Dynacomp another noob question

[Le québécois]

Hi, I have read about OTA in order to understand the CA3080 in dynacomp.

If we make abstraction of the envelope follower that retroactively adjust the gain of the CA3080 (pin 5) to boost small signal more than big ones, (i think I understand this part enough), what set the gain of the OTA? From my reading the bigger is delta V (voltage difference between pin 2 and 3 of the OTA), the bigger will be the output current (pin 6). The voltage of this output current can vary and is determine by the load that pin 6 see ( in this case the 150 k resistor coming from vb and the base of Q2 --- how to calculate that, I have no idea!). In theory, if I manage to increase delta V, I will have more current at pin 6 and it should result in a louder output//more gain right?  If this is correct, does playing with only one of the 2 1M resistor to ground will help me to achieve this goal or will this go directly against the trim pot goal in this circuit?

I have already play with the 150k resistor connected to pin 6 of the OTA without any significant improvement in gain ( in fact I end up socketing almost everything on my dynacomp to boost my signal when at low sustain setting). I would like to know if socketing this 1M resistor is a lost of time?

please help me!!!

 

[PRR]

> voltage difference between pin 2 and 3 of the OTA

It is better to think of the CURRENT into the OTA control port. Output is very proportional to current. The voltage is uncertain and not a good guuide to operation.

> result in a louder output//more gain right?

If you want more output, add a booster after the OTA. There are other ways to do this, but they all need more thought and computation.... booster is obvious and simple and will work.

[R.G.]

If we make abstraction of the envelope follower that retroactively adjust the gain of the CA3080 (pin 5) to boost small signal more than big ones, (i think I understand this part enough), what set the gain of the OTA? From my reading the bigger is delta V (voltage difference between pin 2 and 3 of the OTA), the bigger will be the output current (pin 6). The voltage of this output current can vary and is determine by the load that pin 6 see ( in this case the 150 k resistor coming from vb and the base of Q2 --- how to calculate that, I have no idea!).

Let's think for a minute about the name: Operational Transconductance Amplifier. A resistance is simple - it's how many volts are produced per amp of current flow. A 1K resistor is one volt per milliampere.  Conductance is the inverse of resistance, it's the amps that are let flow per volt across it. In the same way that a 1K resistor is one volt per milliampere, it could equally be said to be one milliampere conducted per volt across it.

Trans-conductance is a little slippier. It's how much current flows in the output of some device per voltage on the input. Power MOSFETs for instance often have a transconductance of about one ampere per volt, meaning that they let through about one ampere on their drain for each change of 1V on their gate. (I'm simplifying here to get the point across.)

Unlike MOSFETs, OTA have a variable transconductance. The National Semiconductor LM13700s happen to have a transconductance of 19.2*Iabc. That is, their transconductance depends directly on the current into the gain control pin. So with 1ma into the bias pin, the transconductance would be 19.2 milliamperes per volt. The "per volt" there is the differential voltage between the two input pins. So for a differential voltage of 10mV, the output current would change 10mV*19.2 ma/V =190uA. And the output is reliably a current, not a voltage.

To get voltage gain, we return to our friend the resistor. A 1K resistor is equal to 1V/ma, so if we load it with a 1K the gain of the example OTA is 19.2ma/V * 1V/ma = 19.2 volts per volt. If we used a 100K resistor for a load, the voltage gain would instead be 19.2ma/V times 100V/ma or 19,200 times.

So the gain of a transconductance amp is the transconductance times the load resistor.

And the OTA has a variable transconductance. It is a constant times the current in the bias pin. And that's how the OTA works in this circuit.

In theory, if I manage to increase delta V, I will have more current at pin 6 and it should result in a louder output//more gain right?

No. What's happening is that the voltage out of pin 6 is run into a level detector that takes several paragraphs to explain, but it's about 0.6V. Anytime the output is bigger than 0.6V peak, the circuit lessens the current into the bias pin and decreases the gain of the OTA. It's a feedback loop that tries to keep the output always the same. If you increase the signal at the input pins, it just reduces its gain.

You can get more output level out of it by increasing the threshold of the feedback loop that sets the gain, but frankly, it's easy to mess things up in this DC coupled gain adjusting loop.

