Transistor Switching for Driving LEDs - Expert Advice Needed

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In summary: LEDs. 3. the transistor is being powered by 3.3v from the 555 timer. 4. input current to the base of the transistor is 0.In summary, you may be able to achieve a brighter LED by adding a transistor in series with the LEDs. However, you may not be able to get the LED to reach its full brightness due to the voltage drop.
  • #1
Dear Experts,

Hope all of you are enjoying the holidays so far.

I am newbie dabbling with circuits to make something for myself.

recently, i made a pulsing signal go into a NPN transistor trying to amplify the output required to drive some LEDs using the transistor as a switch. I know this sounds redundant as i could have used the pulsing signal to act as switch. But somehow, i added the transistor to learn to use it as a switch hoping to avoid voltage drop (Vce = 0 at saturation).

I could not reach close to the source voltage Vcc to drive the LEDs which still works but not as brightly as i liked.

Not sure if its due to the input signal not high enough? Am i right to say that the Vce if zero would give me the Vcc , ie. no voltage loss, needed to drive the LEDs?

If so, should i adjust the input current to the base to reduce it further so that Ib(max) can be reached that would give me the Vce = 0 whereby the Collector is 'shorted' to Emitter closing the circuit? Due to the pulses going to the base I could not really tell what was the highest amount of current that pulsed in. What i measured was roughly 600 microamps.

Oh by the way, the pulsing comes from a 555 timer. I experimented with it and found that increasing the resistance at pins 6-7-8 could up the voltage that drives the LEDs.
But i only have 20K ohms.

Thanks for reading my question.

Peace and love to all!

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  • #2
You only need about 1.5 to 2 volts to light a LED to full brightness. If it is a white one it may need 3.5 volts. Any more any you risk destroying the LED.
In fact, you may know that you have to limit the current in a LED by adding a series resistor.

So, assuming you are using something like this:
LED driver.PNG

you should be able to do this easily. That 270 ohm resistor is a minimum value for a 5 volt supply. For a 12 volt supply you need about 600 ohms to get 20 mA LED current.

If the 270 ohm resistor is not there, the transistor cannot go into saturation without blowing up the LED, because this would mean the LED had more than its rated voltage across it.

Don't forget that the time the LED is not conducting makes the appear LED dimmer.

Also, LEDs vary a lot in brightness. You can get "super-bright" LEDs for only a small extra cost and these give a huge increase in brightness.

A 555 with only 5 volts on it will only give about 4 volt pulses out across your 20 K resistor. This would mean you have only 200 uA base current. (4v / 20K ).
So you would need a transistor with an actual current gain of 100 to get 20 mA of collector current.
So, if you do only have 5 volts on your 555 you will probably need to make this base resistor smaller. Maybe 6.8 K?

If your 555 has 12 volts on it, you would get about 550 uA of base current which means you need a gain in the transstor of about 36. So your 20 K should be OK.
  • #3
Dear vk6kro,

Yes, my circuit is basically the same as you envisioned, except there is no resistor used before the base. Sorry for taking some time to revert as i was taking measurements and finding a spare transistor from old circuit boards. :)

I am using about several parrallel sets of 6 superbright 2v red LEDs in series with each other and with a resistor such that when this lamp is directly connected to a 12V dc supply, the LED is each dropping 1.9v or thereabouts.

I have changed the transistor after it become overheated due to re-soldering. Initially, i used a TL431A NPN and now have replaced this with a 13001 NPN which has a gain ranging from 40 to 80.

However, the voltage drop to the LED lamp is still not able to reach near 12V after the change which only improved very little with a tiny increase of voltage for the lamp , at 11.2v.

measurements as follows:
1. Vce = min 0.9v max 1.1v
2. 555 timer resistance kept at 20K to provide 11.2v for the lamp.
3. transistor base pin is connected to pin 3 of 555 without any resistance in-between.
4. bas current max = 857uA

It appears to me that even without controlling the base current, i am only able to get 11.2v for the lamp which is in series with the collector. I am not able to lower the Vce to zero.

the 555 configuration used is as follows:

1. pin 8 and 7 is where the 20k pot sits.
2. pin 7 and 6 is where a 1k resistor sits.
3. pin 5 and 1 is connected with a 1 uF capacitor.
4. pin 4 and 8 are shorted.
5. pin 3 is connected to the base.
6. pin 1 is connected to pin 2 with a capacitor controlling the flash rate.

For this configuration, i am not sure how to make Vce = 0 such that the lamp gets the Vcc at 12V if not near to this value. Theoretically, this should be possible right?

I don't think the lamp resistance is limiting anything since when directly supplied 12v, this lamp drop about 11.9v from a 12.1v source. So, i think the output from the flashing circuit is not high enough for some reason.

Thanks for reading.

  • #4

It is perfectly normal for there to be a saturation voltage across the transistor. This can be up to 1 volt for small transistors and up to 4 volts for big power transistors carrying a lot of current.

The resistor you have in series with the LEDs will determine the total current and brightness of the LEDs. You didn't say what value this was, but maybe you could reduce it. Monitor the CURRENT through the LEDs. That is more important than the voltage.

You could also try it with just 5 LEDs or try to get 15 volts or so as your supply. Just the LED voltages add up to 11.4 volts leaving 0.7 volts for the resistor and the transistor's saturation voltage. So, the resistor isn't doing much.

Or, you could put the LEDs in two parallel strings of 3 with a series resistor for each string and then being switched by the one transistor.

