정전류회로 for LED Driver

LED driver용 IC가 많이 나오고 있는데, 대부분 정전류 회로를 바탕으로 한 것같습니다.

제가 수배한 DIODES사의 AL5802의 경우 TR 2개를 이용한 정전류 회로를 사용하고 있습니다.

패키지의 모습입니다.

패키지 안을 보면 위와 같이 TR 2개를 이용하고 있습니다.

응용회로 입니다. 응용회로를 보시면 더욱 이해하시기 편하실 겁니다.

5번핀과 6번핀을 Rext를 통해 GND로 연결하는데, 이 저항은 센싱저항이고,

이 저항값에 따라 1번핀에 있는 LED에 흘리는 전류값이 정해지게 됩니다.

=>

저항값은 위 식으로 계산하게 되는데 VBE(Q1)은 대략적으로 0.7V로 계산합니다.

아래는 설명이 잘 되어있는 참고 사이트입니다.

더보기를 누르시면 확인가능합니다.

OVERVIEW

In the following circuit, I've intentionally boxed up the circuit so that it appears to have three pins. There is a good reason for this choice, as will become clearer later on (if not already.)

^{simulate this circuit – Schematic created using CircuitLab}

The circuit requires a GND pin, of course. This needs to be connected as the lowest voltage of the three pins and should be tied a low-impedance node so that the current can sink to that node without moving the node voltage around. This is typically attached to the overall circuit ground reference node.

The SINK pin is controlled such that the current specified by RSET$R_{\text{SET}}$ is sunk at this pin. A load may be attached between this pin and a positive voltage supply (relative to whatever the GND pin is tied to, anyway.) I show an LED here. But this could be other kinds of loads, as well.

The burden voltage for the SINK pin, relative to GND, is at least 1V$1\text{V}$. (The circuit can often be designed to tolerate some saturation of Q1$Q_1$, so that its VCE$V_{\text{CE}}$ can be as little as 300mV$300\text{mV}$.) But for some designs this may be somewhat higher -- perhaps as high as around 1.6V$1.6\text{V}$ in some more extreme cases with high sinking currents set by RSET$R_{\text{SET}}$.

The ENA pin is usually connected to the same positive voltage supply (relative to GND) that the load uses. This "enables" the circuit. But using the positive supply directly need not be the case here, as I'll discuss a little later towards the very end, below. It may also be connected to an MCU and used to turn the circuit ON and OFF. For this reason, I show it as a separate pin and call it ENA to signal that this is an active-HI enable pin, of sorts. You can tie it HI (as shown above) or you can use an I/O pin to control it. (As I said, more on that later on.)

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OPERATION

Let's first look at the circuit to see how it does what it does.

The main idea is to use the fact that a silicon BJT has a relatively stable VBE≈700mV$V_{\text{BE}}\approx 700\text{mV}$. (Assuming the temperature of Q2$Q_2$ isn't moving around -- more on that, later.) All of the current sinking into the SINK pin, plus a little more from the base of Q1$Q_1$ and less a little bit that goes into the base of Q2$Q_2$, is driven through RSET$R_{\text{SET}}$ and therefore develops a voltage across it. If Q1$Q_1$'s collector were to try and increase the current it was sinking (Early Effect?) then this added collector current would cause the voltage across RSET$R_{\text{SET}}$ to exceed the nominal VBE$V_{\text{BE}}$ of Q2$Q_2$. In response, Q2$Q_2$'s collector will sink more current (its collector current is controlled by its base-emitter voltage.) This added Q2$Q_2$ collector current has to come through RB$R_{\text{B}}$ and therefore must increase the voltage drop across it. This lowers the voltage at the base of Q1$Q_1$, which therefore lowers its emitter voltage. In effect, the overall response of the circuit to an attempted increase in the current through the load (LED?) is to oppose that change and return the base voltage of Q2$Q_2$ to the nominal VBE≈700mV$V_{\text{BE}}\approx 700\text{mV}$. Similar logic also applies should the collector current of Q1$Q_1$ decline. The circuit response to a lowering in Q1$Q_1$'s collector current is to raise the base voltage of Q1$Q_1$ and therefore oppose that decline.

