Honestly I spent hours on multiple sources to solve the learnings to this problem. This guy waltzes in, scribble and yells thoughts, 4:56 later, I have my answer finally, Thank you haha
I LOVE your comment! I'm grad I could do in minuts what others couldn't in multiple sources. Let me know what other concepts you're studying and I may have a video for that or I can add the concept into oner of my gfuture videos.
thank you very much! I am trying to make an 8-bit adder from transistors for my IT final paper but we didn’t even learn transistors or even LED’s in physics, this helped me very much
No. All manufacturers' data sheets specify saturation with a base current one-tenth of the collector current. You should drive the base with 0.1mA in this case, which calls for a 4.3K base resistor. The reason is the switching time being limited by the input capacitance, and that needs current to charge and discharge. A 2N3904, for example, specifies its delay time, rise time, storage time and fall time with an Ic=10mA and an Ib=1mA. You can't have fast switching without driving some current. In fact, some switching designs use a small capacitor in parallel with the base drive resistor to further improve the transient response.
Thanks for the feedback, Rex! Always appreciated. I agree with the one-tenth. In this example I mentioned to divide the calculated resistance by 2 two or so, which will be somewhere around 47 k ohms. That will give around 1 mA which is one-tenth of 10 mA. I selected a gain of 200 so I would end up with a base current of around one-tenth, but I suppose I should have unpacked that from the perspective of the datasheet as well. My focus in this video is to teach the operation principles of a BJT, but I am getting the sense from some comments that I should also use Datasheets and specific transistors. ---- ----- You calculated for 0.1 mA which is one-one-hundredth. Your value of 4.7 k is the correct number but is just off by a factor of 10. I fully make those mistake way too much :-) Sometimes in lecture which is funny! Especially when my students catch me on it 😊 Keep up the feedback
@@larsexplains3086 Thanks, Lars, I clearly wrote the wrong base current there, and you're right that it should be 1mA to saturate a transistor sinking 10mA collector current. I gave the correct value for the base resistor, which is 4.3K, but it's not critical in my experience and 4.7K would work fine.
Thank you for your great videos...i have a questiin please : if we put around 40k Rb then the base current will increase , so will the current across CB also increase after multiplied by the gain?..Thank you
No sir I'm asking you, I need to clear my doubts....i request you to make a tutorial about finding fan-out for bjt inverter for N and N-1 inverter if you could do it...
Hello Alvin. I selected 1000 ohms because I needed a collector current that was nice and simple. 10 milliamps is a simple number. The thing about BJTs is that they're typically not designed to control a lot of current. Anything under 1000 ohms would draw possibly more current than the BJT can handle and anything more than 100 ohms would mean that the base current would be so small that I may need a mega ohm resistor for the base resistance. I guess the best way to answer your question is, 1000 ohms for the collector resistor creates a current that's easy to work with and the numbers work out well. Of course, my answer doesn't tell you anything about application. I think typically you will see in a classroom setting a 1000 ohms resistor is typically used so that the example is nice and simple, but depending on the application that collector resistor may be much lower even down to 50 ohms and much higher up to 10,000 ohms. Typically, Darlington transistor (Which can handle higher currents with collector resistors of low values) and FETS (Field Affect Transistors) are used to control higher currents like 100 mA to 1 A and even much higher. BJTs have wildly fluctuating Bata (The Gain of the transistor) with changing and high currents above 100 mA. For this reason, you don’t generally see the collector resistor being lower than 100 ohms. If you want to look at examples that use lower resistances for the collector, I did a motor control video with a BJT. th-cam.com/video/cSTv-Ft_HhU/w-d-xo.html Here is another video where a BJT (NP2222A) is used to control a Pneumatic Solenoid that draws under 100 mA. He is using 12 V for the Vcc so I’m assuming the resistance of the Solenoid is (10 - 0.2)V/56 mA = 175 ohms. th-cam.com/video/8DMZSxS-xVc/w-d-xo.html
so thats what i was missing the load is important to determine the base resistor , the transistor doesnt act like a mechanical switch , anyways thank you so much for this great video
Hi Christian. I just made that example up. I selected a Gain of 200. That example is discussing the general steps to select the base resistor for any BJT. This method works for all NPN Transistors. So, in that case I did not need to be specific to any one BJT. Sorry, the datasheet doesn't exist because I just made up the gain, If you like, I can do a video for you on a specifit BJT. Let me know.
