i have a circuits exam (referral exam) in June and i just hope The Organic Chemistry Tutor uploads more videos related to OpAmps, Diodes, Transistors and Small Signal, otherwise i'm definitely failing. Thank you so much for all of the videos you upload, you've helped me a lot!!
After watching this video,I now understand why you 've got almost 3M subscribers.You make things expicit to grasp,which will take me a long time and enormours energy.
Wow! That sure wore my brain out trying to develop each circuit BEFORE you showed and explained the right circuit. I still haven't developed the intuitive understanding, but this certainly helped me better see the current flow and voltage differences. I've gone through your videos on circuit analysis and those helped me follow your explanations after the fact. My wife walked in while i was looking at the video and she (Masters in Adult Education) exclaimed "Now THAT'S the right way to teach!", being impressed with the clarity of how you had laid things out. Congratulations and thanks. I had to keep re-watching parts to fully understand, but definitely worth the time. I've watched videos which clearly explained how to use logic gates with switches (esp. ElectroBoom), but this is the first showing how the logic gates themselves can be built.
Don't know if you will see this or not, but have you ever considered being a private tutor? I would pay good money for you to teach me this subject (electrical circuits and engineering). Though your videos are very helpful, I still get confused and i'd love to be able to ask you questions in real time.
I think that earlier in the video he meant to say "at least" with respect to the other examples with 0.7 Volts as well. I'm guessing that these aren't the only circuit constructions with represent these truth tables?
5:30 That NOR example isn't a real gate, hasn't a real output that you can use, I wish it work like that, but you need a transistor to invert OR diode gate into a real NOR gate with output a signal.
Hey, thanks a lot for the video, it really helped me for my exam.. just a small doubt, in the NAND gate if we choose R3 such that the voltage drop is enough for the LED(>=2V), will current flow through both R1 and R2 or just R2?
What about all of the other logic gates such as XOR, negative-OR, negative-AND, negative-XOR, etc...Also on the AND-gate towards the end, you said the potential at the point right before the LED was 3.5-V, how is that possible? If the potential is 5-V at A or B and the Diodes drop 0.7-V across each of them , that leaves 4.3-V. Since 4.3 > 3.5, shouldn't that be the potential at that point?
At 6:17, there is electrical potential of 4.3 V after diode and 3.0 V after the LED. Why is electrical potential set at 4.3 V and not 3.0 V at this location? Why is the larger chosen?
The voltage drop across the diode (.7v) is less than that of the LED (2v) so it will take the least resistive path to ground (no potential). In other words, the diode only resists .7v leaving the following point at 4.3v, rather than the led resisting 2v and leaving you with 3v at the following point. The greater potential is from 4.3v to 0v so the larger is chosen.
@@chrisp1582 But will any current actually flow through the LED? I need to set this up on a breadboard and do measurementts. You are saying the general rule is this?: When two or more paths connet to a common node, the voltage from that node will be that of the highest voltage source.
@@typedeaf Yes, current will flow through the led. This is simply the measurement at the connecting node which would be that of the least resistive path. The voltage at the node will be that of the greatest potential to ground which can change based on the volume of voltage from the source. So if you increased the voltage source from another point that value could change at that connecting node.
(1) At 6:17, there is electrical potential of 4.3 V after diode and 3.0 V after the LED. Why is electrical potential set at 4.3 V and not 3.0 V at this location? Why is the larger chosen? (2) At 5:27, what is purpose of diodes in OR gate? It seems the circuit would work without them and they are redundant.
For 2) in actuality the values 0s and 1s will not correspond to 0 and 5v We need some device which will restrict current at a some voltage and diode does this, If A had a greater potential than at other side , the diode will be reverse biased and offer high resistance and no current flows, but a conductor wld allow current in the other direction Also, current can pass through conductors even though there is no potential difference...
1. The voltage drop across the diode (.7v) is less than that of the LED (2v) so it will take the least resistive path to ground (no potential). In other words, the diode only resists .7v leaving the following point at 4.3v, rather than the led resisting 2v and leaving you with 3v at the following point. The greater potential is from 4.3v to 0v so the larger is chosen. 2. The purpose of the diodes is to act as a switch in the circuit. If voltage is applied at either A or B then the LED would be OFF due to the greater potential across the diode. Similar to answer 1, the diode resists less which allows a greater potential and an easier path of resistance.
I’m curious as to how these circuits can be used to derive a +5V trigger or gate signal for use with a modular synth circuit. Where would one tap the signal in order to get a healthy +5V based upon the resulting truth tables? I’m especially interested in the AND circuit. I built one with 2 NPN xstrs but the resulting output is useless. Building it with diodes and a 5 volt rail seems like a good time.
