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Let's revisit the conclusion I made above by doing a back-of-envelope calculation that actually analyzes the circuit. We should do this to see if the above estimate i provided holds up to slightly deeper scrutiny. We'll need a schematic so that I can identify parts in the equations:

Let's revisit the conclusion I made above by doing a back-of-envelope calculation that actually analyzes the circuit. We should do this to see if the above estimate I provided holds up to slightly deeper scrutiny. We'll need a schematic so that I can identify parts in the equations:

Changes in the \$V_\text{BE}\$ due to temperature on the drive BJT are automatically compensated by the feedback BJT, which is measuring the collector current of the drive BJT as it passes through the resistor between the feedback BJT's base and emitter.

So this leaves us with the small-scale approach. If we know the magnitude of the base-emitter voltage at some temperature and can guess that it won't change by more than \$-1.8\:\frac{\text{mV}}{^\circ\text{C}}\le \frac{\Delta V_\text{BE}}{^\circ \text{C}}\le -2.4\:\frac{\text{mV}}{^\circ\text{C}}\$, then we can make a simple statement:

$$\Delta I_\text{LED}=\frac{ \frac{\Delta V_\text{BE}}{^\circ \text{C}}}{R_\text{SENSE}}\cdot \Delta T$$

So, if \$\frac{\Delta V_\text{BE}}{^\circ \text{C}}=-2.2\:\frac{\text{mV}}{^\circ\text{C}}\$ and \$R_\text{SENSE}=33\:\Omega\$ and \$\Delta T=15\:\text{K}\$, then \$\Delta I_\text{LED}=-1\:\text{mA}\$. Assuming \$V_\text{BE}\approx 680\:\text{mV}\$ prior to the temperature change, \$I_\text{LED}\approx 21\:\text{mA}\$. So a rise of \$\Delta T=15\:\text{K}\$ of the feedback BJT temperature would then imply a change to \$I_\text{LED}\approx 20\:\text{mA}\$, in this case. This is likely to be quite acceptable.

But if you are seeking the large-scale equation, which provides you with how things are over many decades of design currents, then you'll probably want the original expression I was suggesting -- the sensitivity equation, itself. This will tell you the percent change in \$I_\text{LED}\$ for a percent change in temperature, at any starting set value for \$I_\text{LED}\$ and \$T\$. But this also requires the combination of several equations and the use of derivatives. If that's what you want, say so. Otherwise, the above small-signal local change equation is probably sufficient.

$$\begin{align*} I_\text{LED}&=\frac{\beta_1}{\beta_1+1}\,I_{\text{E}_1}=\frac{\beta_1}{\beta_1+1}\left(\frac{V_{\text{BE}_2}}{R_\text{SENSE}}+I_{\text{B}_2}\right)\\\\&=\frac{\beta_1}{\beta_1+1}\left(\frac{V_{\text{BE}_2}}{R_\text{SENSE}}+\frac1{\beta_2}\left[\frac{V_\text{DRIVE}-V_{\text{BE}_1}-V_{\text{BE}_2}}{R_\text{DRIVE}}-\frac{I_\text{LED}}{\beta_1}\right]\right)\\\\\text{solving for }I_\text{LED},\\\\ &=\left[\frac{\beta_1\,\beta_2}{\beta_1\,\beta_2+\beta_2+1}\right]\cdot\left[\frac{V_{\text{BE}_2}}{R_\text{SENSE}}+\frac{V_\text{DRIVE}-V_{\text{BE}_1}-V_{\text{BE}_2}}{R_\text{DRIVE}}\right] \end{align*}$$

Even with temperature variations on \$\beta\$, the value of the first factor above will be very close to 1 (slightly less.) So we can remove it from consideration. \$V_\text{DRIVE}\$ is reasonably assumed to be temperature-independent for analysis purposes. So this leaves us with:

$$\Delta I_\text{LED}=\frac{\frac{\Delta V_{\text{BE}_2}}{^\circ \text{C}}}{R_\text{SENSE}}\cdot \Delta T-\frac{\frac{\Delta V_{\text{BE}_1}}{^\circ \text{C}}+\frac{\Delta V_{\text{BE}_2}}{^\circ \text{C}}}{R_\text{DRIVE}}\cdot \Delta T$$

So there's an adjustment term that I'd not included in the original case. However, because for all intents and purposes it will be the case that \$R_\text{DRIVE}\gg R_\text{SENSE}\$ and that term will not matter much.

