Thank you! I was fortunate to have had a very good circuit analysis professor. We used the school's main frame to solve circuit analysis problems. We used a lot of linear algebra using the APL programming language at Syracuse University. 🙂
The video is very well done but it is in ideal world and not in the real world. In this video you chose to find Re 1st, I chose to find the values for R1 and R2 1st. With a Vcc of 12 Volts, I set the base voltag as 3 Volts(1/3) of Vcc. With current 10 times the base current. I then chose a value for Rc, based on Beta of the transistor with a curent of some value to set the Vc (collector voltage) is 1/2 of Vcc. I then find the value of emitter resistor. To effect the gain I spilt the value and put a cap across part it to set the gain of the stage.
Thank you for your thought provoking comment! You are correct about it sitting in an ideal world ... it sits all by itself and no transistor circuit does that. That is why I am following up with a multi-stage transistor circuit to demonstrate the dependency on subsequent loading (e.g. a circuit to follow). As I say, "Anything connected to the input or output of the circuit becomes part of the circuit. It either becomes a current sink or source." Regarding the methodology...yup there are quite a few ways to approach the analysis. The one you speak of is one of the ways that I saw as I was researching the topic. I wanted to keep it simple and a direct move from the simple common-emitter to the beta stabilized version, so I avoided the split emitter resistor. As my emphasis was on beta stabilization, I also purposely avoided dealing directly with the gain issue apart from mentioning that the emitter resistor affected the gain. With that said, both the split emitter resistor AND the whole gain issue will appear in the multi-stage video to follow. What I did find in the research was that most seemed to start with the Thevenin equivalent circuit and the rule of thumb that Rth=(B+1)*Re/10; this includes my EE textbook by Donald Neaman. This means that you cannot calculate Rth without knowing Re, first. This yields I(R2) at about 13 x I(B). So, I went with what seemed most prevalent and what my viewers would most likely see should they look around.
This is a common-emitter circuit which, by its very nature, implies that the input is on the base (likely capacitively coupled) and the output is on the collector (also likely capacitively coupled, but not necessarily). So, yeah, the input and outputs are there. I just didn't overtly show it in the schematic. In the next video release in this series (coming up in a couple weeks), I will be demonstrating the analysis and design of a multi-stage transistor circuit with a 10 KOhm *input* resistance (the base of the first stage) and a 10 Ohm *output* resistance (the emitter of the second stage). I truly hope this dispels any confusion that I may have left behind. 🙂
It would be nice to have a much better explanation of where the rule of thumb at @6:36 comes from, when it applies, and why. It can't apply in every transistor circuit, so why does it apply in this one?
As as far as the origins ... experience and handed down tribal knowledge (I got it from an engineering book which gave no explanation). If we think about it, though, the higher the voltage we allow on the emitter, the less available voltage swing we have on the collector. We want this voltage to have an "insignificant" effect on the available collector voltage swing (and still remain linear). To this end, one of the things we often see is factors of ten used in engineering. If this thing is a factor of 10 greater than that thing, then we have to option to ignore it or the effect of the lesser is "insignificant" as compared to the greater. This rule of thumb applies to the common-emitter circuit. While you would still use the voltage divider biasing routine for the common-collector, we certainly wouldn't apply this rule of thumb to that. And the common-base ... don't even need it. The second rule of thumb (Rth) ... reality is, you can make this anything you want. But, there is a significant trade-off. The lower the value of Rth, the more impervious to beta changes the circuit becomes. BUT this come at a price - input impedance! This second rule of thumb basically sets the current through R2 at about 13 times the base current and the current through R1 about 14 times. Remember what I said about insignificance. So, this rule of thumb is a compromise between maintaining a nice, high input impedance and making the circuit beta stable. It's origins? An engineering book with little explanation ... tribal knowledge passed from generation to generation of engineers. Hope this helps. 🙂
