LED Vf - Page 1
Forum section: electronics / beginners

author: electrolust
author_rank: Unknown
timestamp: February 08, 2017, 07:34:44 am
content: How do I model / calculate the voltage drop across an LED?If I set my PS to 20mA CC 3.3V, and power the LED directly, the display reads 1.3V and I can measure 2.0V drop across the LED. If I add a dropping resistor, everything works textbook. Vd across the LED measures 2.0V and Vd across the resistor measures 1.3V, or rather VCC-2.0V as I vary the supply voltage, and irrespective of resistance value.This is what I'd expect.But 3 online simulators I tried disagree with this. They all show the voltage drop across the LED to go down as the resistor value goes up. As if the LED and resistor form a resistive voltage divider. (Just qualitatively. I didn't test to see if some ratio is preserved as the R value changes.) Two of the simulators have 6-8 complex properties for the LED, which do not include Vf. The 3rd has only a single property, Vf (for that one, there's no LED, only a generic diode). Even for that simpler simulator, if I set Vf to 2.0V, it is not used as a static drop, it varies with R value.

author: Ian.M
author_rank: Super Contributor
timestamp: February 08, 2017, 07:59:48 am
content: Vf varies with If. There should be a graph of it in your LED's datasheet.In your experiment you wont observe much change in Vf if you vary If by a factor of two. If you switch in a bunch of different decade value resistors (e.g. 100R, 1K, 10K, 100K), each time adjusting the supply voltage to get 0.15V then 0.30V then 0.60V across the resistor (for currents of 1.5, 3 and 6 in the decades *1mA,*100uA, *10uA and *1uA), measuring Vf for each, with a DMM with 10Meg input impedance, and plot the results you should get pretty close to the results of the better simulators.You wont get great results if you try to extend this for 1Meg due to your meter shunting the series resistor and the diode junction, but if you have a FET input voltmeter with a high enough input impedance, if you plot Vf against log(If) you'll get a nearly straight line from typically 10% of the max rated If down for five or six decades of If.

author: danadak
author_rank: Super Contributor
timestamp: February 08, 2017, 12:32:23 pm
content: The behavior you see, Vdiode dropping as R increases, very much describedby classic diode equation (load line shown as well).Regards, Dana.

author: grumpydoc
author_rank: Super Contributor
timestamp: February 08, 2017, 01:22:24 pm
content: I am not sure that you are interpreting (or explaining?) your initial experiments with a bench PSU correctly. EDIT - or at any rate I am confused by your description.When you say that you set your supply to 20mA CC 3.3V what do you mean?If you set constant current mode the PSU will adjust voltage to get the specified current to flow, possibly up to a user-set maximum.If you powered the LED directly and the PSU volt meter read 1.3V then this should have been the VFof the LED. I am not clear how you managed to measure a 2 volt drop across the LED if the PSU said 1.3V (although 2.0V is a perfectly reasonable VFfor a LED, 1.3V is too low for anything except an IR LED).You seem to have got things right with the PSU when you add a resistor.Simulators, as you have found, vary in the quality of their simulation.As has been said if you increase the current through a LED the VFwill risea littlebut not much. A resistor and LED combination fed from a constant voltage supply does not act like like a resistive divider but the LED will have a (mostly) fixed VF, Kirchoff's voltage law then says how much voltage is across the resistor (as you say VCC- VF) and then Ohm's law will give you the current through the resistor and hence the LED.If you increase the resistor the current will drop, this will in turn cause a slight decrease in the LED forward voltage.You can do the plot as Ian M suggests. If you have two bench multimeters you can just measure current and voltage at the same time and won't need to keep changing resistors, if you have just one DMM you can measure the voltage across a known resistance to work out the current flow when it is smaller then the resolution of the ammeter on your PSU (rarely do these go below 1mA and are usually pretty inaccurate anyway).

author: electrolust
author_rank: Unknown
timestamp: February 08, 2017, 06:30:21 pm
content: Quote from: Ian.M on February 08, 2017, 07:59:48 amAh-HA. There's my first problem then. The datasheet is a 1 pager and does not have a graph like that. Just a chart of typical values for Vf (and others) @ I=20mA. Makes sense now.Also explains why the DMM diode check only reported 1.7V ... it doesn't source 20mA.

author: electrolust
author_rank: Unknown
timestamp: February 08, 2017, 06:34:18 pm
content: Quote from: grumpydoc on February 08, 2017, 01:22:24 pmYeah, my explanation was wrong. I meant current limiting, not constant current.QuoteAh, ok it must have read 2.0V then, but by the time I got here I misremembered.QuoteCool. I do have 2 DMMs so I'll try this.

author: electrolust
author_rank: Unknown
timestamp: February 08, 2017, 06:49:56 pm
content: Quote from: grumpydoc on February 08, 2017, 01:22:24 pmFirst I needed to understand why VFwas not what I expected, but now this is getting to my real question.For a circuit like this:VCC - R1 - LED - R2 - GNDDo I just calculate current as VCC/(R1+R2)? The reason I started this thread is, the simulators were giving me different voltage drop across the didoe for the same RTOTAL, depending if R2 were absent (R1 = RTOTAL) or if I added R2.How would I calculate the voltage drop across each resistor in this case? In the simulators, in no case can I maintain 2.0V across the LED when I have 2 resistors split across the diode.

