From this schema enter image description here

I don't understand this sentence from here, particulary in section "Basic BJT current mirror"

"...If Q1 and Q2 are matched, that is, have substantially the same device properties, and if the mirror output voltage is chosen so the collector-base voltage of Q2 is also zero..."

My question is : Why collector-base voltage of Q2 is also zero ?

  • 2
    \$\begingroup\$ To ignore the early effect \$\endgroup\$ – G36 Sep 15 '20 at 14:50
  • \$\begingroup\$ The sentence you quote does not say that the collector-base voltage of Q2 is zero. It says if it is zero. \$\endgroup\$ – Justin Sep 15 '20 at 14:58
  • 1
    \$\begingroup\$ @G36 *Early after James M. Early \$\endgroup\$ – Spehro Pefhany Sep 15 '20 at 15:16

Without getting too involved in the details, there are typically two very important and one slightly less important (usually) details to keep track of with respect to a simplified BJT model. (Ignoring temperature differences.) These are the saturation current, \$I_\text{SAT}\$, which associates the relationship between \$V_\text{BE}\$ and \$I_\text{C}\$; the value of \$\beta\$, which associates the relationship between \$I_\text{B}\$ and \$I_\text{C}\$; and a parameter describing the Early effect, \$V_A\$, which modifies the earlier relationship between \$V_\text{BE}\$ and \$I_\text{C}\$ by including an effect due to \$V_\text{CE}\$.

If you can hold just the first two ideas in mind at one time, you can analyze many circuits close enough for usable results. The third idea, the Early effect, can become important in other circuits. This one is an example where it often may matter.

This is a good point to suggest that you also look at the cascode arrangement where one BJT is used in a way to pass along the collector current of another BJT while also mitigating the Early effect in that other BJT. But I won't discuss it here.

So let's look at the schematic, with some added bits to help clarify as I write more:


simulate this circuit – Schematic created using CircuitLab

On the left, I've added two resistors. These are due to the Early effect. Their values can be approximated with \$r_o\approx \frac{V_A}{I_\text{C}}\$. On the right, I've simplified the diagram, slightly. Since \$Q_1\$ is diode-connected, we can replace it with a diode symbol. This may help get some points across more easily. (Or it may do the opposite. We'll see.)

The first part of the quote from your link is:

If \$Q_1\$ and \$Q_2\$ are matched,

For the most part, by matched they mean the first parameter I mentioned: \$I_\text{SAT}\$. This is the most important parameter to match because of the Shockley equation relating \$V_\text{BE}\$ and \$I_\text{C}\$ (lacking the Early effect):


You can solve the above for \$V_\text{BE}\$, as well:

$$V_\text{BE}=V_T\cdot \ln\left(\frac{I_\text{C}}{I_\text{SAT}}+1\right)$$

If you know one, you can calculate the other.

The current mirror depends upon this fact. By letting \$R\$ set \$Q_1\$'s \$I_\text{C}\$, it responds by generating a \$V_\text{BE}\$. It can do this, because the base is tied to the collector and will draw the \$I_\text{B}\$ needed to handle base recombination for \$I_\text{C}\$. (If you assume that \$\beta\to\infty\$, then you can temporarily ignore \$I_\text{B}\$ and therefore not worry about \$\beta\$.)

Since \$Q_1\$'s generated \$V_\text{BE}\$ is passed along to \$Q_2\$'s \$V_\text{BE}\$, this implies that \$Q_2\$'s \$I_\text{C}\$ will be the same.

All this assumes that both BJTs are operating in their active regions, of course.

\$I_\text{SAT}\$ is the most important parameter to match because of the current mirror's ratio of currents is \$\frac{I_{\text{C}_2}}{I_{\text{C}_1}}\approx \frac{I_{\text{SAT}_2}}{I_{\text{SAT}_1}}\$. (You can derive that relatively easily from the above equations and using the given current mirror arrangement above.)

Discrete BJTs from the same device family can have their values of \$I_\text{SAT}\$ vary by as much as a factor of 3 around the average value. This translates to as much as \$\pm 30\:\text{mV}\$ difference in \$V_\text{BE}\$ given the same \$I_\text{C}\$. If you selected poorly enough, getting one BJT from one end of the spectrum and another one from the opposite end, the difference in current could be as much as 8 or 9. Not so good for a current mirror where you want very similar currents. A way to help repair this is to add emitter resistors in discrete designs. But how to design that so it works well is beyond the scope here.

Manufacturers offer BJT pairs. Some are just offered in pairs for convenience and there's no matching intended beyond what accidentally occurs. Some are offered with varying degrees of \$V_\text{BE}\$-matching. Some are offered with very good degrees of \$V_\text{BE}\$-matching and then also reasonably good degrees of \$\beta\$-matching, as well. (The latter costs more, of course.)

