Understanding the CA3080...

[JDoyle]

Before I head into the backwoods for my holiday vacation, I wanted to leave a little gift for the DIY community.

Since I started this hobby, I have found that one of the most common circuits that elicits questions on its operation is that for the Ross or MXR compressors. I have spent a lot of time studying this circuit and have tried to help people where I could, but I think that the main problem is that after struggling through learning about every other component, when it comes time to deal with the CA3080, people freeze because it's operation appears to be a mystery. Its name, Operation Transconductance Amplifier, implies that it is similar to the 'standard' op amps everyone is used to but it rarely appears in a circuit with feedback and the resistor networks around it and their ratio's never seem to make any sense. So most of the time, the 3080 is ignore in favor of the rectifier, because at least that circuit is easy to understand.

Well, in order to try and give a little back to the FX community, I thought that I would do a little write up on the CA3080 to help people get a better handle on it's operation.

Because it turns out that you probably already knew how it worked, but it's not like a 'standard' op amp, it can be thought of as being that lowly originator of all of our noise: the simple bipolar transistor.

So, without further delay, best wishes and health to you and yours in the New Year, and I hope that this helps some people...

Jay

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One way to think about and better understand the operation of the 3080 is to view it as a transistor with an enormous amount of gain, or even better, to think of it as a certain number of transistors in parallel, all with the same, and user adjustable, current amplification properties. The current outputs of each of these transistors are summed together to produce the output signal for our 3080 'transistor. By controlling the main '3080 transistor' we also similarly control the other certain number of transistors at the same time. As we will find out, this 'certain number' is 19.2. (The 3080 is actually a differential amplifier with a huge current gain).

- The two inputs are the 'base', which gives you a choice of input vs. output polarity that you don't get with a 'standard' transistor. The input signal (V_in) is the differential voltage between the two inputs. The differential voltage is the difference between the voltages on the two inputs of the 3080 [ Vin = (V_+in) - (V_-in) ]. For our purposes, it is probably easiest to understand this concept if you tie one of the inputs directly to the bias voltage. Then the signal on the other input will cause that input to fluctuate around Vbias, causing a difference in voltage, or a differential input, between the two inputs; that way it is just like a 'standard' transistor except you truly have a choice of input vs. output polarity.

- The output is the 'collector', but instead of attaching the load resistor (R_load) to V+ as in a 'standard' transistor, you attach it to whatever voltage you want the output to bias to, and, it will. This is possible because the output of the 3080 is a current that varies only with the magnitude of differential voltage between the inputs and is unaffected by the voltage on the output. The output signal of the 3080 appears to whatever follows it as a current that alternates around the bias voltage attached to (R_load). Or in a way that I think will make more sense in the realm of guitar effects: the output is a voltage that is biased to the voltage attached to R_load and has an extremely high output impedance due to the fact that it is mostly a current signal (...because it just happens to be dropping across R_load).

- The Iabc port (pin 5 on the 3080; the 'abc' stands for 'amplifier bias current', emphasizing the current, not voltage, nature of the 3080) is the 'emitter', except instead of attaching the 'emitter' resistor to ground, you attach it to a positive supply voltage. But this resistance doesn't work as it does in an actual transistor, it supplies a steady bias current to the I_abc port, and this current doesn't vary with the input signal like the current through an emitter resistor in a 'standard' transistor. This amount of current into the I_abc port is the same amount of current that is going through the 3080 'transistor', so just like a 'standard' transistor, the current at the 3080's 'emitter' is the total current through the 3080 'transistor'. And also just like a 'standard' transistor, the emitter of the 3080 'transistor' is a diode that has to be biased 'on' with a voltage of approximately 0.6V above ground.  Therefore you can determine the total current through our 3080 'transistor' via Ohm's law: divide the voltage the resistor is attached to (V_abc) less the diode drop of the 'emitter' diode at the I_abc port (approx. 0.6V), by it's resistance (R_abc), and you then have the amount of current you are allowing through the 3080 'transistor'. But UNLIKE a 'standard' transistor, the 'emitter' current of our 3080 'transistor' is not the total current OUT of the entire 'transistor' stage. The output signal, or more accurately, the output current, is the 'emitter' current (I_abc) multiplied by the transconductance (g_m) of the 3080. [So in a way, as the ⁺/_- inputs of the 3080 can be considered a 'base' that amplifies a voltage input, the I_abc port can also be considered as another 'base', but unlike the ⁺/_- inputs which amplify a voltage signal in, the I_abc 'base' amplifies the current signal into it. But because this current is the determinate factor in the gain, I think it is more accurate and is more readily understood if thought of as the 'emitter' in this example.]

So how do we fit this all together? Well, we do need some more information before we can continue...

