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### Topic: driving darlington array (Read 5319 times)previous topic - next topic

#### Runaway Pancake

#15
##### Dec 17, 2011, 11:37 pmLast Edit: Dec 17, 2011, 11:41 pm by runaway_pancake Reason: 1
The 2803 outputs provide a path to ground.

The UDN2981 is a "high-side driver".

They're still available from Jameco.
http://www.jameco.com/Jameco/Products/ProdDS/1762681.pdf

Yes, resistors are (still) required.
"Who is like unto the beast? who is able to make war with him?"
When all else fails, check your wiring!

#16
##### Dec 18, 2011, 12:01 amLast Edit: Dec 18, 2011, 12:05 am by magnethead794 Reason: 1

The 2803 outputs provide a path to ground.

The UDN2981 is a "high-side driver".

They're still available from Jameco.
http://www.jameco.com/Jameco/Products/ProdDS/1762681.pdf

Yes, resistors are (still) required.

I just poked google- Aren't ULN2003/2004 high side drivers?

If I set it up to run at half power (30ma), saturation voltage is 0.9-1.1 volt, to display an 8 is 240mA for the whole unit. Max power is .03*5 = .15 of a watt.

Half-power segment resistor would be ((5-0.9)-4.0)/.03 = 4ohm -> ((5-1.1) - 4.0)/.03 = 4ohm -> (5-4.0)/.03 = 35ohm

So the most conservative resistor would be 35 ohms or greater per segment?
KF5RVR

#### Runaway Pancake

#17
##### Dec 18, 2011, 01:43 am
No, the 2003 is a low-side switch, too, equivalent to a 2803 only the 2003 has 7 "switches" and the 2803 has 8.

Can't you sketch out your circuit and take a picture of that with your cell or something and post that?  You haven't provided a schematic for what you're on about, but I think your maths are off.
"Who is like unto the beast? who is able to make war with him?"
When all else fails, check your wiring!

#18
##### Dec 18, 2011, 07:40 amLast Edit: Dec 18, 2011, 08:56 am by magnethead794 Reason: 1

No, the 2003 is a low-side switch, too, equivalent to a 2803 only the 2003 has 7 "switches" and the 2803 has 8.

Can't you sketch out your circuit and take a picture of that with your cell or something and post that?  You haven't provided a schematic for what you're on about, but I think your maths are off.

Per the datasheet I link, each LED is 2.0V 30mA nominal. So 4V 60ma nominal per segment, or to be safe with how it's set up, I'd like to run each segment at 25-30mA total.

I suppose I could have 2 sets of 7 TO-220 case NPN's, and put them on 3 rows of female headers, basically making my own 3 row/7pin transistor array. Easily mountable on a daughter board. I think I'll do that.

KF5RVR

#### Grumpy_Mike

#19
##### Dec 18, 2011, 10:17 am
You need to post the schematic of how you are driving this, it sounds like you are trying to use the transistor array incorrectly.

Just for anyone reading this thread later note that connecting LEDs in parallel like this is wrong.

#20
##### Dec 18, 2011, 11:56 amLast Edit: Dec 18, 2011, 12:41 pm by magnethead794 Reason: 1

You need to post the schematic of how you are driving this, it sounds like you are trying to use the transistor array incorrectly.

Just for anyone reading this thread later note that connecting LEDs in parallel like this is wrong.

IC current: 0.01A
Vs = 5V
Vsat = 0.7V
Vsegment = 4.0V
Isegment = 0.03A
Rsegment = (4.3-4.0)/0.03 = 0.3/0.03 = 10 ohms
hFE = 5 * (0.03 / 0.01) = 5 * 3 = 15
Rb = (5.0 * 15) / (5 * 0.03) = 75/0.15 = 500 ohms

KF5RVR

#### Runaway Pancake

#21
##### Dec 18, 2011, 03:04 pm

Just for grins, you should be able to work out one "segment", your way, on a breadboard and turn it on/off, that's easy enough, and see how that goes.  Maybe the result will be acceptable to you.

Resign yourself to re-work, modification.

I'd have the LEDs in one segment (4) wired in series, each string having its own resistor, and run that from 12V or so.
"Who is like unto the beast? who is able to make war with him?"
When all else fails, check your wiring!

#### Grumpy_Mike

#22
##### Dec 18, 2011, 03:43 pm
Yes that circuit is right. Note that the transistor shown is a PNP transistor. The arrays you were asking about are darlington drivers made from NPN transistors. These are not the same thing.