The output is where the LED is. Remember that this is diode logic. It has long been extinct in logic circuit design. Transistors had not been invented yet when this was still being used.
@ 10:00 How are A and B considered as inputs of the AND gate with the diode blocking them? A and B would have to be a switch where the NC is to ground and the NO is to 5v? That seems like a big assumption to not have mentioned. The typical circuit would have A and B as a switch with NC being an open circuit. Also, the only reason the current follows D1 or D2 when they are 0v is because it has less resistance than the LED and second resistor, right? You never state the resitance of the diodes and LED. I am asking this because if you dont consider that fact, and instead first calculate the voltage drop across R1 as if it would next follow the LED, it would only drop 1.5v, which would be enough to power the LED.
This is a 5V logic circuit. All inputs are assumed to be either +5V or 0V (ground). Open circuits are undefined in logic circuitry. They should never occur because they can cause errors. Secondly, no. The input diodes have a biasing voltage drop of 0.7V, and we are assuming they are connected directly to ground. This means the voltage at "point C" (as he labeled it later in the NAND circuit) will be pulled down to 0.7 volts. That is not enough to turn on the LED. Voltage drops across resistors do not remain static when the circuit condition changes. When point C becomes 0.7V, the voltage drop across R1 (top resistor) will necessarily become 4.3 volts. If there was a fault in this video, it was the fact that this was not explained clearly.
When analyzing circuit behavior we usually assume we are using "ideal" components (which have perfect characteristics) or we assign them a typical value, which is obviously more like what actually happens in the real world. An ideal diode would have infinite resistance in reverse bias and would have zero resistance in forward bias. That is not what happens in reality, in either case. He is assuming characteristics of a typical diode throughout this video. Real diodes have a leakage current in reverse bias (this characteristic can almost always be safely ignored, which he does), and they have a voltage drop when forward biased. Ideal diodes are almost never used in circuit analysis, but it might be useful for you to try it yourself in this instance (or imagine what would happen if you replaced the diode with a switch and turned it on and off if that's easier for you). If your question wasn't already answered by that rambling, see if you can perform thought experiments by replacing a diode in the circuit with an ideal diode or a switch and think about what would happen on your imaginary volt meter at point C or at other places in the circuit.
Stick to chemistry; your teaching technique for logic functions is worse than poor; it is wrong. By changing how the LED is connected to the input diodes, you are flipping back and forth between active-high and active-low logic conventions without a word of explanation. On and Off have no place in a truth table, which is why 50% of your tables are semantically correct, but logically wrong. If you instead assign 1's and 0' to the outputs, you will see that the NOR and NAND tables are wrong. The problem is the schematics. The schematic for a particular type of gate is what it is, with no regard for what is connected to the output. Your truth tables depend on changing output connections, whereas correct ones do not.
Final Exams and Video Playlists: www.video-tutor.net/
He must be genius for sure as he explains as simple and easy as possible with all points to make me understood in logic gates operations.
i have a circuits exam (referral exam) in June and i just hope The Organic Chemistry Tutor uploads more videos related to OpAmps, Diodes, Transistors and Small Signal, otherwise i'm definitely failing. Thank you so much for all of the videos you upload, you've helped me a lot!!
After watching this video,I now understand why you 've got almost 3M subscribers.You make things expicit to grasp,which will take me a long time and enormours energy.
2 years past, now he has 7M subs
Wow! That sure wore my brain out trying to develop each circuit BEFORE you showed and explained the right circuit. I still haven't developed the intuitive understanding, but this certainly helped me better see the current flow and voltage differences. I've gone through your videos on circuit analysis and those helped me follow your explanations after the fact. My wife walked in while i was looking at the video and she (Masters in Adult Education) exclaimed "Now THAT'S the right way to teach!", being impressed with the clarity of how you had laid things out. Congratulations and thanks. I had to keep re-watching parts to fully understand, but definitely worth the time. I've watched videos which clearly explained how to use logic gates with switches (esp. ElectroBoom), but this is the first showing how the logic gates themselves can be built.
the best teacher in the world... he teaches in very easier way...
thank you very much your explanation is very simple that make me understand very well 👏👏👏👏
OR - 0:11
NOR - 3:40
AND - 8:30
NAND - 14:43
This guy knows I'm his biggest fan
Awesome video, also thank you for doing it with 5V and LEDs making it easy to test out a breadboard with an arduino. Great learning
yes we have exam tomorrow 😂
*today 🙂
Ah thank you so much I've been searching everywhere for a reasonable explanation and finally ended up here.
Damn!! This is a great explanation. Cleared all my doubts.
Best teacher ever💯
Thanks very much for the very simplified explanation,you are a real pro
Don't know if you will see this or not, but have you ever considered being a private tutor? I would pay good money for you to teach me this subject (electrical circuits and engineering). Though your videos are very helpful, I still get confused and i'd love to be able to ask you questions in real time.
