OTA revisited

Remember operationnal transconductance amplifiers? So my question: Is it suitable to use a one megohm resistor at the output of an OTA (before the darlington buffer) ?

LM13700 documentation says "output voltage swing can be set at any level by selecting RL" but fails to mention any recommended upper limit.

Danny

Reply to
D. Graveson
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Remember operationnal transconductance amplifiers? So my question: Is it suitable to use a one megohm resistor at the output of an OTA (before the darlington buffer) ?

LM13700 documentation says "output voltage swing can be set at any level by selecting RL" but fails to mention any recommended upper limit.

Danny Sorry if I posted twice.

Reply to
D. Graveson

D. Graveson schrieb:

Hi Danny

I know another OTA very well, the CA3080, made by fairchild for instance. I guess that national chip works quite similar: The bias current fed into the control input is mirrored to the unbuffered output. If you put 1uA into the control-pin, the maximum output current is restricted to +/- 1uA. So the maximum voltage swing would be +/-1V when loaded with a resistor of 1M.

By changing bias current and load resistance you can set the output level according to to your needs, obvously there is no absolute limit of the output resistor.

joe

Reply to
joe

Tanks Joe! It answers my question. Not many people out there knowing OTAs.

I would have another question or two about linearizing diodes which CA3080 don't have. Are you familiar with this too?

Danny

Reply to
D. G.

hi Danny

Yes, I know about these linearization technique which was part of RCA CA3280 dual OTA which I designed into some audio gain-control-circuitry in the early

80s of the last century. At that time, there existed some fine application notes from RCA and it might be a good idea to search them, perhaps you are lucky and some people have scanned them in. Nowadays I use CA3080 without linearization to achieve a perfectly smooth distorting guitar PreAmp. Anyway, you might put your question and I can try to answer them.

Joe

Reply to
joe

Intersil has marked the CA3280 and CA3280A as "inactive" although they still have web pages up, with datasheets and app notes, see

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But they don't seem to have AN6818 up any longer. Sadly, I don't seem to have saved that one in time.

Gerber has 22 CA3280 in stock, at $2.58 each. There's also lots of aftermarket supply.

--
 Thanks,
    - Win
Reply to
Winfield Hill

I'm just concerned about the input current (Is) calculation to the amplifier input as well as the peak-to-peak voltage at the same input . Correct me if I'm wrong:

With current (Id) injected into the diodes, datasheet says the amplifier now behaves as a current amplifier because the transfert function is: Iout = Is * 2 * Iabc / Id where Is is the input current into OTA input.

Datasheet says Is should not exceed Id/2. I use Thevenin to find out Is and, once simplified, the final equation I use is: Is = Vin / Rin where Rin is the limiting resistor betwwen the OTA input and the voltage source.

I pick Vin peak value to find Is peak value. I assume it is the peak value that is of concern. For example, if Vin=1Vp-p and Rin=10K, then Is peak =

0.5V / 10K or 50uA. There is also a very small value resistor called Rp shunting the input to ground and it does not seem to do anything in that calculation. The value of Is seems to be related to Rin only. According to Thevenin...

Even though they describe the OTA as an current amplifier when diodes are used, it seems to me that we still have to limit the input voltage. No (apparent) word about this in datasheet. I assume this because of a small graph showing harmonic distorsion with and without diodes, related to peak-to-peak input voltage.

My final assumption is that Rin must meet two conditions:

1) limit Is below Id/2 2) limit peak-to_peak input voltage lower than 100mVpp (with the help of Rp to do a voltage division)

Hope my explanation is coherent and correct...! :)

Danny

Reply to
D. G.

Hello Danny

I think your maths is ok, but I will try to give you some brief explanations... Did you ever see the equivalent 'circuit diagram' of the OTA? It is not too complicated and shows the following basic stages:

Both inputs are the base-electrodes of a differential npn transistor pair. The sum of their emitter currents Ie1+Ie2 is fed by a common npn current mirror set by Iabc. With a current amplication of 1A/1A,

Ie1+1e2=Iabc!

-> Emitter-, collector- and base currents of the differential amp are all biased by Iabc.

The collector currents Ic1 and Ic2 are fed into two complementary current mirrors with a current amplication of 1A/1A. Their collectors are tied together and form the current driven output.

All in all, the OTA works similar to an ideal differential amp, where the output current equals the diffence of the collector currents of the input stage, which is an exponential function of the differential voltage between base1 - base2 of the input stage, multiplied by the sum Ie1 + Ie2 = Iabc. Without doing extended maths, it evident that

1) with Vbase1=Vbase2: Ic1=Ic2=Iabc/2 Iout=Ic2-Ic1=0 -> perfectly balanced

2) with Vbase1=Vbase2 +/- (>100mV) Ic1 = Iabc/0 Ic2 = 0/Iabc Iout= +/-Iabc -> 100% overdriven

As with any differential amp, rising the base1-base2 differential voltage beyond 100mV does not make any difference, as the whole Iabc is fed through one single input Transistor. Keep in mind that at that stage output CURRENT of the OTA is controlled in the linear region by an input VOLTAGE in the range of few 10 mVs.

On the other side, if you use the linearizing diodes parallell to the base-inputs, you shunt the inputs by the low impedance of a diode, which decreases linearly as diode bias current increases. Now the OTA is converted to an input current controlled current source!

Combined with the current limiting series resistors, it turns back to a voltage driven current source again.

Without linearization diodes, I recommend a differential voltage limitation done by to antiparallell 1n4148 or similar. This is a good practise that does not affect the transfer function anyway.

Using linearization diodes, there is no need for an input voltage limitation provided that the input is driven by some opa or similar with a restricted voltage drive capability: Match the current limiting input resistor to the diode biasing current and everything should run fine. So I agree to your first condition, the second one being adjusted by me.

Joe

Reply to
joe

This gives me something to digest. But things are clearing up. The next step is to get my protoboard, grap my scope and my audio generator and... to experiment.

What I find quite admirable about this kind of device is the mere simplicity of the circuitry. A differential pair and four current mirrors. That's about it!

Thanks for shading some light! Danny

Reply to
D. G.

That's assuming perfection of the circuits. In the case of the CA3280, this may be close to the truth, but in the previous incarnation, the CA3080, the important current-biasing mirror, not being a Wilson-mirror type, had huge Early-effect errors. "That's about it," did not apply, unless one was very careful.

--
 Thanks,
    - Win
Reply to
Winfield Hill

I should've said the topology of the OTA is quite simple and identical. All the models I know (I don't know about recent models... if any) used the same topology. Indeed, deep inside the chip, there must be a little bit more like internal biases, references and compensations tricks; that is caracteristic to integrated circuits.

Regards! Dan

Reply to
D. G.

Walt Jung described these in his "Audio Ic Op-Amp Applications" 2nd and 3rd editions (and probably 1st too). This book (all editions) has been out of print and used prices have been rather high ($50 and up) for several years. Jung described both the 3280 with its internal linearizing diodes, and a 3080 circuit using external diodes on the input to do the same thing the 3280 does internally.

Also, a 'standard' way of connecting the output is to go directly to the negative input of an opamp, with the positive input grounded and a feedback resistor, scaled for gain, from the output to the negative input. I forget if the output had any extra distortion caused by varying voltage as opposed to going into a zero-impedance summing input, but the op-amp does gives direct conversion of current output to a low-impedance voltage output, which is what's wanted in most cases.

There's a practical upper limit, with the limiting factor probably being stray capacitance and the current output's ability to drive it. Higher resitance will result in a lower max frequency response.

Reply to
Ben Bradley

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