peak current when i open the switch on LEDs

[yassin.kraouch]

Hi,
I have a QDC driver which is driver by NPN transistor, and it will drive LEDs,
when i turn ON the MOSFET i have a peak current, and the LEDs does not support this hight current ( 140mA peak),
the problem is that when the transistor is turned ON, capacitor C4 will give a current with the mains,
can you help me to solve this issue ?

 

[Audioguru]

Your circuit has no switch. It has no AC power supply voltage V1, has no capacitor values and has no LED forward voltage.
Your scope shows the current through D11 but your schematic has no D11.
Then we do not know what you are talking about.

 

[yassin.kraouch]

yes i moved the values in order to make the schematic clear,
V1 is AC source 230VAC/50hz,
Q1 is the NPN transistor used as a switch,
due to the big quantity of LEDs=60 it was difficult to put them all in the picture ,
i have LEDs from D1 to D63,
the capacitor value C4 is 6.8uf
is it clear now ?

 

[godfreyl]

You could put a capacitor in parallel with the LEDs to absorb the current surge.

 

[Audioguru]

I guess you connect the 60 LEDs in series? I have Christmas Tree lights like that.

C4 will charge to the peak voltage which is about +309V but since you did not say the forward voltage of your LEDs then I cannot calculate the current. I see no current surge if the value of C4 makes it a good filter.

 

[yassin.kraouch]

I guess you connect the 60 LEDs in series? I have Christmas Tree lights like that.

C4 will charge to the peak voltage which is about +309V but since you did not say the forward voltage of your LEDs then I cannot calculate the current. I see no current surge if the value of C4 makes it a good filter.

the total forward voltage of the LEDs is 192V, yes all LEDs are in series, and i need a current around 16mA throught the LEDs,

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You could put a capacitor in parallel with the LEDs to absorb the current surge.

how much will be this capacitor value ?

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You could put a capacitor in parallel with the LEDs to absorb the current surge.

i added a capacitor in parallel with the LEDs but it does not help,
i have the same result

 

[Audioguru]

You have a very high current in R4 and the LEDs because the voltage across R4 is very high. The voltage and current will be less when the value of C2 is reduced.

 

[AveztruZ]

I've worked with LEDs before and in all experiences I had with more than 3 LEDs in series I needed to use a current driver.
The LEDs are meant to be exactly the same but real world it's not true... And without the current driver some of them will start losing bright and in some days with LEDs on, you will waste all your time.

Posted via Topify on Android

 

[D.A.(Tony)Stewart]

I usually found the best low cost power supply was a universal laptop charger with tapped voltages. usually around 35 cents per watt.

You then add up the LED voltages to be just below the Vdc (within a 1 LED difference) then simply use an array of series parallel to use the best combination. If all LEDs come from the same batch in a bag then they are well matched within10mV so they can run in parallel.

Then use Ohms law on the current for the voltage drop.

Unregulated supplies will work, but dont rely on the Cpacitor as the ESR is about 1 Ohm for a 1Watt LEd.

 

[yassin.kraouch]

I usually found the best low cost power supply was a universal laptop charger with tapped voltages. usually around 35 cents per watt.

You then add up the LED voltages to be just below the Vdc (within a 1 LED difference) then simply use an array of series parallel to use the best combination. If all LEDs come from the same batch in a bag then they are well matched within10mV so they can run in parallel.

Then use Ohms law on the current for the voltage drop.

Unregulated supplies will work, but dont rely on the Cpacitor as the ESR is about 1 Ohm for a 1Watt LEd.

that's good what did propose, but this wil not help me to solve the problem, i want to find a solution to the problem, not avoid completly the problem and find other solution

 

[D.A.(Tony)Stewart]

If you want a solution then start a question with LED specs for V, I qty and supply voltage DC and AC ripple. If rectified voltage is 100 Hz, it is straight forward for me to tell your optimum matrix series, parallel.

For ripple reduction you need at least 3x T period of ripple of 10ms for 100Hz or 30ms , but warning if you have high current a capacitor may turn into a battery size to yield low ripple.

 

[FvM]

that's good what did propose, but this wil not help me to solve the problem, i want to find a solution to the problem, not avoid completly the problem and find other solution

A better thread title might have been "how to save a bad design?".

I presume that switching C4 footpoint to Q1 collector will fix the LED peak current problem, unfortunately creating a transistor peak current problem instead. You might find a compromise in splitting the total series resistance between R1 and R4 differently.

In any case, I doen't see a particularly reason to have C4 in this place.

 

[Gorgon]

C2 is also part of the series impedance. If you remove the C4 and adjust the value of C2 to a value that is correct for the voltage you want out of the driver circuit. The peak voltage value at start should go down a bit.

Another possibility is to keep the C4 and add a zener diode over it to bleed away the voltage buildup. This will give you a slightly higher start voltage, but nothing like the one you have now. You should still adjust the C2 to the correct value. The wattage and cooling of the Zener is depending on the value of C2, and the time the LEDs are off.

 

[Audioguru]

C4 prevents the LEDs from flickering on and off at 100Hz like my Christmas Tree LEDs do.

