Talk:Current mirror

message: This article is now much more advanced than its talk page. I propose a wishlist comparison table of current mirror architectures in terms of output impedance, operating voltage, and perhaps cleaning up the huge article.  — Preceding unsigned comment added by 132.239.189.204 (talk) 18:40, 8 March 2017 (UTC)
thread_title:  Comments 

message: The simple current mirrors without leg resistors do not increase their output resistance as a result of feedback. However, if a leg resistor is included, feedback comes into play. See the example at two-port network. Brews ohare (talk) 21:59, 9 February 2008 (UTC)
thread_title:  About feedback 

message: Brews ohare and other editors of this page,
I highly appreciate your professional edits about the famous current mirror circuit and its versions. My remark is only that the page looks quite formal, specialized and somewhat "sterile". It gives a lot of details about current mirror versions but it does not show the basic ideas behind them. That is why, I have decided to join this discussion. Circuit-fantasist (talk) 13:04, 19 February 2008 (UTC)
thread_title:  Revealing the philosophy behind the current mirror 

message: Once revealed the general idea we can create as many as we want current mirror circuits (this is the power of this heuristic approach). In all these versions, only the electronic components (BJT, FET, op-amps, etc.) will be different; the general idea behind them will be the same. Well, let's begin with the most popular of them – the basic BJT current mirror.
==== An output part ====
We can drive a bipolar transistor by voltage or by current. If we change the base-emitter  voltage as an input and use the collector current as an output (Fig. 5), we can think of a BJ transistor as a non-linear voltage-to-current converter having exponential characteristic. So, we can use it directly as an output part of our simple BJT current mirror:
The output part of the simple BJT current mirror is just a bipolar transistor acting as an exponential voltage-to-current converter.
Operation. How does the transistor T2 behave in this arrangement? In order to know, let's carry out an experiment – set a constant input voltage VREF = 0.5 ÷ 0.7V and then vary the load resistance RL (with the same success, you can vary the supply voltage V2 or even both the resistance RL and the voltage V2). The result is surprising – the transistor changes its present resistance RT2 between the collector and the emitter so that to keep up a constant total resistance Rtot = RL + RT2 = const (Fig. 6). As a result, the output current remains constant IOUT = V2/Rtot. In this way, the output collector-emitter part of the transistor T2 acts as a current-stable non-linear resistor. They usually say the transistor T2 acts as a simple current sink.
==== An input part ====
Now, we need to make the BJ transistor serve as the opposite current-to-voltage converter. Only, we cannot "reverse" it directly since the transistor is a one-way device, whose base-emitter junction can control the collector current; the opposite is just impossible. What do we do then?
We have already known the remedy – its name is negative feedback. In our case that means to make the transistor adjust its base-emitter voltage so that the collector current to be IREF = VCC/R. For this purpose, we have just to connect the collector to the base, in order to apply a "100% parallel negative feedback" (Fig. 7). As a result, although it seems strange, in this "reversed" transistor the collector current serves as an input quantity while the base-emitter voltage serves as an output quantity!
The input part of the simple BJT current mirror is just a bipolar transistor with 100% parallel negative feedback.
==== Operation ====
But how does the transistor T1 behave in this arrangement? In order to know, let's carry out another experiment – set a constant input (supply) voltage V1 and then vary the input current-setting resistance R (again, with the same success, you can vary the input voltage V1 or even both the resistance R and the voltage V1).
The result is still more surprising than before – now, the transistor changes its present resistance RT1 between the collector and the emitter so that to keep up an almost constant resistance ratio K = RT1/(RT1 + RL) = const. As a result, the output voltage remains almost constant VOUT = VCE1 = VBE = const (Fig. 8). In this way, the output collector-emitter part of the transistor T1 acts as a voltage-stable non-linear resistor. But this is the recipe of making various active diodes ("ordinary", "zener" or "rubber", adjustable...)! It is just a wonder! The parallel negative feedback has made a current-stable resistor (the output part of T1) behave as a voltage-stable one! This is the same transistor but in the first case it serves as a current-stable element while in the second case it serves as a voltage-stable element.
==== A "reversed" transistor ====
At the same time, the input voltage source V1 and the current-setting resistor R form a composed current-source that "want" to produce a current IIN = V1/R through a voltage-stable element (the transistor T1). By the way, some mystic cascode circuits are based on the same arrangement (a current source supplies a voltage-stable element and v.v.). It is interesting that, in this situation, the voltage-stable element changes its present resistance, in order to "help" the current source to establish the desired current. For example, if we decrease the resistance R to increase the current, the transistor T1 will also decrease its present resistance thus helping us to increase the current and v.v. Doing that, the transistor T1 adjusts its base-emitter voltage so that the collector current to be always IIN = V1/R. As a result, although it seems strange, the collector current serves as an input quantity while the base-emitter voltage serves as an output quantity! The negative feedback has reversed the one-way transistor!
==== An equilibrium point ====
It is interesting to discover how the transistor T1 manages to reach the equilibrium point (a system with negative feedback always reaches the equilibrium). For this purpose, we change firstly the magnitude of the input voltage V1 or the resistance R. The transistor responses to this "intervention" by changing its present resistance RT1... but till when? In order to understand, let's carry out an interesting experiment – break the feedback loop and drive the "true" base-emitter transistor input by a separate voltage source "VOUT" (Fig. 9). Then, let's begin increasing the "true" transistor's input voltage "VOUT"; the transistor will begin dropping its collector voltage VC1 by decreasing its RT1 and v.v. Figuratively speaking, the two voltages "move" against each other. In order to imitate the negative feedback behavior, we have to stop changing "VOUT" when the two voltages become equal (a zero indicator connected between the collector and the base can show this moment); this is the equilibrium point. Now, if we short the zero indicator (connect the collector to the base or, as they say, close the feedback loop), the system will remain at rest as it is at the point of the equilibrium.
==== Another example: a transdiode ====
By the way, there is another paradoxical circuit that does the same but it is an almost perfect – the logarithmic converter based on the so called transdiode (a BJT transistor connected in the op-amp feedback loopIntegrated DC Logarithmic Amplifiers). As here, in this odd and also never explained circuit configuration, the collector current serves as an input quantity while the base-emitter voltage – as an output quantity. The only difference is that there an op-amp adjusts the transistor's base-emitter voltage so that its collector current to be exactly equal to the input one. The op-amp does this magic by observing the virtual ground point and keeping it (almost) equal to zero.
==== Assembling the whole circuit ====
Finally, we have only to connect the output of the input part (the base-emitter junction of T1) to the input of the output part (the base-emitter junction of T2) to build the famous BJT current mirror circuit!
  + 
  = 
Fig. 10. Assembling the whole BJT current mirror.
What is the use of applyng this stiff heuristic approach – step-by-step building instead giving the "cut-and-dry" classical circuit? The benefit is that we can build now various current mirror circuits knowing only one powerful general idea! We know what the transistors T1 and T2 do in this circuit; we know the truth about the circuit!
Operation.
Circuit-fantasist (talk) 11:49, 29 February 2008 (UTC)
thread_title:  Implementing the general idea into a BJT electronic circuit 