@@LightningFreezer Hey Lightning. The video is generic and does not specify a transistor, but the math is all the same. If you are interested in using a BC549 then use the Datasheet to establish the gain will be for whichever Collector-Emitter current you need and the maximum voltage applied to the Collector (Or more specifically what the VEC will be) can handle. The BC549 has a maximum VEC of 30 volts and the gain will range from 100 to 800 depending in the current through the transistor. Does that answer your question?
@@LightningFreezer Here is a link to the Wikipedia page on the 549 en.wikipedia.org/wiki/BC548 Here is a link to a Datasheet from Fairchild: www.mouser.com/datasheet/2/149/BC547-190204.pdf
@@larsexplains3086 Oddly enough, the BC547 series is of the few BJTs where the manufacturer specifies the base current as 1/20 of the collector current to guarantee saturation. Almost every other transistor specifies 1/10, except for the Darlingtons, of course, where it's common to see 1/250.
*YOU DIDN'T TELL HOW TO CALCULATE THE GAIN?*---how tp calculate the gain and how got the value 200? What if the base voltage is same as the collector boltage? Is there a simple way to calculate that in that case?
Thank you for your great question. I get your frustration. You are not alone. I had a similar issue when I first started learning transistor theory. The thing is, the Gain is not calculated, but put into the design of the transistor by the manufacturer. The manufacturers decided what the gain is and then include that information on the datasheet. This gets a bit more complicated, because the gain changes with the collector current. If you look at the date sheet of any transistor here is a link: chrome-extension://efaidnbmnnnibpcajpcglclefindmkaj/www.onsemi.com/pdf/datasheet/tip47-d.pdf If you look at the Datasheet, you will see the Gain chart hFE DC Gain Current Gain. The Max Gain is 150 when the collector current is low and the and the minimum gain is 10 when the collector current is at its highest (The highest current is listed as 2 Amps) Knowing the Gain, you can do all the math I demonstrated in the video. I would like to also note that I used a gain of 200 to make the math easier, but it is an uncommon Gain. Typically, smaller Transistor have a lower max gain lie 80- or lower for the max collector current, I hope that helped your frustration.
Great video. To bad it was so hard to find :) One question: do I need a base resistor if I want to control transistor from output of an IC that gives 2V and 5mA on it's output?
@Taurruth Putting 2V that can source 5mA of current into the Base of a BJT is dependent on the transistor. What he didn't show you is how to get: Ic Saturation , Ib Saturation, Vce Saturation, Vbe Saturation, and Beta (Current Gain, the Greek letter that looks like a B). These vary for each transistor type (even among various manufacturers for the same type). Determining those parameters is the "hard part" of this exercise, each manufacturer has their own way of specifying how their part works, so its not always straight forward as some manufacturers may not specify all of them clearly and you have you derive them from studying the data sheet. So, if you have, say, a 2N2222 NPN Transitor, I beleive it can handle 5mA on the Base, but since the the voltage across BE Junction will be about 2V for this transistor (your output's supply voltage) then that output driving the Base is effectively current limiting to the transistor's spec by default (not the norm but possible), but that output is working at its maximum and you should always have about 10% design margin from Ibe Max (check the data sheet) to allow for temperature changes and other tolerances. Pull out a 2N2222 data sheet and check this as I am going from memory. In this case I would put a 10 Ohm resistor in series withe Base as a current limiter so your output is slightly below max output current and not at specified max. You may find that even a 500 Ohm resistor may work because you may only need about 3 mA to get the transistor to saturate for the amount of current you need to flow through the CE pins. Usually, we try to drive a NPN BJT in the Common Emitter configuration with a output that has voltage higher than the Vbe voltage and use a current limiting resistor for better Base current control. Using a 2V output to drive the 2N2222 gives you little control of the current. For example, if your output was 3.3V at 5mA, the current limiting resistor would drop 1.3V and to provide 4mA @2V across the BE - that resistor would be 325 Ohms. This way, you have a design that you can "Tweak" the Base current (selecting the best current limiter resistor value) for optimum operation and not waste power (using more Base current than you will ever need). If you only have 1 tranistor circuit that "wastes" 1mA of current that may be fine, but if you had 10 circuits, that would be 10mA wasted for no reason, remember, more circuits, more power used, thus less power available for other circuits in your design. Does this make sense?
