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ICL8038 Waveform Generator Questions

11/03/2016 4:58 AM

ICL8038 is a monolithic waveform generator IC that can produce sine, square and triangular waveforms with very little distortion. The frequency can be programmed from 0.001Hz to 300 KHz using external timing capacitor and resistor. Frequency modulation and sweeping can be attained by using an external voltage. Other features of the ICL8038 are high linearity, high level outputs, simultaneous sine, square, triangle wave outputs, low external parts count, high temperature stability etc.

The working of Intel ICL8038 is as follows. The external timing capacitor (C2 in the circuit diagram) is charged and discharged using two internal current sources. The first current source is on all the time and second current is switched ON and OFF using a flip-flop. Suppose the second current source is OFF and the first current source is ON, then the capacitor C2 will be charged with a continuous current (i) and the voltage across C2 increases linearly with time. When the voltage reaches 2/3 supply voltage, controlling flip flop is triggered and the first current source is activated. This current source carries double the current (2i) making the capacitor C2 is discharged with a current i and the voltage across it drops linearly with time. When this voltage reaches 1/3 supply voltage, the flip flop is resetted to the initial condition and the cycle is repeated again.

Looking at this circuit I see there are are at least 3 problems:

1 I assume the supply is +10v-0v–10v, where then is the 0v connected?
2 what forms the output of the circuit, I see the +ve sides are formed from the outputs of the IC on either pins2,3 & 9 of IC1, what forms the -ve output?
3 I appears that pin 9 of IC1 is connected via a 15K resistor to the +10v line, are there similar connections to pins 2 & 3?

Audio oscillator circuit

The circuit diagram given above shows a variable audio frequency oscillator using ICL8038. Such a circuit is very useful while testing audio related projects. The frequency range of this circuit is 20Hz to 20KHz. POT R6 can be used for adjusting the frequency while POT R9 can be used for adjusting the distortion. POT R4 can be used for adjusting the duty cycle while POT R7 can be used for nullifying the variations in duty cycle. C2 is the external timing capacitor and R5 is a pull up resistor.

Notes.

  • The circuit can be assembled on a vero board.
  • Use +10 /-10V DC dual supply for powering the circuit.
  • The power supply must be well regulated and filtered.
  • All fixed resistors are rated ¼ W.
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#1

Re: ICL8038 waveform generator questions

11/03/2016 5:06 AM

The -ve connection can also be ground according to the pin outs I've seen by searching for that chip.
Some ICs have poor documentation and application notes/diagrams that simply don't work as described. Others are very good. Personally I've found National Semiconductors and Maxim to be be excellent.
I've also found that complexity of function is inversely proportional to likelihood of success.
You need to study all the appliction notes not just try to build a circuit straight from a diagram. Layout and power supply is often critical with ICs.

You say you "see" there at least 3 problems... have you actually built the circuit? If not, then you don't actually have a problem.
Del

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Re: ICL8038 Waveform Generator Questions

11/03/2016 6:26 AM

1. You can either use +ve and -ve supplies (in which case the outputs will swing above and below 0V), or a single-ended supply, 0V and +ve (where the outputs will be all above 0V).

2. The output is taken between the appropriate output pin (2, 3 or 9) and "ground" (pin11).

3. If you look at the detailed schematic, the square-wave output from pin 9 is a "bare collector"; without a pull-up resistor, no voltage will appear. The sine and triangular ouputs are both driven "up" and "down". Resistors RSINE ans RTRI in your drawing are load resistors for the test circuit, and are not necessary for operation.

Note that this is an obsolete product with no recommended replacement.

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#3

Re: ICL8038 Waveform Generator Questions

11/03/2016 9:30 AM

1 I assume the supply is +10v-0v–10v, where then is the 0v connected?

The flip-flop output (square wave) is an open collector output and requires a resistor to pull it high. The sawtooth and sine wave output return to ground (0v) either by the load or a resistor. The test circuit is shown below. Connect the other side of your load to +V for square wave signal or to ground (0v) for the other two signals.

Here is the datasheet.

https://www.intersil.com/content/dam/Intersil/documents/icl8/icl8038.pdf

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Re: ICL8038 Waveform Generator Questions

11/03/2016 2:34 PM

I had a look at the datasheet and all these questions are answered in it. Do download one if you haven't already.

The only thing that is not clearly explained is the return for the +/- power supply and output signals.

1 - Ground

2 - Ground

3 - No, that is a pull-up resistor needed for operation of the square wave output channel only.

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Re: ICL8038 Waveform Generator Questions

11/04/2016 1:53 AM
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