Bidirectional current source

[JKowalski]

This is for the 555, as a triangle wave generator.

I need a bidirectional current source that is adjustable with just one resistor. And I want it very simple (few components).

The best method I have found so far is two JFET current sources - the source has a variable resistor in series with it, and the gate is connected to the opposite side. Two of these in series, in opposite directions, is exactly what I was looking for in terms of simplicity and function - but, it requires a dual ganged pot to adjust the current symmetrically.

Anybody know of a good design? If nobody knows of anything, maybe I'll just look for a dual ganged pot.

[R.G.]

OTA

[JKowalski]

Huh? What does OTA mean?

EDIT: As is operational transconductance amplifier, correct? I'll look into it. Thanks.

[JKowalski]

Okay. This is how my 555 circuit is set up currently. I am using the output of the 555 in series with my bidirectional current source, going into the capacitor. When the voltage trips the comparator, it flips the output, and then the current source flips too and so on.

So I found this, the output current of the OTA:     

Where the transconductance (Gm) is directly controlled by the bias current.

Seems pretty simple.

So I would have my 4.5V on the noniv input, and the 0-9 square wave output of the 555 on the inv. Then, the output goes to the capacitor? And the current will be controlled by a current source into Ibias? That way, it would be (9v-4.5v)(f(Ibias)), and then on the trigger, (4.5v-9v)(f(Ibias)). So then I would have a +/- 4.5(f(Ibias)) current source to the capacitor? Am I oversimplifying this way too much? This is my first experience with OTA's.



EDIT: Would this mean I could have a adjustable duty cycle for the oscillator without frequency alteration, by altering the 4.5 bias voltage on the noninv input? I had a simple duty cycle addition to the JFET current source that worked decently, with some frequency variation, and it would be great if I could incorporate it a similar or even better duty control with the transconductance approach.


Wow, if all this is correct, then this is exactly what I was looking for in every way! IF.  I'm hopin'

[R.G.]

People just don't get OTAs. They're simple.

The Iabc bias current sets the total current at the output node. There is a total of Iabc/2 flowing from the positive power supply to the negative power supply all the time. That's why you can't increase Iabc over 1ma for the 3080 and 2ma for the 13700 and others - the fixed current in the output overheats the chip.

What the differential input does is apportion the Iabc between the upper driver and lower driver on the output. If the inputs are balanced, the upper driver puts out Iabc/2 and the lower driver 'eats' Iabc/2, so there is zero current left over to go out the output pin. If the input diffamp is shifted one way, the upper driver allows more current through from the power supply than the lower one can eat. The difference has to flow out the output pin. Shift the input the other way and the upper driver supplies less current than the lower one demands, so the lower one sucks the excess current in from the output pin.

The output is zero to Iabc, either going into or coming out of the output pin. The direction and amount are determined by which way the + and - pins are unbalanced, and how much. For a diffamp, the total linear range from unbalanced one way to completely unbalanced the other way is about 25mV. So you can switch the direction of the output current by switching the input diffamp by more than 25mV, and then control how much current flows with Iabc.

This is a semi-secret trick that is only used in one commercial pedal I know of, and even there it's hidden behind a cloaking layer or two. I use it all the time for my private stuff.

[JKowalski]

Ah! Why can't a find such a good explanation on the net? Thanks so much RG.

So the basic operation of what I proposed is correct, including the duty cycle adjustment? I just didn't fully understand how this comes to pass? All I need to adjust is the square wave input differential range around Vref to match the OTA's range, correct?

I took a look at a random OTA datasheet, hoping to find a V/I curve for the diff voltage and Iout with a fixed Ibias. This is the first I found -

http://www.datasheetcatalog.org/datasheet/intersil/fn1174.pdf

At the "effects of diode linearization... etc." part. So, I see that the output current can only go to +/- 600uA with a 650uA bias current, so that seems to agree with your description. But, the graph shows that the voltage in goes from +/- 4 volts to reach each of those peak output currents. I suppose they all differ in this aspect (voltage diff/output current) somewhat?

[R.G.]

The input range of the OTA is extended by using large resistors or resistor dividers to convert voltages to currents or to voltages in the 25mV range.

You can drive the input with more than +/-25mv, but that's as big as the linear range is. If you connect the two inputs with a 1K and drive the inputs through 10ks, the linear range gets to being 10x as large from the 10:1 voltage divider, for example.

[JKowalski]

Ah, okay, so when you said "as big as the linearization range is", you meant the second graph on the datasheet I showed you, the unlinearized one, in the middle around the origin. So it's commonly that kind of a curve for an OTA, and the linearization trick that they use was just a certain special method used to extend the viable range.

Yes, I knew I could use a divider to solve that problem, but I just wanted to clarify my understanding of the input voltage ranges, and see how far the limits went, exactly.


I think I'm okay from here. Thanks again for your help, RG.