Perhaps I am missing a point, but a 5V zener does not provide a 5V voltage drop. Beyond break-down the voltage drop is negligeable.
This situation calls for a resistor. The LED's take 100 mA at 40V, so their resistance at this voltage is 400 Ohms. We want to go to a situation where 45V draws 100 mA, so the total resistance needs to be 450Ohms. We need to add a 50 Ohm resistor to get there. Dissipation over that resistor is 1/2 watt, so keep it cool and take a 3 or 5 W.
vac
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Maybe it will drop more than 0.7v if you turn it backwards so it can drop 5v?
No, forward drop is the standard .7v, backwards resistance above break down is very low, and so is the voltage drop.
Paul,
I believe you mean the dynamic resistance above reverse breakdown is very low. The voltage drop is simply according to its spec, like (about) 6.8V for a 6.8V zener. I'm sure you know this, just clarifying.
jan
Hi,
Or a transistor as an amplified diode, shouldn't be that critical.
rgds, sreten.
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Many thanks to you all for all the replies and great ideas.
I never knew about the high voltage version of the LM317.
I asked for an elegant solution because of the limited space available on the PCB and at the same time a solution that would not waste energy. I cannot use the classic configuration of series resistor and zenner because i can only draw a small amount of current from the trafo ( about 100mA ) and i need almost all of this current to power up 10 1Watt white LEDs. The LEDs are driven by TLE4242 driver at a current somewhere below 100mA. I am using the LEDs to light two big vu-meters in my amp. The original solution was built with incandescent bulbs, 9 volt bulbs in series of 5 and 10 bulbs in total for both vu-meters. The problem is that the original bulbs are in a fuse like 2 cm long package. Because the filament is almost 20mm long it broke almost every time i moved the amp because of the mechanical shock. The bulbs are 2 Euros a piece and i kept replacing them as they failed for about 4 years now. I had enough of this and from the tests i made until now, a combination of white and orange LEDs give almost the exact warm light as the bulbs.
So the cost of the LM317HV (60 volt rating) at about 2 Euros is not an issue if it saves me of the trouble to open up the amp and replace bulbs every time i lift it up from it's 'feet'.
I never knew about the high voltage version of the LM317.
I asked for an elegant solution because of the limited space available on the PCB and at the same time a solution that would not waste energy. I cannot use the classic configuration of series resistor and zenner because i can only draw a small amount of current from the trafo ( about 100mA ) and i need almost all of this current to power up 10 1Watt white LEDs. The LEDs are driven by TLE4242 driver at a current somewhere below 100mA. I am using the LEDs to light two big vu-meters in my amp. The original solution was built with incandescent bulbs, 9 volt bulbs in series of 5 and 10 bulbs in total for both vu-meters. The problem is that the original bulbs are in a fuse like 2 cm long package. Because the filament is almost 20mm long it broke almost every time i moved the amp because of the mechanical shock. The bulbs are 2 Euros a piece and i kept replacing them as they failed for about 4 years now. I had enough of this and from the tests i made until now, a combination of white and orange LEDs give almost the exact warm light as the bulbs.
So the cost of the LM317HV (60 volt rating) at about 2 Euros is not an issue if it saves me of the trouble to open up the amp and replace bulbs every time i lift it up from it's 'feet'.
I may have missed it in all the posts, but if this is just to run LEDs from a too-high voltage, you can just put a resistor and the LEDs all in series. LEDs are current devices anyway. No need to go fancy on lowering the voltage and THEN feed the LEDs.
Example: 60V, 2 LEDs that need 2.5V at 5mA, have a series R that eats the excess 55V at 5mA = 11k. Power in resistor = 55V x 5mA = 275mW...
You can use this for 10 LEDs just scale values, as long as the supply is high enough.
jan didden
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My understanding is that an IC LED driver is being used, and it's the driver that needs the voltage drop.
If a 3-pin regulator is used, IMO it might be more elegant to just configure it as a constant-current source. Otherwise, I think the one-component diode or resistor solution wins going away.
If a 3-pin regulator is used, IMO it might be more elegant to just configure it as a constant-current source. Otherwise, I think the one-component diode or resistor solution wins going away.
Yes Jan. I know this. But it is very important for me to keep the same brightness in the LEDs. For this i need a driver with thermal compensation. I prefer to use a driver for PWM dimming if necessary and also to switch the LEDs off in an instant using an AC detect.
If i use a simple series resistor i know it will work, but the LED's will work in a high temperature environment besides the temperature generated by themselves and they will begin to draw more and more current and shift brightness, causing them to fail in time. I have seen this the first time in a LED version of the Maglite torch and confirmed it when testing the white LEDs.
3v x 15 = 45. Want to power 15 white led's? If the transformer max is below the current max of an LED, the driver, regulator and other power wasting circuitry is useless. But if the transformer is a bit stronger, all you'd need is to come very close to the target voltage with a string of LEDs and then add a tiny value resistor like 20 ohm or so, depending on just how close you get to the target voltage. Don't be shy of using several series power diodes, since those drop voltage rather efficiently. And lastly adjust the resistor for brightness and not more than 42% led current, so they will last. Using an online LED calculator, simply get your resistor wattage requirement down to bare minimum, and necessarily that is maximum efficiency.
P.S.
I shottky trimmed several RV's worth of lighting. The schottky trim was interesting because as the battery runs down, the current demand drops and the LED's do not dim since schottky passes more voltage during lower current demand. Well it isn't particularly efficient plugged in; however, as soon as it unplugged and then onto a freshly charged battery, 12.77vdc, the drainer resistors and light fixtures have no detectible heat output even though they did not dim.
A solar panel will charge the battery easily. These do not need to stay plugged in. Gosh, I'm glad I didn't use any power wasting drivers and regulators.The transformer is AC 38,2 V.
The original layout consisted of two PCBs connected in parallel on the 38,2 AC.
Each PCB held 5 series bulbs similar to these : CML INNOVATIVE TECHNOLOGIES|3820AL|LAMPE, AXIAL, T3.8, 8V, 0.6W | Farnell Deutschland
I can only assume the current rating of the power supply based on the fact that the bulbs drew 200mA. ( measured current is a bit below 200mA ).
But because i will have to parallel the two circuits i cannot rely on the maximum current rating of the trafo. I need to make sure the LEDs are safe. Otherwise i'd be better off with the original bulbs and take extra care when lifting the amp.
The original layout consisted of two PCBs connected in parallel on the 38,2 AC.
Each PCB held 5 series bulbs similar to these : CML INNOVATIVE TECHNOLOGIES|3820AL|LAMPE, AXIAL, T3.8, 8V, 0.6W | Farnell Deutschland
I can only assume the current rating of the power supply based on the fact that the bulbs drew 200mA. ( measured current is a bit below 200mA ).
But because i will have to parallel the two circuits i cannot rely on the maximum current rating of the trafo. I need to make sure the LEDs are safe. Otherwise i'd be better off with the original bulbs and take extra care when lifting the amp.
I don't think you understand how a zener diode works ??
If the resistors are dropping the excess voltage then the current through the zener will be ZERO.
The aim is to pass about 5mA through the zener + the remaining current through the load 100mA in your case.
The series resistor is then calculated to drop the excess 5V with 105mA passing through it ~ 50 Ohms.
The zener is only rated at 5W because it has to pass the full 105mA if the load suddenly ceases to draw any current.
Stating that the transformer isn't big enough to cope with a zener regulator is incorrect.
ALL regulators will, to varying degrees, draw their own power from the supply.
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