Still not equal... IX = (IL+2Ib) its just approximating... KCL not allows us todo that. I frist transistor Ix= ic+2ib and but in 2nd transitor IL = ic so its just approximation
@@incxxxx , it should work. Of course it won't work exactly as calculated because there are some assumptions we made in the calculations that aren't necessarily true. For example, I assumed that the two transistors have the same beta value, but if they don't IB1 and IB2 won't be the same.
Thank you . Even though ive been in the electronics industry for nn years and now very retired, I still like to tinker , and never in the past requred or fully understood current mirrors. This explination video concludes my understanding.
Great explanation! I would add, this specific circuit will only work as a mirror if the load on Q2 has a lower resistance than the resistor on Q1. Because in this circuit, Q1 is at the knee of the saturation curve, (VCE is at the minimum of 0.7V) so Q2 can only regulate to reduce current (and increase VCE). If you add a current control in series with Q1, and move away from saturation, you then have room to regulate Q2 in either direction.
Thanks a lot! Finally I heard the key point - the current in load is managed to be constant by change of voltage E-C of Q2. Even P. Horowitz and W. Hill did not highlight it in "The Art of Electronics" : )
Awesome video, very clear and concise. I really appreciate that you showed the load current. Other videos just leave that node floating which can make it confusing.
love you man.... great explanation... i just got it by watching two times... my professor couldn't do that by explaining this for like 10 times... you are amazing
really good, for those who did not approximate IB1, and IB2 ( where IB1=IB2=IE1/B+1=IE2/B+1 , so IE1=IE2) I got for Ix=IE1(B+2)/B+1 , so since B+2>B+1 always, Ix>IE1 always, but B+2~=B+1, so the same conclusion Ix~=IE1=ILoad
A big reason the transistors are in one package is to keep them thermally coupled, As the output transistor heats up the input transistor heats up too keeping them matched.
why are the emitter currents equal to eachother? i understand that the base currents are the same but doesn't the emitter current also depend on the voltage between collector and emitter? (which is likely different for q1 and q2)
If you use two different discrete transistors, then likely the emitter currents will be different. If the current mirror is fabricated on the same die, the two transistors can be made to have nearly identical characteristics, so the base currents and the emitter currents will be the same. As +Robonza noted, having the transistors in the same package also keeps them thermally coupled.
I was wondering too. Let's put equal resistors R_1 and R_2 just before the base of of Q_1 and the base of Q_2 respectively. We have by Kirchhoff's law V_CC = (I_x)(R_x)+(I_B1)(R_1)+0.7 and similarly V_CC=(I_x)(R_x)+(I_B2)(R_2)+0.7 where 0.7 is the base-emitter voltage. Thus I_B1=I_B2 Now we may argue that the wires have this small resistance, and therefore the currents are equal, however if we considered the resistance of these wires, we should also consider the resistance of the emitter-ground wires, should not we ?! To be honest I have never seen any convincing argument on why they are equal, but using resistors makes sense anyway.
Check out Wikipedia's article "Bipolar junction transistor" -- The emitter current Ie is determined by emitter-base voltage (Vbe) and by emitter saturation current Ies (a constant) -- Ie = Ies [exp ((Vbe/Vt) - 1)] If the transistors are matched, then Ies of transistor 1 = Ies of transistor 2. Since the transistors' emitters are grounded and since their bases are directly connected, then the Vbe of transistor 1 = the Vbe of transistor 2. Hence the Ie of transistor 1 = the Ie of transistor 2.
Avid Williams: As a retired electronics engineer, not an IC designer , but a USER ...with no choice ! I often asked myself how does a current mirror work. An almost correct explanation, BUT YES : Vbe1 - Vbe2 by design , Ie depends upon Vbe, temperature, and characteristics of the junction material in the Q1, Q2 which are similar but NOT IDENTICAL these characteristics depend upon the molecular level of the transistor junction , which coarsely controllable So NO: Ie1 = Ie2 except by can by chance: BUT, YES Ie1 ~ Ie2 : where ~ means approximately, but NOT EQUAL
We are making the assumption that the two BJTs have identical characteristics. Since the bases are connected and the emitters are connected (ground), the base-emitter voltages (VBE) of the the two BJTs are equal. Since the BJTs are identical and the VBEs are the same, the current in to the bases must be the same. Also, current's out of the emitter must also be the same.
