# Finally, a little progress?

Discussion in 'General Electronics Chat' started by charlie_r, Nov 16, 2009.

1. ### charlie_r Thread Starter Member

Jul 12, 2007
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Thanks in large part to Bill_Marsden's comprehensive guides to the 555, I have finally come to understand how to set up a 1/10 duty cycle with a Tm of 1ms.

However, this brings out another oddity that I am having trouble wrapping my head around:

Driving several LEDs with a 555, without using current limiting resistors, I expected to see a total current draw of ~300mA in my circuit.

LEDs are 2.6 Vf, 30 mA If units, wired in parallel, driven with a 2N5088 transistor. Without limit resistors, I expected to see a lot higher total current than just 30mA for the entire circuit.

What gives? I think I understand that the It (how do you do subscripts?) will be an average, but still, shouldn't the circuit draw a lot more than this? I don't have access to a wigglescope [oscilloscope] to see the actual peak instantaneous draw when the output is high.

Can anyone give me a clue here?

2. ### Wendy Moderator

Mar 24, 2008
20,874
2,654
LEDs will burn out without current limiting. So will the 555's. Are these high power LEDs? Let's see the schematic please.

Glad you liked my work, I aims to please.

3. ### charlie_r Thread Starter Member

Jul 12, 2007
16
0
Schematics attached.
LED datasheet attached.

$V_in = 7.5V$
LED color = Yellow

Is a simple test bed for a larger project I've been slowly working on as time and \$\$ are available.

I agree, LEDs are supposed to require a limit resistor. As I am trying to pump the max mA through these for max brightness, I am experimenting around with various ways to achieve that goal. I have burned out many in this attempt. No matter, they are fairly cheap.

I have a set on the breadboard currently that have been running for ~24hrs, with no problems. Not even the slightest hint of dimming yet.

That is why I am curious about the lack of current draw as tested with my cheap multimeter.

Schematics drawn with Eagle 5.4

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4. ### StayatHomeElectronics Distinguished Member

Sep 25, 2008
958
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Your multimeter is reading something closer of the average current draw rather than the peaks. It is not fast enough to catch the 0.1 ms on time. If you increase the duty cycle of the 555 output you will also see an increase in current on your meter...

You also definitely need the current limiting resistors. The data sheet gives absolute maximum ratings for pulsed forward current (with waveform equivalent to yours).

(7.5 V (supply) - 2.6 V (Vf)) / 100 mA (max current) = 49 ohms

It is also best to run the current lower than the absolute maximum rating for longevity of the parts.

5. ### charlie_r Thread Starter Member

Jul 12, 2007
16
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What is messing with me is that the $I_t$ is so low with all three LEDs lit. I expected a lot higher even with it being an average over time.

I'm not worried about the 555 current, that is why I am using the 2N5088 as a current amplifier, limiting the amount of output current from the 555.

I thought that the way the driving current worked is that as the pulses get shorter, the mA could go higher because the die would have time to cool. I assume it would be a log scale as opposed to linear, with a second assumption that the POV apparent brightness would basically remain the same. Maybe I need to rethink?

In spite of appearances, math as applied to electronics is not my strong suite, re: (7.5 V (supply) - 2.6 V (Vf)) / 100 mA (max current) = 49 ohms, which I do understand. If these LEDs that I have running now show signs of failure after a few days, I will add in the limit resistors, being 51Ω 1/8W.

6. ### Audioguru New Member

Dec 20, 2007
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896
Please post the schematic as an image that we can see without using special Eagle software.

The max allowed current in your LEDs is only 50mA.
The max allowed current in the 2N5088 transistor is only 100mA.
Why blow them up without current-limiting??

7. ### charlie_r Thread Starter Member

Jul 12, 2007
16
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Ok, here is a .png of the circuit.

As the circuit name implies, this is a test for how far I can push the components.

I'm not sure of why the 2N5088 hasn't let the magic smoke out yet. From the datasheet on it, it should have many times over. According to the LED datasheet, 100mA pulsed at 1/10 duty, max .1ms is the absolute max.