It's much simpler to put an opamp gain stage after the whole mess and get bigger signal that way.
 

[Le québécois]

Thanks for your answers.

The booster is already done but I feel like cheating if I don't understand the circuit well enough to make it do what I want. Unfortunately, this one seem to be a little complicated in my slowly rising learning curve.

What's happening is that the voltage out of pin 6 is run into a level detector that takes several paragraphs to explain, but it's about 0.6V.

Is this 0.6V threshold is cause by the two diode in front of Q3 and Q4? If so, adding more diode in series would increase the threshold to 1.2V. or LED?

[R.G.]

Thanks for your answers.

The booster is already done but I feel like cheating if I don't understand the circuit well enough to make it do what I want. Unfortunately, this one seem to be a little complicated in my slowly rising learning curve.

It's not cheating. Sometimes understanding the circuit well means understanding what it won't do, or won't do well. It's that old discretion is the better part of valor thing.

Is this 0.6V threshold is cause by the two diode in front of Q3 and Q4?

Close, but not exactly. It's the forward drop of the base-emitter of the transistors that do the sensing. The other diodes clamp the signal to ground on negative excursions.

If so, adding more diode in series would increase the threshold to 1.2V. or LED?

You could put a diode in series with the transistor's emitter or base. That would increase the threshold. It's possible, but the buffer is simpler and more predictable.

[PRR]

> understand the circuit well enough
> this one seem to be a little complicated

Limiters are some of the trickiest audio tools around.

I been studying them 37 years, published a few, and still get mind-clouded.

This one is actually "simple". Deceptively simple. There's a lot going on, but it comes together with very few parts just-barely doing what needs to be done.

Your question certainly is not "noob".

What is the problem? Gain or Level?

It has a ton of gain already. I make it as over 90 full-UP.

The maximum output is around 0.35V rms. This is seven times more than any guitar amp needs to reach full power output. Also I do not see a lot of posts saying the output is low.

So why are you seeking "more"? Two thoughts:

1) Build mistake. Something is not as it should be.

2) Odd use. Unusual sounds in, or not connected at guitar-cord level.

First let's see if it works like it should. Full limiter testing is tedious, start with basic DC checks. Read and follow DEBUGGING - What to do when it doesn't work. Particularly WHICH schematic you used (I found two drawings) and WHAT changes you have made. I don't know the target voltages, but I'd want Q2 Emitter at 2V-3V, Q2 Collector at 6V-7V, Q5 Emitter over 8V on dead silence and going below 2V when SLAMMED with signal.

If possible, compare with another DynaComp.

Odd signals: this limiter has fast attack and slow release. For a very percussive signal, it will "duck". The initial transient spike slams it to low gain, and the slow release does not recover before the note fades. As a quick hack, replace the 10uFd timing cap at Q3 Q4 Q5 with a 10K resistor and 1uFd cap in series. This gives slower attack and faster release. It may be "too much", choppy and won't max-slam. It will brutally slash an initial transient. But it may guide you toward your goal.

[Le québécois]

Here is the checklist to fill out:
1.What does it do, not do, and sound like?
Good compression overall. Add a bit of distortion when the sustain knob is more than 90% up. Under this condition (full sustain) the output gain is more than unity. The problem: when the sustain is at or below 75%, the volume drop below unity gain even with the volume knob at max. The sustain knob then become unusable without a booster in the pedal chain.

2.Name of the circuit = Comparous from Tonepad (identical to the one find at Geofex except for the transistors types)
3.Source of the circuit (URL of schematic or project) = http://www.tonepad.com/getFile.asp?id=9

4.Any modifications to the circuit? Yes, the attack mod but the problem was there before this modification. You probably know, but this MOD consist to replace the 150K resistor with a pots (in my case 250K + 10K resistor in series). This way we change the recharging time of the 10 uF capacitor to ground and therefore limit the time before the next transient can be operated / detected by the limiter
I have put sockets here and there (all transistor and IC + some resistor that I was believing can increase the gain but didn't work). 
It was my second build (first one MXR microamp) on home made PCB carefully check and recheck and , yes rechecked again with DMM for continuity were it needs to and a magnifying glass. This check was done before soldering anything on the board and after every new part soldered to the board. This way, the thing did work right away when I plugged my guitar!   

