Diodes are strange - Page 1
Forum section: electronics / beginners

author: Legion
author_rank: Frequent Contributor
timestamp: March 22, 2014, 07:59:12 pm
content: I'm messing around with diodes right now. The behavior is strange.Some circuits I breadboarded:Circuit 1Expected: All of V_Source is dropped across the diode D1. If V_Source is greater than D1's forward voltage, current flows. The relationship between voltage and current is non linear, so Ohm's law doesn't hold.Actual: If V_Source is greater than D1's forward voltage, the voltage across D1 is always the forward voltage. It'll draw as much current as it can to keep the voltage across it equal to the forward voltage. Good thing I had OCP turned on.Circuit 2Expected: V_Source minus D2's forward voltage is dropped across R1. R1 effectively "linearizes" the circuit once D2's forward voltage is exceeded. Circuit current will be equal to (V_Source - D2_FV)/R1.Actual: Worked as expected. Some variation of the diodes forward voltage depending on the paired resistance value. Lower resistance/higher current increased forward voltage. Is this just movement along the diode's V/I curve?Circuit 3Expected: Similar to circuit 2. R2 and R3 will divide V_Source less D3's forward voltage.Actual: Mostly worked as expected. I played around with the ratio of the resistors and found some odd behavior. Diodes are "greedy". If R3 is 500K and R2 is 100R pretty much all of V_Source should be dropped across R3. There shouldn't be enough voltage left to get the diode to work. But there is a voltage drop across the diode and it's conducting.I experimented further with circuit 3 and mapped the total circuit voltage vs total circuit current. It's way more linear than I thought it would be. I almost wonder if these diodes are bad or something. But I checked them with my DMM and it shows a forward voltage around 550mV. It's just weird that I don't see that in the circuit.Here's the excel file with the graph:https://dl.dropboxusercontent.com/u/91808950/Diode%20V%20vs%20I.xlsx

author: Rerouter
author_rank: Super Contributor
timestamp: March 22, 2014, 08:18:01 pm
content: diodes forward voltage doesn't increase much short of a very large current increase once there conducting a few hundred milliamp (first and second example)while with the third i would greatly ask you to measure your supply voltage for your calculation, diodes cannot steal voltage that is not there, arguably towards the lower currents its exactly following there current curve for the voltage drop to decrease (get really low and you can approach 10's of mV)also just as a tip, try instead comparing the diodes own forward voltage to the current through the circuit, that will give you the IV curve of the diode your looking at, whereas measuring the supply voltage with the series resistor will look very linear as until you get to below 1-2V you are mainly measuring the resistors drop in voltage with minimal influence from the diode itself,

author: IanB
author_rank: Super Contributor
timestamp: March 22, 2014, 08:35:54 pm
content: Quote from: Legion on March 22, 2014, 07:59:12 pmOf course there is a voltage drop across the diode and of course it is conducting. You have a supply voltage and a circuit. Current flows in the circuit and a voltage appears across every element with a current flowing through it.Voltage cannot get "consumed" and there is no such concept as "not enough voltage left". That's not how voltages work.If Vsource provides a voltage, then that voltage will appear across R3, D3 and R2. The voltage will be divided up between those components, but the sum of voltages across all components will equal the supply voltage. R3 will have the most voltage, R2 will have less voltage, and D3 will have a very small voltage. But those three voltages will always add up to the source voltage.

author: babysitter
author_rank: Frequent Contributor
timestamp: March 22, 2014, 08:46:47 pm
content: Diodes are strange when you're a stranger,IV diags look ugly when you're alone...

author: w2aew
author_rank: Super Contributor
timestamp: March 22, 2014, 09:48:04 pm
content: Quote from: Legion on March 22, 2014, 07:59:12 pmOnce you get your head around the "strange" DC properties of the diode, and how they can be used as rectifiers, you might want to try taking a look at these other cool applications for diodes.Here's how they can be used as electronic switches:Here's an application using diodes as current switches:And even stranger yet, these special PIN diodes can switch RF signals which reverse bias the diode, but they don't rectify the signal:

author: Legion
author_rank: Frequent Contributor
timestamp: March 23, 2014, 01:02:55 am
content: Quote from: IanB on March 22, 2014, 08:35:54 pmPerhaps I phrased it poorly. What I'm referring to is the voltage drop across the diode's stiffness. If I had a voltage source and three resistors in series, A, B and C; if I vary the values of A and C while keeping B constant, the voltage drop across B will change in relation to the ratio of B to the sum of A and C. But the diode doesn't. I thought it would have some kind of equivalent resistance value. So if it were in series with a large enough resistor, the resistor would drop the majority of the voltage and the remaining voltage wouldn't be enough to cause the diode to conduct. Like a voltage divider.