Hi, Why IC1 isn't equal to 0? As I see it, the BC junction of Q1 is shorted so IE1 should be equal to IB1, and because Q1 the same as Q2 -> IX=2IB so IE1=IE2=IX/2 and as IC2=IE2-IB2=0 it shouldn't work?
So what happens when the transistors go to saturation mode?? Also I have seen a lot of circuits having a current source instead of Rx - So is this any different from those circuits?
The resistor is a current source in this circuit bc it controls current Ix. So theres nothing different about those circuits besides them using a different type of current source.
How does he assume that all of the current Ix goes to the bases of the two transistors? Shouldn't it branch out to equal: Ix = 2Ib + Ic ? (While calculating Ix using the supply voltage and voltage drop Vbe with Ohm's law)
Because the collector and the base are tied together, the collector voltage will be equal to the base voltage. The emitter is tied to ground, so the voltage at the collector will be Vbe. The voltage across the resistor will therefore be Vcc-Vbe and Ohm's law can be used to calculate Ix.
Okay so I've simulated this circuit in everycircuit...but I put a Vcc on Q2, I've also put a different resistor value for the load, but the currents aren't the same
Sir, could you please explain why you traced the path that way when you were looking for the value of Ix using KVL? Isn't KVL applicable for loops only? How does the path you trace become a loop? Thanks :)
4.98mA is pretty close to 5mA, which should be the value of Ix if you have used Vcc=10V and Rx=10kOhm. So therefore there is nothing wrong with your calculations
Without resistors, between emitter and ground, it does not work, is highly unstable and current in the second transistor rises many times over right value and distroys it.
For the example I gave, I just made the value up. For a given transistor it will be a property of the transistor. The beta will be in the datasheet. Usually it is called Hfe.
They are not actually equal, but are very close. To be precise: IE = IC + IB, but since IC = beta*IB and beta is typically pretty big (let's say around 100 for arguments sake) then IC is 100 times bigger than IB. Meaning IE = IC within about 1%
It's pretty simple... the transistor bases are tied to Rx, which sets the current gain equally for both transistors. This typology is commonly used to balance a Long Tailed Pair.
![BLACK BAG - Official Trailer [HD] - Only in Theaters March 14](http://i.ytimg.com/vi/Du0Xp8WX_7I/mqdefault.jpg)
Man, you have such a good voice to explain this kind of stuff
Thanks for the awesome comment - made me smile.
This was an awesome explanation. I really struggled with amplifiers in college, I’m brushing up on it now. Current mirrors were a nightmare I remember
you re not alone 😄😄
Still not equal... IX = (IL+2Ib) its just approximating... KCL not allows us todo that.
I frist transistor Ix= ic+2ib and but in 2nd transitor IL = ic so its just approximation
Finally I got the idea how the c. mirror works. Thank you for your explanation.
So try it experimentally. It will not work!
@@incxxxx , it should work. Of course it won't work exactly as calculated because there are some assumptions we made in the calculations that aren't necessarily true. For example, I assumed that the two transistors have the same beta value, but if they don't IB1 and IB2 won't be the same.
@@incxxxx ..No, the circuits work
@@nickharrison3748 Witout emiter resistor it doesnot work - I found this experimentally!
@@incxxxx ..ok good. I did not try this one but I was referring to Amplifier circuits that Dave described. Those work for me
Thank you . Even though ive been in the electronics industry for nn years and now very retired, I still like to tinker , and never in the past requred or fully understood current mirrors. This explination video concludes my understanding.
Fantastic explanation. I always find youtube videos to be far more cognizant of their audience than lecturers or text books
This has given the total working needed for band gap references... Thank you brother
Great explanation! I would add, this specific circuit will only work as a mirror if the load on Q2 has a lower resistance than the resistor on Q1. Because in this circuit, Q1 is at the knee of the saturation curve, (VCE is at the minimum of 0.7V) so Q2 can only regulate to reduce current (and increase VCE). If you add a current control in series with Q1, and move away from saturation, you then have room to regulate Q2 in either direction.
Thanks for the comment. Yes, this is the simplest form for a current mirror.
Thanks a lot! Finally I heard the key point - the current in load is managed to be constant by change of voltage E-C of Q2. Even P. Horowitz and W. Hill did not highlight it in "The Art of Electronics" : )
Awesome video, very clear and concise. I really appreciate that you showed the load current. Other videos just leave that node floating which can make it confusing.
Thanks so much for the great comment.
Thank you so much for this explanation! You made it so clear with the assumptions explained and even an example which is exactly what i needed!
