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schematic

simulate this circuit – Schematic created using CircuitLab

Since the voltage at node(A) is 30V, the diode(D1) does not conduct because the voltage at the anode is 20V. What about the current that has been divide due to the parallel branches? At point B is there current = (V/R)/2 (equal currents between branches since both resistance are the same)?

If so, the load would have lower current input?

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  • \$\begingroup\$ A reverse biased diode conducts very little current (ideally zero). There is no resistive divider here -- neither the diode nor the 10V source can be modeled as resistive. \$\endgroup\$ – Null Oct 14 '15 at 3:54
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    \$\begingroup\$ Voltage sources are effectively short circuits when doing current analysis. \$\endgroup\$ – Ignacio Vazquez-Abrams Oct 14 '15 at 3:54
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    \$\begingroup\$ Say I have a battery, a switch and a lightbulb in series. The switch is on. When I turn off the switch, what happens to the current? \$\endgroup\$ – user253751 Oct 14 '15 at 5:56
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Lets look at this circuit in a slightly clearer topology - below is exactly the same circuit, but rearranged to put all the elements in a way that height is signifying voltage - the higher up in the diagram, the higher the voltage. (It's a nice analogy which helps visualise what is going on).

schematic

simulate this circuit – Schematic created using CircuitLab

So, at the top of the diagram is \$20 + 10 = 30 \mathrm{V}\$. At the bottom is \$0 \mathrm{V}\$. So lets see if we can work out the currents \$I_r\$ and \$I_d\$. I should note that the direction I have drawn the current arrows is the direction that conventional current would have to flow (positive to negative).

Well, for \$I_r\$ it is fairly easy. Ohm's law tells us for a resistor (or any device with 'Ohmic' behaviour), that \$V=I\times R\$, so from this we can work out:

$$I_r = \frac{V}{R} = \frac{30}{50} = 0.6 \mathrm{A}$$

Why? Well, at the top of the diagram is \$30\mathrm{V}\$, and at the bottom is \$0\mathrm{V}\$, so there must be that much voltage across the resistor.

Now lets look at the diode. For a diode to conduct the voltage at the Anode must be higher than the voltage at the cathode (*). Now in this diagram we can see that the Anode is at \$20\mathrm{V}\$, and the cathode is at \$30\mathrm{V}\$, so this means the voltage across the diode \$V_ak = 20 - 30 = -10\mathrm{V}\$. So from this we can see that the diode is reverse biased - the voltage at the anode is negative with respect to the cathode. So we know then that \$I_d = 0 \mathrm{A}\$.

The current can't 'split down the parallel branch', because the diode is reverse biased so is blocking the flow of current - it's acting essentially as an open circuit.

If the diode was placed the other way around (not a good idea!), then current could flow down that branch. But the current that flows would not reduce the amount of current that flows through the resistor. Instead it increases the amount of current drawn from the 10V supply (**).


(*) There is some 'reverse leakage current' for a diode, basically the amount of current that it conducts when reverse biased, but this is pretty small so for simplicity we say it is \$0\$.

(**) Assuming an ideal power supply - one which has no limit on how much current you can draw.

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    \$\begingroup\$ nice job with the redrawn schematic. \$\endgroup\$ – tcrosley Oct 14 '15 at 4:36
  • \$\begingroup\$ I agree about the re-drawn schematic, made it much easier. However, if we removed(or flipped) the diode wouldn't that be an short-circuit there? \$\endgroup\$ – Pupil Oct 14 '15 at 9:13
  • \$\begingroup\$ @XCIX if the diode was flipped around, then a very large current would flow through it - if you've seen a diode I-V curve it is clear that to drop 10V it would allow a lot of current to flow. For an ideal power supply this wouldn't matter, but in reality you would either short out the power supply or melt the diode, or both. If you removed the diode and put nothing in its place (i.e. left the wires unconnected), nothing would happen as currently the diode is already open circuit. If you replaced the diode by a piece of wire, then yes that would be a short circuit. \$\endgroup\$ – Tom Carpenter Oct 14 '15 at 14:00
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"When a diode does not conduct, no current flows through it"

An ideal diode is equivalent to an open-circuit if reverse biased. And hence the reverse biased diode placed across the voltage source (10V) plays no role (it carries zero current).

So current 'at B' is zero. The entire current passes through the other path parallel to it, through the voltage source. (In practical cases there will be a reverse saturation current through diode but will be in the order of micro- or nano-amperes and hence neglected here).

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