author: bson
author_rank: Unknown
timestamp: February 08, 2017, 07:46:56 pm
content: Quote from: electrolust on February 08, 2017, 06:49:56 pmNo, you can't, at least not quite this simply. It would be I=(Vcc-Vf)/(R1+R2), but since Vf is a function of I you'd end up with I=(Vcc-Vf(I))/(R1+R2). You need to replace Vf(I) with Shockley's diode equation and solve the resulting equation for I. (Or, more commonly, you have Vcc and I and need to determine R1+R2, which is easier.)

author: electrolust
author_rank: Unknown
timestamp: February 08, 2017, 09:17:49 pm
content: Quote from: bson on February 08, 2017, 07:46:56 pmIndeed, that's what I want to do.

author: electrolust
author_rank: Unknown
timestamp: February 08, 2017, 10:09:24 pm
content: And actually this was now calculated correctly by circuitlab. Not sure what I was doing differently before.

author: radiogeek381
author_rank: Regular Contributor
timestamp: February 09, 2017, 02:04:56 am
content: QuoteClose. But your expression left out the drop across the diode.Remember, the components in the circuit are all going to obey the laws thatgovern their terminal voltages and branch currents.The same current flows through R1, R2, and the LED (D1) (kirchoff's current law)The sum of the voltage drops across R1, R2, and D1 are going to equal VCC (kirchoff's voltage law)Let's call the current Id.R1 and R2 are going to obey ohms law, so the sum of the voltages across them willbe Id * (R1 + R2)The voltage across D1 is going to be Vd and it will satisfy an exponential curve:Id = Is *(eVd / (n*Vt)- 1)where Vt is kT/q -- a temperature dependent voltage that is about 26mV at room temperature.But we don't know Is for most diodes.So we resort to the graph of Id vs. Vd in the diode spec sheet.OR-- if we don't have such a thing, we take a guess. Let's assume we want Idto be 20mA. Let's assume that Vd is 2V. (This is a good guess for many LEDs,a lousy guess for a silicon diode. For a normal silicon diode, we'd use 0.6 or 0.7V)Then VCC = Id * (R1 + R2) + Vd = 0.02 * (R1 + R2) + 2if VCC is 5V then the sum of R1+R2 needs to beR1+R2 = (5 - 2) / 0.02 = 3 / 0.02 = 150 ohms.That smells right. But let's see what happens if the diode voltageat that current is only 1.0 V....Then the current is going to be (5 - 1)/150 = 27mA -- a little high,but not in the incendiary range.What if the diode voltage is 3V..Then the current is going to be (5 - 3) / 150 = 13mA -- high enoughto light the LED.--------------------Finally, if we DO have the curve family, then we can calculate R1+R2 bydoing the following.Place a point on the diode's I vs. V curve that intersects a horizontalline going through I = 20mA. Label that point "Q".Place a point on the X (Vd) axis at VCC. Label that B.Draw a straight line from B through point Q and extend it all the way tothe vertical axis. It will intercept this axis at Im.(R1 + R2) = Vcc/ImThe line that you drew was theload line. Each point (Vx, Ix) on that line correspondedto the resulting voltage across the diode (Vx) , assuming that Ix was flowing throughthe two resistors. If no current is flowing, then the entire voltage is across the diode.If Im is flowing, then all the voltage is dropped across (R1+R2).But, there's a diode in this circuit. For it to satisfy reality, the current through thediode and the voltage across the diode must be somewhere on the diode curve.Both Ohm's law and the diode curve must be satisfied. So, the only combination ofVd and Id that satisfies both requirements is where the load line and the diode curveintersect -- at point Q.If you want to try this on a real datasheet, take a look at figure 1 here:http://www.cree.com/~/media/Files/Cree/LED-Components-and-Modules/HB/Data-Sheets/C503B-RAS-RAN-AAS-AAN-RBS-RBN-ABS-ABN-RCS-RCN-ACS-ACN-1079.pdfNote that you'll have to extend the graph to the right a bit, as it's horizontal axis only goes to 2.8and you'll want to extend it to Vcc to draw point B on your load line.-------This load line stuff is just one way of expressing circuit behavior as the resolution ofmultiple constraints. But it is a meme that is underneath much, if not all, of circuitdesign. In fact, there is a class of calculating engines -- analog computers -- thatare entirely based on this notion. With a fairly simple set of primitives, one can buildcircuits that add, subtract, multiply, divide, take the log, take the exponential, Nthroot, sin, cos, and others.And it all boils down to realizing that an expression likeId = Is(eVd/(nVt)- 1)is not a description of a black box that takes Vd as an input and produces Id as an output.A device equation is arelationthat must be satisfied (subject to certain conditions) whenthe element is in a circuit.So we could just as easily write it asVd = nVt * ln(Id/Is + 1)Because, no matter what, both expressions must be satisfied if this diode is in a circuit.In fact, the latter expression is how widgets that produce ln(x) or log(x) work in analogcomputers and other instruments. Drive a current corresponding to X through a diodeand measure the voltage drop.Good question.

author: Audioguru
author_rank: Super Contributor
timestamp: February 09, 2017, 03:08:27 am
content: The datasheet for your LED should show a graph of its "typical" forward voltage that changes with current and temperature. The datasheet should also list therange of forward voltagefor that LED part number since they cannot make a bunch of LEDs the same. Some red LEDs will be 1.8V and others will be 2.2V.