Glad it was helpful!
love you man.... great explanation... i just got it by watching two times... my professor couldn't do that by explaining this for like 10 times... you are amazing
So what? Try to do it experimentally and you see if it works. Im assure you it will not work!
really good, for those who did not approximate IB1, and IB2 ( where IB1=IB2=IE1/B+1=IE2/B+1 , so IE1=IE2) I got for Ix=IE1(B+2)/B+1 , so since B+2>B+1 always, Ix>IE1 always, but B+2~=B+1, so the same conclusion Ix~=IE1=ILoad
Thank you , I still have some confusion about the source voltage of Q2 but , I never had it explained to me this good
Thank you David ! This is a valuable contribution to make the operation of current mirrors understandable.
Thanks!
A big reason the transistors are in one package is to keep them thermally coupled, As the output transistor heats up the input transistor heats up too keeping them matched.
Good explanation David.
Regards.
Great video thank you! Absolutely loved the music you used, thank you for the credit so I could look it up!
awesome explanation
awesome vid! thanks for the explanation
Thank you so much!! Very Clear, Thanks again!!👍👍👍
Awesome explanation!
Straight to the point. Thank you
I learned a lot from this video, thanks a lot
You're welcome. Thanks for watching
you are the best Sir thank you so much
#RespectFromSouthAfrica
why are the emitter currents equal to eachother? i understand that the base currents are the same but doesn't the emitter current also depend on the voltage between collector and emitter? (which is likely different for q1 and q2)
If you use two different discrete transistors, then likely the emitter currents will be different. If the current mirror is fabricated on the same die, the two transistors can be made to have nearly identical characteristics, so the base currents and the emitter currents will be the same. As +Robonza noted, having the transistors in the same package also keeps them thermally coupled.
I was wondering too.
Let's put equal resistors R_1 and R_2 just before the base of of Q_1 and the base of Q_2 respectively. We have by Kirchhoff's law
V_CC = (I_x)(R_x)+(I_B1)(R_1)+0.7
and similarly
V_CC=(I_x)(R_x)+(I_B2)(R_2)+0.7
where 0.7 is the base-emitter voltage.
Thus I_B1=I_B2
Now we may argue that the wires have this small resistance, and therefore the currents are equal, however if we considered the resistance of these wires, we should also consider the resistance of the emitter-ground wires, should not we ?!
To be honest I have never seen any convincing argument on why they are equal, but using resistors makes sense anyway.
Check out Wikipedia's article "Bipolar junction transistor" --
The emitter current Ie is determined by emitter-base voltage (Vbe) and by emitter saturation current Ies (a constant) --
Ie = Ies [exp ((Vbe/Vt) - 1)]
If the transistors are matched, then Ies of transistor 1 = Ies of transistor 2.
Since the transistors' emitters are grounded and since their bases are directly connected, then the Vbe of transistor 1 = the Vbe of transistor 2.
Hence the Ie of transistor 1 = the Ie of transistor 2.
What a great video!! thank you very much
Solve numerical on current mirror circuit to find current when beta is given 100
Thank you. Finally, everything is clear.
Avid Williams:
As a retired electronics engineer, not an IC designer , but a USER ...with no choice !
I often asked myself how does a current mirror work.
An almost correct explanation, BUT
YES : Vbe1 - Vbe2 by design , Ie depends upon Vbe, temperature, and characteristics of the junction material in the Q1, Q2 which are similar but
NOT IDENTICAL these characteristics depend upon the molecular level of the transistor junction , which coarsely controllable
So NO: Ie1 = Ie2 except by can by chance:
BUT, YES Ie1 ~ Ie2 : where ~ means approximately, but NOT EQUAL
Really good. Hope you come back.
You are a life saver... Thanks a lot!!!
Why does the same voltage imply the same current? Just as a property of the amplifier?
How do we know the base currents are equal to each other and emitter currents are equal to each other?
We are making the assumption that the two BJTs have identical characteristics. Since the bases are connected and the emitters are connected (ground), the base-emitter voltages (VBE) of the the two BJTs are equal. Since the BJTs are identical and the VBEs are the same, the current in to the bases must be the same. Also, current's out of the emitter must also be the same.
well done.exellent🤗
2:59 why Ic and Ie are getting same.??
They aren't exactly the same Ie = Ic + Ib, but since Ib = Ie/(beta) and we are assuming beta is quite big, Ib is small enough to ignore
@@ElectronXLab ... thanks 😄 a lot..one of the best explanation I have ever seen
Thanks man..easy , simple ...appreciated
In my simulator;however, I noticed that if load resistance is greater than that if Rx, it will decrease the current of q2
If, for maintaining IL VCe of load transistor changes, wouldn't that drive the transistor out of linear region?
