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BJT Current Mirror - Enlighten me

by Machinia
Tags: current, enlighten, mirror
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Machinia
#1
Jan14-12, 03:41 AM
P: 6
Attached is an image of a typical BJT current mirror. The reference current is established by the resistor.

We can calculate R with: R = (V.CC - V.BE) / I.REF

Now what I don't understand:

I know I.C = I.S * e ^ (V.BE / V.T). The collector current is quite sensitive to changes in the base-emitter voltage yes? Then why is it I'm able to choose the almost arbitrary value of 0.7V for V.BE and end up with the calculated reference current?

Could explain to me how the circuit ends up with a precise V.BE (causing the correct current to flow through Q1 and Q2) despite our rough calculations? Or maybe my whole thought process is wrong, I'm not really sure at this point.
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Kholdstare
#2
Jan14-12, 04:15 AM
P: 390
Ib = Is exp(Vbe/Vt)

Iref1 = Ic + Ib = (beta + 1) Is exp (Vbe/Vt)
Iref2 = Ic = beta Is exp (Vbe/Vt)

So Iref2 = Iref1 when beta >> 1
Vbe may not be exactly 0.7V. The important thing is whatever is Vbe for Q1 is Vbe for Q2
NascentOxygen
#3
Jan14-12, 06:51 AM
HW Helper
Thanks
P: 5,137
Quote Quote by Machinia View Post
Attached is an image of a typical BJT current mirror. The reference current is established by the resistor.

We can calculate R with: R = (V.CC - V.BE) / I.REF
If Vcc is >> Vbe, then an inaccuracy of 0.1 volt or so in the value used for Vbe won't have much percentage effect on the difference (Vcc - Vbe).

yungman
#4
Jan14-12, 01:23 PM
P: 3,883
BJT Current Mirror - Enlighten me

Quote Quote by Machinia View Post
Attached is an image of a typical BJT current mirror. The reference current is established by the resistor.

We can calculate R with: R = (V.CC - V.BE) / I.REF

Now what I don't understand:

I know I.C = I.S * e ^ (V.BE / V.T). The collector current is quite sensitive to changes in the base-emitter voltage yes? Then why is it I'm able to choose the almost arbitrary value of 0.7V for V.BE and end up with the calculated reference current?

Could explain to me how the circuit ends up with a precise V.BE (causing the correct current to flow through Q1 and Q2) despite our rough calculations? Or maybe my whole thought process is wrong, I'm not really sure at this point.
This is a very typical circuit used in integrated circuits to set up constant current sources. The point is not to get precise Iref, the point is to set up current sources that is proportion to Iref.
You are right, Vbe is not exactly 0.7V, it's only an approximation for calculation. If Vcc is 10V, Iref=(10-Vbe)/R. error is very small if Vbe is off by 0.1V. But yes, if Vcc is low like 2V, the error is much larger. But in the whole picture, it is not important. You should never rely on absolute value on the design.
In fact inside IC, the absolute value of the R is the biggest error. In my days of designing BJT ICs, the resistor is accurate to only 30% ei. if you say 1K, one run can be as low as 700Ω and the next run can be 1.3K!!!! That is where the error really started. Vbe is never even a concern in real life.

Now in addition to your question. This current mirror work only in IC where you can control the dimension of the transistor and design so the two transistors are right next to each other and at the same direction. This will make their characteristics very close and their temperature is closely tracked. You can get good current ratio out of this design. But still if you want precision current mirror, you are going to use emitter resistors. Now you must question about the 30% accuracy of the resistors!!! The answer is the resistors are not accurate, BUT they can be made to track and proportion with each other very precise. You layout the two resistors right next to each other to get a very tight match with each other much more so if you use two discrete 0.5% resistors.

In IC design, it is only about ratio, not about absolute. That's the reason why you see the power supply current has a large range. Look at the data sheet of opamp, they always spec typical say 1mA, but always spec max of something like 2mA or worst. That's mostly because of the 30% error of the resistors when setting up the current mirror.
jim hardy
#5
Jan14-12, 03:21 PM
Sci Advisor
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P: 3,498
this might help..

at low current ,
where voltage drop across resistance of internal parts(silicon wafer and little gold connectiing wire) isn't significant,

current through a junction is pretty close to theory:

i = e^( QV / KT)
where
e is 2.718+++,
Q is charge of electron,
V is voltage,
K is Boltzmann's constant,
T is absolute temperature

so if the two transistors eb junctions are matched, having same doping and construction
and are at same temperature, maybe even both on same silicon die

and they have same voltage across them, their currents will match.
Everything except voltage and temperature is an established physical constant.
Machinia
#6
Jan14-12, 06:29 PM
P: 6
Ok thanks to all who replied, this is making more sense and there are some great tips on circuit design.

My initial problem was when I first looked at this circuit, I was under the impression transistors could only amplify base currents, as in Ic = beta Ib. So I wondered how did Q1 know how much base current to pull? After some studying I realized that as long as the eb junction is forward biased, a base current will automatically flow that is proportional to the collector current which has been established by our resistor. (Correct?)

