LED voltage drop - Electronic Circuits Projects Diagrams Free

Peepsalot · 27th March 2006, 05:26 PM

From what I've read, I used to think that LEDs have a constant votlage drop across them at all times. People or texts will say things like "This diode has a drop of X volts".
But I realized from a little experimenting yesterday that this is not really so.

For example my multimeter on the diode setting measures a 1.58V drop for a generic red LED I have. But when I put it in a circuit and power it up, at only 10ma, I measure a voltage of 1.98 across it. This is a huge discrepency. Is there some equation I can use to predict the voltage drop of an LED when driven at different currents?

For the multiplexed LED display project I am working on, I need to figure out what kind of resistors I will need for pulsing these LED. I am aiming for 160ma pulses. If I get the supply voltage at about the right level, could I do without resistors entirely? Seems like at high enough current, there will be enough drop over the LED to just even things out on it's own, if that makes any sense.

windozeuser · 27th March 2006, 05:35 PM

The diode is a ohmic linear device. Hence you don't need to predict values when you can calculate them using ohms law. The amount of current, directing proportional to voltage will affect your LED's voltage drop.

Say you have a LED with a forward bias voltage drop of 1.6 volts, and a maximum current of 10mA. Using a 9 volt source You would need:

R = E/I = (9v - 1.6v)/0.010 = 740 ohm resistor

Remember that Resistance is additive in series and decreses in parallel.

For example using a 9volt 1A supply with a series 740 ohm resistor, with each LED using 10mA you can have 100 LED's In parallel.

Peepsalot · 27th March 2006, 06:19 PM

I'm not sure I follow you, what is ohmic linear?

If the voltage drop is 1.6V at 10ma, what is the equation for the amount of drop when run at 5ma or 20ma? It is not going to be 1.6V still, my tests have shown that.

Oznog · 27th March 2006, 06:32 PM

A diode is more or less constant voltage. Once the threshold voltage is exceeded current rises rapidly with small increases in voltage.

The curve varies depending on the device and how you choose to measure the temperature effects of heating inside the device. It's often describes as exponential though it's often a fairly linear slope in the important region. The slope is steep and small changes in voltage produce large changes in current.

The thing is the threshold voltage- and the slope of voltage vs current- varies with die temperature. As the part heats up, the threshold goes down and the whole curve slides over to the left. Thus in reality driving with a constant voltage never works. What voltage constitutes 20mA when it's cold might make 40mA when it starts warming up. 40mA then heats the die up even more and it draws 80mA and burns out.

Also there are always differences from LED to LED. Even with the same mfg & part number, and with both at identical temps, a voltage that produces 20mA on one part might produce 30mA on another. Very poor matching.

Roff · 27th March 2006, 06:35 PM

Quote:

Originally Posted by windozeuser

The diode is a ohmic linear device.

No it's not. That would imply that, if your LED's forward voltage was 2.0V at 10ma, it would be 0.2V at 1ma. This is definitely not the case. The current is an exponential function of voltage.

windozeuser · 27th March 2006, 06:51 PM

Oh thanks for clearing that up, but I was referring to the way resistance changes in conjunction with variable voltage

FusionITR · 27th March 2006, 07:14 PM

Constant voltage drop is an approximation.

The real voltage drop is n * VT * ln(Id/Is)

Peepsalot · 27th March 2006, 07:20 PM

Quote:

Originally Posted by FusionITR

Constant voltage drop is an approximation.

The real voltage drop is n * VT * ln(Id/Is)

Thanks for that, could you explain what the variables are though? :roll:

Roff · 27th March 2006, 07:23 PM

Quote:

Originally Posted by windozeuser

Oh thanks for clearing that up, but I was referring to the way resistance changes in conjunction with variable voltage

I'm still not sure that you get it. Below is a voltage vs current curve for 6 different types of LEDs. You can see that the voltage is definitely not a linear function of current.
As was pointed out by Oznog, these curves move to the left as LED temperature goes up.
I'm not picking on you. I'm just trying to help you understand, if you don't already.

Peepsalot · 27th March 2006, 07:34 PM

Well, I think I have somehting similar to that super red in the chart you posted. I wish the chart went higher, since I am going to be pulsing these. Looks like 160ma would be somewhere around 2.5V then. That's a big difference from the 1.58V measured on the multimeter.

FusionITR · 27th March 2006, 07:47 PM

Quote:

Originally Posted by Peepsalot

Quote:

Originally Posted by FusionITR

Constant voltage drop is an approximation.

The real voltage drop is n * VT * ln(Id/Is)

Thanks for that, could you explain what the variables are though? :roll:

n = nonideality factor (n=1 or n=2)
Vt = thermal voltage
Id =current through the diode
Is = saturation current

Roff · 27th March 2006, 08:47 PM

Quote:

Originally Posted by Peepsalot

Well, I think I have somehting similar to that super red in the chart you posted. I wish the chart went higher, since I am going to be pulsing these. Looks like 160ma would be somewhere around 2.5V then. That's a big difference from the 1.58V measured on the multimeter.

The multimeter forces a low current through the diode, probably on the order of a milliamp, and measures the voltage drop.

Oznog · 27th March 2006, 09:36 PM

Do you catch why the resistor is critical?