Simply thank you...finally got it!
So what? Try to do it experimentally and you see if it works. Im assure you it will not work!
Hi, Why IC1 isn't equal to 0? As I see it, the BC junction of Q1 is shorted so IE1 should be equal to IB1, and because Q1 the same as Q2 -> IX=2IB so IE1=IE2=IX/2 and as IC2=IE2-IB2=0 it shouldn't work?
So what happens when the transistors go to saturation mode?? Also I have seen a lot of circuits having a current source instead of Rx - So is this any different from those circuits?
The resistor is a current source in this circuit bc it controls current Ix. So theres nothing different about those circuits besides them using a different type of current source.
What is the small signal model?
Why are the collector and base shorted??
Nice video!
How does he assume that all of the current Ix goes to the bases of the two transistors? Shouldn't it branch out to equal: Ix = 2Ib + Ic ? (While calculating Ix using the supply voltage and voltage drop Vbe with Ohm's law)
What you said (Ix=2Ib+Ic) is correct and is what I stated in the video.
David Williams No, I know that but I think I asked my question the wrong way. Why is Ix=(Vcc-Vbe)/Rx ? The video is great btw, keep up the good work!
Because the collector and the base are tied together, the collector voltage will be equal to the base voltage. The emitter is tied to ground, so the voltage at the collector will be Vbe. The voltage across the resistor will therefore be Vcc-Vbe and Ohm's law can be used to calculate Ix.
Ok, thanks for clearing it out for me.
Okay so I've simulated this circuit in everycircuit...but I put a Vcc on Q2, I've also put a different resistor value for the load, but the currents aren't the same
Try adding a resistive load
Sir, could you please explain why you traced the path that way when you were looking for the value of Ix using KVL? Isn't KVL applicable for loops only? How does the path you trace become a loop? Thanks :)
It's an implied loop. There is a power supply in the loop that is spit in to VCC at the top of the diagram and ground at the bottom of the diagram.
@@ElectronXLab Sir could you explain it a bit further? I'm not sure if I understand :)
@@RanielDG David's explanation was perfect - the 'implied loop' is a valid KVL loop, and that loop completes the circuit.
Can I use diode 1n4148 instead of q1?
Using LTspice and used the same values, if the Load is a 2k resistor, then IL is only 4.98ma, not the same as Ix. What's did I do wrong?
4.98mA is pretty close to 5mA, which should be the value of Ix if you have used Vcc=10V and Rx=10kOhm. So therefore there is nothing wrong with your calculations
What is happening if you have current mirror in mosfets? Could you make a video for them as well?
The circuit is the same, you just do the math using the gate voltage instead of current.
What is that yellow thing at the end?
It's a hexbug :-)
@@ElectronXLab cool, never seen that
Woah this is magic
Ix=(VCC-VCE)/Rx VCE not VBE
Difference between capacitor and level shifter circuit
you didnt take into account the voltage drop on resistor. its gonna affect the current but the calculations are fine i guess?
Thank you
Without resistors, between emitter and ground, it does not work, is highly unstable and current in the second transistor rises many times over right value and distroys it.
Why current mirror circuit use in integrated circuit
This video is awesome!!!!!!!
Totally understood thanks bro
So what? Try to do it experimentally and you see if it works. Im assure you it will not work!
Thanks, dude. seriously. Same explanation for MOSFET right?
Thank you so much
Wher did u get beta
For the example I gave, I just made the value up. For a given transistor it will be a property of the transistor. The beta will be in the datasheet. Usually it is called Hfe.
thanks for the video
Thanks bro .. clear.
thanks
Thanks.
thx m8 understood. subbed :)
Like and subscription
For me, It is genius.
I really appreciate that, thanks.
Perfect thx
How IC and IE are equal
They are not actually equal, but are very close. To be precise: IE = IC + IB, but since IC = beta*IB and beta is typically pretty big (let's say around 100 for arguments sake) then IC is 100 times bigger than IB. Meaning IE = IC within about 1%
It's pretty simple... the transistor bases are tied to Rx, which sets the current gain equally for both transistors. This typology is commonly used to balance a Long Tailed Pair.
❤️❤️❤️
Back at ya!
Simulate some circuits! Look up androidcircuitsolver on google