Aaaaah! Good question! We first have to remember that the BJT is a current operated device and that the current in the emitter (Ie) = the current in the base (Ib) * (1+B). Suppose B=10 and Ib = 1 mA, then Ie = 11mA. If we have a 10 Ohm resistor in the emitter (Re), then the voltage across that resistor would be V(Re)=Ie*Re=11mA*10Ohms=0.11 Volts. But, from the perspective of the base, the current it is experiencing is only 1 mA. So, Re(effective) = Ve/Ib = 0.11Volts/1mA = 110 Ohms = Re*(1+B). Hope this helps. 🙂
@@davidluther3955 Yes, the 10%, you mean the Beta +1 and with the Beta of 10, this +1 looks like 10%! I got it. The emitter current consists of the collector current (Ic) plus the base current (Ib). The collector current is equal to the base current times Beta (Ib*B). So, Ie = Ic + Ib = Ib*B + Ib = Ib*(B+1).🙂
WHY DID YOU NOT USE STANDARD RESISTOR VALUES?IF I WERE DESIGHNING THIS CIRCIUT THAT IS WHAT I WOULD DO.WOULD YOU DO A PRESENTATION ON EARLY VOLTAGE?WHEN I WAS IN COLLEGE I DID NOT HAVE THIS IN MY ELECTRONICS CURRICULIUM MANY YEARS AGO.HOW DO YOU CALCULATE EARLY VOLTAGE?
This video was intended to be a demonstration of the process to arrive at the initial, ideal component values. This is always the first step in design. The obvious next step is to look at what is available in standard values and, potentially, modify the design to accommodate them while still meeting the requirements. I did not go through this part of the process, it is true. Now for the "Early Voltage." When you plot collector current against collector-emitter voltage (in other words, like we do when using a transistor curve tracer) we get line that initially has a very steep slope and then tapers off nearly horizontally. In a perfect world, these horizontal sections would be perfectly horizontal. Instead, they have a slight positive slope. If we were to lay our ruler on these lines and draw a line back beyond the Vce = 0 (Y) axis until it crosses the Ic = 0 (X) axis, this place where this crosses is called the "early voltage." And, yes, it is a negative value. Ideally, this point would be the same for ever value of Ib we chose. But it is not. For a Silicon-based transistor this value averages out to be about 100 volts. I actually used my curve tracer with a real, randomly chosen transistor and extrapolated back to this point for each of the Ib traces. I then averaged the values I got for each line and it came out to be petty close to 100 volts. Yeah ... doing a video on this subject would be an interesting one to do and it *should* be fairly straightforward. I will add this to my queue! 🙂
Did you download the "go along with the video" formula sheet? It is available here: drive.google.com/file/d/1WXAorRSnhwrvGLb5OYMCHItWTVb00QY4/view?usp=drive_link I think you slipped a decimal point on your calculation of V(Rth). 5 uA * 24K = 0.12V, not 0.012 volts
Good question...check out page 4 in this go-along-with-the-video formula sheet: drive.google.com/file/d/1WXAorRSnhwrvGLb5OYMCHItWTVb00QY4/view?usp=drive_link 🙂
Not to worry my friend! We've all made the same mistake medication or not. Inverted signs and slipped decimal points, forgotten variables ... all part of our very human condition. 🙂
No one else has the special effects with thier chalktalk that you do! Well done sir...
Thank you so much! 🙂
WHEN I WAS IN COLLEGE TRYING LEARN ALL THIS STUFF WE NEVER HAD SUCH GIFTS AS THIS.ABSOULTY INTRIGUING!
Thank you! I was fortunate to have had a very good circuit analysis professor. We used the school's main frame to solve circuit analysis problems. We used a lot of linear algebra using the APL programming language at Syracuse University. 🙂
Nice work Ralph, you keep Illuminating us...
You're a treasure!
Thank You
I appreciate that! 🙂
👍Thank you sir.
You are welcome! 🙂
The video is very well done but it is in ideal world and not in the real world.
In this video you chose to find Re 1st, I chose to find the values for R1 and R2 1st. With a Vcc of 12 Volts, I set the base voltag as 3 Volts(1/3) of Vcc. With current 10 times the base current. I then chose a value for Rc, based on Beta of the transistor with a curent of some value to set the Vc (collector voltage) is 1/2 of Vcc. I then find the value of emitter resistor. To effect the gain I spilt the value and put a cap across part it to set the gain of the stage.
