Talk:Current mirror

Comments
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)

The article only discusses BJT current mirrors and makes no mentionof MOSFET at all. Could someone expert in the field add the missing information?

http://paginas.fe.up.pt/~fff/eBook/Imagens/fig_019.gif

http://paginas.fe.up.pt/~fff/eBook/MDA/MDA.html#sec_3.2.

What is the reference current source? Wouldn't it have to be another current mirror? - Omegatron 16:30, Jun 26, 2005 (UTC)

Omegatron--no, the driver wouldn't have to be another current mirror. Many control devices provide a control current--not voltage. The National Semiconductor LM334, for example, is a temperature dependent current source. It puts out a current linearly proportional to temperature (in K). The current mirror prevents loading of such devices. Many DACs, microcontrollers, and other signal conditioning devices also generate a current output, albeit with a finite output inpedence, hence the need for opamp amplifier circuits or current mirrors. Sorry for the long reply.

The "Current source" article contains this line: "Variations to the basic current mirror are the Widlar current source and the Wilson current source." There is an article for the widlar source and the wilson source. Perhaps we want to mention that on the current mirror page.

An extremely important piece of information is unclear – namely, in the article, it says "when both transistors have zero base-collector bias, the two base currents are equal". There is no diagram or any details showing that both transistors have 'zero' base-collector bias. If somebody is going to write this, then they need to show it. And show it properly. For example, does somebody manually set the DC voltage so that it becomes equal to the base voltage of Q2? Or do we need an automatic voltage measurement system for measuring the base voltage of Q2 and use that measurement to control the DC voltage source so that it matches that base voltage? KorgBoy (talk) 05:26, 16 June 2017 (UTC)

What confuses many people is – what are the mechanisms that allow the collector currents to be the same in each collector branch? The issue is nobody appears to explain it properly. It appears that the main aspects are – the collector current is approximately equal to an expression involving Is (Reverse saturation current), base-emitter voltage Vbe, and a 'thermal voltage' VT. Also, the two transistors Q1 and Q2 must be assumed to be identical (so fabricated to be virtually identical, same characteristics). This translates to same Is and same VT value for both transistors. And, if both transistors are connected so that they have the same applied Vbe, then the collector current Ic in both transistors will be the same. This will then also mean that the base currents in both transistors will be the same as well, due to the Ic ~ beta.Ib relation. KorgBoy (talk) 02:43, 21 June 2017 (UTC)

Saturation Point Of a Transistor
The transistors have a property to retain a specific amount of collector current for a corresponding base current at Saturation point.As we increase the voltage Vcc The base collector junction remains forward biased until Vce remains less than 0.7v after which the transistor comes in its linear region of operation.In which the amplifiers are operated.

About feedback
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)

Revealing the philosophy behind the current mirror
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)

My wiki mission. I have a different "mission" in Wikipedia and Wikibooks – to reveal the philosophy behind circuits. IMO, before showing in detail how to calculate an electronic circuit we have first to show what the very basic idea is behind this circuit. In Circuit Idea, I try to do that relying only on my human intuition, imagination and common sense.



The simple but complex current mirror. At first sight, the current mirror looks a simple bare circuit (Fig. 1); but actually, it is an odd, strange and exotic circuit that is never explained (if you know sources that have revealed the philosophy of the basic current mirror circuit, please cite them here; I have not managed to find such sources, I have found only some guesswork). It is a great paradox that everyone knows what a current mirror is but nobody knows how it operates.

As a rule, the classical current mirror explanations begin with the assertion that the transistor Q1 acts as a diode. But it is very primitive and confusing to say "the input transistor Q1 is a diode". No, it is not a diode; it is exactly a transistor operating in the active mode. It would be a diode, if its collector was disconnected. Then all the current IREF = VCC/R (R is omitted on the picture) would flow through the base-emitter junction acting as a diode. But here the output collector-emitter part of the transistor is connected in parallel to the base-emitter "diode". In this way, it serves as a shunting regulating element that diverts the great amount (β/(1 + β)) of the current. But how and why the transistor does this magic?

There is something strange and confusing in this arrangement... As everybody knows, the base-emitter voltage VBE is the input quantity of the bipolar transistor and the collector current IREF is the output quantity. But here all is on the contrary – the collector current is the input and the base-emitter voltage is an output!?!? What an idiocy?

Since there are no satisfactory explanations, I suggest to forget temporarily all the "cut-and-dried" citations and to try revealing the secret of the famous circuit by ourselves. Please, just forget them and begin thinking by yourself.

Questions to be answered. In order to understand this odd circuit, we need to answer dozens of questions that are never answered. Here are some of them.