author: IanB
author_rank: Super Contributor
timestamp: March 23, 2014, 01:11:30 am
content: Quote from: Legion on March 23, 2014, 01:02:55 amResistors only "drop" voltage if there is a current flowing through them. If there is a current flowing through the resistor, then it is also flowing through any diodes in series with that resistor.If there is no current flowing through the resistor, then it drops no voltage. In that case,the entire supply voltage appears across the diode. How could the diode not conduct with the entire supply voltage appearing across it?In short, a resistor has no effect on a circuit unless it has a current flowing through it. And if a current is flowing through the resistor, it is flowing through any other components in series with that resistor. Including diodes.

author: IanB
author_rank: Super Contributor
timestamp: March 23, 2014, 01:15:55 am
content: You need to learn about Kirchhoff's voltage and current laws. And you especially need to understand Ohm's law better than you do now.

author: tszaboo
author_rank: Super Contributor
timestamp: March 23, 2014, 10:06:45 am
content: Quote from: Legion on March 23, 2014, 01:02:55 amIf you need an easy explanation for this:You apply the rules in different order. If you have 12V, first you subtract the 0.6V voltage drop of the diode, THEN calculate the voltage drops on the resistors. So there is always 0.6V on it if there is current flowing. The order of applying the rules matter.Semi hard explanation: solve the circuit in steps. First, you calculated the voltage drop on the resistors, and the diode. Also calculate the current. See, if it is on the I-V curve of the diode. If it is not, and not even close, select a voltage and current which is on the curve, recalculate everything. Repeat as many times as needed. You end up with about 0.6V drop and the actual current.If you need a complicated answer: Differential equalizations, and non-linear characteristics...

author: Legion
author_rank: Frequent Contributor
timestamp: March 23, 2014, 12:58:43 pm
content: Quote from: IanB on March 23, 2014, 01:11:30 amI see what you're saying. There wouldn't even be a voltage drop across the resistors unless current was flowing. Current can't flow unless the diode is conducting. If the diode's conducting there is a forward voltage drop.I think I'm getting tripped up a bit in the transition from ideal diode to real diode. I'm using a 1N4001 and it leaks like a sieve. Even at 0.1V I was able to detect current through the circuit (0.1uA).

author: Legion
author_rank: Frequent Contributor
timestamp: March 23, 2014, 06:02:02 pm
content: Quote from: IanB on March 23, 2014, 01:15:55 amI know them. But knowing how they apply or their implications isn't always clear. Reminds me of chess. The rules are easy to learn, but mastering the game takes years.

author: babysitter
author_rank: Frequent Contributor
timestamp: March 23, 2014, 07:29:35 pm
content: Remember that these things that look counterintuitive to you now, or at least in your starting post, can be exploited for fun and profit. Besides rectifier use you can use that nasty forward voltage drop for pracical things. Where should a LED get the power for emission from if it where a current-only "member" of the circuit without the forward drop, as P=U*I and there where U=0?Also the diode is a good candidate to shunt a current around a part that should not get too much of it - think ESD diodes and everything that is called "clipping" - the voltage to your part in parallel to a diode will only rise slowly as the diode will be a rising load for your power supply for a while.Also, voltage stabilisation is exploiting this - the diode can be basically used as a shunt regulator, and the Ultra Precision Zener reference discussed in a long thread here is related to that too.Also, your forward drop has a temperature dependency - measuring it and comparing to the datasheet can give you clue about the ambient temperature. Vice versa that means that you have to check the temperature of a diode if you want it to serve a certain purpose.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 23, 2014, 10:46:42 pm
content: 1N4000s are the floor scrapings of the diode world; transistors are generally better silicon, and B-C junctions make reasonable diodes if you don't need much current (obviously, keep it under Ib(max)).LEDs also make good diodes (GaAs and friends exhibit quick switching and low resistance... and a higher bandgap = more Vf), but mind the reverse leakage is proportional to ambient light.(This is, or can be, true of glass-encapsulated diodes as well -- 1N914, etc.)You always have a tradeoff between forward drop, reverse bias and switching speed. 1N4001 has a lower voltage drop than UF4001 (high speed type). SB160 (schottky) ironically has about half the voltage drop and no switching speed (schottky diodes are considered instantaneous), but massive capacitance (that acts like switching speed anyway) and has several orders of magnitude higher reverse leakage (at higher temperatures, it's in the mA range!).There are also a few self-contained synchronous rectifiers, which trade reverse leakage (it's actually supply current) for negligible forward drop:http://www.onsemi.com/PowerSolutions/newsItem.do?article=994(This article was 2005, and it seems the product is 'obsolete' now.) The downside is, nothing that runs on microamperes can switch in microseconds. I see an LT part which drives four MOSFETs as a bridge rectifier, good up to only 600Hz. Most switching applications use a proper synchronous rectifier controller, which keeps up with high frequency switching better than parasitic-powered kinds.Tim

author: Richard Head
author_rank: Frequent Contributor
timestamp: March 27, 2014, 07:31:48 am
content: T3sl4co1lI have also found that the massive capacitance of Shottkeys limits its application. Very often a fast recovery diode with its higher Vf is a better option, even if it does get a bit hotter.With resonant converters you can lump the capacitance into the resonant cap which makes a massive difference.Dick