 

[yassin.kraouch]

A better thread title might have been "how to save a bad design?".

sure its a bad design, if its a good design i will not ask question here,

In any case, I doen't see a particularly reason to have C4 in this plac

i have a reason to put C4 there, because if i put it in parallel with LEDs, and if the transistor if switched OFF, and i have ground layer in the bottom layer, i will create capacitive coupling between TOM layer and buttom layer, because the capacitor is not filetering,
so a small eakage current will flow throught the LEDs

 

[D.A.(Tony)Stewart]

yassin, if you want it to work with 60 x 5mm white LEDs on 230Vac 50Hz, it is better to put LEDs in series to reduce the ripple current.

Putting the LEDs in parallel creates two problems One you need a LiPo battery just to filter the ripple the Amps of current. Next, a risk of thermal runaway** and pffft:-(... if they are not perfectly matched from same batch

With series LED's current should be 20mA average and is possible to smooth ripple current with C4 = 100uF

R1 serves no useful purpose unless it is a fuse and will overheat.
C1 must be self-healing plastic film cap rated for >700V (thus low ESR)
C1 value is the current limiting part. = 15uF @>700V rated for 2A ripple current. LED array ESR = 15 Ohms /60 =0.25 Ohms

But if 60 LEDS in series. Vf=190Vdc ESR= 15x60=900 Ohms @ 20mA
C1= 0.5uF approx @ >700V Film only
then C4=100uF across LEDs to reduce ripple but will now take one (1) second to light up (slow)


**Thermal Runaway can occur at high current when self-heating reduces voltage drop and draws more current in parallel mode away from the other LEDs. The Shockley effect is ~-5mV/degC and thermal resistance is 100 degC/100mW , and only series resistance and/ or ESR prevents thermal runaway, if no cooling is possible. I leave this as an exercise to find threshold for mismatch at rated current.

If you really want to use parallel mode for "safety reasons" then a LiPo cell running at no more than 3.4V in parallel with LEDS to act as C4 with separate On/off switch for AC & DC ( DPST) If precharged, then it wont take hours to charge up

EDIT
Sorry, I got side-tracked by "Christmas light string of 60 LEDs.
You only wanted 140mA pk. What LED type and how many? or rather what total power?

To help you understand C4 issue, Ic=C dv/dt thus for dv=0.1V max in dt=10ms at Ic=140mA ,
C=Ic dt/dv = 0.14A *0.01s/0.1V = 0.014 Farads or 14k uF

also R4 serves no useful purpose as C1 is the currently limiter... unles C4 becomes a LiPo battery then reduce R4 to ~ 1 Ohm +/-100%

 

[yassin.kraouch]

yassin, if you want it to work with 60 x 5mm white LEDs on 230Vac 50Hz, it is better to put LEDs in series to reduce the ripple current.

Putting the LEDs in parallel creates two problems One you need a LiPo battery just to filter the ripple the Amps of current. Next, a risk of thermal runaway** and pffft:-(... if they are not perfectly matched from same batch

]

yes i already have only LEDs in series, and the current throught LEDs needed is 20mA, i do not accept 140mA current ripple, and the maximumm ripple current is 40mA, i don't understand why you are talking about battery, but i guess 320 V battery will be so big,
C2 is the current limiter, but the problem come i switch off the transistor, and when i switch ON, a huge current come from C4( because C4 stay charged), and this current will be limited by R4

 

[FvM]

As already mentioned, it's a simple circuit concept with limited performance. When you decided to use this topology, you got the problems for free.

I'm also under the impression that you didn't yet start to modify the circuit creatively and not actually considered the suggestions that have been made, respectively didn't fully understand their substance.

 

[yassin.kraouch]

As already mentioned, it's a simple circuit concept with limited performance. When you decided to use this topology, you got the problems for free.

I'm also under the impression that you didn't yet start to modify the circuit creatively and not actually considered the suggestions that have been made, respectively didn't fully understand their substance.

the only solution i have now is to increase the value of R4, to for example 2k, and also increase the value of C2, in order to keep the current through LEDs the same,
but in this case the system efficiency will decrease because i will dissipate throught R4,
playing around R4 and C2 will solve the problem, but decrease the system efficiency, i am looking for other solution,

 

[Gorgon]

the only solution i have now is to increase the value of R4, to for example 2k, and also increase the value of C2, in order to keep the current through LEDs the same,
but in this case the system efficiency will decrease because i will dissipate throught R4,
playing around R4 and C2 will solve the problem, but decrease the system efficiency, i am looking for other solution,

Take a look at the last paragraph in my post #13. The reason for the over current when switching on the LEDs, is the no load charge of C4 to peak voltage 300+V. If you load it with a Zener diode(s) that is slightly higher in voltage than the normal loaded LED voltage, you will not have the big current peak when switching the LEDs on. You should select the zener voltage to be some volts more than the normal loaded LED voltage.

The zeners will burn around 4W with the LEDs off, so you should combine a number of zeners in a string that will split the load into small enough chunks of the total power. You must also adjust the value of C2 so the current will be 20mA in both the LEDs and the zeners.