No, a "typical" LED (if there is such a thing) usually starts glowing between 5 an 8 mA, and is fully lit by 12 to 15 mA. Its max current is 20 to 25 mA -- above which it burns out. There do exist "low-current" LED's that are full bright at 1 to 3 mA, and burn out at 10 mA. Most "7-segment LED's" are more like the 'typical' example above.
If you have very good control of the pulsing, you can pulse LED's to produce even brighter displays. But otherwise you have a very good point -- 7-segment LED's take 7 times the current of a single LED.
The point in issue here is the method of driving the LED's. In my proposal, there are no "current limiting" resistors that consume most of the power. Instead·an inductor is used to "transform" the wattage needed, say 50 mW per LED segment, and match that to the available supply voltage, say 25 to 30VDC. At high conversion efficiency, and 80% is not an unreasonable expectation, we are therefore looking at roughly 2 milliamps per segment.
Why all this works is that the 2 mA is the average current consumed from the 25/30 V power supply; the inductor charging pulse is much higher, but then the corresponding duty cycle is low. So, for a reasonably bright "average" LED, 7 segment simultaneously "on" would be in the order of 15 mA.
In my tests on this I actually used·a tiny 1 mH inductor, one end wired to the +30 supply, and the other end to a ground switching transistor, and the LED was the back-emf "catch" diode. The current in the LED flows only while the transistor is "off" and stops when the collapsing field is depleted. Be careful to only pulse the inductor for a few microseconds so that it does not saturate. If by accident, or poor pulse programming techniques, the transistor stalls in the "on" condition, smoke WILL appear.
Depending on the inductor, you will need to experiment with the rep-rate and the duty cycle.
Have fun with lighting your LED's this novel and efficient way!
Peter·is right, he is basically PWMing the LED where the current pulses are above the max average allowable for an LED, so that when the PWM current is integrated over time via the inductor the average current is just below max average. Like he said this has the benifit of using nearly all the power consumed towards lighting the LED rather than wasting it in current limiting resistors. One word of warning, if the driver doing the PWM gets "hung up" during the pulse being high, you will burn out the LED segment. It may be possible to use a zener diode to prevent this from happening, but I'm not sure because it is trip value is dictated by voltage which the inductor can cause to fluctuate possibly causing false triggering (and wasting energy). An alternative is to use the self resetting fault fuses since thier triggering is dictated by the current.
Just a quick note on this technique, the current in the inductor will rise at a rate proportional to the voltage on it and inversely proportional to the inductance. V=L di/dt. As the current increases it builds up a magnetic field which stores the energy you are putting into it. When you switch off the source, the collapse of the inductor's magnetic field will cause a reverse voltage across the inductor such as to attempt to maintain the current flowing through it at whatever value it was just before switch off. So you want to time your pulse according to V/L = di/dt or T=i*L/V . If your supply voltage is 30 V and your LED forward voltage is 2.2 then the time it takes for the inductor to discharge will be 30/2.2 times longer than it takes to charge it. The current through the LED will start out at whatever your peak inductor current was, and will ramp down to zero over time, until the inductor's magnetic field is completely collapsed. Since the current will drop almost linearly because of the nearly constant voltage of the LED, you probably want your peak inductor current to be no more than twice the average allowed LED current, depending on how often you pulse the inductor relative to how long it takes to discharge it.
You can generally pulse LEDs at a much higher current than their average, and a bonus to doing that is higher perceived brightness. I can't recall for sure, it is either a case of the LED output being non linear (which seems not likely) or the eye responding in a non linear way to a pulsed source. A quick look at a typical Lumex green 7 segment LED display shows a steady current rating of 25ma, but a peak forward current of 150ma at less than 10 microseconds duration. Max power dissipation per segment 105 milli watts at 25 C. It is the power dissipation that has to be observed along with the max current. The average power dissipation can't be any higher than rated, and it derates at 1.5 milli watts per degree C above 25 C. The average power dissipation can be figured as the LED forward voltage drop times the average current.
Let's do some numbers: T= .150 * .001 / 30 With a 1 milli Henry inductor at 30 volts you would need to turn it on for 5 microseconds to hit 150 ma. It will take T = .150 *.001 / 2.2 or 68 microseconds to discharge into your LED. Please note, your mileage may vary! The average current per pulse can be looked at as half the peak, or 75 ma. To keep the power dissipation in the ballpark, you can't pulse it at more than 1 part in 3 to keep the over all average current at 25 ma. Which translates into a 5 microsecond charging pulse every 68*3 = 204 microseconds.
If you are going to do 2 displays you can multiplex them both off the same set of 7 (or 8 with decimal point) inductors. Still keeping the same 1/3 duty cycle but doing one display during one third of the cycle and the next display during the second third and doing nothing the last third. Or add a third display to use up that open time slot.....