There really should be a resistor between the 555 and the base of the transistor. Isn't the 555 getting really hot?
  • #5
Hi vk6kro,

I tested with another set of LEDs separately and used the same pulsing circuit. This time, am able to get 1.98v per LED (transistor circuit able to provide 11.9v in total ). This concurs with your advice which is spot on.

I stand corrected, the resistance in the lamp appears to be the cause. I had built the lamp before adding the 555 circuit. I plugged in the lamp to the circuit but the current could not reach high enough to bring Vce to near zero.

I would also adopt your advice about having resistor before the base.

Can i find out is it true that Vce normally is not zero but usually only near zero?

Thanks for your time, vk6kro! :)

Appreciate your help. Happy 2010 in advance.

  • #6
The real problem is that you have too many LEDs in series for the voltage you have available.
With 12 volts available, you need 4 LEDs with 8 volts total across them, a resistor with 3 volts across it and a transistor with 1 volt across it. You calculate the resistor to suit the current you want in the LEDs assuming 3 volts across the resistor.
So, if you want 20 mA in the 4 LEDs, you would do this:
R = 3 volts / 0.02 amps = 150 ohms.
The LEDs will each keep 2 volts across them, almost regardless of the current through them.

You can download data sheets from Internet by looking for them on Google.
They should be free, but some sites want to charge for them.

One good free one is this:

Here is a section of one data sheet for a NPN transistor, a TIP35

You can see where I have marked an arrow in red that the voltage drop can be quite large if currents are large.

Mosfets are much better at this, but you would only be saving half a volt at low currents.

The base resistor will depend on the base current and the supply voltage for the 555. If the 555 has 12 volts on it, it will give 11 volts out. There will be 0.6 volts across the B-E junction of the transistor, so say there is 10.4 volts across the base resistor, roughly.

The gain of the transistor is about 50 so the base current is the collector current divided by the gain, so maybe 1 mA if you have 50 mA of collector current?

Resistor value = 10.4 volts / 0.001 Amps = about 10000 ohms.
  • #7
Hi vk6kro,

I really like your advice. Very clearly explained. Thanks! :)

I tried amplifying the current using a darlington made up of the original transistor and another one (different model) i added. The result confirms to me that voltage for the lamps cannot be increased and it must be due to the 6 red LEDs in series, as you explained.

I suppose no matter how many transistors chained in darlington style, I will never get the voltage i want due to the LEDs in series?

I will re-wire the LEDs as advised. But being the novice that i am, i had glued the entire lamp circuit with hot glue when i made the lamp. Is there an alternative way out? I mean, can i step up the voltage to the lamp from about 11v to 14v using a ferrite toroid? Too small a voltage increase to step up? My resistor in series with the 6 LEDs is 20 ohms and that should be able to handle even 14v of voltage. Initially, i set the resistor to handle such voltage fluctuations as i wanted to use this lamp in a car. Would a ferrite toroid lower the current going into the LEDs further? Currently the LEDs are at most getting 12mA each.

Thanks for reading.

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  • #8
I suppose you could step up the voltage but I think you also want to have the LEDs flashing. That would complicate the problem. I'll have a think about it.

Do you know you can melt hot glue with a heat gun? This is like a handheld hair dryer but more powerful.

If the leads of the LEDs are still accessible, you can rewire the LEDs without movng them.

If not, you can melt the hot glue with a soldering iron. It will be messy and smelly, but a cheap way of doing it.
  • #9
The LEDs are diodes you will never get a large voltage across them they will burn out due to extreme high current before you get a high voltage.

Related to Transistor Switching for Driving LEDs - Expert Advice Needed

1. What is a transistor switch and how does it work?

A transistor switch is an electronic component that can turn an electrical signal on or off. It consists of three layers of semiconductor material, each with a different level of conductivity. By applying a small signal to the base layer, the transistor can control the flow of current between the other two layers, acting as a switch.

2. What is the purpose of using a transistor switch for driving LEDs?

A transistor switch is commonly used to drive LEDs because it allows for precise control of the current flowing through the LED. This is important because LEDs are sensitive to overcurrent, which can cause them to burn out. Using a transistor switch ensures that the LED is only receiving the appropriate amount of current.

3. What type of transistor is best for driving LEDs?

The best type of transistor for driving LEDs is a MOSFET (metal-oxide-semiconductor field-effect transistor). MOSFETs have a low on-resistance, which means they can handle higher currents without overheating. They also have a high breakdown voltage, making them suitable for driving multiple LEDs in series.

4. How do I calculate the appropriate resistor value for a transistor switch driving LEDs?

The appropriate resistor value can be calculated using Ohm's law: R = (Vsource - VLED) / ILED. Vsource is the supply voltage, VLED is the voltage drop across the LED, and ILED is the desired LED current. For example, if you have a 9V supply, a 2V LED, and want 20mA of current, the formula would be (9V - 2V) / 0.02A = 350 ohms. It's also important to consider the power rating of the resistor to ensure it can handle the current flowing through it.

5. Are there any precautions to take when using a transistor switch for driving LEDs?

Yes, there are a few precautions to keep in mind when using a transistor switch for driving LEDs. First, make sure to use a transistor with a high enough voltage rating to handle the LED voltage. It's also important to properly heat sink the transistor to prevent it from overheating. Additionally, you should use a current-limiting resistor to protect the LED from overcurrent. Finally, be sure to double-check your circuit and connections before powering it on to avoid any potential hazards.

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