The next part of the quote from your link is:

that is, have substantially the same device properties,

I've already discussed the important property, \$I_\text{SAT}\$. But I've left out \$\beta\$, above. So I'll address it quickly.

\$\beta\$ is a device property and not all of the current supplied by \$R\$ makes it to the collector of \$Q_1\$. If the value of \$\beta\$ is different between the two BJTs, or if it is the same between them, the fact is that this causes an error in the estimate of \$Q_2\$'s \$I_\text{C}\$, since \$R\$ has to supply two base currents as well as \$Q_1\$'s \$I_\text{C}\$. (This can be greatly aided by the addition of another BJT, not shown.)

Assuming the BJTs are matched on \$I_\text{SAT}\$ and if the \$\beta\$ of both BJTs are exactly the same, the current in \$R\$ still must be reduced by \$2\cdot \frac{I_\text{C}}{\beta}\$ in order to figure out what remains for \$Q_1\$'s \$I_\text{C}\$. Again assuming the BJTs are matched on \$I_\text{SAT}\$ and if the \$\beta\$ of the BJTs are different, then the current in \$R\$ must be reduced by \$\frac{I_{\text{C}}}{\beta_1}+\frac{I_{\text{C}}}{\beta_2}\$ in order to figure out what remains for \$Q_1\$'s \$I_\text{C}\$. (Both will still have the same \$I_\text{C}\$ if they are matched on \$I_\text{SAT}\$.) This is just a matter of figuring out \$Q_1\$'s \$I_\text{C}\$ and the values of \$\beta\$ are typically high enough (and can be repaired by the addition of another BJT, anyway), so I don't really count it as an issue here. But do be aware of it, anyway.

The next part of the quote from your link is:

and if the mirror output voltage is chosen so the collector-base voltage of \$Q_2\$ is also zero,

The point here is addressing itself now to the Early effect. (Those resistors I show in the schematic.) \$Q_1\$ is being operated with its base voltage exactly equal to its collector voltage. So this means, by definition, that \$V_\text{BC}=0\:\text{V}\$ for \$Q_1\$. So \$V_\text{CE}=V_\text{BE}\$, also. Since \$V_{\text{CE}_1}\$ is impressed across \$r_{o_{_1}}\$, it follows that there is a small current through it and this adds to what is observed as \$Q_1\$'s \$I_\text{C}\$. The same thing happens with \$Q_2\$. So if you want to observe the same \$I_\text{C}\$ in \$Q_2\$, then \$Q_2\$'s \$V_\text{CE}\$ must also be the same. But this means that \$Q_2\$'s \$V_\text{BC}=0\:\text{V}\$, as well. That's why they wrote that.

In the above schematics, this would mean that \$V_\text{X}= V_\text{BE}\$. In reality, this almost never happens (unless you use one of variations of a Wilson mirror.) So if, say, you set \$V_\text{X}=V_\text{CC}\$ and you place a resistor in between \$Q_2\$'s collector and \$V_\text{X}\$, then that resistor will drop some voltage due to the current through it. That voltage may, or may not, set the collector voltage of \$Q_2\$ to be exactly the same as \$Q_1\$'s collector voltage. If not, then \$r_{o_{_2}}\$ will sink more current than \$Q_1\$'s \$r_{o_{_1}}\$ does. And this will be yet another unaccounted error.

So they are just being cautious in their wording.

In the interest of some completeness, I'll present some common modifications. You should become familiar with them:


simulate this circuit

In the left-most case above, I've added \$Q_3\$. \$Q_3\$ reduces the error current stolen from \$R_\text{SET}\$, used to supply the two base currents for \$Q_1\$ and \$Q_2\$. That's because \$Q_3\$'s emitter current is doing that job and only a fraction of it (divided by \$Q_3\$'s \$\beta+1\$) is taken from the current supplied by \$R_\text{SET}\$. That will be a lot less than before. The price to pay here is that now \$Q_1\$'s collector voltage has to be \$V_\text{BE}\$ higher. This is easily accounted for when determining the value of \$R_\text{SET}\$, so it's not considered a serious problem but rather a useful improvement.

In the middle case, \$Q_3\$ is moved over into a new position, pushed over so that it falls within the load current path, now. But positioned here, it serves useful two jobs at once. It still supplies the two base currents for \$Q_1\$ and \$Q_2\$. But it also arranges things so that \$Q_2\$'s collector voltage is the same as its base voltage. This effectively nullifies \$Q_2\$'s Early effect. But now \$Q_1\$'s collector voltage isn't the same as its base voltage, but instead one \$V_\text{BE}\$ above its base voltage. So \$Q_1\$ and \$Q_2\$ are operating with slightly different \$V_\text{CE}\$ values and the Early effect has a small effect on the collector current differences. But as the difference isn't that much the problem is relatively small. Another impact is that now \$R_\text{SET}\$ is again supplying more substantial base current to \$Q_3\$. But this is still better than in the case with only \$Q_1\$ and \$Q_2\$, where \$R_\text{SET}\$ had to supply two base currents instead of just one, as in this new case. This is the basic Wilson mirror.

