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Assuming we have a 12 V load which needs 1 A current, when we connect this load to 60 Ah 12 V car battery it does not need to resistor and it will ony draw 1 A.

But on the other side, assuming we have a transformerless power supply circuit connected to 220 V so the DC voltage after diode bridge will be 310 V. Why can we not connect 100 LEDs in series (each 3.1 V) directly without using a capacitor or resistor?

I know this capacitor is used as reactance (\$X_C\$) to limit the current, but why is it needed since the voltage of LEDs is equal to the output voltage of the power supply? Shouldn't the LEDs should draw the required current? According to this equation R=(Vin-VLED)/I =(310-310)/I = zero resistor needed

enter image description here

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    \$\begingroup\$ Not directly related to your actual question, but drawing 1A from a 60Ah battery does not 'leave the other 59A'. Ah (Amp-hours) and A (Amps) are different units. 'Ah' are an indication of the energy storage capacity of the battery. 'A' is an indication of the amount of current flowing at an instant in time. \$\endgroup\$
    – brhans
    Mar 16, 2017 at 20:34
  • \$\begingroup\$ I know , i just want to explain my means. 60AH battery can easily give 500A during car starting for a few seconds, i know 60Ah its capacity not its Max. current. \$\endgroup\$
    – M.A.K
    Mar 16, 2017 at 20:38
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    \$\begingroup\$ A small amount of variation in voltage will cause a massive variation in LED current. Stop thinking of a LED as something that can be fed a voltage. You have to fed it a controlled current. \$\endgroup\$
    – pipe
    Mar 16, 2017 at 20:45
  • \$\begingroup\$ "Why can we not connect 100 LEDs in series (each 3.1 V) directly without using a capacitor or resistor?" - This is a misrepresentation. The circuit you stole only had 24 LEDs in it, not 100. circuitstoday.com/wp-content/uploads/2010/06/mains-optd-led.jpg \$\endgroup\$ Dec 2, 2019 at 21:30
  • \$\begingroup\$ @BruceAbbott There is a ... at the bottom of the image to symbolize the other LEDs not shown. It is unnecessary to show all 100. \$\endgroup\$
    – ccolton
    Dec 2, 2019 at 23:40

5 Answers 5

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How confident are you that your rectified mains voltage will be exactly 310V?

How confident are you that 100 LEDs in series will have a voltage drop of exactly 310V?
Have you read the datasheet? What is the Forward Voltage tolerance specification?

What if your mains is actually 315V and/or your string of LEDs adds up to 308V?

LEDs need current limiting - they are not voltage-driven devices.

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  • \$\begingroup\$ Yup, in order to guarantee it, he needs to figure out the worst case (Lowest.. forward voltage) and the max supply voltage and do the math. It will be way more than 100 LEDs. \$\endgroup\$
    – Trevor_G
    Mar 16, 2017 at 20:47
  • \$\begingroup\$ @brhans-changing voltage between 308v & 315v not very important because my white LEDs according to datasheet can run with 3 to 3.4v so voltage range between 300 and 340v acceptable. \$\endgroup\$
    – M.A.K
    Mar 16, 2017 at 20:54
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    \$\begingroup\$ @mousa the LED can't "run with" 3 to 3.4V. It operates at some voltage within that range, dependent weakly on the current available, as well as temperature, its age, and exactly how well that batch was made. Any voltage in excess of whatever it happens to take will cause a destructively large current through it. And THAT's why you need current limiting - as the current increases, more volts across the capacitor, bringing the voltage down to a safe value. \$\endgroup\$
    – user16324
    Mar 16, 2017 at 21:01
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    \$\begingroup\$ Odd explanation. The OP didn't claim to be confident of anything particularly. Just didn't know diode theory and therefore advocating the data sheet is pointless. Could've been a decent explanation instead, OP's not lazy. \$\endgroup\$
    – TonyM
    Mar 16, 2017 at 21:24
  • \$\begingroup\$ @TonyM the OP is making a lot of assumptions about a lot of things - implying some overconfidence with regard to at least some of the things I mentioned. \$\endgroup\$
    – brhans
    Mar 16, 2017 at 21:36
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If you short the capacitor the total resistance in the circuit will be about 100 \$\Omega\$.

The total impedance with the capacitor not shorted at 50 Hz is $$|100 + {1\over j \omega C}| =|100 + j~3185| = 3186~\Omega.$$

At 60 Hz it will be about \${5\over6} \times 3186~\Omega = 2653~\Omega\$.