One point to notice is that your current will vary proportionally to the supply voltage. Your average current will be 75 ma times the duty cycle of 5 microseconds /204 microseconds * number of segments driven. So for two 7 segment displays that is .075 * 14 * 5 / 204 = 25.7 milliamps per display worst case from a 30 V supply. Brings you to a reasonable 6.5 A total for 255 displays. Not counting the power consumption of the scenix, but a switching buck regulator will give the same kind of benefit there.
In the gotchas department, you will need to set this up as a buck regulator, which means the switching element and a flyback diode to ground has to be at the 30V end of the inductor with the led at the opposite end of the inductor. Assuming you have common cathode LEDs. You also want to design your switches to be OFF while the scenix powers up. Not difficult to do since the scenix powers up with all its I/O pins in the input state.
I was going to suggest a smallish cap to ground at the inductor LED junction to prevent spikes, but thinking about the circuit configuration, the inductor charging current is going to go through the led too, so you won't have the discontinuity I was thinking of when the charging switch turns off.
Note that if you try to use the LED in the position of the flyback diode, which would be a really nice thing if it worked, you will pop the LED when the switch turns on, because LEDs don't have very high reverse voltage capability. TANSTAAFL. (There Aint No Such Thing As A Free Lunch)
Thinking about the configuration of the circuit a bit more, you have to subtract the LED forward voltage from the charge voltage in the calculations above, because the LED will be in series with the inductor when it is charging. For the sake of completeness, you also need to take into account the forward drop of the flyback diode in the discharge side of things, as it is in series with the discharge current, and so increases the voltage across the inductor during discharge, shortening the time. And wasting some fraction of the power as well. Hmm, actually a large chunk if you are to use a normal diode at .7 V forward you will loose .7/2.2 31% or the power to the flyback diode. Even with a schottky at .3 volts or so, you will blow .3/2.2 or 13% of the power in the flyback diode. That may be acceptable, or else you can use a totem pole sort of arrangement with mosfets, using what amounts to an active flyback switch, with a much lower voltage drop.
Another alternative. which only works if you have common anode LEDs, is to put the LED at the flyback position with the flyback diode in series with the LED. The LED will still leak some current, but the diode in series with it will limit it to a safe value.
Gosh, it seems the more questions you try to answer on something like this, the more you find to ask. Oh, well, had to put in my two cents worth. Hope it was more helpful than confusing.
Looking at all the additional components over using simple current limiting resistors, I think you might be better off with a single switching regulator on the board to supply the LEDs and the scenix while running your power bus at the 30V or so to get the bus current down by the ratio of the input to the output voltage of the regulator. Note that you will need a biggish cap at the input of the regulator to compensate for the inductance of the power bus wiring. The average current drawn off the bus is lower, but the peak is still going to be in the vicinity of the peak current on the output. If you were going to use the power bus to communicate to the modules as well, this isn't a big issue, you'll need it anyway, along with a filter inductor in series with it, as mentioned in a previous post.
It was exactly my intention to use the LED segment as the flyback diode, and I was aware that a series reverse protection diode would be required to protect the LED.
In this scenario, the inductor charging current does not flow through the LED, only the discharge current does. Also it does not lend itself well to multiplexing.
Considering that this setup is very easy and inexpensive to implement with an SX, and it does eliminate all the power wasting current limiting resistors, I thought it was a better (more efficient and less aveage consumed current) way to light the LEDs than a single node switching regulator and again using those power wasting limit resistors.
Clearly, for just a few nodes this is inconsequential, but multiplying by 255 it is an issue when one wishes to use 18 Ga lampcord vs bigger wire.
Comments
If you have very good control of the pulsing, you can pulse LED's to produce even brighter displays. But otherwise you have a very good point -- 7-segment LED's take 7 times the current of a single LED.
The point in issue here is the method of driving the LED's. In my proposal, there are no "current limiting" resistors that consume most of the power. Instead·an inductor is used to "transform" the wattage needed, say 50 mW per LED segment, and match that to the available supply voltage, say 25 to 30VDC. At high conversion efficiency, and 80% is not an unreasonable expectation, we are therefore looking at roughly 2 milliamps per segment.
Why all this works is that the 2 mA is the average current consumed from the 25/30 V power supply; the inductor charging pulse is much higher, but then the corresponding duty cycle is low. So, for a reasonably bright "average" LED, 7 segment simultaneously "on" would be in the order of 15 mA.
In my tests on this I actually used·a tiny 1 mH inductor, one end wired to the +30 supply, and the other end to a ground switching transistor, and the LED was the back-emf "catch" diode. The current in the LED flows only while the transistor is "off" and stops when the collapsing field is depleted. Be careful to only pulse the inductor for a few microseconds so that it does not saturate. If by accident, or poor pulse programming techniques, the transistor stalls in the "on" condition, smoke WILL appear.
Depending on the inductor, you will need to experiment with the rep-rate and the duty cycle.
Have fun with lighting your LED's this novel and efficient way!