message: Finally, we have only to connect the output of the input part (the base-emitter junction of T1) to the input of the output part (the base-emitter junction of T2) to build the famous BJT current mirror circuit!
  + 
  = 
Fig. 10. Assembling the whole BJT current mirror.
What is the use of applyng this stiff heuristic approach – step-by-step building instead giving the "cut-and-dry" classical circuit? The benefit is that we can build now various current mirror circuits knowing only one powerful general idea! We know what the transistors T1 and T2 do in this circuit; we know the truth about the circuit!
Operation.
Circuit-fantasist (talk) 11:49, 29 February 2008 (UTC)
thread_title:  Assembling the whole circuit 

message: 'References' 3, 10 and 11 should really be footnotes or the suchlike, as they do not refer to other sources of information, rather they are editors' factual comments. I'll try to amend this soon, if I get chance to, if someone else can do this quickly, please do so :-)  Zangar (talk) 00:11, 9 April 2009 (UTC)
replies: text:  - I managed it! Zangar (talk) 11:07, 9 April 2009 (UTC)
thread_title:  Article References 

message: As far as I can see, Q1 is not needed in Figure 3.  All it does is introduce a voltage drop that's dependent on Iref.  So if Q1 is removed then Iref will change, but the circuit will still cause Iout to be the same as Iref.
If there are no objections I'll make this change.
Occultations (talk) 10:53, 19 April 2010 (UTC)
replies: text: Maybe, some resistor has to be connected in series to VCC as well? Circuit dreamer (talk, contribs, email) 22:20, 19 April 2010 (UTC)
replies: text: There is already a resistor in series with VCC.  Its value is unspecified, because the only thing that matters in this part of the circuit is Iref. Occultations (talk) 13:20, 20 April 2010 (UTC)
replies: text: Sorry, I haven't noticed it. I second your suggestion; Q1 is not needed at all. Its role in the simple current source is to convert the input current into voltage; it acts as a non-linear current-to-voltage converter. In the op-amp version, the left RE acts as such but already linear current-to-voltage converter. The op-amp, Q2 and the right RE constitute voltage-to-current converter with series negative feedback. BTW, the left RE introduces some error in the input circuit and the right RE decreases the compliance voltage of the output circuit. Circuit dreamer (talk, contribs, email) 17:22, 20 April 2010 (UTC)
thread_title:  Simplification to Feedback assisted current mirror 