Thank you for your thought provoking comment! You are correct about it sitting in an ideal world ... it sits all by itself and no transistor circuit does that. That is why I am following up with a multi-stage transistor circuit to demonstrate the dependency on subsequent loading (e.g. a circuit to follow). As I say, "Anything connected to the input or output of the circuit becomes part of the circuit. It either becomes a current sink or source."
Regarding the methodology...yup there are quite a few ways to approach the analysis. The one you speak of is one of the ways that I saw as I was researching the topic. I wanted to keep it simple and a direct move from the simple common-emitter to the beta stabilized version, so I avoided the split emitter resistor. As my emphasis was on beta stabilization, I also purposely avoided dealing directly with the gain issue apart from mentioning that the emitter resistor affected the gain. With that said, both the split emitter resistor AND the whole gain issue will appear in the multi-stage video to follow.
What I did find in the research was that most seemed to start with the Thevenin equivalent circuit and the rule of thumb that Rth=(B+1)*Re/10; this includes my EE textbook by Donald Neaman. This means that you cannot calculate Rth without knowing Re, first. This yields I(R2) at about 13 x I(B).
So, I went with what seemed most prevalent and what my viewers would most likely see should they look around.
Thank you!
You are very welcome! 🙂 Just wait 'till I get to the multi-stage transistor circuit. So much FUN! 😁
Without an input signal, what's the use of the circuit?
This is a common-emitter circuit which, by its very nature, implies that the input is on the base (likely capacitively coupled) and the output is on the collector (also likely capacitively coupled, but not necessarily). So, yeah, the input and outputs are there. I just didn't overtly show it in the schematic.
In the next video release in this series (coming up in a couple weeks), I will be demonstrating the analysis and design of a multi-stage transistor circuit with a 10 KOhm *input* resistance (the base of the first stage) and a 10 Ohm *output* resistance (the emitter of the second stage).
I truly hope this dispels any confusion that I may have left behind. 🙂
It would be nice to have a much better explanation of where the rule of thumb at @6:36 comes from, when it applies, and why. It can't apply in every transistor circuit, so why does it apply in this one?
As as far as the origins ... experience and handed down tribal knowledge (I got it from an engineering book which gave no explanation). If we think about it, though, the higher the voltage we allow on the emitter, the less available voltage swing we have on the collector. We want this voltage to have an "insignificant" effect on the available collector voltage swing (and still remain linear). To this end, one of the things we often see is factors of ten used in engineering. If this thing is a factor of 10 greater than that thing, then we have to option to ignore it or the effect of the lesser is "insignificant" as compared to the greater.
This rule of thumb applies to the common-emitter circuit. While you would still use the voltage divider biasing routine for the common-collector, we certainly wouldn't apply this rule of thumb to that. And the common-base ... don't even need it.
The second rule of thumb (Rth) ... reality is, you can make this anything you want. But, there is a significant trade-off. The lower the value of Rth, the more impervious to beta changes the circuit becomes. BUT this come at a price - input impedance! This second rule of thumb basically sets the current through R2 at about 13 times the base current and the current through R1 about 14 times. Remember what I said about insignificance. So, this rule of thumb is a compromise between maintaining a nice, high input impedance and making the circuit beta stable. It's origins? An engineering book with little explanation ... tribal knowledge passed from generation to generation of engineers.
Hope this helps. 🙂
WHERE DOES Re×(BETA+1) COME FROM WHEN YOU CALCULATED R1OR,
IS SOMETHING YOU SHOULD JUST ACCEPT?
Aaaaah! Good question! We first have to remember that the BJT is a current operated device and that the current in the emitter (Ie) = the current in the base (Ib) * (1+B). Suppose B=10 and Ib = 1 mA, then Ie = 11mA. If we have a 10 Ohm resistor in the emitter (Re), then the voltage across that resistor would be V(Re)=Ie*Re=11mA*10Ohms=0.11 Volts.
