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I have a problem working on Lab 3-5 from Horowitz's Students Manual for The Art of Electronics. The circuit is like that:

Circuit in question

The author suggests compare the RC time constant of two version of this circuit: the original one and with 2.2K resitor removed. The problem (for my understanding) comes when removing the resistor: the RC becomes about 100 times (or even more) bigger. To see what happens with initial differentiator (capacitor and 1K resistor) I've removed the diode also and the RC was as expected. Supposing that something could we wrong with my diode I've tested is with a multimeter and the diode was correct. So, I can't understand why rectifying the differentiator output changes so drastically the RC constant.

I'm very grateful if somebody could help. Thanks

Here's what I'm doing: circuitandscopephotos

Should it behave like this when adding a diode and why? Then adding a 2.2K resistor between the diode output and the ground (i.e. the original lab's circuit) fixes the problem (which also I don't understand)

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  • \$\begingroup\$ Did he say where the signal was to be measured and which edge? \$\endgroup\$ – EM Fields Mar 31 '16 at 13:13
  • \$\begingroup\$ He doesn't give much instructions. Here is the lab's text: Use a diode to make a rectified differentiator, as in the figure above. Drive it with a square wave at 10kHz or so, at the function generator's maximum output amplitude. Look at input and output, using both scope channels. Does it make sense? What does the 2.2k load resistor do? Try removing it. Hint: You should see what appear to be RC discharge curves in both cases-with and without the 2.2k to ground. The challenge here is to figure out what determines the R and C that you are watching. \$\endgroup\$ – Ruslan Mar 31 '16 at 15:06
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Which capacitor is charging in this circuit, the 560pF or the 'scopes input capacitance? And what is the discharge path?

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Simulating it with LTspice, here's what I get:

enter image description here

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  • \$\begingroup\$ Yes, the RC here looks like it should be but measuring the real (not simulated) circuit I get a very long discharge curve at VOUT2. I can't understand what happens and where the problem could reside. \$\endgroup\$ – Ruslan Mar 31 '16 at 19:29
  • \$\begingroup\$ Since the only thing (maybe) not accounted for in the sim is your scope probe's capacitance, That's more than likely the culprit. \$\endgroup\$ – EM Fields Apr 1 '16 at 0:08
  • \$\begingroup\$ If my understanding is right the input capacitance of the scope is made to not to affect the mesured cirsuit too much. More than that, I can't figure out how just inserting a diode can change the discharge time. I took photos of what I was doing in order to someone could see if I'm doing it right (they are in the updated question) \$\endgroup\$ – Ruslan Apr 3 '16 at 16:19

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