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Newbie Question of Resistance

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JEBster

New Member
Greetings!

I put together the circuit from the attached schematic, and it works fine. The circuit uses a 9V source and the resistor called for is 4.7K.

What I can't figure out is how I would know what resistor value to use if I were to build this circuit on my own, or change the voltage or number of LEDs in use. Bear with me as I walk thru my understanding, and I hope you'll let me know where I'm going wrong. Sure it wont take long. :D

I built a circuit with just the 5 LEDs on 9V, and measured .006A between the last cathode and the negative terminal. I've read that LEDs (red) typically draw .02A, so first wondering if my measurement is incorrect.

From the a datasheet I read that the transistor is rated at .15A.

On the attached circuit, I measured .7V between the resistor and the base of the transistor which seems correct (as I've read a transistor requires .6V to switch on.) So, I figured the voltage drop from the resistor is 8.3V.

So, to determine the ohm rating required to get the .7 volts to the base, I figured (voltage souce - voltage drop) / amps would equal the ohm value of the resistor. (9-8.3)/.156 = 4.48.

4.48 makes no sense, especially considering the difference between 4.48 and the 4.7K in the schematic. But for kicks, I dropped the resistor value to 1K, then 470 and the circuit still worked. But down to 100ohms, and the transistor is no more.

I have a feeling I'm in way over my head on this, but hoped there might be a relatively simple idea I'm not grasping. I greatly appreciate your attempts to enlighten me! Thank you.
 

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giftiger_wunsch

New Member
I've read that LEDs (red) typically draw .02A, so first wondering if my measurement is incorrect.

20mA is a typical max current rating for a standard LED, but that doesn't mean that's what you'll get through them. You should always have a series resistor when using LEDs to ensure that you don't blow them out.

As for the transistor, I'm new to electronics myself but my understanding is that the current through the transistor's base activates it, not the voltage. FETs are activated by voltage, while BJTs ("standard" transistors) are activated by current (someone feel free to correct me if I'm wrong).

So if the recommended resistor value is 4.7k and the voltage supply is 9V, the base current required to turn the transistor on should be 9/4700 ≈ 2mA. Decreasing the resistor value to 100Ω would cause a current of ~900mA to pass through the base, which is well above the rating of the transistor that you mentioned.

As for the LEDs, the only thing limiting the current through them is the resistance between the collector and emitter of the transistor; as bountyhunter said, it's not a good design.
 
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Janis1279

Member
I built a circuit with just the 5 LEDs on 9V, and measured .006A between the last cathode and the negative terminal. I've read that LEDs (red) typically draw .02A, so first wondering if my measurement is incorrect.
Hi
When you built the similar schem with the LEDs, you need remember about 2 (2,2 )volts LED forward voltage, too. For the 5 LEDs summary forward voltage is ~ 10 volts.
 
Hi, i would not bother with a transistor at all because if your just using 5 LED`s in series then thats only a current draw of 0.02A, i would replace the transistor with a resistor to limit the current instead, and say you run it on 10V it will only draw 0.2W, the small resistors are usually 1/4W, - thats why i dont think you need the transistor.
 

giftiger_wunsch

New Member
I think the circuit diagram is designed to just be a very simple demonstration of how transistors function. Removing the transistor would defeat the purpose of the demonstration.
 

JEBster

New Member
>>So if the recommended resistor value is 4.7k and the voltage supply is 9V, the base current required to turn the transistor on should be 9/4700 ≈ 2mA. Decreasing the resistor value to 100Ω would cause a current of ~900mA to pass through the base, which is well above the rating of the transistor that you mentioned.

:D Yep. That makes sense. Thanks very much, giftiger_wunsch!
 

RCinFLA

Well-Known Member
I assume you are using red or green LED's which are about 2.1 vdc each.

White or blue LED's have a larger voltage drop, 2.8-3.8 vdc

Even with red/green you are summing up to over 10 vdc drop so you don't have enough voltage from a 9v battery. A brand new alkaline might get you higher current but it won't last long before dropping in voltage.

Assuming red or green LED's, get rid of one so you only have four in the stack.

Put a 27 ohm resistor in the emitter lead to ground (negative terminal of battery). This resistor will set the LED current.

Drop the base resistor value to about 1.8K to 2.2K ohms. In switch mode, where you want transistor to be fully on, you should assume base current needs 20% of collector current. This is called 'forced beta' and the 20% base current assumes the transistor at saturation (fully on, lowest collector voltage) has only a D.C. beta of 5. The beta drops because the collector to emitter voltage is very low.