But then what didn't make sense is what I posted, to do with how Vbe is established. My basic thought was Vbe establishes Ib and not the other way around.

What I've come up with is for Q1: Ic (~Iref) causes Ib which establishes a specific Vbe which is then of course applied across the e-b junction of Q2 causing the same Ib to flow into Q2. Yes/no?
yungman
#7
Jan15-12, 12:59 AM
P: 3,883
Q1 has the base and collect shorted, it is connected as a diode. In this case, all I care is Ie=Ic + Ib. That is the current that mirror to Q2. Yes Ic=βIb. So if you assume Vbe is 0.7V, then you can calculate Ie=(Vcc-Vbe)/R. The formula for Vbe is

[tex]V_{BE}=V_T\; ln(\frac{I_C}{I_S})\;\hbox { where } V_T=\frac{kT}{q}[/tex]

[tex] I_C= I_S \;e^{\frac {V_{BE}}{V_T}}[/tex]


[Quote]What I've come up with is for Q1: Ic (~Iref) causes Ib which establishes a specific Vbe which is then of course applied across the e-b junction of Q2 causing the same Ib to flow into Q2. Yes/no?[Quote]

Not really, it is the Vbe matching that matter. In IC, if the two transistor are matched, If you draw Iref and Vbe of Q1 is certain amount, if you put that voltage across Vbe of Q2, you will make Q2 sink Iref if Q1 and Q2 is exactly the same size and dimension. In fact for precision current mirror, we put another transistor Q3 with the base at collector of Q1, base of Q1 no longer connect to the collector. Q3 emitter the emitter drive the base of both Q1 and Q2. With this, we don't even care about the base current as it is buffered by Q3. We only worry about the Vbe.
Kholdstare
#8
Jan15-12, 01:06 AM
P: 390
See, If you want to fix Iref you can only control two parameters R and Vbe (Your Vcc is fixed by power supply.) Again, Vbe in turns depends on Ib and Ic is related to Ib. Thus Iref is a function of Ib and Ic or simply Ib. So,
Vbe = Vt * ln( Iref / ( Is * (beta+1) ) )
Put this in,
R = ( Vcc - Vbe ) / Iref
and you get,
R = ( Vcc - fn(Iref) ) / Iref
So the solution is not direct where you fix some parameter and get the desired output. (I guess the equation does not have a direct solution). You have to iteratively solve to find out what will be Iref for given parameters. I say Iref only depends on R, Vt, Is and beta.
Kholdstare
#9
Jan15-12, 01:08 AM
P: 390
What I've come up with is for Q1: Ic (~Iref) causes Ib which establishes a specific Vbe which is then of course applied across the e-b junction of Q2 causing the same Ib to flow into Q2. Yes/no?
Yes
Machinia
#10
Jan15-12, 01:48 AM
P: 6
1 not really, 1 yes. Ok, getting close.

@yungman: I understand the importance of transistor matching. It's easy to understand how the whole thing works: If Q2 and Q1 are the same, Q2 mirrors Q1 due to their equal Vbe (With a bit of error due to base currents). But me having no real experience with BJTs in general started asking questions about its operation in more detail. I like to think I figured it out with my bolded statement a few posts up. Also the bolded statement is just for my understanding.
yungman
#11
Jan15-12, 04:41 AM
P: 3,883
Quote Quote by Machinia View Post
1 not really, 1 yes. Ok, getting close.

@yungman: I understand the importance of transistor matching. It's easy to understand how the whole thing works: If Q2 and Q1 are the same, Q2 mirrors Q1 due to their equal Vbe (With a bit of error due to base currents). But me having no real experience with BJTs in general started asking questions about its operation in more detail. I like to think I figured it out with my bolded statement a few posts up. Also the bolded statement is just for my understanding.


Here is the detail equation of transistors.

http://en.wikipedia.org/wiki/Bipolar...ion_transistor
jsgruszynski
#12
Jan18-12, 11:01 AM
P: 276
Quote Quote by Machinia View Post
Attached is an image of a typical BJT current mirror. The reference current is established by the resistor.

We can calculate R with: R = (V.CC - V.BE) / I.REF

Now what I don't understand:

I know I.C = I.S * e ^ (V.BE / V.T). The collector current is quite sensitive to changes in the base-emitter voltage yes? Then why is it I'm able to choose the almost arbitrary value of 0.7V for V.BE and end up with the calculated reference current?

Could explain to me how the circuit ends up with a precise V.BE (causing the correct current to flow through Q1 and Q2) despite our rough calculations? Or maybe my whole thought process is wrong, I'm not really sure at this point.
The very, very simplest explanation about the Vbe: there is a KVL loop that assures the Vbe must be the same. Nothing more is required to explain it.

The other issue, however, is whether the transistors are "well matched" in performance which can introduce errors in the mirroring.


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