Hi. I have a problem. Here I live in Africa, here is either difficult do get hold of DI-boxes for the music or they are way too expencive. So I dercided I will try to build them myself to sell them here. I struggle also to get hold of metal boxes to build them in. A DI box is a Direct box which will isolate and create a better sound from the instrumnent through the mixer to the amp and to the speakers. I Followed all the instruction I could find on internet. But still I got a buzz. And I building passive boxes there is only signals from mostly guitars going through. Can it be because the boxes is all plastic or what can it be? Would it also help if I cut some metal pieces to cover the inside of the boxes to create a good grounding?
@@larsexplains3086 I got stuck on the same thing. I tried finding in the datasheet (using MPS751), and closest thing I could find is hFE but then only gives a Min. based on 4 different test conditions. Thank you BTW for making this. :)
@@joshhaughton1893 You'll see that the gain changes with the Ic. So, select the rigth gain for whatever Ic you have. For BJT as a switch, you can balpark the base resistor so using an approx. gain woks fine.
@@larsexplains3086 Thanks for response! To make sure I understand how to put what you're teaching into practice. If I'm using a MPS751 transistor for instance, and only need to switch a load of up to 1 Amp, although it's rated for up to 2 Amps - I'd reference the datasheet. Here: datasheet.lcsc.com/lcsc/2008241904_onsemi-MPS751_C606060.pdf - Then lookup the hFE Symbol, and use 75 as "B" in your formula BECAUSE that was the minimum DC current gain in a test condition where Ic was >= my calculated load of 1 Amp. So although, per this particular BJT, the min gain drops down to 40 with 2 Amps on the collector, I wouldn't need to use that. Do I understand that correctly?
I do a video on using a BJT as a DC amplifier if that's what you mean. Making a variable DC source to run experiments and different circuits is WAY more complicated than using a BJT. This is because a BJT is configured with a given collector current and a specific collector voltage. A variable DC power source has changing voltage and changing current and is a much more complicated circuit that just BJT.
Each BJT comes with its own B and can b e found on the data sheet. I made up this gain, but could have also taken it from the Data sheet. The Gain is defined by the manufacturer. The size and type of transistor also has a little bit to do with the game, but essentially the manufacturer chooses what the gain would be in the way they manufacture the transistor. This particular gain of 200 is quite large and actually is not very typical for this application. I think I should have probably used A-100 instead or even something lower, but to answer your question, I just pulled it out of my head because I didn't use an actual transistor with its own data sheet and code. The number 200 allowed me to do the math easily. I hope that answers your question.
So, if base current should be one-tenth of the collector current, that means if I want 1A collector current the base current should be 100mA, am I right? (Or completely wrong?).
@@larsexplains3086 OK, thanks! 👍 I just want to drive a 3W LED light panel. It consumes 1A at 3V. I already tried to drive it with one transistor but the heat is too much. I'm thinking of using few transistors in parallel.
It still works. When driving relays with an NPN (2N4401) I usually just use a 1k (or less), and it works well, I'm probably driving lower resistance loads, but not massively lower.... This is with 3.3V microcontrollers, draws maybe 2-3mA from their pins... all good....
I can answer questions via this chat and I may be able to do a video for you. I can't fit in tutoring because I'm taking 18 hours/week this semester. Keeping me CraZy busy! Ill try to fit in a video if your question is in alignment with my classes. If I had more time, I’d do a video on anything short of politics 🤣
NB: Everyting in your explanation is very good but not the beta Value. The beta in Saturation will never be that high. beta is rather about 10 in Saturation.
