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I need square wave without negative voltage. enter image description here

Considering a source which produces square wave (could be function generator or from other circuits) has no ability or I don’t have the control required to do this. So I need to use an extra circuit which can block the negative part.

What circuit would you use?

My simplest idea is to use a diode in series to block the negative voltage. However it doesn’t work. I have tried high and low frequency. Simulation in multisim’s oscilloscope doesn’t show any difference. I have no idea why this is happening. Being relatively new to semiconductor devices, I am hoping to learn something new from this as it goes against my basic understanding of diodes.

Why the diode can’t block negative voltage? Is it because the diode’s p-n junction works like a little capacitor?

Multisim schematic and oscilloscope result: enter image description here

Edit: After applying Olin Lathrop’s answer and Venny’s diode suggestion oscilloscope shows exactly what I want.

enter image description here

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  • \$\begingroup\$ Try connecting the power supply negative terminal to the middle terminal. And if you want only the positive cycle flip the diode around. Also, check the "oscilloscope" it might be set on AC coupling, set it to DC. \$\endgroup\$
    – Mike
    Aug 25, 2014 at 18:18
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    \$\begingroup\$ The ringing is probably caused by the pretty lousy reverse recovery time of the 1N4001. Try replacing it with something faster like the 1N4149 (worked for me). \$\endgroup\$
    – venny
    Aug 25, 2014 at 19:13
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    \$\begingroup\$ I'm not sure you should have edited your question to show the modification that Olin suggested - I've just spent 4 and a half minutes of my life trying to figure out what was going on with your circuit. I would consider it sensible to return your question to how it was originally phrased. \$\endgroup\$
    – Andy aka
    Aug 25, 2014 at 21:03
  • \$\begingroup\$ @Andyaka - Sorry for the mess I have created. I am sure that I shouldn’t have done that. I will not do this to any of my next question. Your time is always appreciated. \$\endgroup\$
    – Amit Hasan
    Aug 26, 2014 at 3:41

4 Answers 4

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A diode is series is one way, but your diagram shows two problems with that setup. First, you have the diode backwards. Second, you have provided no load for the diode. The scope is has high impedance, so the negative part is either being capacitively coupled thru the diode or via its leakage.

Here is how to use a diode correctly:

All negative input voltages will result in 0 output voltage with a impedance of 1 kΩ. Note that due to real diodes not being perfect, during the positive part of the input, the output will be about 700 mV less. Put in mathematical terms:

Out = max(0, In - 700mV)

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  • \$\begingroup\$ Accepted and upvoted? Seems strange, since all you've done is merely duplicate his circuit, so any problems he's experiencing with his will also be experienced with yours. BTW, your diagram should include polarity signs on the input, just to allay any confusion their lack might cause. :-) \$\endgroup\$
    – EM Fields
    Aug 25, 2014 at 20:09
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    \$\begingroup\$ @EMFi: This is different from his original circuit. Note that the OP edited the question after I posted this, apparently changing it to what I suggested. As for polarity, I have already shown that by marking a net as ground. The convention is that unless specified differently, ground is at 0 V and all other voltages are therefor relative to it. Besides, if he's got a symmetric +- signal, then the polarity of the input doesn't matter. \$\endgroup\$
    – Olin Lathrop
    Aug 25, 2014 at 20:47
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To rectify a signal without introducing error from diode forward voltage drop, one has to use ideal rectifier op-amp circuit.

cir9

Both feedback branches must be present even if only one is used, to prevent the op-amp output from saturating.

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At 50HZ, that diode had better block the negative excursion of that square wave! Either MultiSym is badly wrong, or else it's scaled itself up by 2x and readjusted the offset so the simulated "trace" looks the same but the p-p voltage is different.

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Try a bridge rectifier. With a bridge rectifier you can save the whole wave, positive and negative. Or, if you want to completely get rid of the negative portion, just use a diode.

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