Peter (pjv)
·
You can generally pulse LEDs at a much higher current than their average, and a bonus to doing that is higher perceived brightness. I can't recall for sure, it is either a case of the LED output being non linear (which seems not likely) or the eye responding in a non linear way to a pulsed source. A quick look at a typical Lumex green 7 segment LED display shows a steady current rating of 25ma, but a peak forward current of 150ma at less than 10 microseconds duration. Max power dissipation per segment 105 milli watts at 25 C. It is the power dissipation that has to be observed along with the max current. The average power dissipation can't be any higher than rated, and it derates at 1.5 milli watts per degree C above 25 C. The average power dissipation can be figured as the LED forward voltage drop times the average current.
Let's do some numbers: T= .150 * .001 / 30 With a 1 milli Henry inductor at 30 volts you would need to turn it on for 5 microseconds to hit 150 ma. It will take T = .150 *.001 / 2.2 or 68 microseconds to discharge into your LED. Please note, your mileage may vary! The average current per pulse can be looked at as half the peak, or 75 ma. To keep the power dissipation in the ballpark, you can't pulse it at more than 1 part in 3 to keep the over all average current at 25 ma. Which translates into a 5 microsecond charging pulse every 68*3 = 204 microseconds.
If you are going to do 2 displays you can multiplex them both off the same set of 7 (or 8 with decimal point) inductors. Still keeping the same 1/3 duty cycle but doing one display during one third of the cycle and the next display during the second third and doing nothing the last third. Or add a third display to use up that open time slot.....
One point to notice is that your current will vary proportionally to the supply voltage. Your average current will be 75 ma times the duty cycle of 5 microseconds /204 microseconds * number of segments driven. So for two 7 segment displays that is .075 * 14 * 5 / 204 = 25.7 milliamps per display worst case from a 30 V supply. Brings you to a reasonable 6.5 A total for 255 displays. Not counting the power consumption of the scenix, but a switching buck regulator will give the same kind of benefit there.
In the gotchas department, you will need to set this up as a buck regulator, which means the switching element and a flyback diode to ground has to be at the 30V end of the inductor with the led at the opposite end of the inductor. Assuming you have common cathode LEDs. You also want to design your switches to be OFF while the scenix powers up. Not difficult to do since the scenix powers up with all its I/O pins in the input state.
I was going to suggest a smallish cap to ground at the inductor LED junction to prevent spikes, but thinking about the circuit configuration, the inductor charging current is going to go through the led too, so you won't have the discontinuity I was thinking of when the charging switch turns off.
Note that if you try to use the LED in the position of the flyback diode, which would be a really nice thing if it worked, you will pop the LED when the switch turns on, because LEDs don't have very high reverse voltage capability. TANSTAAFL. (There Aint No Such Thing As A Free Lunch)
Thinking about the configuration of the circuit a bit more, you have to subtract the LED forward voltage from the charge voltage in the calculations above, because the LED will be in series with the inductor when it is charging. For the sake of completeness, you also need to take into account the forward drop of the flyback diode in the discharge side of things, as it is in series with the discharge current, and so increases the voltage across the inductor during discharge, shortening the time. And wasting some fraction of the power as well. Hmm, actually a large chunk if you are to use a normal diode at .7 V forward you will loose .7/2.2 31% or the power to the flyback diode. Even with a schottky at .3 volts or so, you will blow .3/2.2 or 13% of the power in the flyback diode. That may be acceptable, or else you can use a totem pole sort of arrangement with mosfets, using what amounts to an active flyback switch, with a much lower voltage drop.
Another alternative. which only works if you have common anode LEDs, is to put the LED at the flyback position with the flyback diode in series with the LED. The LED will still leak some current, but the diode in series with it will limit it to a safe value.
Gosh, it seems the more questions you try to answer on something like this, the more you find to ask. Oh, well, had to put in my two cents worth. Hope it was more helpful than confusing.
Looking at all the additional components over using simple current limiting resistors, I think you might be better off with a single switching regulator on the board to supply the LEDs and the scenix while running your power bus at the 30V or so to get the bus current down by the ratio of the input to the output voltage of the regulator. Note that you will need a biggish cap at the input of the regulator to compensate for the inductance of the power bus wiring. The average current drawn off the bus is lower, but the peak is still going to be in the vicinity of the peak current on the output. If you were going to use the power bus to communicate to the modules as well, this isn't a big issue, you'll need it anyway, along with a filter inductor in series with it, as mentioned in a previous post.
It was exactly my intention to use the LED segment as the flyback diode, and I was aware that a series reverse protection diode would be required to protect the LED.
In this scenario, the inductor charging current does not flow through the LED, only the discharge current does. Also it does not lend itself well to multiplexing.
Considering that this setup is very easy and inexpensive to implement with an SX, and it does eliminate all the power wasting current limiting resistors, I thought it was a better (more efficient and less aveage consumed current) way to light the LEDs than a single node switching regulator and again using those power wasting limit resistors.
Clearly, for just a few nodes this is inconsequential, but multiplying by 255 it is an issue when one wishes to use 18 Ga lampcord vs bigger wire.
Anyway, it is just another way to skin the cat.
Cheers,
Peter (pjv)
Ryan