Nope. This was perfectly fine! Beta drops in saturation simply because for a given Ib, there isnt enought Vce to sink the active region's Ic. The idea here is that the with 50uA Ib, the BJT is is at the edge of Sat and Active. Since this is bot Sat and Active, the same Beta applies. Thus for a higher Ib, it will definatetly get pushed to Saturation.
WOW! You're right! I researched the proper pronunciation and I've been saying it wrong for decades! I think I'm going to blame that on my high school electronics teacher 😂 I've been saying it like that since grade 11. Finally, someone set me straight! Thanks!
![[TH] VALORANT Champions Seoul - Semi-Final - SEN vs TH](http://i.ytimg.com/vi/IewhVY9cAmQ/mqdefault.jpg)
That's the right way to teach the complex topic in a simple way ... thanks a lot sir :)
KISS! "Keep it Simple S...." I like straight forward too
This is so good explanation, short fast and straight forward!
I love fast!
Greatly explained, thank you!!!
Superb style of TEACHING...Real Guru
Honestly I spent hours on multiple sources to solve the learnings to this problem. This guy waltzes in, scribble and yells thoughts, 4:56 later, I have my answer finally, Thank you haha
I LOVE your comment! I'm grad I could do in minuts what others couldn't in multiple sources. Let me know what other concepts you're studying and I may have a video for that or I can add the concept into oner of my gfuture videos.
Wow finally I figured it out. You teach perfectly for me thanx sir.
Thank you for this video with a great explanation. Just in practical situations Beta(saturation) = 10 NOT 200.
Love the energy.
thank you very much! I am trying to make an 8-bit adder from transistors for my IT final paper but we didn’t even learn transistors or even LED’s in physics, this helped me very much
Good clarifiation.I am a beginner. Thanks for your help
No. All manufacturers' data sheets specify saturation with a base current one-tenth of the collector current. You should drive the base with 0.1mA in this case, which calls for a 4.3K base resistor. The reason is the switching time being limited by the input capacitance, and that needs current to charge and discharge. A 2N3904, for example, specifies its delay time, rise time, storage time and fall time with an Ic=10mA and an Ib=1mA. You can't have fast switching without driving some current. In fact, some switching designs use a small capacitor in parallel with the base drive resistor to further improve the transient response.
Thanks for the feedback, Rex! Always appreciated. I agree with the one-tenth. In this example I mentioned to divide the calculated resistance by 2 two or so, which will be somewhere around 47 k ohms. That will give around 1 mA which is one-tenth of 10 mA. I selected a gain of 200 so I would end up with a base current of around one-tenth, but I suppose I should have unpacked that from the perspective of the datasheet as well. My focus in this video is to teach the operation principles of a BJT, but I am getting the sense from some comments that I should also use Datasheets and specific transistors. ---- ----- You calculated for 0.1 mA which is one-one-hundredth. Your value of 4.7 k is the correct number but is just off by a factor of 10. I fully make those mistake way too much :-) Sometimes in lecture which is funny! Especially when my students catch me on it 😊 Keep up the feedback
@@larsexplains3086 Thanks, Lars, I clearly wrote the wrong base current there, and you're right that it should be 1mA to saturate a transistor sinking 10mA collector current. I gave the correct value for the base resistor, which is 4.3K, but it's not critical in my experience and 4.7K would work fine.
Your explanation is so perfect
nice explanation
Love the passion, bro.
Thanks AgrX!
Thank Lars, clear explanation. Love it.
Ya ya ya ya ya ya!
Best explanation ever
Awesome!
Thank you. It's great explanation and to the point.
Excellent explanation
Thank you for your great videos...i have a questiin please : if we put around 40k Rb then the base current will increase , so will the current across CB also increase after multiplied by the gain?..Thank you
Hello thank you very much for your explanation. but do you think a over drive factor (ODF) of about 2 for this matter is really enough?
I think that's IB(eos) = Ic(Sat)/beta
And Ib(Sat) should be equal to KIb(eos)
K being overdrive factor
Thanks vkh2412. I'm going to dive into this and re-work my examples with Edge of Saturation
No sir I'm asking you, I need to clear my doubts....i request you to make a tutorial about finding fan-out for bjt inverter for N and N-1 inverter if you could do it...
why 1000 ohms? All tutorials never explain how they come to decide on the resistance.