5.Any parts substitutions? If yes, list them. Only one: I had no 2 k trim pot so I wired a 2.7K resistor in parallel with a 10K trim pot and check the new ohm prior to installation. It was 2ohm

6.Positive ground to negative ground conversion? N

7.Turn your meter on, set it to the 10V or 20V scale. Remove the battery from the battery clip. Probe the battery terminals with the meter leads before putting it in the clip. What is the out of circuit battery voltage? =>9.39V
Now insert the battery into the clip. If your effect is wired so that a plug must be in the input or output jack to turn the battery power on, insert one end of a cord into that jack. Connect the negative/black meter lead to signal ground by clipping the negative/black lead to the outer sleeve of the input or output jack, whichever does not have a plug in it. With the negative lead on signal ground, measure the following:
Voltage at the circuit board end of the red battery lead = 9.39
Voltage at the circuit board end of the black battery lead = 0

Q1
C = 9.38
B = 2.09
E = 1.58

Q2
C= 5.56
B= 4.43
E= 3.88

Q3
C= 8.85 (this value drift from 8.97 down to 8.85) I think this is the 10 uF cap discharging in my DMM)  / around 7volts with recorded guitar in
B= 0                / 0.07  with guitar in
E= 0                / 0

Q4
C= 8.85 (same drift here)  -  around 7 with recorded guitar in
B= 0               / -0.03 with guitar in *** my DMM really show negative value (I did go on the 2 V scale to be certain)
E= 0               / 0

Q5
C= 9.38                                         / same
B= 8.85 (same discharging here)   / 6.7 with guitar signal
E= 8.32                                        / 6.1 with guitar signal

IC1 (or U1)
P1= 0
P2=4.21
P3=4.21
P4=0
P5=0.7                                / 0.61 up to 0.7 V while turning the sustain knob. (sorry the current measurement is to much work for now!)
P6=4.44                              / 2.86 to 3.7V with guitar signal and with sustain from 0 to 100% open
P7=9.37
P8=0


D1
A (anode, the non-band end) = 0
K (cathode, the banded end) = 9.37

D2
A = 0                   /
K = 0                   / 0.56 with guitar signal

D3
A= 0
K= 0                     / 0.06 with guitar signal

Battery at end : 9.37

To me the only two thing that are weird are the D3 diode and base of Q4 when a guitar signal is in. The signal is only my guitar (it's not music).

I think this is it! Thanks for helping me.

[Le québécois]

Ok, there is an update.

D2 was cold jointed on one side. Now the Voltage whit guitar signal is 0.6 like the other diode
I don't know why but this also replace Q4 base voltage to 0.07 like Q3.

This have not solve my volume problem unfortunately.

[PRR]

There's a reason there's a slight negative voltage on the bases when driven. Ignore it.

> IC1 (or U1)
> P6=4.44 / 2.86 to 3.7V with guitar signal and with sustain from 0 to 100% open

Should be more like 2.9V when silent. This error offsets the phase splitter, it won't make full output, isn't really limiting correctly. This is reflected by the bias-shift toward 2.9V when driven hard (it is trying to find its own center, but should be centered already), the relatively low shift in control voltage (Q5 E= 8.32 / 6.1 with guitar signal.... should go WAY down), and the touch of distortion.

Since the observed IC1 P6 voltage is nearly half the supply voltage, but the schematic shows it coming from a 56K:27K divider, you should look here:

[Le québécois]

Thanks to your helpfull comment and good direction Paul, my dynacomp Voltage are now correct and the little thing have a better usability range now 40 to 100 % sustain.  I resolder some of the bad looking joint around pin 6.   

Just for fun now.... For me, the perfect dynacomp should have : a unity gain when the sustain pots is turn all the way down with the level knob all the way up. The more the sustain is increase, the more we can drop the level knob to maintain unity gain.

But don't misunderstand me. The thing sound really good as is!

thank again for helping me. 

[PRR]

> maintain unity gain.

The whole idea of a compressor is VARIABLE gain. More on small signals, less on large signals. You use a compressor when you are not happy with any specific gain. Asking for "unity gain" misses the point.