First, the current through the I_abc port (I_abc) is controlling the transconductance (g_m) of our 3080 'transistor'. In other words, the current into the I_abc port determines the ratio of the change in output current caused by a change in the differential input voltage; or:

(1) -[ g_m = (I_out)/(V_in) ]

...which is the current gain of our 'transistor' (I_out=(g_m)(V_in).

Second, the exact value of the transconductance of the 3080 'transistor' is extremely easy to determine:

Figure out via Ohm's law the current into the I_abc port:

(2) - [ I_abc = (V_abc - 0.6V)/(R_abc) ]

...and multiply that by 19.2 and the result is the transconductance of the 3080 'transistor':

(3) - [ g_m = (19.2)(I_abc)]

Third, the gain of the 3080 transistor is the transconductance determined in '(3)' above, multiplied by the resistance value of the load resistor (R_load) on the 'collector', or output pin, of the 3080:

(4) - [ Gain = (g_m)(R_load) ]

With this information we can get even closer to an approximation of a transistor:

We know that the gain of a 'standard' transistor stage is essentially its collector load divided by its emitter load:

(a) - [ 'Standard' Transistor Gain = (R_c)/(R_e)]

And from (4) above, in the case of our '3080 transistor', the gain is the transconductance multiplied by the load resistance:

(b) - [ '3080 Transistor' Gain = (g_m)*(R_load) ]

We also know from (3) above that the transconductance is the current into the Iabc port multiplied by 19.2:

(c) - [ g_m = (19.2)*(I_abc) ]

If we substitute (c) into the formula in (b) we get:

(d) - ['3080 Transistor' Gain = (19.2)*(I_abc)*(R_load)]

Continuing, we also know from (2) above that Iabc is determined via Ohm's Law by the resistor on the I_abc pin (R_abc) attached to some positive voltage less one diode drop (V_abc - 0.6V):

(e) - [ I_abc = (V_abc - 0.6V)/(R_abc)]

And if we continue by substituting (e) into (d) we get:

['3080 Transistor' Gain = (19.2)*(V_abc - 0.6V)/(R_abc)*(R_load)]

Which is still a little cumbersome to help all that much; but as I said, the output of the 3080 can be thought of as the 'collector' and the I_abc port the 'emitter', so by substituting the nomenclature for 'standard' transistor resistors, from (a) above, into the formula for the ['3080 Transistor' Gain] above:

['3080 Transistor' Gain = (19.2)*(V_abc - 0.6V)/(R_e)*(R_c)]

...and then rearrange it a bit for clarity:

['3080 Transistor' Gain = (19.2)*(V_abc - 0.6V)*(R_c)/(R_e)]

...we can see that the formula is the same as it is for a 'standard' transistor except the gain is a LOT bigger, a factor of '(19.2)*(V_abc - 0.6V)' bigger! And we can also see that by varying I_abc, or either V_abc or R_c, we can vary the gain. So by viewing the 3080 as a transistor we can then view R_load as the collector resistor, and R_abc and the emitter resistor, and if we multiply their ratio by the voltage on R_abc and 19.2 we will get the total gain.

Said another way, it is like we are using (19.2) transistors in parallel with each other, each with a gain of [(V_abc - 0.6V)/(R_e)*(R_c)] and we are summing the currents together as our output signal. Plus, by controlling the current bias state of one '3080 transistor' we control all of the transistor's bias.

- Important Caveats:

1 - The cost of sinking more than 2ma of current into the I_abc port is that of a new 3080. Simply - above 2ma I_abc, you burn up the chip. While there are still a ton of 3080s out there, they are out of production and the cost per piece has pretty much quadrupled, so be sure to attach a resistor to the I_abc port that will limit the current into I_abc to 2 ma with the supply voltage you are using (R = V+/2 ma) before you power up any circuit with a 3080 in it.

2 - Because the output is a current, it has an enormous output impedance, and therefore the output has a lot of potential to create noise. So make sure you use the smallest resistance possible on the output and it would be wise to make it metal film. Also, because of this enormous Zout, the output signal will have to be buffered in some way to make it at all useful. In fact, if you buffer the output, with a JFET common source buffer for example, and then take THAT buffered version of the 3080 output as the output (the signal at the source of the JFET buffer) instead of the output of the 3080, you now have a semi-'standard' op amp because you have turned the high Zout current into a low Zout voltage.

3 - Just as in a 'standard' transistor the input impedance of the 3080 is a reflection of the 'emitter' impedance. Therefore, the Zin of the 3080 varies with I_abc. The formula for intrinsic emitter resistance in a silicon transistor is r_e=I/26 mV, but because the input stage of the 3080 is a differential amplifier, each input is directly connected to the base of a transistor, and both of the input transistors share the total emitter current, the resistance value is doubled as the emitter current is split in two between the input transistors. So the input impedance of each input is equal to [I_abc/52 mV] and varies around that point with the input signal. Therefore the input impedance varies.