Hello Alvin.
I selected 1000 ohms because I needed a collector current that was nice and simple. 10 milliamps is a simple number. The thing about BJTs is that they're typically not designed to control a lot of current. Anything under 1000 ohms would draw possibly more current than the BJT can handle and anything more than 100 ohms would mean that the base current would be so small that I may need a mega ohm resistor for the base resistance.
I guess the best way to answer your question is, 1000 ohms for the collector resistor creates a current that's easy to work with and the numbers work out well.
Of course, my answer doesn't tell you anything about application. I think typically you will see in a classroom setting a 1000 ohms resistor is typically used so that the example is nice and simple, but depending on the application that collector resistor may be much lower even down to 50 ohms and much higher up to 10,000 ohms.
Typically, Darlington transistor (Which can handle higher currents with collector resistors of low values) and FETS (Field Affect Transistors) are used to control higher currents like 100 mA to 1 A and even much higher. BJTs have wildly fluctuating Bata (The Gain of the transistor) with changing and high currents above 100 mA. For this reason, you don’t generally see the collector resistor being lower than 100 ohms.
If you want to look at examples that use lower resistances for the collector, I did a motor control video with a BJT.
th-cam.com/video/cSTv-Ft_HhU/w-d-xo.html
Here is another video where a BJT (NP2222A) is used to control a Pneumatic Solenoid that draws under 100 mA. He is using 12 V for the Vcc so I’m assuming the resistance of the Solenoid is (10 - 0.2)V/56 mA = 175 ohms.
th-cam.com/video/8DMZSxS-xVc/w-d-xo.html
so thats what i was missing the load is important to determine the base resistor , the transistor doesnt act like a mechanical switch , anyways thank you so much for this great video
You're very welcome!
Great! I would have loved to see the datasheet of your transistor though, so at least part number would have been nice.
Hi Christian. I just made that example up. I selected a Gain of 200. That example is discussing the general steps to select the base resistor for any BJT. This method works for all NPN Transistors. So, in that case I did not need to be specific to any one BJT. Sorry, the datasheet doesn't exist because I just made up the gain, If you like, I can do a video for you on a specifit BJT. Let me know.
@@larsexplains3086 Hey thanks for the video how about a BC 549 Transistor?
@@LightningFreezer Hey Lightning. The video is generic and does not specify a transistor, but the math is all the same. If you are interested in using a BC549 then use the Datasheet to establish the gain will be for whichever Collector-Emitter current you need and the maximum voltage applied to the Collector (Or more specifically what the VEC will be) can handle. The BC549 has a maximum VEC of 30 volts and the gain will range from 100 to 800 depending in the current through the transistor. Does that answer your question?
@@LightningFreezer Here is a link to the Wikipedia page on the 549 en.wikipedia.org/wiki/BC548
Here is a link to a Datasheet from Fairchild: www.mouser.com/datasheet/2/149/BC547-190204.pdf
@@larsexplains3086 Oddly enough, the BC547 series is of the few BJTs where the manufacturer specifies the base current as 1/20 of the collector current to guarantee saturation. Almost every other transistor specifies 1/10, except for the Darlingtons, of course, where it's common to see 1/250.
beautifully explained
thank you
*YOU DIDN'T TELL HOW TO CALCULATE THE GAIN?*---how tp calculate the gain and how got the value 200? What if the base voltage is same as the collector boltage? Is there a simple way to calculate that in that case?
Thank you for your great question. I get your frustration. You are not alone. I had a similar issue when I first started learning transistor theory.
The thing is, the Gain is not calculated, but put into the design of the transistor by the manufacturer. The manufacturers decided what the gain is and then include that information on the datasheet. This gets a bit more complicated, because the gain changes with the collector current. If you look at the date sheet of any transistor here is a link: chrome-extension://efaidnbmnnnibpcajpcglclefindmkaj/www.onsemi.com/pdf/datasheet/tip47-d.pdf
If you look at the Datasheet, you will see the Gain chart hFE DC Gain Current Gain. The Max Gain is 150 when the collector current is low and the and the minimum gain is 10 when the collector current is at its highest (The highest current is listed as 2 Amps)
Knowing the Gain, you can do all the math I demonstrated in the video.