''What does the transistor Q1 do in this circuit? What is its function there? What is the difference between it and a diode? Can we replace it by a bare diode? Why and how the collector current IREF serves as an input quantity and the base-emitter voltage as an output one (we thought the base-emitter voltage VBE was the input quantity of the bipolar transistor and the collector current IREF was the output quantity)? What does the transistor Q2 do in this circuit? What is its function there? How do collector voltages vary when we change the input current IREF (the input voltage VREF, the programming resistor RREF and the load resistance RL)?''

The general idea behind the whole circuit
Below I have begun exposing my original viewpoint at the basic current mirror circuit. It is not based on "verifiable sources" just because there are not such sources; so, one can say it is an original research. Only, I have endeavored to explain the famous circuit so simply, clearly and evidently that there is no any need to verify these assertions (see my advices to creative thinking wikipedians). By the way, the very NOR stipulates such a possibility – to use sources, "the accuracy of which is easily verifiable by any reasonable adult without specialist knowledge".

So, I will first show the truth about current mirror here in the talk page. Then, you will decide, if we should show it to the visitors of the main article or we should leave them to continue thinking of this circuit as a black magic:) Only, in the latter case, if they (Heaven forbid) peep into this talk page, they will exclaim, "The King is naked!":)

IMO, we might at least insert in the introductory article part a concise text showing the general circuit idea. Meanwhile, I will copy this discussion to the according Circuit idea page (talk) where I will disclose completely the mystery of the famous current mirror circuit.

The problem: current direction inverter


In circuitry, especially in the area of microelectronics (e.g., inside op-amps), sometimes we need to invert the direction of a current (to make the flowing in current a flowing out one and v.v. - the flowing out current a flowing in one) without changing its magnitude. In this way, the output current just follows, "copies" the input one but this copy is inverted, "mirror" one. In other words, the input current "programs" the output one and the whole circuit serves as a programmed current source (current-controlled or dependent current source).

Obviously, in order to do that, such a current direction inverter (having a more popular name – current mirror) has to be at least a 3-terminal device (Fig. 2). The reason of that is because the two entering currents have to "go out" from somewhere (Fig. 2a) and v.v., the two exiting currents have to "go in" somewhere (Fig. 2b). In this arrangement, the one lead serves as an input, the other – as an output and the third – as a common terminal (usually connected to the positive or negative rail and more rarely – to the ground).



How to create a current mirror
Basic structure. In electronics, we have two kinds of circuits producing or, more frequently, just controlling current – current sources and current sinks. The difference between them is that, regarding to some circuit point, sources "inject" current while sinks "absorb" current. Obviously, in order to implement the general arrangement above (Fig. 1) into a concrete current mirror, we need both the circuits. As a rule, these circuits are voltage-controlled; so, in order to control them by current, we need to connect current-to-voltage converters before them. Let's now draw the block diagram of the two possible current mirror arrangements.



If we have a current source, we convert the input "entering" current into a voltage and then use this voltage to control a current sink; as a result, we obtain a current sink (Fig. 3a). Conversely, if we have a current sink, we convert the input "exiting" current into a voltage and then use this voltage to control a current source (Fig. 3b); as a result, now we obtain a current source. We can already generalize this basic current mirror structure in a conclusion:

A current mirror consists of two consecutively connected current-to-voltage and voltage-to-current converters.



Characteristics. It is interesting that the two converters may be linear (then IOUT = V/R = IIN.R/R = IIN) but it is not obligatory. They may be non-linear devices having whatever transfer or I-V characteristics that might even depend on another quantity (e.g., temperature); the only requirement is their characteristics be inverse. In this way, if the one converter implements a function y = f(x) and the other represents the inverse function x = f-1(y) the whole function is y = f(x) = f-1(f(x)) = x (the two converters might be connected in any order – Fig. 4). At first sight, this result looks strange and nonsensical but in this way the main problem – reversing the current direction, is solved. So, we can formulate the next conclusion:

A current mirror consists of two consecutively connected converters having inverse characteristics.

Simple reversing. Usually, there is not a direct/inverse converter pair, and we have only one kind of converter. If this is a reversible converter, we can use it as a direct converter and then connect another such converter in an opposite direction to get a reverse converter.

Reversing by negative feedback. The problem is when the converter is not reversible and it is a bare one-way device. In this case, we cannot swap the circuit input and output ports; we cannot apply an input quantity to the circuit output and to get an output quantity from the circuit input (unfortunately, this is our case). The only way to "reverse" the circuit is applying the ubiquitous negative feedback. Only systems with negative feedback have the unique property to reverse the cause and effect relations between the input and output quantities of the objects. They adjust their internal input quantity so that their internal output quantity to become equal to the "true" external input quantity. As a result, the internal input quantity follows, depends on the external input quantity; actually, the internal input quantity serves as an external output quantity. In this way, they can "reverse" the objects.