The right side shows what I call a full Wilson mirror. If you look closely, you'll see now that the collectors of both \$Q_1\$ and \$Q_2\$ are identical. (This is very good, as the Early effect will be nearly identical for them.) \$R_\text{SET}\$ now has to again supply two base currents. But as that can be accounted for when setting a value for it, the benefits are considered worth the costs in some circumstances. You could, if you wanted to (and had the extra voltage headroom for it and liked slapping down more BJTs), place a new \$Q_5\$ in a similar position to \$Q_3\$ shown in the left-most schematic to alleviate the base currents drawn away from \$R_\text{SET}\$. But I don't recall seeing it used.

Please note that \$Q_3\$ in the middle and right schematics is operated as a cascode BJT. It passes along \$Q_2\$'s collector current, which is no longer impacted by collector voltage variations as it is isolated from them by \$Q_3\$. You might think that \$Q_3\$ should still suffer from the Early effect, as it's collector voltage can vary depending on the load and the current setting. But since it doesn't control its collector current, as this is set by \$Q_2\$'s collector current, it doesn't matter anymore. It simply passes long the current to the load, regardless.

There are some remaining nuances which would disagree (but only slightly) with what I presented above. But it would draw this out way beyond where I should. Feel free to think about it on your own.


The sentence you are talking about is explaining that current in each transistor is equal when the collector base voltage of Q2 is zero. This means you are setting Vout equal to VB to achieve the same current in each leg of the circuit. This is done in the example to explain the math behind the current mirror.

Usually the collector-base voltage of Q2 is not zero in practical circuits, but the math gets more complicated because you have to take into account the Early effect. The sections after the initial BJT section in your source (output resistance, compliance voltage, and extensions and complications) go into what happens when the collector base voltage is not equal to zero for Q2


The main problem here is confusing the premises and the conclusions. \$V_{BC}= 0\$ is a premise here, not a conclusion. The (full) statement that confuses you is:

If Q1 and Q2 are matched, that is, have substantially the same device 
properties, and if the mirror output voltage is chosen so the collector-base 
voltage of Q2 is also zero, then the VBE-value set by Q1 results in an emitter 
current in the matched Q2 that is the same as the emitter current in Q1.

What this sentence is telling us, is that IF the following happens:

  • Q1 and Q2 are matched
  • \$V_{OUT}\$ is such that \$V_{BC2}= 0\$

Then, \$I_{E1}= I_{E2}\$.

So, you are right in doubting if \$V_{BC2}= 0\$. It is not \$0\$ in every case, but the analysis made is telling you that IF that voltage is zero, emiter currents are equal. We can conclude, that if the case is \$V_{BC} \neq 0\$, then \$I_{E1} \neq I_{E2}\$.

  • \$\begingroup\$ Effectively, I have confused "so" like "then". Like it is spoken about for \$V_{BC2}\$, i understand that is \$V_{BC2}\$. But from LTspice' graphic response (see the modified schema at the end), for \$V_{BC2} = V(A, vbe) = 10V\$, then \$I_{E1} = I_{E2} \$ \$\endgroup\$ – user7058377 Sep 16 '20 at 15:51
  • \$\begingroup\$ @user7058377 you need to remove R2 in your simulation \$\endgroup\$ – MPA95 Sep 16 '20 at 17:29

It means that only if \$V_{BE1} = V_{BE2} = V_{CE2}\$ we will have \$I_{C1} = I_{C2}\$

enter image description here

enter image description here And the main idea behind current mirrors is that the output transistor (\$Q_2\$) behaves just like a VCCS (Vbe control-current source). So by changing the \$V_{BE2}\$ voltage we changing (setting) the output (\$I_{C2}\$) current. And we don't care about base current we assume \$ \beta = oo\$.

Because the only things that matter here is \$V_{BE}\$ vs \$I_c\$ characteristic and a Shockley equation \$I_C = I_S e^{\frac{V_{BE}}{V_T}}\$.

The first transistor (being called diode-connected transistor) job is to convert the input current into the "input voltage" \$V_{IN} = V_{BE1}\$ and nothing more.

And the output transistor job is to convert this voltage "input" voltage \$ V_{BE1} = V_{BE2}\$ back again into the output \$I_{C2}\$ current.