We can replace the \$\frac{\Delta V_{\text{BE}_i}}{^\circ \text{C}}\$ variables in the above equation with the Shockley expansion that also includes the full temperature-dependent equations for \$I_\text{SAT}\$. A closed solution will involve the use of the product-log function and take a lot of room below. But it can be done.

For now, I think it is enough to see that a basic circuit analysis does confirm the original equation as "close enough" when using reasonable estimates for the variation of \$V_\text{BE}\$ with temperature.

Those are typical curves. From these, it looks as though I can be pretty sure that over a very wide range of temperatures, and so long as \$V_\text{CE}\ge 1\:\text{V}\$, that \$\beta_1\gt 100\$.

This provides a worst-case reading. It's for \$I_\text{C}=2\:\text{A}\$, which is 100 times what I'm considering. But if you look again at the above curves, you'll see that the positions are about the same in either case. So let's design this for \$\beta_1=60\$. We are rock-solid safe with that choice.

This means \$I_{\text{B}_1}\le 333\:\mu\text{A}\$. Different D44H11 devices may vary, but we can be pretty sure the base current won't exceed this value range. Taking worst-case and best-typical as the extremes, \$100\:\mu\text{A} \le I_{\text{B}_1}\le 333\:\mu\text{A}\$.

For \$Q_1\$, I actually don't care too much right about about its operating \$V_{\text{BE}_1}\$ because it's the job of \$Q_2\$ to make adjustments there. So I'm not going to think about it. The circuit will handle it.

Let's move on to \$Q_2\$. It's the device that is doing the measuring function and there is the following relationship between its all-important \$V_{\text{BE}_2}\$ and its \$I_{\text{C}_2}\$ (for this device, \$\eta=1\$):

$$V_{\text{BE}_2}=V_T\cdot\ln\left({\frac{I_{\text{C}_2}}{I_{\text{SAT}_2}}+1}\right)$$

This is crucial because \$V_{\text{BE}_2}\$ essentially determines \$Q_1\$'s collector current and therefore the LED/LOAD current. So setting the \$Q_2\$ collector current is important. Part and temperature variations in the D44H11, \$Q_1\$, will cause variations in its base current and these variations will cause variations in the collector current of \$Q_2\$ and that will cause variations in \$V_{\text{BE}_2}\$, directly impacting the controlled current sink.

$$\begin{align*}\frac{\%\, V_{\text{BE}_2}}{\%\,I_{\text{C}_2}}=\frac{\frac{\text{d}\, V_{\text{BE}_2}}{V_{\text{BE}_2}}}{\frac{\text{d}\,I_{\text{C}_2}}{I_{\text{C}_2}}}&=\frac{\text{d}\, V_{\text{BE}_2}}{\text{d}\,I_{\text{C}_2}}\cdot \frac{I_{\text{C}_2}}{V_{\text{BE}_2}}=\frac{V_T}{V_{\text{BE}_2}}\\\\&\therefore\\\\\%\,I_{\text{C}_2}&=\%\, V_{\text{BE}_2}\cdot\frac{V_{\text{BE}_2}}{V_T}\end{align*}$$

Let's say that we want to allow only \$\%\, V_{\text{BE}_2}\approx 0.05\$ (or 5%.) This means for thermal and part variations, we want to keep \$19 \:\text{mA}\le I_{\text{C}_1}\le 21\:\text{mA}\$. We should use the largest \$V_T\$ that we are likely to encounter for \$Q_2\$. (Since \$Q_2\$ will drift with ambient temperature and hopefully isn't coupled to \$Q_1\$, this means that perhaps the highest temperature we consider is \$55^\circ\text{C}\$, or \$V_T\le 28.3\:\text{mV}\$.)

First, note that this is for \$V_\text{CE}=1\:\text{V}\$. Luckily, we'll be operating \$Q_2\$ at only a little more than this (two \$V_\text{BE}\$'s), so the chart is close enough for our use.

