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schematic

simulate this circuit – Schematic created using CircuitLab

I'm a bit confused by a question which asks me to plot current through R1 as a function of Vin using a constant voltage model for D1. I'm a bit confused by this. Given the model, even if I considered Vin to be the voltage over the current source, wouldn't it be V_Diode_on once I_in*R1 = V_Diode_on? Does this make sense or is this likely a typo and I should be plotting V_R1 as a function of I_in?

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  • \$\begingroup\$ But if Vin's voltage doesn't go higher than that ... do I just stop the graph at Iin*R1 = V_diode_on? \$\endgroup\$ – Daniel B. Oct 2 '15 at 4:05
  • \$\begingroup\$ But that doesn't make sense does it? There is no value of i_R1 for Vin > V_diode_on because there is no voltage higher than V_diode_on? Am I making sense? \$\endgroup\$ – Daniel B. Oct 2 '15 at 4:07
  • \$\begingroup\$ That doesn't make sense ... there's voltage higher than V_diode_on from the current source when v_out can never be higher than v_diode_on? I get that increasing current from the source would only increase diode current, but I don't see how the input voltage could change without the output voltage changing. \$\endgroup\$ – Daniel B. Oct 2 '15 at 4:12
  • \$\begingroup\$ I agree with you - it's a typos and should say Iin. \$\endgroup\$ – Kevin White Oct 2 '15 at 4:15
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If you have the I vs V for the Diode, all you need to do is subtract Id (Diode Current) from Iin. because:

VR = VD = Vin (parallel configuration), and IR = Iin - Id

Note that Resistor I/V plot has two axis (Horizontal axis for Vr and Vertical axis for Ir) as Diode I/V has two axis (Horizontal axis for Vd and Vertical axis for Id). Because Vr and Vd are equal, to can simply take the Diode I/V plot and use its horizontal axis as Vr and for each point of it (i.e each Voltage value) calculate the Ir as Iin - ID

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