it's about filter caps (35v 470uF found stock in the dbx 160VU).

I know increasing uF value can improve filtering and improve sound (tighten up the bass).  So I'm looking at using a Panasonic FM 35v 1800uF 105 degC with a very high ripple current (3820 mA) or a larger Panasonic NHG 35V 4700uF 105 degC with a ripple current of 1900 mA.  Is there an advantage using the 4700uF filter cap in the 160VU?

Thanks in advance,

DY

 

[JohnRoberts]

Sorry I don't engage in the silly stuff...

JR

 

[damnyankee]

Hi John,

I figured as much.  I told AC earlier today I wish I could ask Jim Williams as he does this sort of thing, but he hasn't been on the other board for quite a spell.  Well, thanks any way - I do appreciate the time you took last fall to answer questions about your Loft Flanger (SAD1024's, stereo linking, etc) and now on this dbx VCA.  8)

Take care,

DY 

 

[JohnRoberts]

Jim Williams is around, I've seen posts from him on another forum within last few days.

I believe he modifies old products for money, so he may just not want to give away for free what he makes his living from. That is understandable.

JR

 

[damnyankee]

Jim Williams is around, I've seen posts from him on another forum within last few days.

I believe he modifies old products for money, so he may just not want to give away for free what he makes his living from. That is understandable.

JR

Hi JR,

Actually, Jim has told posters a few times on the other board to yank the stock 35v 470uF caps and replace with the largest size 35v that can be stuffed on the board (for the 160, 119, 118).  He said he uses 35v 4700uF caps.  The bigger caps "removes noise and makes the bass more punchy". 

I guess I can understand the noise part, but I don't understand why it would affect the audio as they aren't directly in the audio path.  "Smoother" supply to the rails???  More reduction of hum within the sound???  Does increasing uF affect frequencies tho the caps aren't in the audio path?

I've researched this on the web and to be honest, there isn't a good scientific explanation why filter caps impacts audio.  Just subjective comments.  Some of those comments recommend filter caps with high ripple currents, filter caps with low impedance, and even pro/con of certain caps such as Nichicon Muse caps.  One person claimed the Muse are the greatest thing since sliced bread yet another person stated the Muse caps are overpriced, re-labeled Nichicon KWs in a pretty black/gold jacket.  I've also read too high of a uF rating may overtax the trafo.

So I'm trying to research to see if I should just purchase 35V 470uF Panasonic FMs for $0.44 or get something like a Nichicon PW 35v 4700uF rated at 105 degC and a ripple rating of 3800 for a couple of bucks each or get something in between.  That's where I'm at because I'm just not educated about filter caps other than they filter/smooth out AC ripple in a power supply.

 

[JohnRoberts]

Hi JR,

Actually, Jim has told posters a few times on the other board to yank the stock 35v 470uF caps and replace with the largest size 35v that can be stuffed on the board (for the 160, 119, 118).  He said he uses 35v 4700uF caps.  The bigger caps "removes noise and makes the bass more punchy". 

I guess I can understand the noise part, but I don't understand why it would affect the audio as they aren't directly in the audio path.  "Smoother" supply to the rails???  More reduction of hum within the sound???  Does increasing uF affect frequencies tho the caps aren't in the audio path?

I've researched this on the web and to be honest, there isn't a good scientific explanation why filter caps impacts audio.  Just subjective comments.  Some of those comments recommend filter caps with high ripple currents, filter caps with low impedance, and even pro/con of certain caps such as Nichicon Muse caps.  One person claimed the Muse are the greatest thing since sliced bread yet another person stated the Muse caps are overpriced, re-labeled Nichicon KWs in a pretty black/gold jacket.  I've also read too high of a uF rating may overtax the trafo.

So I'm trying to research to see if I should just purchase 35V 470uF Panasonic FMs for $0.44 or get something like a Nichicon PW 35v 4700uF rated at 105 degC and a ripple rating of 3800 for a couple of bucks each or get something in between.  That's where I'm at because I'm just not educated about filter caps other than they filter/smooth out AC ripple in a power supply.

PS caps...  if the capacitance is adequate to prevent regulators from dropping out of regulation at low line voltage and ripple minimums, they should be "completely" adequate.

If the circuit PS rails are not regulated, it is more complex analyzing PSRR (power supply rejection ratio) of the individual circuits.

Punchy bass...  sounds a little optimistic to expect as an improvement from an otherwise competent circuit. If it's a cheap unregulated circuit, this brute force solution could probably be better accomplished by adding $0.25 regulators. 

JR

 

[damnyankee]

That's what I have to learn, John: how to calculate capacitance.  I know the general rule of thumb is to double voltage values over the voltage of the power rails (e.g., if power rail is 9v, the filter cap should be minimally 18V or the selected cap in this case would have a 20V value).  But the mystery is, I don't know how to calculate what capacitance value would be acceptable.  Take our 20V electrolytic example: uF values can run from 33uF to 15,000uF.  I have to keep reading so I can figure this out.

 

[JohnRoberts]

Few people actually calculate reservoir caps for modest current rack gear but as long as I am playing answer man there is fairly simple math to determine ripple voltage. Half wave rectification charges up the caps 60x sec, full wave rectification 120x per sec. Ignoring the actual charging time, the caps have either 16.6 mSec to decay or 8.3 mSec to decay. The rate of change for cap voltage per second is I/C or Amps divided by Farads.

For the sake of example and round numbers, lets assume 100mA of current and a 1,000 uF cap. Plugging in I/C gives us .1/.001=100v/sec rate of change. Now factoring for our recharge rate or decay time between recharges, we get either 1.6V of ripple voltage half wave, or .83 V of ripple full wave.

You can scale this simply. 200 mA or twice the current gives you twice the ripple, 10x the capacitance gives you 1/10 the ripple, 1000x the capacitance gives you 1/1,000th ripple.

There are surely web pages about designing a simple power supply.

JR

 

[damnyankee]

Few people actually calculate reservoir caps for modest current rack gear but as long as I am playing answer man there is fairly simple math to determine ripple voltage. Half wave rectification charges up the caps 60x sec, full wave rectification 120x per sec. Ignoring the actual charging time, the caps have either 16.6 mSec to decay or 8.3 mSec to decay. The rate of change for cap voltage per second is I/C or Amps divided by Farads.

For the sake of example and round numbers, lets assume 100mA of current and a 1,000 uF cap. Plugging in I/C gives us .1/.001=100v/sec rate of change. Now factoring for our recharge rate or decay time between recharges, we get either 1.6V of ripple voltage half wave, or .83 V of ripple full wave.

You can scale this simply. 200 mA or twice the current gives you twice the ripple, 10x the capacitance gives you 1/10 the ripple, 1000x the capacitance gives you 1/1,000th ripple.

There are surely web pages about designing a simple power supply.

JR

Alright!  You are officially annoited, "The Answer Man"!  ;D

I think I understand simple power supplys: transformers, diodes to convert or rectify AC to DC, the transformer & diodes sends the power to the filter caps in spurts so the job of the filter/smoothing caps is smooth out the spurts or ripples.  Thank you for the mathematical discussion - it really does help.  Ya know, I can slap one of these kits together, but this is my first non-9V batter project and they are much more complicated than overdriving a J201 opamp.  I want to understand these power supplies, how they work, and why particular values are used.  Same goes for VCA and RMS circuits.  I am very, very grateful of your time and expertise, John!

Take care - I've got reading to do...I'm reading how the transformer/rectifiers send the input power to the caps in "blips" every x-milliseconds.

Steve