Brilliant explanation for this topic!
★★★★★
as always thanks for the hard work your putting out.
It's only God that will bless you
Thanks so much
Amazing video and very clear and detailed explanations. Thank you sir.
the first time ive left from a toct video without understanding anything :/
Nice. Very comprehensive. The picture helped a lot so the concept is more visual than being abstact. Thanks.
Cant thank enough! RESPECT ++
Super explanation 🎉sir
thanks a lot bro!!
good explanation gates
Great videos. You explain very well. Are you going to add videos on FETs and more? Thank you
Awesome job!
Thank you!
I think that earlier in the video he meant to say "at least" with respect to the other examples with 0.7 Volts as well. I'm guessing that these aren't the only circuit constructions with represent these truth tables?
Wow, this is amazing. Do you have one for Transistor Logic Gates?
yes we have mids tomorrow
Really great presentation, thanks!
mum me lee
Thanks a lot. ❤️
5:30 That NOR example isn't a real gate, hasn't a real output that you can use, I wish it work like that, but you need a transistor to invert OR diode gate into a real NOR gate with output a signal.
17:00 why does the current flow through the R2 and the led instead of the path with R1?
It will flow through both paths.
It's been wonderfully taught that even a moron can get it so easily...
It's been wonderfully taught that even a moron can get it so easily...
@Killer Boy It's been wonderfully taught that even a moron can get it so easily...
jesus, im worse than a moron
Hey, thanks a lot for the video, it really helped me for my exam.. just a small doubt, in the NAND gate if we choose R3 such that the voltage drop is enough for the LED(>=2V), will current flow through both R1 and R2 or just R2?
Need Digital Elctronics please
Which software did you use? I whan to make the same
Diodes connected in parallel would act like which gate???
Kindly explain me.
What about all of the other logic gates such as XOR, negative-OR, negative-AND, negative-XOR, etc...Also on the AND-gate towards the end, you said the potential at the point right before the LED was 3.5-V, how is that possible? If the potential is 5-V at A or B and the Diodes drop 0.7-V across each of them , that leaves 4.3-V. Since 4.3 > 3.5, shouldn't that be the potential at that point?
At 6:17, there is electrical potential of 4.3 V after diode and 3.0 V after the LED. Why is electrical potential set at 4.3 V and not 3.0 V at this location? Why is the larger chosen?
The larger value takes dominance at one certain point at a time
The voltage drop across the diode (.7v) is less than that of the LED (2v) so it will take the least resistive path to ground (no potential). In other words, the diode only resists .7v leaving the following point at 4.3v, rather than the led resisting 2v and leaving you with 3v at the following point. The greater potential is from 4.3v to 0v so the larger is chosen.
@@chrisp1582 But will any current actually flow through the LED? I need to set this up on a breadboard and do measurementts.
You are saying the general rule is this?: When two or more paths connet to a common node, the voltage from that node will be that of the highest voltage source.
@@typedeaf Yes, current will flow through the led. This is simply the measurement at the connecting node which would be that of the least resistive path. The voltage at the node will be that of the greatest potential to ground which can change based on the volume of voltage from the source. So if you increased the voltage source from another point that value could change at that connecting node.
(1) At 6:17, there is electrical potential of 4.3 V after diode and 3.0 V after the LED. Why is electrical potential set at 4.3 V and not 3.0 V at this location? Why is the larger chosen?
(2) At 5:27, what is purpose of diodes in OR gate? It seems the circuit would work without them and they are redundant.
For 2) in actuality the values 0s and 1s will not correspond to 0 and 5v
We need some device which will restrict current at a some voltage and diode does this, If A had a greater potential than at other side , the diode will be reverse biased and offer high resistance and no current flows, but a conductor wld allow current in the other direction
Also, current can pass through conductors even though there is no potential difference...
1. The voltage drop across the diode (.7v) is less than that of the LED (2v) so it will take the least resistive path to ground (no potential). In other words, the diode only resists .7v leaving the following point at 4.3v, rather than the led resisting 2v and leaving you with 3v at the following point. The greater potential is from 4.3v to 0v so the larger is chosen.
2. The purpose of the diodes is to act as a switch in the circuit. If voltage is applied at either A or B then the LED would be OFF due to the greater potential across the diode. Similar to answer 1, the diode resists less which allows a greater potential and an easier path of resistance.
Great
I’m curious as to how these circuits can be used to derive a +5V trigger or gate signal for use with a modular synth circuit. Where would one tap the signal in order to get a healthy +5V based upon the resulting truth tables? I’m especially interested in the AND circuit.
I built one with 2 NPN xstrs but the resulting output is useless. Building it with diodes and a 5 volt rail seems like a good time.
