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Topic: resistor power (Read 4155 times)previous topic - next topic

#15
Dec 22, 2011, 11:05 pm
I'll get some readings for you tonight

#16
Dec 23, 2011, 09:35 amLast Edit: Dec 23, 2011, 10:26 am by magnethead794 Reason: 1

So if it's dropping 2.9 volts, I is 19.3 mA ,
Can you verify that this is the voltage being dropped, using a multimeter?

Quote
Passing 2.1 volts to the LED resistor, which is then dropping current to 14mA. So how many volts are being dropped by the second resistor/how many volts are the LED's seeing?
The LEDs should drop about 2 volts but that may not be the case at such small currents. I don't know what LED you're using and even if I did the manufacturer may not test it for such low currents.

Can you measure the voltage across one of the LEDs when it's in the dim condition and the voltage across "its" resistor?

If you're actually getting 19.3mA through the single "dimming" resistor, this will be split between all 39 LEDs and their resistors, so it should be about 0.5mA each. This would give a volt drop across the 39 LED resistors of about 75mV. Again any actually voltages you can measure would help.

Iain

Vs = 5.050V
Vdimmer = 3.219V
Vprimary = 160.7mV
VLED = 1.667V

21.35 mA flowing between Vs and the dimmer resistor

Pdimmer = 0.02135^2 * 150 =  0.068 watt

Vs = 4.96V
Vprimary = 2.927V
VLED = 1.945V

341.6 mA flowing between Vs and primary resistors.

measurements taken with 20 LED's turned on. Each LED is getting 1.667V @ 1mA dimmed and 1.945V @ 17.08mA at full brightness.

---------

Vs = 3.310V (NOTE: Will be 3.8 volts in application)
Vdimmer = 1.391V (NOTE: 150 ohms, same as 5V circuit)
Vprimary = 207.6mV
VLED = 1.709V

9.22 mA flowing between Vs and the dimmer resistor

Pdimmer = 0.00922^2 * 150 = 0.0127 watt

Vs = 3.310V
Vprimary = 1.403V
VLED = 1.889V

59.34 mA flowing between Vs and primary resistors.

measurements taken with 4 LED's turned on. Each LED is getting 1.709V @ 2.305mA dimmed and 1.889V @ 14.835mA at full brightness.

Strangely, both sets of LED's were about the same brightness at both full brightness and reduced brightness. I found that somewhat peculiar since I used a 150 ohm resistor on the lower-voltage set of LED's.

sixeyes

#17
Dec 23, 2011, 10:57 am
Vs = 5.050V
Vdimmer = 3.219V
Vprimary = 160.7mV
VLED = 1.667V

21.35 mA flowing between Vs and the dimmer resistor
Thanks for the readings. When taking into account the lower VLED of 1.667V and only 20 LEDs we get similar results:

Vdimmer = (5.05 - 1.667) * 20 / 21 = 3.221V
Vprimary = (5.05 - 1.667) / 21 = 161.0mV

However the problem with all of this is the brightness variance as you change the number of segments illuminated. The 39 LEDs which are always illuminated will appear dimmer than the ones controlled via the ULN2803A's. As I don't have any actual figures for your LEDs controlled by the ULN2803A I'm going to assume that they're ideal switches. I believe this is valid because you've used lower resistors in practice for these LEDs to account for the voltage drop across the ULN2803A. This will allow me to demonstrate the issue you face with the current design.

If you're showing a "1" (segments b & c) you would get:

Vdimmer = (5.05 - 1.667) * 4 / 5 = 2.673V
Vprimary = (5.05 - 1.667) / 5 = 668.2mV
ILED = 0.6682 / 150 = 4.45mA

(For this example I'm assuming VLED hasn't changed although in practice it will rise slightly because of increased current)

But if you show an "8" (segments a - g) you would get:

Vdimmer = (5.05 - 1.667) * 28 / 29 = 3.266V
Vprimary = (5.05 - 1.667) / 29 = 116.7mV
ILED = 0.1167 / 150 = 0.778mA

I hope you can see that the current when all the LEDs are illuminated has changed dramatically. This change will be noticeable. And when I tried something similar with a matrix display I found it very irritating that the brightness varied with the displayed character. I notice you've shown some figure with a 3.3v supply.

Can hook up a ULN2803A and 1 LED (or more if you wish) to 3.3v (via your 90 ohm resistor) and see if it's dim enough. If so that would solve the variable brightness issue for the LEDs controlled by ULN2803A's. Then we could find a matching resistor for the 39 LEDs using 5v. So the DPDT switch would switch between 5v and 3.3v for the groups of 28 LEDs and the between 5v and 5v + resistor for the group of 39 LEDs

BTW your schematic didn't show 39 LEDs being on permanently but 20 permanently on and 8 switched (in addition to the 2 groups of 28 controlled via the ULN2803A).

Iain

sixeyes

#18
Dec 23, 2011, 11:03 am

Strangely, both sets of LED's were about the same brightness at both full brightness and reduced brightness. I found that somewhat peculiar since I used a 150 ohm resistor on the lower-voltage set of LED's.
What's really going to irritate is the difference (when dimmed) of 16 LEDs (i.e. "1 & 1") and 56 LEDs (i.e. "8 & 8").

I'm not sure if you've can hook up your Arduino to the 56 LEDs yet but if you could write some code to toggle between 1 & 8 on both digits you'd see what I mean.

Iain

#19
Dec 23, 2011, 11:07 am
The varying brightness per digit isn't too big an issue, I don't think.

the 3.3 figure was to simulate the 3.8 volt remainder of the darlingtons.

the first digit is a 5. A toggle switch turns on 2 extra segments to make it an 8.

Usually the second digit will be a 4, 5, or 6.

The third digit will vary and cannot be predicted.

The 2 LED (it was 4, i removed 2) decimal point will be tied to the default 5, for all intensive purposes.

22 permanently on + 8 switched + 56 controlled by darlingtons.

86 LED's total.

sixeyes

#20
Dec 23, 2011, 11:13 am
OK. Given your selection of digits it may not be much of an issue.

So are you happy with the power calculations?

Iain

#21

#22
Dec 23, 2011, 11:22 amLast Edit: Dec 23, 2011, 11:29 am by magnethead794 Reason: 1

OK. Given your selection of digits it may not be much of an issue.

So are you happy with the power calculations?

Iain

Yes I am happy with the calculations. Just had to wrap my brain around the convention.

The 5 will be constant, the second will be 4/5/6 which is one extra segment per number, the third can be anything from 1 to 8 as far as current draw. So it will see the most current swing. But really, if you think about it...

0: 6 segments
1: 2 segments
2: 5 segments
3: 5 segments
4: 4 segments
5: 5 segments
6: 6 segments
7: 3 segments
8: 7 segments
9: 6 segments

So other than 1 and 7, every digit should have a similar brightness to other digits.

I dont know for sure yet, but the target is going to be between 42 and 68....probably tighter (48 and 62?). So the highest contrasts will be 51, 57, and 61. Not a problem.

sixeyes

#23
Dec 23, 2011, 11:26 am
Once you said the second digit wasn't showing a 1 or 7 I didn't think there would be much of an issue. You won't see a 1 next to an 8.

Iain