I was looking at driving some LEDs with this ic, and I was wondering how much current it could handle with all outputs on:
The equation for the power dissipation is on page 18, and it is:
Pd = (Vcc * Icc) + (Vout * Iolc * N * dPWM)
Where:
· Vcc = device supply voltage
· Icc = device supply current
· Vout = OUTn voltage when driving LED current
· Iolc = LED current adjusted by RIREF resistor
· N = number of OUTn driving LED at the same time
· dPWM = duty ratio defined by GS value
I assume Vcc should be 5v.
I don't know what I should put for Icc.
I'm guessing Vout should be the forward voltage of the led? Or would that be Vcc - Fv?
Iolc I'm pretty sure is whatever I decide to set the maximum current per channel to with RIREF.
N would be 24 since all 24 outputs may be on at once.
And dPWM would be 1, under the assumption that all LEDs are turned fully on.
So, how do I determine Icc? Is that N * Iolc?
And why is that being multiplied by Vcc, when I would think that the right side of the equation would tell me how much power needs to be dissipated for all leds combined?
And which method do I use to calculate Iolc?
TI has a type right after Icc. It should be minus sign. This formula simply tells you the energy conservation. The chip takes in power from input and sends most of it down it's 24 channels. The rest of the power is consumed by the chip. So it makes sense to have a minus sign. The output voltage is the voltage on your output pin, not the forward voltage. The formula considers when all channels are on and have the same voltage and current and same PWM ratios, which is probably when the chip consumes most power. The current on each channel is controlled by a resistor. So what else is bothering you ?
Okay, so what you're telling me is the datasheet is wrong, and that + sign in the middle should be a - sign.
I'm still confused.
The output voltage is the voltage on your output pin, not the forward voltage.
But what is that voltage? Is it 5v? Is it 5v minus the led's forward voltage?
And how do I determine Icc? If I assume all the LEDs are fully lit and each is drawing 20mA, is Icc 480mA?
If so, how do I know if the Vcc pin can handle that much current? I know there is a limit for what the package can dissipate but usually datasheets list limits on how much current is allowed on the inputs, and it's usually a lot lower than 480mA. I think TI's high current shift register can only handle something like 120mA total output. So 480mA seems awfully suspect to me.
scswift:
But what is that voltage? Is it 5v? Is it 5v minus the led's forward voltage?
The power supply - the LED's forward drop = voltage across OUTn.
e.g.
5V - 2V = 3V
scswift:
And how do I determine Icc?
Page 4: "Supply current"
liudr:
TI has a type right after Icc. It should be minus sign.
That isn't a typo. It should be a plus sign as it is shown in the datasheet. The power dissipation is the combination of what the chip internally dissipates PLUS power dissipated as current flows through the I/O pins.
The chip is just a current sink. It only controls the amount of current flowing through the "output" pins. Voltage (and power) for the LEDs are applied externally from chip, as shown in the diagram on the first page, Vled.
This chip is a sophisticated replacement for a resistor...
So the power dissipates by the chip is the remaining voltage drop, current flowing through the pins, and the power the chip dissipates itself.
It's not terribly clear from that how to calculate the supply current. I see a bunch of test conditions. But if those conditions are for Vout = 1V it seems like that is for a best case batch of LEDs with a 4v drop across them. Not sure what Latch and blank low conditions mean. Or what the reference resistor values specify for output current, but I'll look into that in more detail this evening and see if I can figure it out.
Not sure what Latch and blank low conditions mean. Or what the reference resistor values specify for output current, but I'll look into that in more detail this evening and see if I can figure it out.
Latch and blank really are not that cryptic. Latch is like any latch found in a shift register. Blank is also known as "output enable". Those test conditions are saying, once data is clocked in and the outputs are enabled.
The reference resistor sets the max current the channels can provide. There is a formula in the data sheet for calculating what reference resistor you need.
The 1V listed in the test conditions concerns me though. It looks like the Typical condition tests were done at 3.3v. That would mesh with 1V on the output pins, if the LED had a 2.3v voltage drop.
However, if the maximum is also calculated for 1V on the output pins, and that is assuming 5v on Vcc, then that's like driving an LED with a 4v voltage drop, which while maybe possible
with white leds, is hardly a worst case scenario like driving a red led with a 2.1v drop from 5v would be.
So I'm still at a loss for how to calculate Icc. I guess maybe if I find the right ohms law equation I can input the values from the test conditions and maybe that will give me a clue as to how to calculate for a different voltage drop.
Ohms law formulas aren't helping. I did notice however that Icc Typical is the same as the max current for each output when using the associated Riref value.
So in addition to what current the chip outputs to any one pin, it also uses itself? When Vcc is 3.3v anyway? Regardless of how many pins are being used at once?
That doesn't seem right.
Assuming:
Vcc = 5v
Icc = ? but going with 90 cause it's the max listed as an example in datasheet
Vout = 3.5v cause lowest voltage drop I could find for an LED was 1.6v so I just went with 1.5v since I need a little safety margin anyway
Iolc = 20mA
N = 24
dPWM = 1.0 (leds on 100% of the time)
I get Pd = 450 + (84 * Iolc)
So for Iolc of 20mA I get 1680mW, for a total Pd of 2130mW.
And the chip can supposedly handle 2500mW when the heat sink isn't even soldered down, and 5000mW when it is.
That's sinking 480mA from the LEDs. That is a little less than the TPI6B595 is supposedly capable of sinking, and the datasheet for that is pretty clear about how much it can sink unlike this one. So I guess I might be in the ballpark.
Yes, but it's unclear what the Icc is when Vcc is 5v. I'm not sure if that is what the maximum listed represents. And if it is what it represents, is that for 1V at the pin? If so, the voltage drop on the LED could leave as much at 3.5v on the pin, which I would assume would increase Icc.