There is negative feedback in this system to oppose changes in the sinking current at the SINK pin. And the sinking current is nominally:

ISINK=700mVRSET$$I_{\text{SINK}}=\frac{700\text{mV}}{R_{\text{SET}}}$$

This entire circuit works to control the voltage at the collector of Q1$Q_1$ by observing the current sinking into the collector of Q1$Q_1$. Other than the voltage across RSET$R_{\text{SET}}$ and across the load (the LED in this case), all of the rest of the voltage difference between the positive supply and GND is taken up by the VCE$V_{\text{CE}}$ of Q1$Q_1$. But this detail is under management by the circuit so that the load's voltage is just what is necessary for the current set by RSET$R_{\text{SET}}$.

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A DESIGN

You must be sure that you have enough extra voltage for the positive supply. It must be at least the burden voltage of the circuit's SINK pin plus whatever burden voltage is required by the load. For a typical LED that uses somewhere from 2V$2\text{V}$ to 3.5V$3.5\text{V}$, this means that the positive supply voltage must be at least 5V$5\text{V}$. But if you are using a red LED with, say, 1.9V$1.9\text{V}$ and if you can accept a design with a little saturation in Q1$Q_1$, then you could get away with as little as 3V$3\text{V}$ for the positive supply. (There is no hard and fast rule here and there are many tradeoffs to consider.)

So let's assume that VCC=5V$V_{\text{CC}}=5\text{V}$ and that this is a red LED that is to be operated at ILED=20mA$I_{\text{LED}}=20\text{mA}$. (Also, obviously, we will use VENA=VCC$V_{\text{ENA}}=V_{\text{CC}}$.)

RSETRB=VBE2ILED⋅(1+1β1−110β2)=10⋅VENA−VBE1−VBE2ILED$$\begin{array}{rl}{R}_{\text{SET}}& =\frac{V_{{\text{BE}}_2}}{I_{\text{LED}}\cdot (1+\frac{1}{{\beta }_1}-\frac{1}{10{\beta }_2})}\\ \\ R_{\text{B}}& =10\cdot \frac{V_{\text{ENA}}-V_{{\text{BE}}_1}-V_{{\text{BE}}_2}}{I_{\text{LED}}}\end{array}$$

We could assume that VBE1=VBE2=700mV$V_{{\text{BE}}_1}=V_{{\text{BE}}_2}=700\text{mV}$ and that is what most folks would do. Feel free to do that, in fact. But if you want to be a little "techy" about it, you might realize that I'm going to set the collector current of Q2$Q_2$ to be 110$\frac{1}{10}$th the value of ILED$I_{\text{LED}}$. (You can see that much in the above equations, if you are sly about reading them.) That means there will be a 60mV$60\text{mV}$ difference in the base-emitter voltages between the two BJTs. Also, because we are dealing with 20mA$20\text{mA}$ and since I happen to know that VBE$V_{\text{BE}}$ in small signal BJTs is roughly 700mV$700\text{mV}$ when the collector current is somewhere in the vicinity of 3mA$3\text{mA}$, I can work out the fact that:

VBE1VBE2=700mV+26mV⋅ln(20mA3mA)≈750mV=700mV+26mV⋅ln(20mA10⋅3mA)≈690mV(See appendix)$$\begin{array}{r}\text{(See appendix)}& V_{{\text{BE}}_1}& =700\text{mV}+26\text{mV}\cdot \ln \left(\frac{20\text{mA}}{3\text{mA}}\right)\approx 750\text{mV}\\ V_{{\text{BE}}_2}& =700\text{mV}+26\text{mV}\cdot \ln \left(\frac{20\text{mA}}{10\cdot 3\text{mA}}\right)\approx 690\text{mV}\end{array}$$

I'm doing this to illustrate a few things. One is about the fact that VBE$V_{\text{BE}}$ isn't actually constant. (We are still ignoring temperature.) Instead, it does depend on the currents involved. So if you choose very different currents than I show here, you may need to know this detail. Another is to show how little it really matters in the case of dealing with an LED. We could just assume and be "close enough." It helps to see it done both ways to make that point.

Designing this for a low burden voltage of 1V$1\text{V}$ (Q1$Q_1$ going into saturation, so let's say β1=20${\beta }_1=20$) and assuming a semi-worst case β2=100${\beta }_2=100$, we find that

RSET≈33ΩRB≈1.8kΩ$$\begin{array}{r}{R}_{\text{SET}}\approx 33\mathrm{\Omega }\\ \\ R_{\text{B}}\approx 1.8\text{k}\mathrm{\Omega }\end{array}$$

And you will find that both those values are true whether or not you use my fancy calculated values for the base-emitter voltages or not. Either way, the results are very much the same. (So this points out one reason why the base-emitter voltages of BJTs can often be taken as a given value.)

