02-05-2012, 08:39 AM
(01-05-2012, 09:22 PM)johnmcgivern Wrote: Hi everyone, received circuit.
Checked voltages on IC's all good apart from IC3 AND IC4 should be +20v and =20V both are 3.5v higher +23.5 =23.5.
These supplies are unregulated, so I wouldn't worry too much about that...
Quote:Struggling with how circuit actually works any suggestions would be most grateful.
Despite the apparent complexity, taken at a "block diagram" level, it's actually pretty generic...
The way the triangle waves are made involves putting constant currents into a capacitor. First the capacitor is charged, and when it reaches a predetermined threshold, the polarity of the current is reversed, and the voltage heads down towards the negative threshold, at which point the current is once again flipped and the cycle repeats.
The comparator that determines when the positive and negative thresholds are reached outputs a square wave, which controls the current to the capacitor. This square wave is also available as an output.
To make a sine wave, the triangle wave is deliberately distorted, taking advantage of the exponential relationship of VBE and IC in a transistor.
Looking at the diagram, IC3 is the positive current source and IC4 is the negative. The op-amps before these control them, summing the frequency control and the external input. These currents are fed into a diode bridge, which is used to determine whether the positive or negative current reaches the timing capacitor.
You'll note that the frequency decade switches are able to change the capacitor being charged (C1 to 4), and also the value of the resistors in the current sources. By altering the currents, you need fewer capacitors (large ones get expensive, so reduce the charging current instead). IC5 is almost certainly being used as a capacitance multiplier instead of a 10u non-electrolytic device. You'll also note the duty-cycle control, that alters the balance between the currents...
The output from the capacitor goes to the FET (TR5). A high impedance is required here, as any current taken will distort the triangle wave. The following transistors in the CA3086 form a conventional amplifier, the output of which is on TP9. It goes to 3 places; the comparator (IC7), the sine converter (bottom right) and the waveform selector switch.
The comparator is a conventional enough circuit, with the output being amplified and buffered and fed back to the diode bridge (arrow A). The output from the chip is also fed to the TTL buffer and the wavechange switch.
The amplifier takes the "mixsig" from the switch and deals with it in a conventional manner, and has constant-impedance attenuators dealing with the output. Similarly, the PSU is entirely conventional, but perhaps not at glance if you're more familiar with 78xx regulators...
Incidentally, I built one of these some years back - probably with more transistors and discrete circuitry than this one, which is how I'm so familiar with it
Quote:I think there has been a new revision on this drawing as it calls for 2 off TL072.
There is only one on the board unless I am reading it wrong and there is a distinct possiblity that I am.
No, definitely two shown on the diagram...
Time to take some measurements to determine where the fault lies. Any sign of the TTL output? Any signal on the capacitors? Is there a DC output - a faulty comparator would probably stick at one rail, so you should have a voltage on the timing capacitor...
As you say the power supplies are OK, this might not be completely straightforward, but look for common-sense stuff first.
Good luck,
Mark
