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I have this simple circuit with a diode (with voltage drop \$V_d\$) and resistor in parallel, and another resistor in series with them both. The circuit can be seen in the diagram below.

schematic

simulate this circuit – Schematic created using CircuitLab

What is the voltage, \$V_o\$, across \$R_2\$? I think that when \$V_i<V_d\$, the diode does not conduct, so the circuit behaves live a potential divider, with \$V_o=V_i\frac{R_2}{R_1+R_2}\$. Otherwise, the diode behaves like a short circuit, so \$V_o=V_i-V_d\$. If this is true, the signal would look quite strange. That's what confuses me.

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  • \$\begingroup\$ That is not a clip circuit. The diode should go across the output resistor. \$\endgroup\$ – Andy aka Jan 24 '14 at 14:52
  • \$\begingroup\$ @Andy aka: That's another way to do it from what I understand, but what I have above should also do the same thing (if the diode is reverse-biased). \$\endgroup\$ – someguy Jan 24 '14 at 14:58
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    \$\begingroup\$ no it won't. Think about it. On small voltages the output is divided by the resistor attenuator and on bigger signals the output is input minus 0.7 volts. \$\endgroup\$ – Andy aka Jan 24 '14 at 15:04
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    \$\begingroup\$ on opposite half cycles the signal is just subject to the resistor divider. \$\endgroup\$ – Andy aka Jan 24 '14 at 15:05
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    \$\begingroup\$ Or, maybe R1 is the output resistor. Nothing is really labeled, but if you tip your head to the left... \$\endgroup\$ – Phil Frost Jan 24 '14 at 17:05
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I think that when \$V_i < V_d\$, the diode does not conduct, so the circuit behaves live a potential divider, with \$V_o=V_i\dfrac{R2}{R1 + R2}\$. Otherwise, the diode behaves like a short circuit, so \$V_o=V_i−V_d\$.

One problem (maybe the only one) with your analysis, is that the diode behavior doesn't depend on \$V_i\$ but on the voltage across the diode itself, which is not the same thing in this circuit.

If you consider that the diode is "on" when \$V_i\dfrac{R_1}{R_1+R_2} > 0.7 \mathrm{V}\$ instead of just \$V_i > 0.7\mathrm{V}\$, the behavior in the "on" and "off" states will match up better at the transition.

Also, for this kind of simple modeling, when the diode is "on" we don't consider it as a short circuit, but as a fixed voltage of about 0.7 V.

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The behavior of the circuit depends on where your 0V (ground) reference is located. Your analysis is correct. The math, and the resulting waveform, will look different depending on where ground is located.

If ground is at the node common to \$V_i\$ and \$R_2\$, the waveform will roughly sinusoidal in shape with \$V_o = V_i-V_d\$ when the diode is conducting and \$V_o=V_i\frac{R_2}{R_1+R_2}\$ when it is not.

If ground is at the node common to \$Vi\$, \$R_1\$, and the diode, the waveform will look like a half-wave, rectified sine wave with \$V_o=V_i\frac{R_2}{R_1+R_2}\$ when the diode is not conducting, and \$-V_d\$ when it is.

Edit:

Now that there is a ground reference in your schematic, the analysis is different. That is, if the sine wave produces negative voltages, the diode will be completely ignored for the negative portion of the sine wave. During this time, \$Vo\$ will always be greater than \$Vi\$ (as in less negative, not a larger amplitude). Then \$V_o=V_i\frac{R_2}{R_1+R_2}\$. During the positive half of the sine wave, still \$V_o=V_i\frac{R_2}{R_1+R_2}\$ until \$Vi - Vo\$ is greater than the cutoff voltage of the diode (typically ~0.7V). Doing the math, this happens when \$V_i(1-\frac{R_2}{R_1+R_2}) > 0.7V\$. After, \$V_o = V_i - V_d\$, until \$V_i(1-\frac{R_2}{R_1+R_2}) < 0.7V\$.

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  • \$\begingroup\$ "The behavior of the circuit depends on where your 0V (ground) reference is located" What? Not more than any other circuit. "ground" and "0V" aren't concepts the circuit knows about, and if I say "hey, ground is my left big toe", this circuit just keeps working as it always does. \$\endgroup\$ – Phil Frost Jan 24 '14 at 17:01
  • \$\begingroup\$ You are right. If Vi is simply an ideal, sine-wave voltage source, Vo will be equal to (Vi*R2)/(R1+R2) until Vi - Vo is greater than the cutoff voltage of the diode. Then Vo will simply be Vi - Vd. However, in a simulation such as CircuitLab, if a -5V to 5V sine wave is used, the ground symbol is referenced to the center of that waveform. Then, it will matter where the ground reference is located. Also, in practice, if this circuit is part of a larger circuit with multiple power supplies sharing a common reference, unexpected waveforms could be produced. \$\endgroup\$ – Justin Trzeciak Jan 24 '14 at 18:38

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