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Diodes - A very boring question which someone might know the answer to...
tbloxham
Registered User regular
Registered User regular
OK, so, I'm using a diode and almost all sources say that the voltage drop across a silicon diode is about 0.7 V. However is this true for any current from zero up to the destruction point of the diode? So, if I apply a few micro Amps, will I still see a 0.7 V drop across the terminals?
I suppose what I'm asking is, at a few microamps of current, does a diode effectively behave like a Mega Ohm resistor along it's correct flow path. Otherwise the whole, current drop of 0.7 V thing I see everywhere would make no sense.
I suppose what I'm asking is, at a few microamps of current, does a diode effectively behave like a Mega Ohm resistor along it's correct flow path. Otherwise the whole, current drop of 0.7 V thing I see everywhere would make no sense.
"That is cool" - Abraham Lincoln
tbloxham on
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http://en.wikipedia.org/wiki/Diode#Current.E2.80.93voltage_characteristic
I'm really, really, really not an expert at all but it says at the bottom of the section that more current results in a greater voltage drop. And that if you don't have high enough voltage, you won't see current pass through at all. And I really wish I stuck with electronics because way too much of this is way over my head.
CUZ THERE'S SOMETHING IN THE MIDDLE AND IT'S GIVING ME A RASH
When reverse-biased, it will conduct slightly more current when below breakdown voltage, then break down and conduct as a wire once you exceed that (the difference, however, is that when this happens in the "wrong" direction it will usually be toast afterwards...it'll no longer function correctly).
The voltage drop will vary slightly across the usable range of the diode, but for most applications yes you can just treat it as either a 0.7 voltage drop or an open circuit (depending whether it's forward biased or not).
That's an ideal diode (0 current when a negative voltage is present, full current when a positive is present), plus a voltage source (approximating the 0.7 drop) plus a current-limiting resistor (which will give you a slight slope to approximate the I-V curve when forward biased).
Though obviously that model doesn't deal with reverse biasing at all. And for simpler functions, you can omit the resistor (which is what I did in the previous post).
once you've passed that amperage (usually in the 10's of thousands of micro amps) Voltage drop across the diode will become .7, or roughly that.
(In forward, or correct, bias)
Hey, just to be sure, do you mean mA or uA? Because mA is a milliamp, and uA is microamps, and 1mA = 1000uA - i.e. a microamp is really, really tiny.
So tiny, that for most diodes, a few microamps is about the amount of leakage current through the diode. Think of a diode like a valve which can't close properly - there'll always be a very small amount of current leaking through the diode. For a vast majority of designs, leakage current is so small that we just ignore it.
On the flip side, a fair few (small signal) diodes happily operate as you'd expect at 1mA.
For a first order approximation:
* If you can only supply a few microamps through the diode, then the circuit is set up so that you have insufficient voltage to forward bias the diode. You should treat the diode as an open circuit.
* If you can supply a few milliamps, then you can treat a standard small signal silicon diode as having a forward drop of 0.7V (e.g. 1N4148).
For a slightly more detailed second order approximation:
* If the diode only has a few microamps, then you can apply Ohm's law to gain the equivalent resistance in megaohms (or hundred of kilo ohms). However, because the voltage/current characteristic of a diode is extremely non-linear, the equivalent R will change dramatically depending on the amount of current that's leaking through.
* If the diode has a few milliamps, then you can look at the datasheets for your diode to determine which voltage you'll actually run at, for a given current. For example, Figure 2 in this datasheet for a 1N4148 shows that 1mA will typically operate the diode at about 0.4-0.5V.
Yes, it can act like an equivalent megaohm resistor if you want to do a more detailed approximation.
The whole 0.7V voltage drop is just a first order approximation for standard small signal silicon diodes operating at rated current <- this is the important assumption that is often left unstated - as you can see from the datasheet I pulled above, operating at non-rated currents changes the voltage drop.
OK, thanks a lot for everyones help. It's certainly been useful for trying to solve the problem I'm having.
What the problem is that an engineer I work with refurbished this device that uses a diode to monitor it's operating temperature. You just get the voltage drop at the correct current and then as the temperature changes, the voltage drop changes. You can then solve the diode equation to find T.
The problem is, that in all tests this device doesn't behave like a diode! In fact, I don't know what it behaves like! It has a negative thermal coefficient (it increases it's resistance as it cools) like a semiconductor, but the rate of change is really wierd, only changing a small amount over what we believe is a large temperature range.
Blasted refurbed devices, covered in mysterious doodads added by previous refurbishers...
The forward bias voltage isn't going to vary much over the entire operating range, maybe –2 mV/°C change riding on 600 mV.
A front-end inverting aplifier to adjust gain and offset could be useful, a realtively simple amp could get you an order of magnitude increase in sensitivity.
Edit: Wait a second
When you say "device" and when you say "diode", are they the same thing?
All the behaviour that's been described here is geared towards the "diode", but if what you mean by "device" includes all the circuitry around the diode, then that might include the amplifier (probably does, actually) that darkgrue suggested.
I'd need to know what you mean by "small amount" and "large temperature range." As mentioned, the voltage across the junction (for a given current) won't vary all the much with temperature...it'll be measurable, but not dramatic.