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TEMPERATURE AND THE EARLY EFFECT

I've avoided these topics until now. But they are both important enough that it is time to bring them up. One of the huge benefits of this circuit is that it solves both these problems (to a reasonable degree) by adding Q2$Q_2$.

Q1$Q_1$ suffers from the Early Effect in a circuit like this, because the VCE$V_{\text{CE}}$ voltage can vary widely depending upon variations where the load drops different voltages. BJTs have, in effect, a resistor that goes from their collector to their emitter. This resistor is sometimes called ro$r_o$ and its value is ro=VA+VCEIC$r_o=\frac{VA+V_{\text{CE}}}{I_{\text{C}}}$, where VA$VA$ is the Early Voltage for the device. This is often around 100V$100\text{V}$ but it can be quite a bit less (the D45H11 can be as little as 10V$10\text{V}$.) For ILED=20mA$I_{\text{LED}}=20\text{mA}$, this value might be as little as 5kΩ$5\text{k}\mathrm{\Omega }$ and, unchecked, would increase the current in the LED.

Q1$Q_1$ also suffers from heating. It's the work-horse device here and it may have to dissipate power and, in the process, heat up. If we were depending upon its VBE$V_{\text{BE}}$ voltage as a way of programming the LED current, we could be in some trouble because this voltage changes by somewhere between −1.8mV∘C$-1.8\frac{\text{mV}}{{}^{\circ }\text{C}}$ to −2.4mV∘C$-2.4\frac{\text{mV}}{{}^{\circ }\text{C}}$. With some BJTs in TO-92 packages with thermal resistances of 200∘CW$200\frac{{}^{\circ }\text{C}}{\text{W}}$, you can easily see a fairly large change in the base-emitter voltage due to heating caused by their normal operations.

But this circuit moves both problems away from Q1$Q_1$.

First, Q2$Q_2$ operates with a nearly fixed VCE$V_{\text{CE}}$ in this circuit and therefore the Early Effect on Q2$Q_2$ doesn't change as the circuit operates. But more importantly, the Early Effect on Q1$Q_1$ is neutralized because Q2$Q_2$ will adjust its operations to automatically compensate for such variations by observing the voltage drop across RSET$R_{\text{SET}}$. If Q1$Q_1$'s Early Effect attempts to make any changes in the LED current, Q2$Q_2$ will immediately adjust its own collector so as to counter it. The result is that the Early Effect in Q1$Q_1$ is neutralized by the same negative feedback loop that is controlling the LED current in the first place.

Second, since Q2$Q_2$ is the BJT measuring the LED current, it's important that Q2$Q_2$ base-emitter voltage remain as constant as possible. But also, because Q2$Q_2$ is operating with far less current (on the order of the base current needed by Q1$Q_1$) and because it has a very small VCE$V_{\text{CE}}$ voltage, the dissipation in Q2$Q_2$ is very modest. This means it is only barely affected by heating and this helps to ensure that its base-emitter voltage is only depending on the ambient temperature for operation; without the additional problems caused by having to dissipate the far greater power being handled by Q1$Q_1$. So this helps to minimize temperature effects.

It doesn't completely eliminate them, since the ambient temperature of the circuit will still have an effect. But it does help to separate these problems into two devices.

So using the 2nd BJT here is helping in two ways: nullifying the Early Effect in Q1$Q_1$ and removing the self-heating portion of the problem in Q1$Q_1$ from impacting the programmed current.

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FINAL NOTES

There is another use for this circuit I'd mentioned earlier. The ENA input can be managed by an MCU I/O pin, instead of tying it to the positive supply. If so, the value of RB$R_{\text{B}}$ might be adjusted per the VCC$V_{\text{CC}}$ of the MCU itself (which may be different than that used by the load.) This allows ON/OFF control of this circuit.

So it's really quite a useful circuit pattern to know about and use.