4 - As I quickly noted at the beginning and above the input is a differential amplifier. While not in the scope of this tutorial, the common mode input range, or input headroom, is such that a differential voltage of ⁺/_- milliVolts causes one of the transistors in the differential amplifier to be passing 98% of the current and effectively clipping the output. Therefore, keeping the input signal smaller than 100mV is essential for clean operation when running open loop (without feedback). This is normally done by dividing down the input with a resistive divider. While this essentially converts the input signal to a current, it is the amount of differential voltage cause by the input signal, and not it's current component, that determines the output signal.

5 - Other than the obvious advantage of direct control over the entire stage's transconductance, the other major advantage of the 3080 when compared to 'standard' op amps is speed. Current amplification is inherently faster than voltage amplification (see Miller Capacitance). BUT, as great as that sounds, the ability to amplify 'fast' in electrical engineering terms is the equivalent of being able to play scales faster than the speed of sound, that's great, but it doesn't matter and in the end is useless because we can't hear it. And actually it is a hindrance because that extended range extends the risk of high frequency oscillations that will unpleasantly distort the output with intermodulation distortion. This can be solved with a cap in parallel with Rload, with which it forms the 3dB corner frequency.

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I hope that this has been a help to those of you who find yourself confuse about the 3080 and needed a better model to view it in order to better comprehend its operation.

Have a safe and happy holiday season and my your New Year find every other guitarist envious of your tone, every time.

Regards,

Jay Doyle

[jlullo]

you're the man, jay.. there's a ton of info to take in here, and we all appreciate it!

[drewl]

I'll never sleep tonight knowing that all of that is going on in my MXR compressor.....in fact it's staring at me right now....mocking me....stop it compressor.....stop mocking me...

[R.G.]

That's a more accessible explanation that I have in my head. Good one!

My internal explanation is more complicated, but less satisfying.

First you have to know current mirrors. You don't need to know the inner workings of them, only that they do work.

Let's say that the 9V power supply is the ceiling. 0V is the floor. Here on the floor is a clot of transistors attached to the floor but with two free leads, marked "input" and "output". I hook the lead marked "output" to the 9V ceiling through a resistor and watch the current through the resistor. Nothing happens. So I take the lead marked input and suck current out of it. Nothing happens, it appears to be blocked.

But then I pour a little current INTO the input lead, and suddenly current flows through the resistor to the ceiling. Reading the meter, I see that the amount of current that is pulled out of the ceiling is the same as I'm pouring into the input lead. The current mirror on the floor converts current poured into its input into current sucked from a higher voltage. It pulls out of the ceiling the same amount I pour into the input.

Looking up, there is a similar, but different colored current mirror attached to the 9V ceiling, also with two leads, input and output. Guessing what might happen from my floor mounted current mirror experiment, I hook a resistor between the "output" lead and the 0V floor, and then suck current out of the input lead. Sure enough, a current equal to what I suck out of the input lead comes out of the output lead. So I can have ceiling mounted (PNP) and floor mounted (NPN) current mirrors. If I pull current down from a PNP mirror, current flows down from its output lead. If I pour current into the intput lead of an NPN mirror, current gets pulled into its output lead.

Here's how OTAs are made:
1. There is a differential amplifier. The diffamp is powered by a current source in its emitter. Diffamps have no current gain at all. They can only steer their emitter current, whatever it is, into one or the other transistor,  in proportion to the two base voltages. 25mv difference is enough to steer the current entirely to the transistor with the higher base. So the diffamp can steer the emitter current one way or another.

2. There is a current mirror providing the emitter current. This is the Iabc input. It sets the diffamp total current. Only the amount of current poured into Iabc can possibly flow through the diffamp, and Iabc is the maximum either transistor can conduct.

3. The two diffamp collectors drive two ... PNP current mirrors on the ceiling. Each of these conducts the same amount of current pulled out of its input by the diffamp, and in the same proportion.

4. ONE of the PNP mirrors on the ceiling has its output connected to a NPN mirror on the floor, reflecting its current from the floor. The other PNP mirror has its output connected to the output of the NPN mirrror. The place where the PNP output connects to the NPN output is the output pin of the OTA.

OK, got all these mirrors. What's important to realize is that if the diffamp is balanced, 1/2 of Iabc flows in each of the diffamp transistors. That means 1/2 of Iabc flows through the PNP output mirror and 1/2 flows through the NPN output mirrror. The OTA is balanced.

If I unbalance the diffamp, the Iabc all flows through one of the diffamp transistors. It goes through either the output PNP mirror and lets Iabc flow out of the output pin, or if it's the other transistor, it goes through a PNP mirror, then the NPN output mirror and sucks Iabc into the input pin.

Iabc ...IS... the output current, just mirrored either once or twice to get the direction right.

The 19.2 gm stuff is all about how the diffamp deals with the unbalance of the bases.

But as I said, it's less intuitive.

[antojado]

Here's some interesting info about availability:

http://www.till.com/blog/archives/2005/06/last_of_the_ota.html