I would like to also note that I used a gain of 200 to make the math easier, but it is an uncommon Gain. Typically, smaller Transistor have a lower max gain lie 80- or lower for the max collector current,
I hope that helped your frustration.
Thanks sir
You're very welcome!
good
Great video. To bad it was so hard to find :)
One question: do I need a base resistor if I want to control transistor from output of an IC that gives 2V and 5mA on it's output?
@Taurruth Putting 2V that can source 5mA of current into the Base of a BJT is dependent on the transistor. What he didn't show you is how to get: Ic Saturation , Ib Saturation, Vce Saturation, Vbe Saturation, and Beta (Current Gain, the Greek letter that looks like a B). These vary for each transistor type (even among various manufacturers for the same type). Determining those parameters is the "hard part" of this exercise, each manufacturer has their own way of specifying how their part works, so its not always straight forward as some manufacturers may not specify all of them clearly and you have you derive them from studying the data sheet.
So, if you have, say, a 2N2222 NPN Transitor, I beleive it can handle 5mA on the Base, but since the the voltage across BE Junction will be about 2V for this transistor (your output's supply voltage) then that output driving the Base is effectively current limiting to the transistor's spec by default (not the norm but possible), but that output is working at its maximum and you should always have about 10% design margin from Ibe Max (check the data sheet) to allow for temperature changes and other tolerances. Pull out a 2N2222 data sheet and check this as I am going from memory. In this case I would put a 10 Ohm resistor in series withe Base as a current limiter so your output is slightly below max output current and not at specified max. You may find that even a 500 Ohm resistor may work because you may only need about 3 mA to get the transistor to saturate for the amount of current you need to flow through the CE pins. Usually, we try to drive a NPN BJT in the Common Emitter configuration with a output that has voltage higher than the Vbe voltage and use a current limiting resistor for better Base current control. Using a 2V output to drive the 2N2222 gives you little control of the current. For example, if your output was 3.3V at 5mA, the current limiting resistor would drop 1.3V and to provide 4mA @2V across the BE - that resistor would be 325 Ohms. This way, you have a design that you can "Tweak" the Base current (selecting the best current limiter resistor value) for optimum operation and not waste power (using more Base current than you will ever need). If you only have 1 tranistor circuit that "wastes" 1mA of current that may be fine, but if you had 10 circuits, that would be 10mA wasted for no reason, remember, more circuits, more power used, thus less power available for other circuits in your design. Does this make sense?
@Paul Romsky thank you for your reply.
You Make learning this stuff fun.
Thanks Richard!
Hi. I have a problem. Here I live in Africa, here is either difficult do get hold of DI-boxes for the music or they are way too expencive. So I dercided I will try to build them myself to sell them here.
I struggle also to get hold of metal boxes to build them in. A DI box is a Direct box which will isolate and create a better sound from the instrumnent through the mixer to the amp and to the speakers.
I Followed all the instruction I could find on internet. But still I got a buzz. And I building passive boxes there is only signals from mostly guitars going through.
Can it be because the boxes is all plastic or what can it be? Would it also help if I cut some metal pieces to cover the inside of the boxes to create a good grounding?
clear, thank!
very good
Thank you sir!
Thank you so much but could you what is the B = 200 again? i mean is it a resistance , a standard or what?
B stands for the Gain of the Transistor. The formula for Gain is B = IC/IB where IC is the collector current and IB is the base current.
@@larsexplains3086 I got stuck on the same thing. I tried finding in the datasheet (using MPS751), and closest thing I could find is hFE but then only gives a Min. based on 4 different test conditions. Thank you BTW for making this. :)
@@joshhaughton1893 You'll see that the gain changes with the Ic. So, select the rigth gain for whatever Ic you have. For BJT as a switch, you can balpark the base resistor so using an approx. gain woks fine.