But, from the perspective of the base, the current it is experiencing is only 1 mA. So, Re(effective) = Ve/Ib = 0.11Volts/1mA = 110 Ohms = Re*(1+B).
Hope this helps. 🙂
WHERE DOES THE 10% COME INTO IT?
@@davidluther3955 Yes, the 10%, you mean the Beta +1 and with the Beta of 10, this +1 looks like 10%! I got it. The emitter current consists of the collector current (Ic) plus the base current (Ib). The collector current is equal to the base current times Beta (Ib*B). So, Ie = Ic + Ib = Ib*B + Ib = Ib*(B+1).🙂
WHY DID YOU NOT USE STANDARD RESISTOR VALUES?IF I WERE DESIGHNING THIS CIRCIUT THAT IS WHAT I WOULD DO.WOULD YOU DO A PRESENTATION ON EARLY VOLTAGE?WHEN I WAS IN COLLEGE I DID NOT HAVE THIS IN MY ELECTRONICS CURRICULIUM MANY YEARS AGO.HOW DO YOU CALCULATE EARLY VOLTAGE?
This video was intended to be a demonstration of the process to arrive at the initial, ideal component values. This is always the first step in design. The obvious next step is to look at what is available in standard values and, potentially, modify the design to accommodate them while still meeting the requirements. I did not go through this part of the process, it is true.
Now for the "Early Voltage." When you plot collector current against collector-emitter voltage (in other words, like we do when using a transistor curve tracer) we get line that initially has a very steep slope and then tapers off nearly horizontally. In a perfect world, these horizontal sections would be perfectly horizontal. Instead, they have a slight positive slope. If we were to lay our ruler on these lines and draw a line back beyond the Vce = 0 (Y) axis until it crosses the Ic = 0 (X) axis, this place where this crosses is called the "early voltage." And, yes, it is a negative value. Ideally, this point would be the same for ever value of Ib we chose. But it is not. For a Silicon-based transistor this value averages out to be about 100 volts.
I actually used my curve tracer with a real, randomly chosen transistor and extrapolated back to this point for each of the Ib traces. I then averaged the values I got for each line and it came out to be petty close to 100 volts.
Yeah ... doing a video on this subject would be an interesting one to do and it *should* be fairly straightforward. I will add this to my queue! 🙂
I THINK YOU MADE ERROR IN YOUR CALCULATION OF THE THEVENIN VOLTAGE. I GET .012V FOR VRTH AND,1.912V FOR VTH.
Did you download the "go along with the video" formula sheet? It is available here: drive.google.com/file/d/1WXAorRSnhwrvGLb5OYMCHItWTVb00QY4/view?usp=drive_link
I think you slipped a decimal point on your calculation of V(Rth). 5 uA * 24K = 0.12V, not 0.012 volts
BEING A CURIOUS PERSON,HOW DID YOU COME UP MATHEMATICALY WITH THE FORMULAS FOR FINDING Rb1 AND Rb2?
Good question...check out page 4 in this go-along-with-the-video formula sheet: drive.google.com/file/d/1WXAorRSnhwrvGLb5OYMCHItWTVb00QY4/view?usp=drive_link
🙂
THANKS SOME HOW I OVER LOOKED THOSE DERIVATIONS.
@@davidluther3955 You are very welcome! 🙂
WHO THAT LITTLE MAN WITH THE QUESTION MARK?😊
LOL! 😀
TO ME IT LOOKS LIKE THE PILSBERRY DOE BOY.
@@davidluther3955 LOL! 🙂
@@eie_for_youI LIKE THAT LITTLE MAN.HES REALLY COOL DUDE.
@@davidluther3955 True that! 🙂
I MUST APPOLIGIZE I WAS IN ERROR.MY CLINICAL DEPRESSION MEDICATIONS CAUSE ME TO NOT THINK CLEARL CLEARLY.
Not to worry my friend! We've all made the same mistake medication or not. Inverted signs and slipped decimal points, forgotten variables ... all part of our very human condition. 🙂