Second, note that this is a typical chart. And that we do NOT have a way of working out the minimum and maximum between parts within a bag. We are looking to avoid changes due to temperature since that's the whole point of this exercise, but we do need to have an idea what to expect for device variations. The main factor determining \$V_\text{BE}\$ is the saturation current for a device and as this depends on the exact area of contact between the emitter and the base, you can easily find devices varying between 50% to 200% of the nominal 100% figure in the same bag. Due to the log function involved, this works out to about \$\pm 20\:\text{mV}\$.

We don't yet know the collector current for \$Q_2\$, but let's eyeball the \$25^\circ\text{C}\$ curve here and pick off a value of \$660\:\text{mV}\$. We can now estimate that \$640\:\text{mV}\le V_{\text{BE}_2}\le 680\:\text{mV}\$ for part variation alone. From here, we find that \$\%\,I_{\text{C}_2}=0.05\cdot\frac{680\:\text{mV}}{28.3\:\text{mV}}\approx 1.2=120\,\%\$ and \$\%\,I_{\text{C}_2}=0.05\cdot\frac{640\:\text{mV}}{28.3\:\text{mV}}\approx 1.13=113\,\%\$. The (barely) tighter spec is this last one, so that's the one to meet. (Note that the sensitivity equation pretty much tells us that we can accept quite a lot of variation in \$Q_2\$'s collector current, which allows us to set its collector current much closer to the needed base current of \$Q_1\$.)

Solving \$I_\text{DRIVE}-100\:\mu\text{A}=\left(1+1.13\right)\cdot\left(I_\text{DRIVE}-333\:\mu\text{A}\right)\$ provides \$I_\text{DRIVE}=540\:\mu\text{A}\$.

Now we return to the fact that \$640\:\text{mV}\le V_{\text{BE}_2}\le 680\:\text{mV}\$. Let's use \$R_\text{SENSE}=33\:\Omega\$. This means that we expect \$19.4\:\text{mA}\le I_\text{SINK} \le 21\:\text{mA}\$, with an geometric mean (to center things so the plus/minus part is evenly distributed) \$I_\text{SINK}=20.18\:\text{mA}\pm 4\,\%\$.

Going back, we find the tighter spec is \$\%\,I_{\text{C}_2}=0.01\cdot\frac{640\:\text{mV}}{28.3\:\text{mV}}\approx 0.226=22.6\,\%\$. And then \$I_\text{DRIVE}-100\:\mu\text{A}=\left(1+0.226\right)\cdot\left(I_\text{DRIVE}-333\:\mu\text{A}\right)\$ provides \$I_\text{DRIVE}\approx 1.4\:\text{mA}\$. Note that we increased the collector current that \$Q_2\$ will have to handle by a fair bit in order to keep this variation down to a minimum.

Let's assume that \$V_\text{CC}=V_\text{DRIVE}=30\:\text{V}\$. This means \$R_\text{DRIVE}=\frac{V_\text{CC}-V_{\text{BE}_1}-V_{\text{BE}_2}}{I_\text{DRIVE}}\approx 20.5\:\text{k}\Omega\$. We can select either the next lower or next higher value and be "pretty good." Since I want to tighten up a little more to account for some of that resistor variation, I'll select \$R_\text{DRIVE}=18\:\text{k}\Omega\$.

Changes in the \$V_{_\text{BE}}\$ due to temperature on the drive BJT are automatically compensated by the feedback BJT, which is measuring the collector current of the drive BJT as it passes through the resistor between the feedback BJT's base and emitter.

So this leaves us with the small-scale approach. If we know the magnitude of the base-emitter voltage at some temperature and can guess that it won't change by more than \$-1.8\:\frac{\text{mV}}{^\circ\text{C}}\le \frac{\Delta V_{_\text{BE}}}{^\circ \text{C}}\le -2.4\:\frac{\text{mV}}{^\circ\text{C}}\$, then we can make a simple statement:

$$\Delta I_{_\text{LED}}=\frac{ \frac{\Delta V_{_\text{BE}}}{^\circ \text{C}}}{R_{_\text{SENSE}}}\cdot \Delta T$$

So, if \$\frac{\Delta V_{_\text{BE}}}{^\circ \text{C}}=-2.2\:\frac{\text{mV}}{^\circ\text{C}}\$ and \$R_{_\text{SENSE}}=33\:\Omega\$ and \$\Delta T=15\:\text{K}\$, then \$\Delta I_{_\text{LED}}=-1\:\text{mA}\$. Assuming \$V_{_\text{BE}}\approx 680\:\text{mV}\$ prior to the temperature change, \$I_{_\text{LED}}\approx 21\:\text{mA}\$. So a rise of \$\Delta T=15\:\text{K}\$ of the feedback BJT temperature would then imply a change to \$I_{_\text{LED}}\approx 20\:\text{mA}\$, in this case. This is likely to be quite acceptable.

