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FreeTrack Forum > FreeTrack : English Forum > Support : Tracking System > Possible to connect 3 LEDs to 3V battery without resistor?
Sartoris | #1 07/05/2012 - 18h37 |
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Would it be possible to connect 3 LEDs to a 3V battery without resistors? Would such a battery be enough to power the LEDs?
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Standby | #2 07/05/2012 - 21h36 |
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You always need resistors when you are using led whatever your battery
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Sinister | #3 07/05/2012 - 23h47 |
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Doesn't it kind of depend on the voltage of the LEDs and how you wire them i.e. Series vs Parallel?
For Example, 3 LEDs rated at 1.2v wired in series would have a max of 3.6v. I don't see why running this off of a 3v battery would be a problem. |
Steph | #4 08/05/2012 - 11h31 |
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Read:
http://en.wikipedia.org/wiki/LED_circuit Especially "Power source considerations". Even if the LED is not destroyed immediately because of the internal resistance of battery, both won't resist a long time. Always drive LEDs with resistors.
Edited by Steph on 08/05/2012 at 11h31.
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Gian92 | #5 09/05/2012 - 22h45 |
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I haven't read anywhere that you have to use resistors compulsorily, especially when using DC sources (accumulators, batteries, etc.).
From the same article:
And that's pretty logical, straightforward and self-explanatory.
Edited by Gian92 on 09/05/2012 at 22h51.
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Steph | #6 10/05/2012 - 14h08 |
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1. You do not know the right voltage drop of the LED at any moment.
2. Batteries are not constant-voltage sources, as there voltage output could vary from 1.7V to 0. A way to drive LEDs without resistors is to use a constant current regulator. More expensive than a resistor and a simple switched-mode power supply. |
Gian92 | #7 11/05/2012 - 16h50 |
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1. LEDs are rated with a nominal forward voltage drop.
2.
3. The function of a resistor is to augment electrical resistance, not to balance.
Thus, following the relation I=V/R, if I have an electric current of 1 V flowing through a resistor of 1 Ohm it will have an intensity of 1 A; 2 A in case of doubled voltage. In this example, did the resistor balance the current? Following your thought, limiting the voltage will suffice instead of using a resistor and power consumption will be cleverer. Please cite your sources where it is explicitly written that resistors are always compulsory for LEDs circuits, and that they will balance voltage peaks.
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Gian92 | #8 11/05/2012 - 17h15 |
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Forgot to say, also the underlying reasons are likewise important.
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Steph | #9 11/05/2012 - 22h43 |
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And...? We are talking about voltage not current.
As I said above, you can drive LEDs without a resistor using a constant current regulator. Another way is to drive them with pulse width modulation, requiring at least a micro-controller. But why going the expensive way, when a resistor does the same job? Why using a resistor? Well, let's look at that closer. LEDs are semiconductors that behave different than resistors. If you apply a specific voltage to a resistor, you can describe the resulting current with: I = V / R That does not work with LEDs because they don’t behave like a linear resistor. To explain look for example at the data-sheet of the SFH485. It gives you two important curves: One gives you the max. forward current for DC which is: I(f)= 100mA The other gives you the voltage-current characteristic line. You can rise the voltage from 0 Volt to ~1.1 Volt without resulting in noticeable current. Apply a bit more voltage and there is current and the LED lights up. We have reached the forward voltage which is needed to open the pn-gate. From now on small changes in the voltage produce large effects on the resulting forward current. As you can see with a step off 1.38V to 1,44V the current rise up from 70mA to 100mA, our max. forward current for the diode. Apply a little bit more, and you quickly shorten lifetime off your LED. The reason why the LED does not blow up with more than 1.5V of a fresh batterie is because of the internal resistances of batteries. So, yes, you can drive LEDs without resistors, but this require current limitation in an other way, as you can see the voltage regulation is not appropriated. |
Gian92 | #10 11/05/2012 - 23h26 |
Class : Habitué Off line |
I=V/R V=IR If R is the same in the two equations then V and I are strictly and solely dependent to each other. In fact:
That's quite logical. Given R the same, If I varies then V varies proportionally to I, and vice-versa. LEDs are non-ohmic circuit, right, yet the whole circuit, except the LEDs, is ohmic. Since we have to control the circuit in order not to blow our LEDs up (and since we can't change the way LEDs behave), ohmic law or linearity still applies. We didn't need to calculate or extrapolate anything: according to Osram datasheet, nominal voltage is 1.5 V and maximum voltage is 1.8 V for SFH485. Our circuit would be composed as follows: Source (battery) -> Resistor -> Electrical utilizer (LED) As said before, if the source voltage is equal to the LED voltage drop and the current - and, therefore, the voltage - is constant (which is the case, since a battery is a DC source), then a resistor or current regulator is unneeded. Sir, pardon me, but still you didn't explain me the reason why resistors are required in the above-mentioned conditions.