message: 1. "To maintain matching, the temperature of the transistors must be nearly the same. In integrated circuits and transistor arrays where both transistors are on the same die, this is easy to achieve. But if the two transistors are widely separated, the precision of the current mirror is compromised." Why would you want to keep them separated. Either way, in a chip, they can't be too far from each other. I don't understand the usefulness of this claim. Will it make a huge difference if the two branches of a current mirror are microns apart from each other?
2. How was the expression for Id obtained from Id=Kn(VGS-VT)2(1+λ(VDS-(VGS-VT))) from the MOSFET article? Yes, I do understand the VGS-VT part is not included in this article but consider using it. If I assume VDS=VGS for both sides through diode-connection and right side proper bias, the equation reduces to Id=Kn(VGS-VT)2(1+λVT) which is not the same as Id=Kn(VGS-VT)2(1+λ(VDS)).  How would this simplification affect the equation for the current?
ICE77 (talk) 21:35, 22 May 2011 (UTC)
thread_title:  Branches separation and current equation 

message: The transistor Q1 in this arrangement is a diode. The collector and base are shorted by a metal wire, so the base-collector junction is not functioning anymore. The electrons will take the easiest path, thru the metal wire used for shorting the terminals. A diode can be used, but the results are harder to get. Q1 and Q2 must be the same model, from the same manufacturing batch. This will ensure identical characteristics for base-emitor junctions for both transistors. In fact this is the trick, to have identical junctions in two sepparate devices.
Back to schematic, biasing Q1 as a diode, one will get a constant current thru it and a certain diode voltage drop. The Q2 will start conducting, using its beta to get a collector current. But soon enough, its base-emiter junction will reach the same voltage as Q1 (considered a diode). (to be continued)  — Preceding unsigned comment added by 84.161.158.170 (talk) 18:18, 27 August 2014 (UTC)
thread_title:  Circuit explanation 

message: Why/when are they needed, and are they essential to monolithic integrated circuits ? - Rod57 (talk) 04:05, 25 February 2015 (UTC)
thread_title:  Significance not clear 

message: 1. The collector current for the BJT is given as
:: I_\text{C} = I_\text{S} \left( e^{\frac{V_\text{BE}}{V_\text{T}}} - 1 \right) \left(1 + \frac{V_\text{CB}}{V_\text{A}}\right) ,
As far as I know and according to the Early effect article, VCB is usually VCE and VCB is only used for SPICE modeling. The BJT article also uses VCE.
Why use VCB then?
2a. The drain current for the MOSFET is given as
::\begin{align}
  I_{d} &= f(V_{GS},V_{DG}) \\
        &= \frac{1}{2}K_{p}\left(\frac{W}{L}\right)(V_{GS} - V_{th})^2 (1 + \lambda V_{DS}) \\
        &= \frac{1}{2}K_{p}\left[\frac{W}{L}\right][V_{GS} - V_{th}]^2 \left[1 + \lambda (V_{DG} + V_{GS})\right] \\
\end{align}
What is the point of going from () to [] but, most of all, what is the point of splitting VDS into VDG+VGS. It just seems pointless to me.
2b. The output resistance for the MOSFET is given as:
:: R_N = r_o = \frac{\frac{1}{\lambda}r + V_{DS}}{I_D} = \frac{V_{E} L + V_{DS}}{I_D}
I've never seen this equations. What is r? What is VE?
Shouldn't the output resistance simply be 1/λID?
ICE77 (talk) 23:28, 14 October 2015 (UTC)
thread_title:  Questionable equations 

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thread_title:  External links modified 

message: There is mention of "R2" Extensions and complications. That is the first mention of R2 in the article, and R2 is not in diagram. Recommend author removing.
Bryanfw (talk) 17:00, 31 January 2020 (UTC)
thread_title:  Issues within the article 

message: Sub section Extensions and reservations: Why Vgs (lowercase "gs")? For small signal analysis? Could it be made clearer? Or is it VGS that is meant (like in the previous subsection)? --Mortense (talk) 20:08, 17 February 2022 (UTC)
replies: text: Upper case is traditionally reserved for DC or bias point values, whereas lower case includes DC. Constant314 (talk) 23:11, 17 February 2022 (UTC)
thread_title:  Why Vgs (lowercase "gs")?