same doubt here!

I think it was arbitrary and the solution made sense at Ve = 2 Volts.

I think it may be the value to drive the LED ie RED approx 2v@@anirudhshandilya4032

Transistor has to be in active mode. We will need at least 0.7v to turn on diode. If ve is 2v, that leaves 2.3v across the transistor. Rule of thumb, approximately half supply across vce.

There is a voltage drop across the CC, and with the Vf of White LEDs there is a limit of 3 in series

Your two questions are actually related. We wanted to pick an RB2 value so that the current flowing into the base is very small compared to what flows through RB2 - then RB1 and RB2 can be treated as having approximately the same current flowing through them. Then, since we're assuming the base current is small enough to ignore, we can use the approximation that VE is (VB-0.7) and then use VE/RE to calculate IE. We could have done a more complicated analysis and thevenized the VCC/RB1/RB2 part of the circuit then use that to calculate IB, and use beta to calculate IE. This would give a more accurate result, but is more work.

I think R2 is suppose to be 850 ohms and r1 is 1000

I calculate like this beta=100 , led red =2v and 0.02A ----- Re=3v/0.02A =150ohm , Rb2=3.7v/0.002A (1/10 beta) =1850ohm , Rb1=1.3v/0.0022A (1/10 beta+extra 1/100beta = 0.002A+0.0002)=590ohm .

constant drop i.e. 2.8V

yes use the 2 BJT version

No problem as long as the current is under the LEDs max current

At 3:33, I showed that VB needs to be 2.7V, then at 4:15, showed that the ratio of VRB2/VRB1 is 1.17. VRB2/VRB1 = RB2/RB1 = 1.17, so with a little algebra we can see RB1 = 0.85RB2.

AT 6:58 in the case where the forward voltage of the LED is 3.3V, the voltage at the emitter cannot be 2V, because 3.3V + 2V = 5.3V which is more than VDD

@@ElectronXLab : I'm trying to learn here. You mentioned that led 3.3v will saturate the transistor. I agree. My understanding of saturation is that the transistor won't pass any more current via the collector. The maximum current will be that what is set by Re(100 ohms) resistor, which is 20mA? If I am wrong would you clarify further? Thanks.

@@shazmiah let's approach this a different way. In the last example, I said that the LED turned on at 3.3V and that it needed 20mA to reach that voltage. If 20mA goes through the 100 ohm resistor, the voltage across it will be 2V. The power supply is at 5V, so that means after you take off the 2V across the 100 ohm resistor, you only have 3V left. If you have 3V across the LED, then it will not have 20mA through it, so the actual current must be something less than 20mA.

@@ElectronXLab : Thanks for your explanation and quick reply.

Oops, that was a bit of an oversight - thanks for the comment.

I think that you need an active circuit to have a constant current driver; I'm not sure how you'd do it with just an RC circuit. With just RC, the capacitor will help hold a steady voltage for a short period of time...depending on its capacitance and the load.

No. Correct calculation should be: Re = (5V - VLed - Vce) / ILoad. VLed takes different values according to the LED type.

shamrai 4.2v 700mA 3Watt 3Ω that is you LED

If you use resistive voltage divider to supply base current, can not keep the current fix when Vcc changed. But, if use Zener diode, you can keep it fix.

@@futuhcoklu1396 are the changes significant

yesssssssssssssssssssssssssssssssssssss

It won't provide constant current.

@@tianapyre2837 Yes it will. IF used in CC configuration. Read the DATA SHEET

Problem is that you need a big wattage low value resistor and it dissipates a bunch of heat.

Извини. Я не знаю

Get your basics right. *Current flows in the opposite direction of electrons*

@@the1aboveall483 You said it, "CURRENT FLOWS". What is flowing? water? air? heat? How about Current is the flow of ELECTRONS. Unless ELECTRONS are swimming like salmon and going against the "flow" so as to go upstream. And if you use that analogy then current "flow" is water? and electrons are salmon. I'll accept that if you can explain what is "flowing" in electric current if it's not electrons. kzfaq.info/get/bejne/b66cd5dhuLLbomQ.html Here is a basic discription of how a battery works. Weird that it states that the energy in the battery goes from the - pole around and back into the + pole of the battery. The battery is an energy storage device. It has a chemical reaction taking place inside it that wants to push electrons OUT the - pole while the + pole has a hole trying to attract an electron back in to take that one's place. I'll help you out a bit. Electromotive force (volts) exist in the battery and in any circuit hooked up to that battery. Electrons moved from the negative pole to the positive pole until there were no electrons to drive the chemical action in the battery of sufficient quantity to move current through the circuit attached to the battery. So whatever current is, it's not moving in the opposite direction of the electrons flowing in a circuit. This stuff was a whole lot easier in the old days. Everyone could see the filaments of an electron tube glow a pretty bright orange and it made perfect sense that the glow came from electrons moving through the filament heated it up. Gee wonder what heats up light bulbs and makes them glow? I bet it's electrons. Electrons that are being PUSHED by VOLTS NOT PULLED by VOLTS. There in lies the explanation . Electrons passing through a resistance generates heat. Proof that there are actually electrons flowing. In regards to the video above.. current flows up through the transistor, NOT down through it.

@@trongod2000 In short. You're right. Electrons flow in the opposite directions. But scientists in the old days thought otherwise and it's not until lately that we discover current flows in the electron flow's direction. The problem is everybody is used to the old way (AKA CONVENTIONAL CURRENT) that nobody bothers to change. So calculation and explainantions all use the opposite direction of electron flow.

@@An-wd9kk One way to show people how current flow "works" is to show them how current flowing through a conductor can pass energy to do work in something completely detached from the wire. Show them an electro-magnet for instance. Perhaps seeing a solenoid in action will help also. Both these examples however do not show the "direction" or "flow" of the energy within the conductor. At least not usually. It might open their eyes to the idea that there is "direction" if you reverse the leads going to a solenoid and see that the piston portion of the solenoid is pushed in the opposite direction. That alone though does not visually show which way electrons are flowing and even less enlightening on the subject of current flow. Current is a very hard concept to show visually and direction of current flow is even harder. One of the best ways to show the effects of current flow (electrons flow) is to take an uncharged capacitor and measure the voltage on it. Since it not charged there won't be an excess of electrons on either of it's plates. Hook the capacitor up to a source of electrons and then disconnect it and see what happened to the capacitor. Surprisingly, almost nothing. In order for electrons to build up on a plate the opposite plate needs to shed electrons. This adding and shedding electrons is pretty instructive in determining direction of flow.

think again