Edited by Gian92 on 11/05/2012 at 23h56.
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Steph | #11 12/05/2012 - 16h57 |
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You said it yourself:
Ohmic law yes, but not linearity. The current flowing in an LED is an exponential function of voltage across the LED. You can’t say that LEDs have “resistance.” Resistance is defined as the constant ratio of voltage to current in a resistive circuit element. LEDs have diode like non-linear I/V characteristics. This picture will show it better than the logarithmic diagram from Osram datasheet. The curve may also varies among different colors, different sizes, temperature and even tolerance in manufacturing.
Depending on what your DC source can handle, either you load down the supply until its voltage drops to Vf or current becomes so high that using E=IR, the voltage drop across the low resistance of the wires drops the remaining voltage and you still have Vf across the LED. Now, batteries are not regulated power sources. I have three new batteries here where I can measure: Duracel Alkaline - 1.59V No name Alkaline - 1.48V Panasonic Oxyride - 1.6 |
Gian92 | #12 12/05/2012 - 17h39 |
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Linearity happens because of Ohm's law compliance (this concept was in my examples, but perhaps I'm a dreadful expositor). I was talking about the circuit, minus LEDs. Moreover, the electrical resistance I was referring to is the one of the resistor. I took LEDs out of the computation because, as already said, in a DC environment, if the source voltage is equal to the LED forward voltage drop, no resistor is needed.
Then, if you match the maximum voltage output of the utilized battery (the maximum value is generated when brand new or just recharged) to the forward voltage drop of the LED, no resistor is needed. A resistor, as above-mentioned, just limits the amount of voltage, and therefore, current intensity. Do you finally agree with that?
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Steph | #13 12/05/2012 - 21h29 |
Class : Moderator Off line |
Call me Simson, but I don't understand this argumentation. Sorry, perhaps I missed something. Regarding the current-voltage curve of the sfh485 1.5V let more than 100mA current flow. But anyway, this is theory. Put a sfh485 on a 1.5 Alkaline and you never will measure more than 80mA in circuit. The internal resistance of battery prevents current rising. So I agree with you, you can drive a LED on battery without a resistor. |
dewey1 | #14 13/05/2012 - 06h11 |
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Gian92;
You need to understand that the IR LEDs may have a Vf from 1.50 to 1.79 at 100mA. This is what the specs indicate for the SFH485P. In other words these are tolerances for Vf at 100mA. The specs do not even show a minimum Vf as some manufacturers do. I can tell based on your references to wikipedia, which has a lot of mistakes and misconceptions that you have a very limited knowledge of Electronics. There is the theoretical world and the real word applications. When you design for real world applications you must take into account worst case scenarios which includes all the variable tolerances that can occur. If you are disagreeing just for the sake of argument, then these diagrams will not make any sense to you. In the attached diagram you can see the difference what happens with a regulated 1.8 VDC regulated supply capable of 10 Amps. This is basically no resistors as you seem to think will work. The second diagram shows the resistor values to get 100mA. I used 4 LEDs that had different Vf at 100mA which will occur. This should help you understand it better. |
Zaphod | #15 13/05/2012 - 09h41 |
Class : Apprenti Off line |
dewey1, the need for resistors in parallel circuit is well established, the original question is about series circuit.
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