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Take a look at this schematics and tell me what happens

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Ezugudor

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Someone should pls help me with this schematics
 

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It appears the transistor is conducting and the voltages appear to be correct. Is this a school test question?
 
Assuming thaty is an LED?

the 100k is way to big and IF that is an LED then it will never light up.
Is that an LED or supposed to be?
if so the resistors are all wrong
 
Maybe he does not know what does it means...
 
Please tell us what this part of the circuit is supposed to do.....more info and stuff needed.

It just looks like.....what's it's purpose? Unless of course the Diode is a supposed to be a LED....which gets turned on when the transistor switches on and grounds the +12V through the "diode" and the 2.2K resistor. 2.2K resistor is still waaay to high to light up any LED from a 12V supply.

I don't get it.
 
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Ib= (12V-0.6V)/100K = 0.114mA
Ic= (10.6-.2)/2.2K = 4.72 mA
4.7ma will easily light an LED
current gain = Ic/Ib = 4.72mA/0.114mA = 41.46 easily within the HFE of common transistors.

Ken
 
simulation says no way on the 4.28 ma

Using TINA it says 4.28 not enough for LED to light but TINA may be wrong?
I inserted a switch across the transistor to verify saturation. With switch closed the current shows very little change.
 

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It depends on the transistor.

The transistor will probably not saturate with a forced beta of 40 so the voltage drop will be higher causing a lower current to flow.

The LED will light but both the base and series resistor should be lower, I'd say 22k and 1k respectively.
 
Ib= (12V-0.6V)/100K = 0.114mA
Ic= (10.6-.2)/2.2K = 4.72 mA
4.7ma will easily light an LED
current gain = Ic/Ib = 4.72mA/0.114mA = 41.46 easily within the HFE of common transistors.

Ken

Sorry Ken....4.7ma through an LED will not light it.

No matter what color the LED is.

15 to 30ma does the job.
 
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Sorry Ken....4.7ma through an LED will not light it.

No matter what color the LED is.

15 to 30ma does the job.


Sorry but you're wrong, 4.7 mA will light an LED, although not very brightly, 30mA will also overheat some LEDs.
 
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Well...on a real bench...2N3904, 100K, 2.2K, 12V, red LED

LED lights and Vce=0.147V :)

Ken
 
if i have an led that needs to be lit from 12v i would use a 2.2k ohm resistor. it is also what velleman suggests to use on the back of their 80 led packet for 12v and use a 470 ohm if its 5v.

The led will light up. that is all that will happen with that circuit. you will probaly find though that the 12v input will be switched on and off as a signal

edit: KMoffet, you just beat me too it:D
 
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Sorry but you're wrong, 4.7 mA will light an LED, although not very brightly, 30mA will also overheat some LEDs.

This is funny. Great forum BTW.

I like my LED's to be driven properly. Either on or off. Not in "low power mode". LED's are supposed to indicate the status of a system. And warn you with bright flashing lights.

The LED reports the status is either good or bad guys....

Common

Till next time. You folks are great.
 
I like my LED's to be driven properly. Either on or off. Not in "low power mode".
So, at what current is "on" for an LED. I frequently use, and see, status LEDs running at currents >10mA. I've used super bright LEDs for status indicators at 1ma on battery devices. Definately shows on or off status. Why waste power. ;)

ken
 
So, at what current is "on" for an LED. I frequently use, and see, status LEDs running at currents >10mA. I've used super bright LEDs for status indicators at 1ma on battery devices. Definately shows on or off status. Why waste power. ;)

ken

I stand under correction Ken. You are way wiser than me.

1ma. With a LED. And it is bright.

I said earlier I was a kid @47 years in this industry. Teach me please.

Best regards
 
1ma. With a LED. And it is bright.
No, it's not "bright". But it definitely shows an on/off status. I've also used low brightness as a status..."the LED works!", "these is power!"...high brightness then indicates a change in status.

Something like the attached.

Ken
 

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Interesting, never saw a transistor with a resistor on the collector/emitter.
I left the switch in the circuit and current draw is almost identical.
 
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