# triangular and square wave generator

the circuit below is a triangular and square wave generator. With the first stage (the integrator) I can generate a triangular wave, and with the second stage (comparator) I can generate a triangular wave with a limited amplitude.

I've never seen this circuit with the net circled in red. What does this net does? The text of the exercise says: Analyze the circuit assuming ideal op-amps. ...the circuit is an astable multivibrator in which zener diodes are ideal with Von =0,6V and Vbkd = 5.6V. 1)Describe the circuit and what kind of waveform can be found at Vout and Vs 2) find R1 and R2 in order to obtain an amplitude of Vout of Vpeak = 10V, and f = 10kHz. • I expect the opamp (or comparator) part numbers are the key to the answer. – Brian Drummond Oct 31 '16 at 13:20
• For my opinion, R5 - together withe Z-diodes - constitutes a kind of output voltage stabilization at a fixed level. This makes sense because it is this voltage which feeds the integrator. – LvW Oct 31 '16 at 13:39

The right hand op-amp looks like a comparator (with an open collector output). The emitter I suspect is pin 1 and this is tied to -15 V. Open collector outputs require a pull-up resistor hence the need for R5 in your circuit.

If you could link where you found this circuit this could be confirmed. I think the comparator matches the pin out of the LM311: - Collector = pin 7 and emitter = pin 1.

• There's no link. This is a university exercise in which I have to find R1 and R2 under some conditions. The operational amplifier are to be considered ideal, so there is no open collector output. – FataMadrina Oct 31 '16 at 13:21
• @FataMadrina No, you are probably wrong on this - it looks like it uses an LM311 comparator as the triangle to square generator. Of course if you use ideal op-amps then you don't need R5. Here's a link to plenty of circuits using two op-amps that might be useful: google.co.uk/… – Andy aka Oct 31 '16 at 13:24
• Edited with the full text translated (from italian). That net have to count in some way. – FataMadrina Oct 31 '16 at 13:33
• @FataMadrina I've explained what is happening. The original circuit uses a comparator (with a pin-out the same as the LM311) and they are asking you to assume the comparator is an ideal op-amp. So, omit R5 and proceed with the ideal op-amp solution. – Andy aka Oct 31 '16 at 13:37
• The original circuit cannot work without R5 because a comparator like the LM311 needs a pull-up resistor because of the nature of its "floating" output transistor. – Andy aka Oct 31 '16 at 13:49