But if you are seeking the large-scale equation, which provides you with how things are over many decades of design currents, then you'll probably want the original expression I was suggesting -- the sensitivity equation, itself. This will tell you the percent change in \$I_{_\text{LED}}\$ for a percent change in temperature, at any starting set value for \$I_{_\text{LED}}\$ and \$T\$. But this also requires the combination of several equations and the use of derivatives. If that's what you want, say so. Otherwise, the above small-signal local change equation is probably sufficient.

$$\begin{align*} I_{_\text{LED}}&=\frac{\beta_1}{\beta_1+1}\,I_{_{\text{E}_1}}=\frac{\beta_1}{\beta_1+1}\left(\frac{V_{_{\text{BE}_2}}}{R_{_\text{SENSE}}}+I_{_{\text{B}_2}}\right)\\\\&=\frac{\beta_1}{\beta_1+1}\left(\frac{V_{_{\text{BE}_2}}}{R_{_\text{SENSE}}}+\frac1{\beta_2}\left[\frac{V_{_\text{DRIVE}}-V_{_{\text{BE}_1}}-V_{_{\text{BE}_2}}}{R_{_\text{DRIVE}}}-\frac{I_{_\text{LED}}}{\beta_1}\right]\right)\\\\\text{solving for }I_{_\text{LED}},\\\\ &=\left[\frac{\beta_1\,\beta_2}{\beta_1\,\beta_2+\beta_2+1}\right]\cdot\left[\frac{V_{_{\text{BE}_2}}}{R_{_\text{SENSE}}}+\frac{V_{_\text{DRIVE}}-V_{_{\text{BE}_1}}-V_{_{\text{BE}_2}}}{R_{_\text{DRIVE}}}\right] \end{align*}$$

Even with temperature variations on \$\beta\$, the value of the first factor above will be very close to 1 (slightly less.) So we can remove it from consideration. \$V_{_\text{DRIVE}}\$ is reasonably assumed to be temperature-independent for analysis purposes. So this leaves us with:

$$\Delta I_{_\text{LED}}=\frac{\frac{\Delta V_{_{\text{BE}_2}}}{^\circ \text{C}}}{R_{_\text{SENSE}}}\cdot \Delta T-\frac{\frac{\Delta V_{_{\text{BE}_1}}}{^\circ \text{C}}+\frac{\Delta V_{_{\text{BE}_2}}}{^\circ \text{C}}}{R_{_\text{DRIVE}}}\cdot \Delta T$$

So there's an adjustment term that I'd not included in the original case. However, because for all intents and purposes it will be the case that \$R_{_\text{DRIVE}}\gg R_{_\text{SENSE}}\$ and that term will not matter much.

We can replace the \$\frac{\Delta V_{_{\text{BE}_i}}}{^\circ \text{C}}\$ variables in the above equation with the Shockley expansion that also includes the full temperature-dependent equations for \$I_{_\text{SAT}}\$. A closed solution will involve the use of the product-log function and take a lot of room below. But it can be done.

For now, I think it is enough to see that a basic circuit analysis does confirm the original equation as "close enough" when using reasonable estimates for the variation of \$V_{_\text{BE}}\$ with temperature.

Those are typical curves. From these, it looks as though I can be pretty sure that over a very wide range of temperatures, and so long as \$V_{_\text{CE}}\ge 1\:\text{V}\$, that \$\beta_1\gt 100\$.

This provides a worst-case reading. It's for \$I_{_\text{C}}=2\:\text{A}\$, which is 100 times what I'm considering. But if you look again at the above curves, you'll see that the positions are about the same in either case. So let's design this for \$\beta_1=60\$. We are rock-solid safe with that choice.