Implementing the general idea into a BJT electronic circuit
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 loop ). 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)

Past discussion about the philosophy behind the current mirror
It is very primitive and confusing to say "the input transistor Q1 is a diode" (see Fig. 1). No, it is not a diode; it is a transistor operating in the active mode. It will become a diode, if we disconnect its collector. Then all the current IREF = VCC/R (R is omitted on the picture) will flow through the base-emitter junction acting as a diode. But here the output collector-emitter part of the transistor is connected in parallel to the base-emitter "diode". In this way, it serves as a shunting regulating element that diverts the great amount (β/(1 + β)) of the current. But how and why the transistor does this magic?

There is something strange and confusing in this arrangement... As everybody knows, the base-emitter voltage VBE is the input quantity of the bipolar transistor and the collector current IREF is the output quantity. But here all is on the contrary – the collector current is the input and the base-emitter voltage is an output!?!? What an idiocy?

A transistor (as well as a voltage divider) is a one-way device, in which the base-emitter junction can control the collector current; (as far as I know) the opposite is just impossible. The only way to make it possible is to apply the ubiquitous negative feedback. Only systems with negative feedback have the unique property to reverse the cause and effect relations between the input and output quantities of the objects. In this way, they can "reverse" the objects. Imagine a block diagram where a converter implementing a function y = f(x) is connected in the feedback loop; as a result, the whole nfb system will implement the reverse function Y = F(X), where Y = x, X = y and F = 1/f.



We frequently use this technique in circuitry to swap the converter's ports (especially of the one-way devices) by connecting them into the feedback loop. From this viewpoint, an active voltage-to-current converter is actually a reversed passive current-to-voltage converter, an op-amp non-inverting amplifier is a reversed voltage divider (see the picture on the right), an op-amp integrator is a reversed differentiator and v.v., an op-amp logarithmic converter is a reversed antilogarithmic converter and v.v., an analog-to-digital converter is a reversed digital-to-analog converter, etc.

Another very interesting example (close connected to current mirror circuit) is a transdiode that is actually a transistor connected in the op-amp feedback loop. In this odd and also never explained circuit configuration, the collector current is the input quantity (?!?) while the base-emitter voltage is the output quantity (?!?) The op-amp does this magic by adjusting the base-emitter voltage so that the collector current becomes equal to the input one.

Eureka! It is clear that the same trick is used in the input part of the current mirror: a transistor with 100% parallel negative feedback adjusts its base-emitter voltage so that the collector current to be IREF = VCC/R. 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!

If you would like to know more about this negative feedback phenomenon, join the talk page of this future Circuit Idea module where I have placed a part of correspondence between me and Gordon Deboo (the inventor of Deboo integrator).

(...to be continued...) Circuit-fantasist (talk) 17:59, 10 February 2008 (UTC)
 * Although it is desirable to keep the device in mind when doing the circuit, in the confines of an encyclopedia article that may be too many words. For the purposes of designing the mirror the notion of a diode suffices, and will not get the designer into any difficulties.
 * In this particular example, the base-emitter junction does act as a diode, and obeys the diode law between current and VBE. So it seems to me your interest is in what role the collector-base junction plays when its VCB = 0 V. To obtain the diode model for the transistor in this case, the CB junction simply is ignored, which requires some explanation, although introducing the idea of feedback does not seem to me to add to the clarity of the explanation.
 * Brews ohare (talk) 18:18, 10 February 2008 (UTC)


 * Thanks for reply! It seems intriguing; I need some time to assimilate this wisdom:) Think of this text only as a discussion and as an idea exchange, not as an article text. Circuit-fantasist (talk) 18:28, 10 February 2008 (UTC)

Article References
'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)
 * :✅ - I managed it! Zangar (talk) 11:07, 9 April 2009 (UTC)

Simplification to Feedback assisted current mirror
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)


 * Maybe, some resistor has to be connected in series to VCC as well? Circuit dreamer (talk, contribs, email) 22:20, 19 April 2010 (UTC)


 * 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)


 * 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)

Branches separation and current equation
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)

Circuit explanation
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)

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

Questionable equations
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)

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Issues within the article
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)

Why Vgs (lowercase "gs")?
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)
 * Upper case is traditionally reserved for DC or bias point values, whereas lower case includes DC. Constant314 (talk) 23:11, 17 February 2022 (UTC)