# Basic question about diode voltage drop and resistor position

I'm new here and also new to electrical engineering so no fancy stuff yet, still tinkering with basic components such as resistor, led, caps and batteries. I think I caught up with the bad side of conventional + and - so I'd like to impose a few question.

I was reading this book "All New Electronics Self-Teaching Guide, 3rd Ed" by Kybett and Boysen to remembers bits what I've learnt so far. On Ch 2 about diode, I stumbled on the idea of the voltage drop in diode. In a series of resistors, it's pretty straight forward, it's a ratio of resistors' value, higher value drops more voltage, etc.

But when a diode is introduced after the resistors, you have to decrease the source voltage with the voltage drop of the diode first before you can measure the new voltage drop of each resistor. Why is this happening? Is it pretty bizarre to count backward first as the current move from + to -.

Also I've seen several examples where a resistor is introduced before anode, which I think is logical and introduced after cathode which is plenty weird as resistor is used to limit the current going into the led but in actual reality, both do just fine(I tried both ways). Is this because of positive charge and negative charge? Because P-N junction, holes and p stuff?

Update (request by Anindo) : Here is the page of the book in googlebook

Previous pages explaining in simple series are gone. For full book you can see it on page 47 (or 71/450, ch 2 starts at 59/450) here (Please edit this, if it's against the law)

• Not everyone has access to the book you mention. Hence, direct quotes of relevant, specific excerpts, rather than your paraphrasing thereof, would be helpful. – Anindo Ghosh Jun 14 '13 at 3:33
• @IgnacioVasquez-Abrams, why delete your answer? --- I thought it made a very important point that lots of beginners miss out on. – The Photon Jun 14 '13 at 4:15

Junction diodes are constructed from a single crystal of semiconductor material that has been altered to form a PN junction. Semiconductors fall somewhere between the conductors (metallic elements) and non conductors (non metallic elements). Generally speaking pure (intrinsic) semiconductor is an element with 4 electrons in its outer shell and is pretty useless electrically. It is neither a good conductor or a good insulator. The first semiconductors used Germanium. Devices today use Silicon.

The reason semiconductor materials are useful is that we can easily alter their electrical properties (especially conductivity) by adding or DOPING them with (very) small amounts of other elements. These doping atoms fit into the crystal lattice but their different electron structure alters the way electrical current can flow through the material.

Making P type and N type semiconductors.

N type has lots of 'extra' electrons because the dopant had 5 electrons in its outer shell - 1 more than (intrinsic) semiconductor.

Similarly P Type has gaps or HOLES in the outer electron shells because the dopant only as 3 electrons compared to 4 of the (intrinsic) semiconductor.

When the PN junction is made the material in the 'middle' is neither P or N type as all the free charge carriers are swept to one side or the other. This is known as the DEPLETION layer. (a bit like no-man's land between two opposing armies) This depletion layer is the source of the voltage drop across the diode.

To get current (flow of charge) through the diode the charge has to 'jump over' this barrier (its more technical than that but let's keep it simple). It needs an extra bit of energy to do this.

Now energy is charge x voltage. The value of the charge is fixed - its simply the electronic charge - 1.602 X 10^-19 so the only charges that can cross the barrier must have have energies of more than the barrier. As the charge is fixed and unchangeable we simply talk about the barrier voltage. For Silicon this is about 0.6 volts. For Germanium this is about 0.2 volts.

The barrier acts like a small battery of 0.6V connected in the OPPOSITE direction to the current flow. (Conventional current - positive to negative). You can only measure this when current is flowing through the diode.