Say you need to run an LED @ 2.0v, 20mA off a 6v battery. So you use a 200ohm resistor to drop 4v.

The actual device decides it wants to run at 30mA @ 2.0v either due to mfg differences or heat. But the resistor won't let it do that. The resistor would drop 6v at 30mA leaving no voltage for the LED at all. In reality the LED might run at 1.95v @ 20.25mA so it's essentially stable.

Also consider what happens if the source voltage is nonideal. If we tried to drive 2.0v without a resistor but were off by +10%, the LED would see 2.2v and would pull like 60mA and burn out. With the resistor situation, +10% meaning the supply puts out 6.6v, you'd get an LED current of around 23mA.

audioguru · 28th March 2006, 12:06 AM

This is the typical forward voltage drop of the very bright red LEDs that I use. The absolute max momentary current rating is 200mA. Notice how the voltage drop isn't a straight line.

windozeuser · 28th March 2006, 05:27 PM

Thanks audioguru and Ron H, you cleared a lot of things up for me!

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27th March 2006, 05:26 PM	(permalink)
Peepsalot	LED voltage drop From what I've read, I used to think that LEDs have a constant votlage drop across them at all times. People or texts will say things like "This diode has a drop of X volts". But I realized from a little experimenting yesterday that this is not really so. For example my multimeter on the diode setting measures a 1.58V drop for a generic red LED I have. But when I put it in a circuit and power it up, at only 10ma, I measure a voltage of 1.98 across it. This is a huge discrepency. Is there some equation I can use to predict the voltage drop of an LED when driven at different currents? For the multiplexed LED display project I am working on, I need to figure out what kind of resistors I will need for pulsing these LED. I am aiming for 160ma pulses. If I get the supply voltage at about the right level, could I do without resistors entirely? Seems like at high enough current, there will be enough drop over the LED to just even things out on it's own, if that makes any sense.

27th March 2006, 05:35 PM	(permalink)
windozeuser	The diode is a ohmic linear device. Hence you don't need to predict values when you can calculate them using ohms law. The amount of current, directing proportional to voltage will affect your LED's voltage drop. Say you have a LED with a forward bias voltage drop of 1.6 volts, and a maximum current of 10mA. Using a 9 volt source You would need: R = E/I = (9v - 1.6v)/0.010 = 740 ohm resistor Remember that Resistance is additive in series and decreses in parallel. For example using a 9volt 1A supply with a series 740 ohm resistor, with each LED using 10mA you can have 100 LED's In parallel.

27th March 2006, 06:19 PM	(permalink)
Peepsalot	I'm not sure I follow you, what is ohmic linear? If the voltage drop is 1.6V at 10ma, what is the equation for the amount of drop when run at 5ma or 20ma? It is not going to be 1.6V still, my tests have shown that.

27th March 2006, 06:32 PM	(permalink)
Oznog	A diode is more or less constant voltage. Once the threshold voltage is exceeded current rises rapidly with small increases in voltage. The curve varies depending on the device and how you choose to measure the temperature effects of heating inside the device. It's often describes as exponential though it's often a fairly linear slope in the important region. The slope is steep and small changes in voltage produce large changes in current. The thing is the threshold voltage- and the slope of voltage vs current- varies with die temperature. As the part heats up, the threshold goes down and the whole curve slides over to the left. Thus in reality driving with a constant voltage never works. What voltage constitutes 20mA when it's cold might make 40mA when it starts warming up. 40mA then heats the die up even more and it draws 80mA and burns out. Also there are always differences from LED to LED. Even with the same mfg & part number, and with both at identical temps, a voltage that produces 20mA on one part might produce 30mA on another. Very poor matching. __________________ I thought what I'd do was I'd pretend I was one of those deaf-mutes.

27th March 2006, 06:51 PM	(permalink)
windozeuser	Oh thanks for clearing that up, but I was referring to the way resistance changes in conjunction with variable voltage

27th March 2006, 07:14 PM	(permalink)
FusionITR	Constant voltage drop is an approximation. The real voltage drop is n * VT * ln(Id/Is)

27th March 2006, 07:34 PM	(permalink)
Peepsalot	Well, I think I have somehting similar to that super red in the chart you posted. I wish the chart went higher, since I am going to be pulsing these. Looks like 160ma would be somewhere around 2.5V then. That's a big difference from the 1.58V measured on the multimeter.

27th March 2006, 09:36 PM	(permalink)
Oznog	Do you catch why the resistor is critical? Say you need to run an LED @ 2.0v, 20mA off a 6v battery. So you use a 200ohm resistor to drop 4v. The actual device decides it wants to run at 30mA @ 2.0v either due to mfg differences or heat. But the resistor won't let it do that. The resistor would drop 6v at 30mA leaving no voltage for the LED at all. In reality the LED might run at 1.95v @ 20.25mA so it's essentially stable. Also consider what happens if the source voltage is nonideal. If we tried to drive 2.0v without a resistor but were off by +10%, the LED would see 2.2v and would pull like 60mA and burn out. With the resistor situation, +10% meaning the supply puts out 6.6v, you'd get an LED current of around 23mA. __________________ I thought what I'd do was I'd pretend I was one of those deaf-mutes.

28th March 2006, 05:27 PM	(permalink)
windozeuser	Thanks audioguru and Ron H, you cleared a lot of things up for me!