This means \$I_{_{\text{B}_1}}\le 333\:\mu\text{A}\$. Different D44H11 devices may vary, but we can be pretty sure the base current won't exceed this value range. Taking worst-case and best-typical as the extremes, \$100\:\mu\text{A} \le I_{_{\text{B}_1}}\le 333\:\mu\text{A}\$.

For \$Q_1\$, I actually don't care too much right about about its operating \$V_{_{\text{BE}_1}}\$ because it's the job of \$Q_2\$ to make adjustments there. So I'm not going to think about it. The circuit will handle it.

Let's move on to \$Q_2\$. It's the device that is doing the measuring function and there is the following relationship between its all-important \$V_{_{\text{BE}_2}}\$ and its \$I_{_{\text{C}_2}}\$ (for this device, \$\eta=1\$):

$$V_{_{\text{BE}_2}}=V_T\cdot\ln\left({\frac{I_{_{\text{C}_2}}}{I_{_{\text{SAT}_2}}}+1}\right)$$

This is crucial because \$V_{_{\text{BE}_2}}\$ essentially determines \$Q_1\$'s collector current and therefore the LED/LOAD current. So setting the \$Q_2\$ collector current is important. Part and temperature variations in the D44H11, \$Q_1\$, will cause variations in its base current and these variations will cause variations in the collector current of \$Q_2\$ and that will cause variations in \$V_{_{\text{BE}_2}}\$, directly impacting the controlled current sink.

$$\begin{align*}\frac{\%\, V_{_{\text{BE}_2}}}{\%\,I_{_{\text{C}_2}}}=\frac{\frac{\text{d}\, V_{_{\text{BE}_2}}}{V_{_{\text{BE}_2}}}}{\frac{\text{d}\,I_{_{\text{C}_2}}}{I_{_{\text{C}_2}}}}&=\frac{\text{d}\, V_{_{\text{BE}_2}}}{\text{d}\,I_{_{\text{C}_2}}}\cdot \frac{I_{_{\text{C}_2}}}{V_{_{\text{BE}_2}}}=\frac{V_T}{V_{_{\text{BE}_2}}}\\\\&\therefore\\\\\%\,I_{_{\text{C}_2}}&=\%\, V_{_{\text{BE}_2}}\cdot\frac{V_{_{\text{BE}_2}}}{V_T}\end{align*}$$

Let's say that we want to allow only \$\%\, V_{_{\text{BE}_2}}\approx 0.05\$ (or 5%.) This means for thermal and part variations, we want to keep \$19 \:\text{mA}\le I_{_{\text{C}_1}}\le 21\:\text{mA}\$. We should use the largest \$V_T\$ that we are likely to encounter for \$Q_2\$. (Since \$Q_2\$ will drift with ambient temperature and hopefully isn't coupled to \$Q_1\$, this means that perhaps the highest temperature we consider is \$55^\circ\text{C}\$, or \$V_T\le 28.3\:\text{mV}\$.)

First, note that this is for \$V_{_\text{CE}}=1\:\text{V}\$. Luckily, we'll be operating \$Q_2\$ at only a little more than this (two \$V_{_\text{BE}}\$'s), so the chart is close enough for our use.

Second, note that this is a typical chart. And that we do NOT have a way of working out the minimum and maximum between parts within a bag. We are looking to avoid changes due to temperature since that's the whole point of this exercise, but we do need to have an idea what to expect for device variations. The main factor determining \$V_{_\text{BE}}\$ is the saturation current for a device and as this depends on the exact area of contact between the emitter and the base, you can easily find devices varying between 50% to 200% of the nominal 100% figure in the same bag. Due to the log function involved, this works out to about \$\pm 20\:\text{mV}\$.

We don't yet know the collector current for \$Q_2\$, but let's eyeball the \$25^\circ\text{C}\$ curve here and pick off a value of \$660\:\text{mV}\$. We can now estimate that \$640\:\text{mV}\le V_{_{\text{BE}_2}}\le 680\:\text{mV}\$ for part variation alone. From here, we find that \$\%\,I_{_{\text{C}_2}}=0.05\cdot\frac{680\:\text{mV}}{28.3\:\text{mV}}\approx 1.2=120\,\%\$ and \$\%\,I_{_{\text{C}_2}}=0.05\cdot\frac{640\:\text{mV}}{28.3\:\text{mV}}\approx 1.13=113\,\%\$. The (barely) tighter spec is this last one, so that's the one to meet. (Note that the sensitivity equation pretty much tells us that we can accept quite a lot of variation in \$Q_2\$'s collector current, which allows us to set its collector current much closer to the needed base current of \$Q_1\$.)

