Can someone please show me this circuit?

Thread Starter

Dleberfinger

Joined Dec 18, 2011
49
If anyone feels up to it, I would like to see a 555 timer circuit to compare with mine. Unfortunately, I seem to have messed something up, seeing as I blew an IC and a transistor this morning.

I need a time of 5 μs, so the timing capacitor is .01 μF and the resistor is 460 ohms. The timer is used for an array of LEDs. The LEDs draw 140mA, and eventually I need to get them to 1000mA. I had no resistor between the NPN transistor and the output of the 555, so I'm assuming that's why my transistor literally blew up and my 555 fried.

Please don't give me a link to the 555 section or LED section, I've read them enough to know that reading them again isn't going to fix it. I think I just need to see the circuit made properly before I ruin any other components.

Thank you in advance to anyone willing to help.
 

nerdegutta

Joined Dec 15, 2009
2,676
Some would say that blowing IC, transistors and capacitors is part of the learning process.:)

How about posting your circuit first? Hard to say without seeing it.
 

Thread Starter

Dleberfinger

Joined Dec 18, 2011
49
Some would say that blowing IC, transistors and capacitors is part of the learning process.:)

How about posting your circuit first? Hard to say without seeing it.
Working on getting it all down. I really just want to see one that done right, I would enjoy finding them problems between mine and a proper one, but I'll post it here once I finish it in Eagle.
 

DerStrom8

Joined Feb 20, 2011
2,390
I think I've got it right now. Here's the circuit I've been working with.



The 3 LEDs are these.

http://www.mouser.com/Search/ProductDetail.aspx?R=LD_CN5M-1R1S-35-1-Zvirtualkey62510000virtualkey720-LDCN5M1R1S351Z


Whoops. Ignore the 10k values on the LED resistors. I forgot to change them. They are all 91 ohm resistors.
Yeah, that circuit is definitely wrong. For one thing, the LEDs are backwards. Also, that looks like a monostable (one-shot) timer, but the resistor and capacitor values are extremely low. It would only be on for a few milliseconds (sorry, getting late and I'm too tired to do the math right now :p ). I also don't understand what the point of the LC oscillator on pin 2 is all about. Where'd you get this circuit?

This is the basic monostable timer circuit:



I hope this helps!
Regards
 
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Thread Starter

Dleberfinger

Joined Dec 18, 2011
49
Yeah, that circuit is definitely wrong. For one thing, the LEDs are backwards. Also, that looks like a monostable (one-shot) timer, but the resistor and capacitor values are extremely low. It would only be on for a few milliseconds (sorry, getting late and I'm too tired to do the math right now :p ). I also don't understand what the point of the LC oscillator on pin 2 is all about. Where'd you get this circuit?

This is the basic monostable timer circuit:



I hope this helps!
Regards
The circuit is from my understanding of a 555 monostable circuit. The time is 5 μs like I said in the original post, if you read it.. I need that short pulse for my application. As for the LC Oscillator, I have no idea what that even is. The symbol to the left of the capacitor is a resistor, not a coil. The capacitor is used to make sure that the 555 timer doesn't activate for longer than the pulse determined by the timing capacitor and resistor. The 10k resistor is to make it so the circuit can be pulsed without manually shorting the capacitor after every run.

According to AllAboutCircuits, diodes point in the opposition of electron flow. Since electrons flow from negative to positive, with ground the positive, they are facing the correct way.
 
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Thread Starter

Dleberfinger

Joined Dec 18, 2011
49
I also noticed the LEDs.

How did you get to 91 Ohms?
If my LEDs each draw 140mA, and my source is 12v, I need approximately 85 Ohms. Right? And I just happen to have a few 91 ohm resistors to use.

12v = Resistance x .140A
Resistance = 12v/.140A
Resistance = 85.7 Ohms

Unless I screwed something up..

Ah yes, you're right. That was my mistake. I think I need to go to bed now :D:p
Haha.. Thanks for the input though!
 

nerdegutta

Joined Dec 15, 2009
2,676
I don't think the value of the LED resistor is the reason for the IC and transistor to blow, but:

R = V / I
R = (12-3.3) / 0.14
R = 8.7 / 0.14
R = 62Ohm

Have you breadboarded your circuit?
 

Thread Starter

Dleberfinger

Joined Dec 18, 2011
49
I don't think the value of the LED resistor is the reason for the IC and transistor to blow, but:

R = V / I
R = (12-3.3) / 0.14
R = 8.7 / 0.14
R = 62Ohm

Have you breadboarded your circuit?
Yeah, that's how I blew the components. Oh, I forgot about the voltage drop. I don't have another transistor, so I can't try the circuit again. Is there anything else you can see that could cause the problem?
 

DerStrom8

Joined Feb 20, 2011
2,390
If my LEDs each draw 140mA, and my source is 12v, I need approximately 85 Ohms. Right? And I just happen to have a few 91 ohm resistors to use.

12v = Resistance x .140A
Resistance = 12v/.140A
Resistance = 85.7 Ohms

Unless I screwed something up..
You didn't take into account the voltage drop of the LED. According to the site, the voltage drop is 3.3 volts. Subtract that first from the source voltage, and you get 8.7 volts. This is what you divide by .140 amps. You get about 62 ohms. That even happens to be a standard resistor value.

Anyway, the 555 can source up to 200mA. My guess is that with the 3 LEDs you have, you're trying to draw too much current from the chip, and that's why it keeps frying. Like I said though, I'm tired, so my brain isn't working very well at the moment. I'm sure somebody will let me know if I'm wrong :D

EDIT: Darn it. Nerdy beat me to it. I guess I took to long to write this :p

EDIT EDIT: Just realized you have a transistor. I definitely need to go to bed :D Cheers!
 