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8. ### StayatHomeElectronics Distinguished Member

Sep 25, 2008
958
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The DMM does not give good results from short pulses. You can not trust the readings as the real and average current are most certainly different.

It is true that the shorter pulse allows you to use a higher current. The data sheet shows a DC forward current max of 30 mA and a peak pulsed forward current max of 100 mA.

The math is simply V = I * R, voltage equals current multiplied by resistance. Equivalent equations are I = V / R and R = V / I.

The pulsing of the current has probably saved the 2N5088, for now. However, a different part should be chosen if you want to continue with these high currents in the LEDs.

9. ### SgtWookie Expert

Jul 17, 2007
22,189
1,741
What are you using for a power supply?

A rough calculation shows that you should be getting about 15mA base current.

A 2N5088 isn't a very good choice for this test. A 2N2222 would be much better.

Your LEDs will likely have a short life.

10. ### charlie_r Thread Starter Member

Jul 12, 2007
16
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Power supply is an adjustable voltage wall wart, with smoothing caps, Max output 1500mA.

The 2N5088 is just what I had in my junk box, I will be using UDN2803 Darlington arrays on the final circuit, with limit resistors on the LEDs. For now, the LEDs seem to be handling the extra mA, but I have noted a slight color change between with and without limiters. So yes, the lifespan of the LEDs are obviously being shortened, and time will tell by how much.

Big picture here is that I am trying to relearn some things from high school (long long time ago, transistors were still very new), and then progress into more modern components such as microcontrollers and fpga etc.

11. ### SgtWookie Expert

Jul 17, 2007
22,189
1,741
I see. Well, you should know that when the LEDs change color, that's a sign of significant overcurrent, and if operated like that for even a short period of time, their brightness will be permanently reduced.

ULN2803's are designed to be driven by logic level (5v) signals. They have a 2.7k base current limiter per channel. If you wish to use them at higher input levels than logic, add a 7.5k Ohm resistor per channel; that will be adequate for operation from about 6v to 16v.

Note that the maximum practical continuous sink current for the ULN2803's is about 350mA; and at that current you will see a Vce of around 1.1v. At minimal sink currents, Vce will be around 0.6v. Remember to include the Vce in your current limiting resistor calculations.

12. ### charlie_r Thread Starter Member

Jul 12, 2007
16
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Thanks! I'll remember to figure that into my calculations as I finalize my project.

13. ### Wendy Moderator

Mar 24, 2008
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2,654
Hate to say this, but without the resistor the short test you've run have already damaged the LEDs and shortened their lifespan significantly. A lot of people new to LEDs have problem with this, but the resistor is not optional. It can be substituded with a current regulator, but never eliminated. That transistor ain't none too happy neither.

There was a period of time manufacturers experimented with built in current regulators for LEDs, but it was a flop. They burned out way too fast. The reason this was done is because it is such a necessity though.

I have not done the math, but the value shown in post #4 sounds reasonable.

14. ### charlie_r Thread Starter Member

Jul 12, 2007
16
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The LEDs I have been testing this with are leftovers from another project, and I was not planning on using them in the final build. I expected them to burn out. Interesting to note, it is going on 36 hrs as of this writing, and I have had only 1 failure, which was one that I had rejected from use on a different project due to being a bit off color.

At \$0.10 each for the 2N5088, if it blows during test phase, no problem. I would have expected it to have at least gotten hot by now, but it is still only a few degrees above ambient. Hardly noticeable by touch.

To me these are very strange developments in this testing phase, when I would have expected massive failure by this time.

15. ### Wendy Moderator

Mar 24, 2008
20,874
2,654
LEDs can be unpredictable how tough they are. The big point is you want a light that never goes out, which can't be said about an old bulb arrangement.

Used to fustrate me, I used LEDs to test laser current sources (solid state lasers) a an equipment maintenance tech. I'd leave the lab for a few minutes and the production techs would pop the LED off the power supply just to watch the flash. That and see how long it was before I caught on. Least they didn't lie about it. If you look at my picture you can see a lot of my hair is gone, it never happens all at once you know.