As if that weren't enough already, the above circuit can also be modified to use a MOSFET:

^{simulate this circuit}

Here, the value of RSET$R_{\text{SET}}$ is set up much the same way as before. However, as the MOSFET doesn't require any base current, the value of RB$R_{\text{B}}$ can be much higher -- say 22kΩ$22\text{k}\mathrm{\Omega }$ or still more, if desired. One impact of setting RB$R_{\text{B}}$ higher or lower will be the collector current of Q2$Q_2$ and therefore will have some effect on the voltage across RSET$R_{\text{SET}}$. Also, there is a threshold voltage required for the MOSFET (NFET), and this voltage requirement must be met by whatever voltage is supplied to ENA.

For example, the BSS123 has a threshold voltage of about 1.7−2.0V$1.7-2.0\text{V}$. This is added to the already needed base-emitter voltage for Q2$Q_2$. So the voltage difference across RB$R_{\text{B}}$ will be ≈VENA−700mV−1.7V$\approx V_{\text{ENA}}-700\text{mV}-1.7\text{V}$. From that, you can work out the collector current for Q2$Q_2$.

So there are a few slightly different calculations involved in this new pattern using an NFET with an NPN BJT. But this also shows the versatility of this pattern, too.

Appendix

Start with the usual equation for a BJT collector current when in active mode:

IC=ISAT(eVBEηVT−1)$$\begin{array}{rl}{I}_{\text{C}}& =I_{\text{SAT}}(e^{{}^{\frac{V_{\text{BE}}}{\eta V_T}}}-1)\end{array}$$

Solve for VBE$V_{\text{BE}}$:

VBE=ηVTln(ICISAT+1)$$\begin{array}{rl}{V}_{\text{BE}}& =\eta V_T\ln (\frac{I_{\text{C}}}{I_{\text{SAT}}}+1)\end{array}$$

At this point, we can simplify the above equation, removing the +1$+1$ term, because ISAT$I_{\text{SAT}}$ is many orders of magnitude smaller than any practical IC$I_{\text{C}}$. So, without any harm we can re-write it as:

VBE=ηVTlnICISAT$$\begin{array}{rl}{V}_{\text{BE}}& =\eta V_T\ln \frac{I_{\text{C}}}{I_{\text{SAT}}}\end{array}$$

Now. Suppose we happened to take a reference measurement of VBE0$V_{{\text{BE}}_0}$ at a collector current IC0$I_{{\text{C}}_0}$. What would we then expect for a new VBE1$V_{{\text{BE}}_1}$ at a new collector current, IC1$I_{{\text{C}}_1}$?

ΔVBE=[ηVTlnIC1ISAT]−[ηVTlnIC0ISAT]=ηVT[lnIC1ISAT−lnIC0ISAT]=ηVT[lnIC1−lnISAT−(lnIC0−lnISAT)]=ηVT[lnIC1−lnIC0]=ηVTlnIC1IC0$$\begin{array}{rl}\mathrm{\Delta }{V}_{\text{BE}}& =[\eta V_T\ln \frac{I_{{\text{C}}_1}}{I_{\text{SAT}}}]-[\eta V_T\ln \frac{I_{{\text{C}}_0}}{I_{\text{SAT}}}]\\ \\ & =\eta V_T[\ln \frac{I_{{\text{C}}_1}}{I_{\text{SAT}}}-\ln \frac{I_{{\text{C}}_0}}{I_{\text{SAT}}}]\\ \\ & =\eta V_T[\ln I_{{\text{C}}_1}-\ln I_{\text{SAT}}-(\ln I_{{\text{C}}_0}-\ln I_{\text{SAT}})]\\ \\ & =\eta V_T[\ln I_{{\text{C}}_1}-\ln I_{{\text{C}}_0}]\\ \\ & =\eta V_T\ln \frac{I_{{\text{C}}_1}}{I_{{\text{C}}_0}}\end{array}$$

So, this suggests that:

VBE1=VBE0+ΔVBE=VBE0+ηVTlnIC1IC0$$\begin{array}{rl}{V}_{{\text{BE}}_1}& =V_{{\text{BE}}_0}+\mathrm{\Delta }{V}_{\text{BE}}\\ \\ & =V_{{\text{BE}}_0}+\eta V_T\ln \frac{I_{{\text{C}}_1}}{I_{{\text{C}}_0}}\end{array}$$

For a small signal BJT, it is almost always the case that η≈1$\eta \approx 1$ and at room temperature VT≈26mV$V_T\approx 26\text{mV}$. This is where I got the "See appendix$\text{See appendix}$" equation, above.