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  • \$\begingroup\$ you are right, but here output voltage is 310 and 100LEDs need to 310v so why i need resistor or capacitor to current decreasing? \$\endgroup\$
    – M.A.K
    Mar 16, 2017 at 20:26
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    \$\begingroup\$ You will still need an impedance. You might be able to get away with less, but the forward voltage drop of a diode is not stable. (So be carefull if you do not want to see the smoke escaping from all the diodes. :-) \$\endgroup\$
    – skvery
    Mar 16, 2017 at 20:31
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    \$\begingroup\$ The impedance at 60 Hz is LESS than at 50 Hz ......2652 Ohms @ 60Hz \$\endgroup\$ Mar 16, 2017 at 20:38
  • \$\begingroup\$ @JackCreasey Oops, my bad bud! Will fix. \$\endgroup\$
    – skvery
    Mar 16, 2017 at 20:41
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It is called "capacitor dropper".
Note that you are talking about AC current (230V AC). Capacitors and inductors has some impedance, it decreases current.
Lower capacitance = higher impedance = lower current.
So, current is limited. If you jump it, it means that you remove that impedance and current jumps very high and burns LEDs.

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  • \$\begingroup\$ you are right, but here output voltage is 310 and 100LEDs need to 310v so why i need resistor or capacitor to current decreasing? \$\endgroup\$
    – M.A.K
    Mar 16, 2017 at 20:26
  • \$\begingroup\$ Without that capacitor you would have over 325V and it could be too much for those LEDs. Perhaps. \$\endgroup\$ Mar 16, 2017 at 21:11
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Assuming we have a 12 V load which needs 1 A current, when we connect this load to 60 Ah 12 V car battery it does not need to resistor and it will ony draw 1 A.

You started out with a fallacy. To have a load on 12v drawing 1 Amp you would have to have a 12Ω load. Because the load would be resistive.

You could not connect a string of LEDs directly to a battery with out them burning, so why would you think you can connect them directly to an AC source and have a different result?

LEDs have a dynamic resistance. They required something to limit the current.

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  • \$\begingroup\$ you are wrong. because resistor don't calculated as you think. you have to use ohm's law as follow: R=(Vin-VLED)/I SO R=(12-12)/1= 0 ohm \$\endgroup\$
    – M.A.K
    Mar 17, 2017 at 6:49
  • \$\begingroup\$ check this link: ohmslawcalculator.com/led-resistor-calculator \$\endgroup\$
    – M.A.K
    Mar 17, 2017 at 7:34
  • \$\begingroup\$ @mousa, you lack some fundamental concepts in electronics, like thinking that it is OK to drive LEDs with no resistance in series because of your misinterpretation of an online calculator... so, if you know so much, why ask here? Or maybe if so many people tell you that LEDs are CURRENT DRIVEN devices (and not voltage), could you be nice and start listening instead of proving a point you don't have? Or do whatever you thing is OK, burn your LEDs and come back when you are ready to listen. \$\endgroup\$ Mar 17, 2017 at 9:21
  • \$\begingroup\$ @ Claudio Avi Chami, you means this equation( R=(Vin-VLED)/I ) is wrong? \$\endgroup\$
    – M.A.K
    Mar 17, 2017 at 9:26
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    \$\begingroup\$ @mousa The equation you used is an approximation. It cannot return correct results in all cases. For example, if your input voltage is lower than Vled then that equation will tell you to use a resistor with a negative resistance. Ain't no such beast. When Vin is close to Vled, you can't calculate a resistor. The equation is inaccurate, and Vled is NOT fixed. It varies with temperature, current, manufacturing tolerance, and other things. \$\endgroup\$
    – JRE
    Mar 17, 2017 at 11:35
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Your mistake is assuming that a raw LED is a "normal load", it is not.

Normal loads have a relatively gentle voltage/current relationship. Small changes in conditions, either voltage of the power supply or temperature of the load, result in small changes in the current.

LEDs have a very steep voltage/current relationship, a small change in voltage can result in a very large change in current (so much so that a common first approximation of a LED is a component that maintains a fixed voltage for any forward current). They also have significant variability with temperature and manufacturing.

When LED manufacturers give a range of forward voltages they are NOT talking about a range of voltages that the LED can be fed with blindly and operate correctly, they are talking about a range of voltages you might encounter when you drive the LED with a specified current.

Even with the capacitor the design of your light is awful. Leaving asside the problems of flicker that come from using rectified mains, the current drawn will vary massively over the range of mains voltages you are likely to encounter and likely also with the temperature of the LEDs.

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