Solving \$I_{_\text{DRIVE}}-100\:\mu\text{A}=\left(1+1.13\right)\cdot\left(I_{_\text{DRIVE}}-333\:\mu\text{A}\right)\$ provides \$I_{_\text{DRIVE}}=540\:\mu\text{A}\$.

Now we return to the fact that \$640\:\text{mV}\le V_{_{\text{BE}_2}}\le 680\:\text{mV}\$. Let's use \$R_{_\text{SENSE}}=33\:\Omega\$. This means that we expect \$19.4\:\text{mA}\le I_{_\text{SINK}} \le 21\:\text{mA}\$, with an geometric mean (to center things so the plus/minus part is evenly distributed) \$I_{_\text{SINK}}=20.18\:\text{mA}\pm 4\,\%\$.

Going back, we find the tighter spec is \$\%\,I_{_{\text{C}_2}}=0.01\cdot\frac{640\:\text{mV}}{28.3\:\text{mV}}\approx 0.226=22.6\,\%\$. And then \$I_{_\text{DRIVE}}-100\:\mu\text{A}=\left(1+0.226\right)\cdot\left(I_{_\text{DRIVE}}-333\:\mu\text{A}\right)\$ provides \$I_{_\text{DRIVE}}\approx 1.4\:\text{mA}\$. Note that we increased the collector current that \$Q_2\$ will have to handle by a fair bit in order to keep this variation down to a minimum.

Let's assume that \$V_{_\text{CC}}=V_{_\text{DRIVE}}=30\:\text{V}\$. This means \$R_{_\text{DRIVE}}=\frac{V_{_\text{CC}}-V_{_{\text{BE}_1}}-V_{_{\text{BE}_2}}}{I_{_\text{DRIVE}}}\approx 20.5\:\text{k}\Omega\$. We can select either the next lower or next higher value and be "pretty good." Since I want to tighten up a little more to account for some of that resistor variation, I'll select \$R_{_\text{DRIVE}}=18\:\text{k}\Omega\$.

\$Q_1\$ (the drive transistor) can also be replaced with a MOSFET. In fact, the idea is often jumped-at. But doing may also mean considering the possibility of somewhat wider Early Effect variations for \$Q_2\$, as the \$V_{_\text{TH}}\$ of discrete MOSFETs for \$Q_1\$ may be wider than the \$V_{_\text{BE}}\$ variation of discrete BJTs they replace there. Just a note to keep in mind if tempted to make comparisons.

If tighter tolerances over ambient temperature are desired, emitter degeneration should be added to \$Q_2\$. A resistor that is predicted to drop anything more than about \$150\:\text{mV}\$ should help. (More is better.) This comes at exactly that price, though. So doing this takes away from the voltage compliance range of the circuit.

The degeneration also improves the behavior over part variations, too. But emitter degeneration is more important for managing operating temperature variation, as significant improvement can be had with a small loss of voltage compliance range. More sacrifice is needed to get much with respect to part variation. So it's less often used for this purpose.

If tighter tolerances over ambient temperature are desired, emitter degeneration is an option that may be added to \$Q_2\$. A resistor that is predicted to drop anything more than about \$150\:\text{mV}\$ should help. (More is better.) This comes at a price, though:

• It takes away from the voltage compliance range for the circuit, since more voltage overhead is now required.
• Variations in the collector current of \$Q_2\$ now cause an Ohmic voltage drop variation across the degeneration resistor. So a tighter focus on restricting variations of \$Q_2\$'s collector current come into play. Base current variations in \$Q_1\$ from all sources of those variations must now have a smaller impact on \$Q_2\$'s collector current (which sets most of its emitter current and affects the current measurement duties of \$Q_2\$.) So the resulting circuit may be somewhat less efficient.

The degeneration does also improve behavior over part variations. But emitter degeneration is more important for managing operating temperature variation, as significant improvement can be had with a small loss of voltage compliance range. The sacrifices needed to get much with respect to part variation are partly why it's less often used. There are other, better topologies to consider before embarking too far down this path.