If you are only interested in small scale changes rather than large, non-linear changes, then you can take a differential approach instead (sometimes called "local linearization about a point.") Below, I'll use a differential operator approach:

ICdIC=ISAT(eVBEηVT−1)=d[ISAT(eVBEηVT−1)]=ISAT⋅d[eVBEηVT−1]=ISAT⋅eVBEηVT⋅d[VBEηVT]=ISAT⋅eVBEηVT⋅dVBEηVT(non-linear Shockley)(linearized)$$\begin{array}{r}\text{(non-linear Shockley)}& I_{\text{C}}& =I_{\text{SAT}}(e^{{}^{\frac{V_{\text{BE}}}{\eta V_T}}}-1)\\ \text{d}{I}_{\text{C}}& =\text{d}\left[I_{\text{SAT}}(e^{{}^{\frac{V_{\text{BE}}}{\eta V_T}}}-1)\right]\\ \\ & =I_{\text{SAT}}\cdot \text{d}[e^{{}^{\frac{V_{\text{BE}}}{\eta V_T}}}-1]\\ \\ & =I_{\text{SAT}}\cdot e^{{}^{\frac{V_{\text{BE}}}{\eta V_T}}}\cdot \text{d}\left[\frac{V_{\text{BE}}}{\eta V_T}\right]\\ \\ \text{(linearized)}& & =I_{\text{SAT}}\cdot e^{{}^{\frac{V_{\text{BE}}}{\eta V_T}}}\cdot \frac{\text{d}{V}_{\text{BE}}}{\eta V_T}\end{array}$$

At this point, you may notice that the first two factors in linearized$\text{linearized}$ are very similar to the right-side of non-linear Shockley$\text{non-linear Shockley}$. The difference, the −1$-1$ term, isn't important enough to worry about, as the exponential value is almost always huge (≫1$\gg 1$.) So this means we can re-write (without any loss of generality) linearized$\text{linearized}$ as:

dIC=IC⋅dVBEηVT$$\begin{array}{rl}\text{d}{I}_{\text{C}}& =I_{\text{C}}\cdot \frac{\text{d}{V}_{\text{BE}}}{\eta V_T}\end{array}$$

Solving for dVBE$\text{d}{V}_{\text{BE}}$:

dVBE=ηVT⋅dICIC(small signal VBE change)$$\begin{array}{}\text{(small signal }{V}_{\text{BE}}\text{ change)}& \text{d}{V}_{\text{BE}}& =\eta V_T\cdot \frac{\text{d}{I}_{\text{C}}}{I_{\text{C}}}\end{array}$$

The interpretation of dICIC$\frac{\text{d}{I}_{\text{C}}}{I_{\text{C}}}$, its meaning in effect, is the "infinitesimal percent variation of the collector current." Compare the above small signal VBE change$\text{small signal }{V}_{\text{BE}}\text{ change}$ with what was developed still earlier above:

ΔVBE=ηVT⋅lnIC1IC0(finite VBE change)$$\begin{array}{}\text{(finite }{V}_{\text{BE}}\text{ change)}& \mathrm{\Delta }{V}_{\text{BE}}=\eta V_T\cdot \ln \frac{I_{{\text{C}}_1}}{I_{{\text{C}}_0}}\end{array}$$

There is a difference, as you can see. If all you want are nearby change estimates, all you need to worry about is estimating the %-change in the collector current and you can easily get the estimated change in the base-emitter voltage without having to worry about taking the logarithm of the current ratios.

If you remember 1st year calculus, then you may remember seeing that dlnx=1xdx$\text{d}\ln x=\frac{1}{x}\text{d}x$. This means that dlnIC1IC0=IC0IC1⋅dIC1IC0=dIC1IC1$\text{d}\ln \frac{I_{{\text{C}}_1}}{I_{{\text{C}}_0}}=\frac{I_{{\text{C}}_0}}{I_{{\text{C}}_1}}\cdot \frac{\text{d}{I}_{{\text{C}}_1}}{I_{{\text{C}}_0}}=\frac{\text{d}{I}_{{\text{C}}_1}}{I_{{\text{C}}_1}}$. And now you can easily see, in broad daylight, the relationship between small signal VBE change$\text{small signal }{V}_{\text{BE}}\text{ change}$ and finite VBE change$\text{finite }{V}_{\text{BE}}\text{ change}$.