Thread Starter

Dleberfinger

Joined Dec 18, 2011
49
I just got all of my parts today, I plan to put it together on breadboard. Does anyone have any suggestions to prevent it from blowing up? Or did I probably just short something on accident?
 

Adjuster

Joined Dec 26, 2010
2,148
I guess that depends on the transistor. If it's a BC547 or similar, I think that it wouldn't hurt to put a 1K on the base.
Although a base series resistor is necessary for the usual common-emitter driver, ince this is an emitter follower driver, a large base series resistor will be distinctly unhelpful. (The emitter follower is a bad choice, in my opinion, which I will explain later),

Let's get our facts straight here. The OP is trying for a total output current of 3*140mA = 420mA. BC547 is only a 100mA device, so it is doomed from the outset ! http://www.fairchildsemi.com/ds/BC/BC547.pdf

If we hope that the transistor will turn on fully, the base current requirement would normally be taken to be Ie/11* =38mA.

*A saturated gain of 10 to the collector is 11 at the emitter. :p

In practice, the emitter follower cannot saturate, because the 555 output cannot reach the 12V rail. let alone exceed it by a Vbe as would be required for full saturation. The transistor will therefore drop more voltage, dissipate more power, and get hotter.

Let's assume we have a decently rated transistor which can manage the current. Connected as an emitter follower and not saturated, it might have a current gain of 50, so the base current might be 420mA/51 = 8.2mA. If a 1k series resistor passed this current, this would drop 8.2V.

Since the emitter follower has not quite unity voltage gain, the output voltage would fall by quite a large amount: not quite 8V in practice, as the output current would be lowered. The transistor would drop some volts, and most likely get very hot. Note that the big base resistor is making things much worse.

The right choice in my opinion would be an NPN common-emitter driver using a transistor rated for at least half an amp collector current. The LEDs and their resistors would be put in series with the collector, and the emitter would be returned straight to negative.

A base series resistor is then required: it should be chosen to give about 420mA/10 = 42mA of base current, maybe 240Ω.
 

Thread Starter

Dleberfinger

Joined Dec 18, 2011
49
Although a base series resistor is necessary for the usual common-emitter driver, ince this is an emitter follower driver, a large base series resistor will be distinctly unhelpful. (The emitter follower is a bad choice, in my opinion, which I will explain later),

Let's get our facts straight here. The OP is trying for a total output current of 3*140mA = 420mA. BC547 is only a 100mA device, so it is doomed from the outset ! http://www.fairchildsemi.com/ds/BC/BC547.pdf

If we hope that the transistor will turn on fully, the base current requirement would normally be taken to be Ie/11* =38mA.

*A saturated gain of 10 to the collector is 11 at the emitter. :p

In practice, the emitter follower cannot saturate, because the 555 output cannot reach the 12V rail. let alone exceed it by a Vbe as would be required for full saturation. The transistor will therefore drop more voltage, dissipate more power, and get hotter.

Let's assume we have a decently rated transistor which can manage the current. Connected as an emitter follower and not saturated, it might have a current gain of 50, so the base current might be 420mA/51 = 8.2mA. If a 1k series resistor passed this current, this would drop 8.2V.

Since the emitter follower has not quite unity voltage gain, the output voltage would fall by quite a large amount: not quite 8V in practice, as the output current would be lowered. The transistor would drop some volts, and most likely get very hot. Note that the big base resistor is making things much worse.

The right choice in my opinion would be an NPN common-emitter driver using a transistor rated for at least half an amp collector current. The LEDs and their resistors would be put in series with the collector, and the emitter would be returned straight to negative.

A base series resistor is then required: it should be chosen to give about 420mA/10 = 42mA of base current, maybe 240Ω.
Ok, so I just need a larger transistor that can handle the current and a 240 ohm resistor. The transistor I plan to actually use is rated for 8A, so I think I'm covered.

I actually plan to power more than 3 LEDs if I get this to work. How could I go about modifying this for perhaps 10-20 LEDs, overdriven to 1000mA?
 

Adjuster

Joined Dec 26, 2010
2,148
Before going any further, be sure of the difference between the common-emitter and common-collector connections? This is not just a question of a resistor value. I recommend that you get the circuit working properly on a small scale first, rather than making it more complicated.

If you do plan to use a lot of diodes, it may be better to have them in strings of two or (maybe) three in series, with suitably adjusted series resistors. The maximum number of LEDs possible per series string would depend on their forward voltage, which may increase somewhat at high drive current.

Using a simple bipolar transistor as a driver, you may only reduce the base drive resistor down to the limit where the base current was 200mA.
The total current guaranteed available to drive LEDs would then be ten times this figure, or about 2A. An optimist might risk a very little more, but not much as the transistor may come out of saturation and overheat.

If you really want to drive a lot of LEDs to a high current, a simple bipolar will not really cut the mustard. A suitably rated logic-level MOSFET would be a more suitable as a driver device: a bipolar Darlington-type power transistor would be second-best.
 

strantor

Joined Oct 3, 2010
5,333
According to AllAboutCircuits, diodes point in the opposition of electron flow. Since electrons flow from negative to positive, with ground the positive, they are facing the correct way.
and if your LEDs were in opposition to current flow, they would be dark emitting diodes. LEDs "go with the flow" to produce light. This is even more confusing because the symbol for them came before wide acceptance of electron flow; it is representative of conventional flow. You have drawn them backwards

EDIT sorry didn't realize there was a page 2 here
 
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