#23
##### Dec 18, 2011, 07:10 pm

Yes that circuit is right. Note that the transistor shown is a PNP transistor. The arrays you were asking about are darlington drivers made from NPN transistors. These are not the same thing.

I know it's PNP, read my post above an I realized I had to switch substrates.
KF5RVR

#24
##### Dec 18, 2011, 08:14 pm
So I checked how much one LED pulls without a resistor. 2.5V 30ma LED unrestricted on 5 volts pulled 235ma and didn't burn out.

So with a PNP transistor, I'd be looking at (5.0-0.7)/0.03 = 4.3/0.03 = 143 ohm resistor.

(5-.07)/330 = 4.3/330 = 13ma using test resistors.
KF5RVR

#### dc42

#25
##### Dec 18, 2011, 08:33 pm

So I checked how much one LED pulls without a resistor. 2.5V 30ma LED unrestricted on 5 volts pulled 235ma and didn't burn out.

The useful lifetime of an LED depends on the junction temperature, which depends on ambient temperature and power dissipation. The rated current is quoted for typically 100,000 hours useful lifetime (time before the light output drops by more than 10% or so). At 5v and 235mA I suspect the useful lifetime would be measured in hours or days at most.

So with a PNP transistor, I'd be looking at (5.0-0.7)/0.03 = 4.3/0.03 = 143 ohm resistor.

(5-.07)/330 = 4.3/330 = 13ma using test resistors.

No, you need to take 5v, less the voltage drop of the LED (which you imply is 2.5v), less the saturation voltage of the PNP transistor (around 0.2v if you choose the right transistor and drive it appropriately). So about 75 ohms.
Formal verification of safety-critical software, software development, and electronic design and prototyping. See http://www.eschertech.com. Please do not ask for unpaid help via PM, use the forum.

#### Grumpy_Mike

#26
##### Dec 18, 2011, 08:36 pm
Quote
So I checked how much one LED pulls without a resistor. 2.5V 30ma LED unrestricted on 5 volts pulled 235ma and didn't burn out.

So what!
You subject an LED to nearly 8 times the recommended current and it hasn't vaporised, well woop - de - dee. You have shafted that LED good and proper.
Two points:-
1) Did you actually measure it had 5V across it when you drew the current or was the power supply impedance limiting the current to only 235mA.
2) Did you measure the brightness of the LED at 30mA before and after your abuse?

Overdriven LEDs don't always die immediately they can:-
a) Fail a lot sooner than they would have done, say in thee to six months.
b) Drop in output brightness. In fact the life of an LED is defined by its half life, that is the time it takes the light output to drop by half.

What you are failing to take into account in your calculations is the forward voltage drop of the LED. This must be added to the Vsat of the transistor before subtracting it from the supply voltage.

#27
##### Dec 19, 2011, 12:15 am
I wasn't stating anything useful with that test- it was to see just what current flowed through the raw LED. The PSU is a good \15 year old 160W ATX from an old Win95 desktop.
KF5RVR

#28
##### Dec 19, 2011, 12:36 amLast Edit: Dec 19, 2011, 01:38 am by magnethead794 Reason: 1
I think I came up with a way to make my own 3 layer PCB using 3 of the pre-printed ones. Gonna cost an arm and a leg (At \$7 a layer) plus \$20 for new LED's. I'll just put the minimum resistor LED on each and do it the right way with a grounding darlington. A layer for LED's, a layer for Resistors, and a layer for the Arduino, Darlingtons, Voltage Regulator, filtering caps and diodes, ect. I have to have brightness control, so a 0-20K potentiometer will join the minimum resistors to limit current.

So doing it right, I would still have to account for the 1.2V saturation voltage of the darlington?

I'm going to parallel all of the LED's in each segment to make a 2V 120mA segment array. That'll be 840mA per segment to display an 8- not sure the Darlington's can push that much?? Unless I use 4 arrays and bridge every 2 together?

(This is going on a race car with a 16VDC 180A charging system, so that plus vibrations are gonna be fun to work with, but not worth \$5 per square inch to have made in PCB form.)
KF5RVR

#### Grumpy_Mike

#29
##### Dec 19, 2011, 04:29 am
Quote
So doing it right, I would still have to account for the 1.2V saturation voltage of the darlington?

Yes

Quote
That'll be 840mA per segment to display an 8- not sure the Darlington's can push that much

While each individual Darlington in the array can switch up to 500mA you can't have more than 650mA flowing through the whole chip at any one time or it will over heat.

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