I was able to design the 7 logic gates using only diodes, without any transistors
It's awesome
Bravoo
I use these in Minecraft alot.
❤❤❤
Thaaaanks
you are damn genius !!
In the case of AND gate, how can I achieve that 2 ground wires will only produce one ground wire if both are grounded? Cheers
Please do biochem videos.
Where is the OUTPUT node of these gates? With respect to inputs A and B, there must be an usable Output pin to drive, e.g., the base of a transistor.
The output is where the LED is. Remember that this is diode logic. It has long been extinct in logic circuit design. Transistors had not been invented yet when this was still being used.
why google say there are only two types of diode logic ? one is diode logic OR gate and diode logic AND gate .
That maybe because the nand nor xor and xnor logic gates are dervied from those and are not specific
Whats the purpose of those resistors
have an hour for exam😂
please i have a question, for the And gate why can't current flow from the 5v to led?
Diode blocks it
talented
How to make xor gate using diodes
@ 10:00 How are A and B considered as inputs of the AND gate with the diode blocking them? A and B would have to be a switch where the NC is to ground and the NO is to 5v? That seems like a big assumption to not have mentioned. The typical circuit would have A and B as a switch with NC being an open circuit.
Also, the only reason the current follows D1 or D2 when they are 0v is because it has less resistance than the LED and second resistor, right? You never state the resitance of the diodes and LED. I am asking this because if you dont consider that fact, and instead first calculate the voltage drop across R1 as if it would next follow the LED, it would only drop 1.5v, which would be enough to power the LED.
This is a 5V logic circuit. All inputs are assumed to be either +5V or 0V (ground). Open circuits are undefined in logic circuitry. They should never occur because they can cause errors.
Secondly, no. The input diodes have a biasing voltage drop of 0.7V, and we are assuming they are connected directly to ground. This means the voltage at "point C" (as he labeled it later in the NAND circuit) will be pulled down to 0.7 volts. That is not enough to turn on the LED.
Voltage drops across resistors do not remain static when the circuit condition changes. When point C becomes 0.7V, the voltage drop across R1 (top resistor) will necessarily become 4.3 volts. If there was a fault in this video, it was the fact that this was not explained clearly.
@@aviandragon1390 Can you explain more about how the diodes "pull down" the voltage to 0.7 at C point.
Is it a characteristic property??
When analyzing circuit behavior we usually assume we are using "ideal" components (which have perfect characteristics) or we assign them a typical value, which is obviously more like what actually happens in the real world.
An ideal diode would have infinite resistance in reverse bias and would have zero resistance in forward bias. That is not what happens in reality, in either case. He is assuming characteristics of a typical diode throughout this video.
Real diodes have a leakage current in reverse bias (this characteristic can almost always be safely ignored, which he does), and they have a voltage drop when forward biased. Ideal diodes are almost never used in circuit analysis, but it might be useful for you to try it yourself in this instance (or imagine what would happen if you replaced the diode with a switch and turned it on and off if that's easier for you).
If your question wasn't already answered by that rambling, see if you can perform thought experiments by replacing a diode in the circuit with an ideal diode or a switch and think about what would happen on your imaginary volt meter at point C or at other places in the circuit.
face reveal at 2 million subs???
👏👏👏
Those dialike must be from our teachers
🙏❤️
who r u
Well explained - but my man.. TH and F do not carry the same sound Both is not pronounced BOF
What about EXOR and EXNOR?
what makes it 4.3 v aorund 8:32
the germanium diode has a voltage of 0.7 v for working, so after passage of 5v, only 4.3 v leaves the diode.
edit- i meant SILICON DIODE*
@@SoumyaMehtalove *silicon. Germanium has a Voltage drop of 0.3V
@@AchiragChiragg thank you, cannot believe i made such a blunder and no one else pointed it out.
This is class 12 syllabus in India
I thought LEDs needed 12-20mA to light up
My professor has easier explanation of NAND gate
Dr punit bajπ ( HOD physics dept)
Link to his video. Oh, he doesn’t have one?
개추
Understand nothing
Stick to chemistry; your teaching technique for logic functions is worse than poor; it is wrong. By changing how the LED is connected to the input diodes, you are flipping back and forth between active-high and active-low logic conventions without a word of explanation. On and Off have no place in a truth table, which is why 50% of your tables are semantically correct, but logically wrong. If you instead assign 1's and 0' to the outputs, you will see that the NOR and NAND tables are wrong. The problem is the schematics. The schematic for a particular type of gate is what it is, with no regard for what is connected to the output. Your truth tables depend on changing output connections, whereas correct ones do not.
Oh boy the yapping convention is in town
@@BradleyCodes nah let him cook; what hes saying is actually pretty important
Booo man