Photodiodes can generate actual voltage but that's another matter. This means that for every diode in the circuit we will lose 0.6V when they are conducting (forward biased). (This increases slightly with current value)

In a series circuit with resistors it does not really matter if the resistor comes before or after the diode. The current passing through resistor and diode is the same. The total voltage drop across the resistor and diode will be the same. The LIGHT EMITTING DIODE has a much larger voltage drop (about 1.5V - 3.0 V) than a 'normal' diode. It uses this extra energy to output light. • Crystal clear explanation, sir. It answered most of my questions (I think it might had been vague and hard to grasp the meaning). So this reduce-this-'voltage drop'-first phenomenon is caused by depletion layer? – Try Jun 15 '13 at 10:06
• @Try thank you, its a very simplified model without the maths. What happens in the depletion layer basically controls PN junction action (that also includes transistors etc.) – JIm Dearden Jun 15 '13 at 15:09
• @JImDearden Just want to clarify that in above figure if we supply 10V to circuit which have resistance R and silicon diode 0.6V so do we finally get the 10-0.6V =9.3 V in the end ? What about the voltage drop in resistance if we dont know the current in resistor ? – Yogus Sep 26 '15 at 6:13
• @Yogus (10-0.6 = 9.4) The voltage drop will be 0.6 + I*R as stated. The 'on resistance' of the junction is fairly small and in some models it is even taken as zero (e.g. perfect diode model). If you look at the graph of a 'real' diode (I/V) you will see that the voltage drop increases with current but the (near linear part of the) is quite steep (large change in forward current, small change in voltage drop). The inverse of this slope is the effective forward resistance of the diode (dV/dI). This small increase in voltage drop with current tends to be ignored. – JIm Dearden Sep 28 '15 at 14:19

One way to think of a diode is that a diode is just a special kind of resistor. If you apply a low voltage to it (below about 0.6 V) then its resistance is very high. If you apply a high voltage to it (above 0.6 V) then its resistance is very low.

So when you put it in series with a resistor, you have a resistor divider but you don't know yet whether the diode is in its high-resistance mode or its low-resistance mode. If it's in high resistance mode, then the current through it is low...so the current through the other resistor is also low. So the drop across the other resistor is low...so the drop across the diode must be high (because the two voltages have to add up to the source voltage)... ooops, it can't then be in the high-resistance mode.

If the diode is in low resistance mode then the current through it is high, so the current through the other resistor is high. So the voltage across the other resistor is high, meaning the voltage across the diode must be low. Ooops, the diode can't be in low-resistance mode either.

Since the diode can't be in either mode, it must be right on the edge between them...sitting right at 0.6 V with whatever current through it is allowed by the series resistor.

As for whether to connect the resistor on the anode side of the diode or the cathode side, it doesn't matter. Just like with any two resistors, the current flowing through them only depends on the sum of the two resistances, not what order they're hooked up in.

• That would be "Putting the diode across the knee, and spanking it with current"? – Anindo Ghosh Jun 14 '13 at 3:56
• I think it's fairly misleading/confusing to describe a diode as a resistor, especially when you start talking about a "resistor divider." – Skaevola Jun 14 '13 at 4:09
• @Skaevola, in a low speed circuit, a diode is a nonlinear resistor, whether you think that is confusing or not. – The Photon Jun 14 '13 at 4:10
• Thanks for your answer. In a light buld example, the equation is very simple like this video youtube.com/watch?v=ggKnH-95ty0 .But I'm confused, on diode it's a whole new world. There's must be some explanations for this why is the voltage is quantified by the book's way I mentioned (I tried on breadboard and the reading in DMM is within the range too) – Try Jun 14 '13 at 4:47
• @Try, have a look here: en.wikipedia.org/wiki/Load_line_%28electronics%29 – The Photon Jun 14 '13 at 5:21

The voltage drop across resistors varies depending on the current running through it. The voltage drop across a diode (in the forward biased state) is close to constant regardless of the current running through it.

Here is a graph of the voltage-current relationship of a diode. Voltage is graphed on the x axis, and current is graphed on the y axis. The graph is an exponential curve. The key idea to take away from this graph is that no current flows until a small voltage occurs across the diode, at which point virtually any current produces close to the same voltage across the diode.

For example, look at the voltage produced when 3 A of current goes through the diode. From this graph it looks close to 3 V. Increasing to 4 A of current only increases the potential difference to ~3.05 V. In reality, most diodes reach forward bias at about .7 V. Because the voltage changes so little with current, a good approximation for their function is just to assume that the voltage drop is always around .7 V when they are forward biased. If you want to dig deeper, this effect is a consequence of the non-linearity of semiconductors. Diodes are also called a "p-n junction", searching that term should provide you information about the mechanism by which this occurs.