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I can't seem to find much information online regarding resistance values and diode behaviour.

Am I right in saying that the IV curve of a diode is independent of any external resistance? For example say you set up a standard Si-diode and a 100Ohm resistor in series and plot the IV curve. Would you expect the same IV curve if you repeated the test using a 1kOhm resistor in place of the 100Ohm?

Kevin Reid
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Maitiu
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    I-V curve for diode alone or I-V curve with the resistor? – G36 May 14 '17 at 13:45
  • If you measure the current through/voltage across the diode, while it is in a series circuit with a resistor. – Maitiu May 14 '17 at 14:36
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    In this case, the resistor doesn't have any effect on I-V curve. – G36 May 14 '17 at 15:08
  • G36 is right, the resistor doesn't affect the IV curve of the diode. But it does affect what voltage will be applied to the diode. Voting to close because you could have just fired up LT Spice and answered this question for yourself. – The Photon May 14 '17 at 15:26
  • I'm voting to close this question because the question is easily answered by a simulator with minimal effort. – The Photon May 14 '17 at 15:27
  • Why would I simulate it when I conducted the test using the real components? I was trying to find out if what I obtained was in line with what is expected and what the reasons are for the results. I don't think a simulator would provide much by the way of intuition, if you have come across such as simulator, please link. – Maitiu May 14 '17 at 19:51
  • a simulator will do fine; go for it; above 100uA through diode + 1kOhm, you'll measure a strong linear response instead of the exponential diode response – analogsystemsrf May 14 '17 at 22:47

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Detailed diode theory is well-documented in any microelectronics textbook. But the simplest diode equation that results from such detailed analysis is the Shockley diode equation:

$$I_D=I_{S\left(T\right)}\cdot\left[e^\frac{V_D}{n\cdot V_T}-1\right]$$

\$V_T=\frac{k\cdot T}{q}\$ is based upon statistical thermodynamics and is called the thermal voltage. \$T\$ is the absolute temperature, \$k\$ is Boltzmann's constant, and \$q\$ is the charge on an electron. It is typically about \$26\:\textrm{mV}\$ at room temperatures. \$n\$ is a model parameter called the emission co-efficient and will usually be around 1.7 to 3 for diodes, but may be as little as 1 and can be (infrequently) higher than 3. \$I_{S\left(T\right)}\$ is the y-axis intercept (log chart) of the diode behavior and is also significantly quite temperature-dependent (on the order of \$\propto T^3\$ to \$\propto T^4\$.)

You will note that this equation does NOT include any external, added circuitry elements. It provides a simple relationship between \$I_D\$ and \$V_D\$. In your comment, where you say you are only trying to observe these two parameters, you will expect this behavior without regard to an external resistor.

It's that simple.

(Well, not really. Diode behavior is a little more complex than one equation provides. There are limitations in reverse voltage, current crowding issues, various and multiple surface and volume re-combination issues, physical bonding wire and structure related issues (Ohmic resistance, for example), etc. All of which make even a simple diode more interesting at times.)

Don't be confused into thinking without nuance about this, by the way. Similar equations model the BJT and are similarly independent of the external circuit elements surrounding a BJT. But not all concepts important to the use of a BJT are independent of the surrounding circuit elements. For example, BJT switch-saturation IS a function of external circuit elements and not purely a function of the BJT itself. If you look at BJT models, you will not see any discussion of switch-saturation. That's because it's not a function of the BJT itself, but of the BJT as part of a circuit.

In short, you always have to engage your brain. You never get to just take a free ride or ignorantly apply pre-conceived "bright lines." You must apply element and/or circuit behavior, appropriately. To do that you have to think with good mental models appropriate to each circumstance.

jonk
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