And we can do this thanks to the Shockley equation. And this is why we don't care about IB vs Vbe (at first glance). Also, the first transistor behaves like a diode because its only has two terminals and conduct current only in one direction. And this is the definition of a diode. And this is why we can call it a diode-connected transistor.

  • \$\begingroup\$ From LTspice's graphic for \$ V(a) = V_{CE2} = 10V \implies (V_{B} = V_{BE1} = V_{BE2}) \$ => the word "only" has no place in the tautology \$\endgroup\$ – user7058377 Sep 16 '20 at 16:28
  • \$\begingroup\$ @user7058377 But your circuit is wrong because for the most of a time Q2 in your simulation spending time in saturation region. \$\endgroup\$ – G36 Sep 16 '20 at 16:33
  • \$\begingroup\$ Do You have the circuit for LTspice validation ? \$\endgroup\$ – user7058377 Sep 16 '20 at 16:39
  • \$\begingroup\$ @user7058377 Try it yourself, simply remove R2 and repeat the simulation. \$\endgroup\$ – G36 Sep 16 '20 at 16:44
  • \$\begingroup\$ @user7058377 Additional notice that in LTspice if you are using the default BJT's model (NPN) then you don't be able to see the Early effect influence because the default model does not include Early effect. Thus to see the Early effect influence on the current mirror you need to use a different model for example 2N2222 in your simulation. \$\endgroup\$ – G36 Sep 16 '20 at 17:06

2) from the schema without load :

enter image description here

For x=0.152 :

Ic(Q1) = Ic(Q2) = 13.72mA

Ie(Q1) = Ie(Q1) = -13.97mA

3) Measure with a more realistic transistor that include early effect

With a G36's remark the use of 2N2222 give, for V(2) = 0.75V, it obtains an

only value of Ie(Q1) = Ie(Q2) = 14.12mA- it is explained towards the end in the link here.

enter image description here

Note : with spice directive : ".dc V2 0.1 20 0.01" the simulation run time is much faster than with

.step param V 0.1v 20v 0.01v

4) With 2 times R2

4.1) With 2 times R2 : better (G36 remark : "...resistor is between 0.1V....0.6V...")

Iref realistic = 1mA (for example).

So, $$ R1 = {{15-0.7-V_{RE} (=0.1v \to 0.6v)} \over {1~mA}} $$ $$ R1 = {{15-0.7-0.3} \over {1~mA}} = 14~kΩ $$ $$ Linear \implies I_{C1} \approx I_{R1} \approx I_{E1} \implies R2 = {V_{RE} \over 1~mA} = {0.3~v \over 1~mA} = 300~Ω $$

enter image description here

This is the circuit to do for an implementation with discrete components to avoid thermal runaway.

5) Widlar current source

5.1) Without a load resistance

From this link, and with the schema before it gives the beta. So the calculus can be made :

enter image description here

enter image description here

enter image description here

Note :

  • It can be observed that for Vout = 0.5 V : Ic(Q2) = IC2 = 0.8 mA. Good !
    That makes a good operating rank between Ic(Q1) and Ic(Q2) because the \$ Ic_{1,2} \$ lines do not intersect ?
  • I had to choose Ic(Q2) <= Ic(Q1) because the ln fonction : ln (Ic(Q1) / Ic(Q2)) will give a negative resistance.


In the study of this current mirror: as P(Q1) = VCE1 * IC1 = VBE1 * IC1 and P(Q2) = VCE2 * IC2, and that the curve (from Vout to the Ground) of IC2 will give P(Q2) > P(Q1) it will result in a thermal runaway in the theoretical case. Then the manufacturer makes a pairing so that the component arrives at a good operating point. In this case, for the current mirror, IC1 = IC2 and this for more than one value of Vout finally otherwise there would be no input margin.
  • \$\begingroup\$ this is not an answer; please, remove this answer and add this content to your question \$\endgroup\$ – MPA95 Sep 16 '20 at 17:27
  • \$\begingroup\$ Change the BJT type to see the Early effect influence. Because now Ic1 = Ic2 is always true as long as Q2 is working on active region. \$\endgroup\$ – G36 Sep 16 '20 at 17:48
  • \$\begingroup\$ Now, try to add the resistors into Q1 and Q2 emitters (50R for a start). \$\endgroup\$ – G36 Sep 16 '20 at 19:16
  • \$\begingroup\$ Why did you use such large emitter resistors \$5000k\Omega = 5M\Omega\$? Are you mad? Have you noticed that the Q2 collector current is negative? Do you know what that even means? A typical example of Garbage In, Garbage Out. And we have a similar story with Widlar current source. \$\endgroup\$ – G36 Sep 17 '20 at 14:49
  • \$\begingroup\$ Typical the voltage drop across the emitter degeneration resistor is between 0.1V....0.6V. \$\endgroup\$ – G36 Sep 17 '20 at 14:53

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