@@larsexplains3086 Thanks for response! To make sure I understand how to put what you're teaching into practice. If I'm using a MPS751 transistor for instance, and only need to switch a load of up to 1 Amp, although it's rated for up to 2 Amps
- I'd reference the datasheet. Here: datasheet.lcsc.com/lcsc/2008241904_onsemi-MPS751_C606060.pdf
- Then lookup the hFE Symbol, and use 75 as "B" in your formula BECAUSE that was the minimum DC current gain in a test condition where Ic was >= my calculated load of 1 Amp. So although, per this particular BJT, the min gain drops down to 40 with 2 Amps on the collector, I wouldn't need to use that.
Do I understand that correctly?
A big question answered. How about making a variable source?
I do a video on using a BJT as a DC amplifier if that's what you mean. Making a variable DC source to run experiments and different circuits is WAY more complicated than using a BJT. This is because a BJT is configured with a given collector current and a specific collector voltage. A variable DC power source has changing voltage and changing current and is a much more complicated circuit that just BJT.
The real question is, how much Ib do you need to set the transistor on fire?? 😂 You know, so to avoid doing that 😅
Where did the value of 200 for B come from?
Each BJT comes with its own B and can b e found on the data sheet. I made up this gain, but could have also taken it from the Data sheet.
The Gain is defined by the manufacturer. The size and type of transistor also has a little bit to do with the game, but essentially the manufacturer chooses what the gain would be in the way they manufacture the transistor.
This particular gain of 200 is quite large and actually is not very typical for this application. I think I should have probably used A-100 instead or even something lower, but to answer your question, I just pulled it out of my head because I didn't use an actual transistor with its own data sheet and code. The number 200 allowed me to do the math easily. I hope that answers your question.
So, if base current should be one-tenth of the collector current, that means if I want 1A collector current the base current should be 100mA, am I right? (Or completely wrong?).
Yup! you are 100 % correct.
That is for a BJT as a switch.
For amplifiers, it's quite different.
@@larsexplains3086 OK, thanks! 👍
I just want to drive a 3W LED light panel. It consumes 1A at 3V. I already tried to drive it with one transistor but the heat is too much. I'm thinking of using few transistors in parallel.
Good
But what happens if I put a 1K resistor there?
the entire universe collapses and it’s all your fault
@@lucasc5622 OMG! I was just about to do it!
It still works. When driving relays with an NPN (2N4401) I usually just use a 1k (or less), and it works well, I'm probably driving lower resistance loads, but not massively lower.... This is with 3.3V microcontrollers, draws maybe 2-3mA from their pins... all good....
@@geoninja8971 does your controller heats up and what kind of microcontroller do refer to thanks
@@stevegoodjob5902 Arduino Nano, ESP32, ATtiny85.... no heating up noted, wouldn't be expected at these low currents....
It's not pronounced "kirch-off" -> it's pronounced "kirk-off"
hi, do you have a way to reach out to you for some help?
I can answer questions via this chat and I may be able to do a video for you. I can't fit in tutoring because I'm taking 18 hours/week this semester. Keeping me CraZy busy! Ill try to fit in a video if your question is in alignment with my classes. If I had more time, I’d do a video on anything short of politics 🤣
Teaching* 18 hours/week
No problem thanks for the help, i think i found my solution
NB: Everyting in your explanation is very good but not the beta Value. The beta in Saturation will never be that high. beta is rather about 10 in Saturation.
Nope. This was perfectly fine! Beta drops in saturation simply because for a given Ib, there isnt enought Vce to sink the active region's Ic.
The idea here is that the with 50uA Ib, the BJT is is at the edge of Sat and Active. Since this is bot Sat and Active, the same Beta applies. Thus for a higher Ib, it will definatetly get pushed to Saturation.
Kirrtschoffs Law... :D No... Kir CH hoff...
WOW! You're right! I researched the proper pronunciation and I've been saying it wrong for decades! I think I'm going to blame that on my high school electronics teacher 😂 I've been saying it like that since grade 11. Finally, someone set me straight!
Thanks!
Happy day shinzon0!
Kirchhoff was German and so was my high school physics teacher. That’s how it’s pronounced in German, so I guess I’m also correct.