Ya was going to say. guy was flying, kept having to hit rewind and pause a bunch.

Same feeling here...

But I still watch it in 2x speed for time-saving purpose. Although I agree that it is quite fast for some tutorial video.

Someone who doenst know nothing, knows everything right?

Same here way too fast to allow me time to understand the info

You're welcome

dont know if anyone gives a damn but if you are stoned like me during the covid times then you can stream all the new movies on instaflixxer. I've been binge watching with my brother for the last months :)

@Eric Ty definitely, have been using InstaFlixxer for since november myself =)

er spricht englisch

nix deutsch

@@ademshabani626 er is deutscher du vogel hörst du sogar an seinem Terminator akzent

Hahahhahha

you can rewatch and pause it as much as you want

@@alfineranai6952 It doesn't help. He basically didn't explain anything, just rushes over the calculations like they are easy. I have a degree in EE and I barely understood what he was doing

Then you firstly need to learn the more basic things.

Same. I learned the basics, but there are a lot of things left unexplained. It's not a tutorial for beginners.

You go too fast through your videos. No time to consider what you did or why it works. I’ve stopped watching your videos to learn and now just watch to review.

@@RobertLopez66 Yeah, this is for someone who already knows this and just wants to brush up, or something. You cannot learn from this video. No way.

@@pumbo_nv yes brother

Look at the video at 2:03 and check the datasheet for the conditions that are used for measuring the collector-emitter saturation voltage. Those are Ic=500mA and Ib=50mA. You can ignore the "DC current gain" section because that is only specified for Vce of 2V or greater, i.e, linear mode. When the transistor is operating as a switch, the crucial design consideration is normally to ensure that Vce is as small as possible when the transistor is switched on. This is called "saturation mode" and Vce(sat) for the BC637 family is 0.5V at Ic=500mA and Ib=50mA. So you want to set the base current around one-tenth of the maximum expected collected current. That will ensure that the transistor stays in saturation when switched on, otherwise if the collector voltage were to rise, it would run the risk of exceeding the maximum transistor dissipation of 625mW (Ic x Vce). In the video, he should have used Ib = Ic/10 for a switching application. Ic is given as 310mA, so Ib = 31mA. Then Rb = (Vcc - Vbe)/Ib = (3.1 - 1.0)V / 0.031A = 68 Ω, although this doesn't have to be exact. The dissipation in the base resistor would be less than 100mW (less than 3.1V x 31mA), so any 125mW or higher dissipation resistor would be fine.

What you mean " Ic is given as 310mA"? Is max current on LED or max current in transistor collector? When he test LED with source generator i didnt see any value of 310 mA. He had max 253 mA in LED.

@@anaromana8183 @RexxSchneider I'm also confused about the source of 310mA for the Ic...

It was in the saturation stage. Of course you can use a slightly bigger current to make sure that you reach saturation but simply multiplying with 5 or 10 it not the best idea. My transistor would not have enjoyed a base current of 2A instead of 400mA.

For IB of 450mA and a worst beta of 25, the IC would be 11.25A which I have never seen BJT rated for such current..!!! It seems I am stupid, but that is what I know. I am not professional to make argument here though.

@@greatscottlab The point is that manufacturers specify a base current of one-tenth of the collector current when quoting the maximum Vce(sat) for their devices. If you drive the base with less than that, you have no way of being certain that the voltage across the transistor is below a specified value. If you're passing significant current through the transistor, you can't calculate the maximum power dissipated and therefore have to use guesswork about heatsinking, etc. If the load current is 4A, you design for a base current of 400mA, or you use a power MOSFET instead, but that's a different story.

@@majdinj When the transistor is in saturation, the collector current is determined by the load, not by β times the base current. The concept of β is really only applicable to a transistor operating in its linear mode, i.e. as an amplifier, not as a switch.

Mihajlo Medjedovic Wah beta wah

Dmitry Ilyin Wow! Great achievement

not so great like your comment

Dmitry Ilyin :D I was kidding

me too ;)

Soon

When using a transistor as a switch, the rule-of-thumb is to make the base current one-tenth of the collector current when turned on. That will ensure that almost any transistor is in saturation. Any further increase in the base current will not increase the collector current, so there is no point in using more base current than that. Use a multimeter to measure the current taken by your 5V LED when 5V is applied to it. Let's say it's 30mA. So you need 3mA base current. If you're driving the base from a 5V microcontroller via a resistor, then that resistor has to be around 5V/3mA ≈ 1.7K. You can use a 1.5K or 2.2K base resistor. It's not critical. If you're using a 3.3V microcontroller, you'll need around 1K to 1.2K for your resistor. Hopefully you can see how to work out the resistor you need now.

@@RexxSchneider I see transistor datasheets that show DC Current Gain of 100-300, and down to ~30 depending on Ic. There is a big difference in 10x versus 100x. I'm confused about what gain to use in my calculations. 🤔 (EDIT: I understand now 😏)

@@abandonedcranium6592 There's an important difference between a transistor operating in its linear region with a current gain of 100-300 and the same transistor operating in saturation where you want to make sure that Vce is as small as possible. To achieve that, it is necessary to overdrive the base current giving an effective current gain much less than what is shown for linear operation. Take an example of the 2N222A, a very common switching transistor. The DC current gain at Ic=150mA and Vce=10V (i.e. linear operation) is specified as 100-300. However, the specification for Vce(sat) - the voltage between collector and emitter in saturation, i.e. as a switch - is given as 0.3V max at Ic=150 mA and Ib=15mA. That is an effective current gain of 10 and is the most common ratio of Ic/Ib in datasheets when examining Vce(sat). That's why I give the rule-of-thumb of setting the base current at one-tenth of the required collector current in switching applications. The collector current will be limited by the collector load, not by the base current and current gain. You may find some very high gain transistors, e.g. BC547 family where the maximum Vce(sat) is specified in the datasheet with the base current at one-twentieth of the collector current. But note that the Vce(sat) is 0.6V max for the BC547 family at Ic=100mA, Ib=5mA, which is twice the Vce(sat) of the 2N2222A at similar collector currents, even if the BC547 would be using half the base current.

I'm wondering the same. Probably it's within the specs for the LED but it would be nice to get an answer ^^

Yeah it's probably the LED and the Light bulb's operating current.

You're right about the error. Normal circuit design for switching transistors is to set the base current at one-tenth of the maximum expected collector current. Note that the datasheet for the BC637 family specifies the maximum Vce(sat) at Ic=500mA and Ib=50mA. I would certainly design for Ib = 31mA when Ic=310mA in this application. The only quibble I have with your calculation is that Figure 4 shows _typical_ Vbe(sat), rather than maximum, which would be worst case. The only other place that Vbe(on) is given is in the "On characteristics" section, where a maximum value of 1.0V is given. Plugging that in implies Rb = (3.1 - 1.0)V / 0.031A = 68 Ω. In most cases, the circuit will work with higher values for Rb, but if you have to design for worst case - as you would in production, for example - then 68 Ω would be the value to use in order to guarantee saturation in any worst case scenario.

It's the typical forward current for that 1-W LED at its rated forward voltage of 3.1V (310mA x 3.1V = 961mW).