You should have about 0.55 vdc across emitter resistor (depending on battery voltage), 1.1 vdc at base of transistor.
 

giftiger_wunsch

New Member
If you want to keep the 5 LEDs, you could group them into 2 and 3 in series and put the groups in parallel, but leaving the transistors in series with all of them. That way you'd lose current, which should be in excess anyway, but have high enough voltage to drive both groups of LEDs.
 

colin mac

New Member
So, to determine the ohm rating required to get the .7 volts to the base, I figured (voltage souce - voltage drop) / amps would equal the ohm value of the resistor. (9-8.3)/.156 = 4.48.

4.48 makes no sense, especially considering the difference between 4.48 and the 4.7K in the schematic. But for kicks, I dropped the resistor value to 1K, then 470 and the circuit still worked. But down to 100ohms, and the transistor is no more.
The LEDs should have a resistor in series with them. That way you can calculate the collector current. Then you know what base current is needed
to drive the transistor into saturation. But the idea of a transistor switch is that it switches on and off. A transistor isn't needed if the LEDs are always
on.
 

giftiger_wunsch

New Member
The LEDs should have a resistor in series with them.

Note that you're the fourth person to mention that now colin ;) hopefully he's got the message now :p


But the idea of a transistor switch is that it switches on and off. A transistor isn't needed if the LEDs are always
on.

As I also mentioned before, I've seen similar example circuits for beginners, and the purpose of the circuit he is using is simply to demonstrate the effect of transistors. Though perhaps the functionality would be better illustrated if a switch was added in series with R1 and the transistor's base in his schematic.
 
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giftiger_wunsch

New Member
Try this schematic. I quickly threw it together so I apologise if there's a mistake in it, I'm sure someone will point that out though ;)
 

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colin mac

New Member
Note that you're the fourth person to mention that now colin ;) hopefully he's got the message now :p




As I also mentioned before, I've seen similar example circuits for beginners, and the purpose of the circuit he is using is simply to demonstrate the effect of transistors. Though perhaps the functionality would be better illustrated if a switch was added in series with R1 and the transistor's base in his schematic.
The OP decides when he gets the message. So please keep your comments to yourself. Thanks.
 
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colin mac

New Member
This is an open forum, I will deliver my comments to all those who may benefit from them. That sort of remark is uncalled for.
As will I. Which I did. And you're trying to dictate what and when I comment, and I'm asking you not to.

Btw, it isn't harmful when someone repeats a point that may already have been made.
It just supports the point.
 
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giftiger_wunsch

New Member
Hope the schematic proves useful JEBster. It should provide a better demonstration of how a transistor can act as an electronically-controlled switch. You may wish to replace R2 with a resistor on each branch of the LEDs though, as otherwise damage to D1 may cause D4 and D5 to be damaged by too much current, for example.
 
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audioguru

Well-Known Member
Most Helpful Member
Drop the base resistor value to about 1.8K to 2.2K ohms. In switch mode, where you want transistor to be fully on, you should assume base current needs 20% of collector current. This is called 'forced beta' and the 20% base current assumes the transistor at saturation (fully on, lowest collector voltage) has only a D.C. beta of 5.
Not for a collector current of only 20mA.
The "forced beta" for a 2N3904 little transistor is 10 and the forced beta for a BC547 little transistor is 20 on their datasheets.

A 2N3055 power transistor has a forced beta of only 3 on its datasheet (10A collector current) and it still saturates poorly.
 
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JEBster

New Member
Wow. This is great! I apologize for the poor example, but I'm very glad I put it out there. I learned a great deal... well, I'm sure I will after some more quality time with the breadboard! :D

Thank you all!
 

JEBster

New Member
Rebuilt my Fisher Price "My First Circuit" as per RCinFLA suggestions (no offense giftiger... just afraid parallel anything might make my head asplode). New schematic is attached.

Didn't have exact components, but now have R1 @ 2.2K and R2 @ 33. And 4 red LEDs. :D

I'm measuring .9V across R2 and 1.7V @ base, so I believe I'm on reasonably close to RCinFLA's specs. But best I can figure, there is no minimum current that needs to be applied to the base to close the transistor collector to emitter (I used a 100K R1 providing .07mA to base and still the LEDs light.) Is this correct? I'm afraid I don't understand RCinFLA's post.

And here's another thing thats got me confused: the voltage measured at the base with R1 @ 2.2K is 1.7V. How is it that the resistor is dropping 7.3V? If this is a result of a two resistor voltage drop on the circuit, the math doesn't work out: drop across R1=R1/(R1+R2)*V=2200/2233*9=8.9. There's still 1.6V unaccounted for..?
 

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Nigel Goodwin

Super Moderator
Most Helpful Member
Rebuilt my Fisher Price "My First Circuit" as per RCinFLA suggestions (no offense giftiger... just afraid parallel anything might make my head asplode). New schematic is attached.

R2 is in the wrong place, it needs to be in the collector, the emitter should go directly to the battery.
 
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