I have a question:

What does the diode and the capacitor do in this circuit? What's their role? And how will \$V_{out}(t)\$ graph look like, if we know, that \$V_C(0)=-5V\ , \ SR=0.5V/\mu s\$ and \$V_{ppOUTmax}=13.5V\$?

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  • \$\begingroup\$ Think about when diode starts conducting -> anode is 0.7 more than cathode. \$\endgroup\$ – R.Joshi Jul 11 '18 at 11:54

We don't just answer homework questions here.

Look at the circuit carefully. You will see two parts. The first is a current source, capacitor, and diode making some voltage. The second simply amplifies that voltage. Therefore, the way to analyze this circuit is to analyze each part independently.

Think about what a capacitor does when being charged by a steady current. This is really something you should be able to figure out yourself. Then what is the effect of the diode on the current source and cap?

After that, the voltage just gets multiplied by a constant. The opamp is wired in classic non-inverting amplifier configuration. You really should be able to see for yourself what the gain is.

You should also consider limits of what the opamp can do. Unfortunately, the power supply to the opamp is not shown. We also don't know what opamp it is, so we don't know things like common mode input range, and output swing range. However, those will be issues in any real circuit.

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  • \$\begingroup\$ +1 for mentioning the power supply. The Vpp spec is meaningless unless you assume it is centered around zero. Granted, that's a pretty reasonable assumption for this level of class, but still.... \$\endgroup\$ – WhatRoughBeast Jul 11 '18 at 13:12

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