+5V DC in series with a 100ohm resistor & diode (anode on the resistor, cathode to the - terminal of supply. Diode forward voltage for turn on is 0.7V. I understand this circuit. The cathode is already grounded, so it's very easy to see that all you need is 0.7V on the anode of the diode to get it to turn on.
Now, let's say you have the above circuit, but TWO diodes in series with each other. I know that you need 1.4V measured from Anode #1 to ground to get the diodes to turn on and complete the circuit.
However, my conceptual problem is this: Each diode is essentially an open switch. The first diode needs 0.7V across it to turn on. How do you reference the voltage of cathode #1 to ground? To me, it essentially looks like it's floating, so how do you ever get 0.7V across the diode?
Your idea of how a diode works is a bit oversimplified. The concept that they suddenly turn on when the voltage across them reaches .7 volts is just not quite right. They conduct some current at all forward voltage, just not an amount that is proportional to the voltage. Take a look at this data sheet for a diode:
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On page 2 there is a graph called Forward Characteristics. This is a diode rated for 1 ampere of forward current, and the forward voltage drop that corresponds to the 1 amp point on this curve is about 0.95 volts (not 0.7 volts, in this case). But at 0.8 volts, it will pass 0.2 amps, and at 0.6 volts it will pass 0.01 amps. The curve just keeps going all the way to zero volts. Diodes are not switches, but very nonlinear resistors.
A diode is not an ON/OFF device. It is ON all the way up to 0.7V and beyond. It is just that at the so-called 'turn-on point', it is more 'ON' than at 0.5 volts. In 'Math-Speak' the resistance of the diode is said to be non-linear. You can show this by increasing the value of the series resistance to a high value until the voltage across the diode drops and where you will find that current is still flowing; plot the values then and look at the curve.
BTW, the actual diode voltage also depends on the current through it and the area of the die among other things. Think about the voltage across a bunch of them in parallel at the same total current as a single one. It will be slightly lower.
You have to stop thinking about the diode as just a component to be used as a one way valve - you have to understand the atomic theory.
Both P and N material conduct electricity - not well, but they do conduct.
Ordinarily the silicon crystal is a good insulator. The manufacturer adds controlled amounts of contaminants to unbalance the insulating properties and this causes it to have a either a surplus of electrons (N type material) or a deficit of electrons (P type material) - these "carriers" allow the silicon to pass current.
No big deal so far.
Now, when a junction is formed with N/P types of material, an interesting phenomena takes place for a very thin plane right at the junction. Some of the "majority carriers" of the N material are attracted to the majority carriers of the P type material and they swap places and form a sort of curious insulator because once again the atoms are in a balanced state.
This area where they swap, is called the "barrier region." It is an insulator, but not a very strong one. For the diode to conduct you merely have to overcome the attraction that is forming the insulator (barrier region) with enough VOLTAGE to get the electrons to move across the barrier.
The barrier can be thought of as a small battery with a .6.7 volt output.
Put two diodes in series, and they both respond to the same VOLTAGE, each will begin conducting CURRENT when both barriers are overcome.
And diodes do conduct electricity even before the barrier voltage is satisfied - a voltage of just a few millivolts will cause some conduction but is just a leakage amount. It isn't an on - off thing or not perfectly on or off - diodes are non linear in the amount they conduct with potential difference.
If you look at the curves for diodes, a silicon diode will show a slight current that will increase as voltage increases until the barrier voltage is overcome then the conduction takes off very sharply as voltage increases further. Same thing happens with a reverse biased diode current increases slowly until the diode breaks down (and as long as the current is limited - breakdown doesn't destroy the diode)
When a diode is reversed biased the depletion region grows and becomes thicker - but there's still some small leakage current.
Any book on basic electronics could probably explain it better than I just did . . .
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On Fri, 06 Jul 2007 08:03:27 -0700 in sci.electronics.basics, TVisitor wrote,
It's not all that completely off. That's a simplified model.
Slightly better: the diode is almost all off with slight leakage, which is enough to equalize the voltage on both diodes (same leakage current through both.)
For a better model, look at the actual graph in the datasheet for the actual diode you are using. For example Figure 3. from
Actually I think everyone did a good job of explaining my "fundamental" problem - thinking of it as a simple on-off switch. With it always conducting some current through it, it now makes more sense; there is some drop across it, and my head is a lot clearer on this now. It also helps as I was having the same fundamental problem with transistors.
(I'll go back and consult my texts for some finer reading, this is the problem when you did electronics a long time ago, but since then have only been a software guy!)
I might as well add something else that may help a little. The junction voltage barrier is proportional to a logarithm of current through it plus an amount related to the 'saturation current' (which is just where an extrapolated line intersects the y-axis.) So the voltage across goes something like 60mV/10X-change-in-current or some not so distant multiplier of that. (10X is a 'decade' change.) That's often close for small signal transistors, for example. Diodes are a little different from the base-emitter junction of a transistor, in that there is more commonly something closer to double that rate -- maybe 100mV/decade or more. There is a number of factors leading to the difference, but these are often combined into a single number called the 'emission coefficient.' For BJTs, it's often near 1. For diodes, often closer to 2, though it can vary to maybe 3-ish.
Anyway, without necessarily knowing the exact figure of change, it gives you a rough handle to just know that there is something on the order of a 100mV change in voltage across a diode for each 10-fold change in current through it. If you know that there is, for example,
0.6V at 1mA, then you can guess that there will be about 0.5V at
100uA, 0.4V at 10uA, 0.3V at 1uA, and 0.7V at 10mA. Something like that. If you continue that process into lower currents for the example diode I'm talking about, you will see 0V at 1nA. That would be the 'saturation current' of the diode, the concept I mentioned earlier as the y-axis intercept.
For BJTs, that will probably be closer to around 60mV/10-fold, but that gets the idea across. Also, they will usually have a somewhat higher voltage at 10uA base current, maybe 0.6V or so. So with smaller steps (60mV) for each decade down, you can see that it would take 10 decades of current to get to 0V, which implies a smaller 'saturation current' of maybe 10^-15 (1fA) or so.
You usually don't worry so much about all this. You just use diodes and can roughly guess their voltage drops for some ballpark current.
100mA would be about 0.8V, as a guess. And that usually is good enough. If you are off by a factor of 2 in your guess about the current, it won't make all that much difference in the voltage you guessed at.
There is another confounding factor at high currents, which is a kind of current crowding. When you start pushing a lot of current through a small signal diode, maybe more than 100mA or so (maybe a factor of
100-1000 less than this for small signal BJT base currents), you start to see the voltage rise more rapidly than you'd expect. So if you were talking about an amp or more, I'd guess that the rate of voltage change would be significantly higher than 100mV/decade for a small signal diode. More current-capable diodes will be uniform for much higher currents, with current crowding taking place at much higher current densities.
So, don't crowd your current (pick a diode or transistor designed for the currents you are working with) and the rules work pretty good.
I'm a hobbyist, by the way, and have _NO_ formal training at all. If a professional corrects me on the above, listen to them.
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