4 added 271 characters in body edited Apr 3 '18 at 7:00 jonk 38.1k11 gold badge3131 silver badges8282 bronze badges $$\V_\text{BE}\$$ varies by about $$\60\:\text{mV}\$$ for each 10-fold change in $$\I_\text{C}\$$. It increases when the collector current increases and decreases when the collector current decreases. This can be computed entirely from the $$\\ref{ic}\$$ equation. $$\V_\text{BE}\$$ varies from about $$\-1.8\:\frac{\text{mV}}{^\circ C}\$$ to about $$\-2.4\:\frac{\text{mV}}{^\circ C}\$$. This must be derived by examining all three equations above. Note here that the $$\\ref{isat}\$$ equation overwhelms the implications of the $$\\ref{vt}\$$ equation and reverses the sign of the behavior. For small signal silicon BJTs, the commonly accepted operating voltage for $$\V_\text{BE}\$$ is about $$\700\:\text{mV}\$$. Broadly speaking, this can be assumed to be nearby with a few milliamps of collector current. You can adjust this value up or down based on the estimated magnitude of change for the operating collector current. For small signal BJTs, there is about a 60% increase in $$\I_\text{SAT}\$$ for each 1% change in absolute temperature. There is a small "dynamic resistance" called $$\r_e=\frac{V_T}{I_\text{E}\approx I_\text{C}}\$$ that can be derived directly from the $$\\ref{ic}\$$ equation. In active mode, the value of $$\\beta\$$ can be treated as remarkably "flat" for typically at least 3 orders of magnitude change and often for 5 orders or more. It does vary with temperature (and it varies somewhat with $$\V_\text{CE}\$$ due to the Early Effect.) You can reflect resistances at the emitter backwards to the base by multiplying them by $$\\beta+1\$$. You can reflect resistances at the base forwards to the emitter by dividing them by the same factor, $$\\beta+1\$$. $$\\beta\$$ is a function of temperature and will usually increase with increasing temperature. You cannot compute this variation over temperature with the equations I've provided. Well, the BJT's $$\V_\text{BE}\$$ is too high, so the collector current responds by increasing -- exponentially so. This increase has several immediate effects: (1) The collector voltage drops because the collector load experiences a larger voltage drop; (2) the emitter current increases, as well, and so the emitter voltage rises, too; and (3) the base current rises (assuming $$\\beta\$$ remains unchanged -- but increasing temperature does increase $$\\beta\$$ so this effect complicates this part of the answer, though the KVL equation will still work out close if you can approximate the new value for $$\\beta\$$ at the new temperature), which increases the voltage drop across the Thevenin equivalent of the biasing and therefore lowers the base voltage. In short, the response is that both the $$\V_\text{BE}\$$ and $$\V_\text{CE}\$$ get pinched a bit. And this fact counters (opposes) the original change due to temperature. $$\V_\text{BE}\$$ varies by about $$\60\:\text{mV}\$$ for each 10-fold change in $$\I_\text{C}\$$. It increases when the collector current increases and decreases when the collector current decreases. This can be computed entirely from the $$\\ref{ic}\$$ equation. $$\V_\text{BE}\$$ varies from about $$\-1.8\:\frac{\text{mV}}{^\circ C}\$$ to about $$\-2.4\:\frac{\text{mV}}{^\circ C}\$$. This must be derived by examining all three equations above. Note here that the $$\\ref{isat}\$$ equation overwhelms the implications of the $$\\ref{vt}\$$ equation and reverses the sign of the behavior. For small signal silicon BJTs, the commonly accepted operating voltage for $$\V_\text{BE}\$$ is about $$\700\:\text{mV}\$$. Broadly speaking, this can be assumed to be nearby with a few milliamps of collector current. You can adjust this value up or down based on the estimated magnitude of change for the operating collector current. For small signal BJTs, there is about a 60% increase in $$\I_\text{SAT}\$$ for each 1% change in absolute temperature. There is a small "dynamic resistance" called $$\r_e=\frac{V_T}{I_\text{E}\approx I_\text{C}}\$$ that can be derived directly from the $$\\ref{ic}\$$ equation. In active mode, the value of $$\\beta\$$ can be treated as remarkably "flat" for typically at least 3 orders of magnitude change and often for 5 orders or more. It does vary with temperature (and it varies somewhat with $$\V_\text{CE}\$$ due to the Early Effect.) You can reflect resistances at the emitter backwards to the base by multiplying them by $$\\beta+1\$$. You can reflect resistances at the base forwards to the emitter by dividing them by the same factor, $$\\beta+1\$$. Well, the BJT's $$\V_\text{BE}\$$ is too high, so the collector current responds by increasing -- exponentially so. This increase has several immediate effects: (1) The collector voltage drops because the collector load experiences a larger voltage drop; (2) the emitter current increases, as well, and so the emitter voltage rises, too; and (3) the base current rises, which increases the voltage drop across the Thevenin equivalent of the biasing and therefore lowers the base voltage. In short, the response is that both the $$\V_\text{BE}\$$ and $$\V_\text{CE}\$$ get pinched a bit. And this fact counters (opposes) the original change due to temperature. $$\V_\text{BE}\$$ varies by about $$\60\:\text{mV}\$$ for each 10-fold change in $$\I_\text{C}\$$. It increases when the collector current increases and decreases when the collector current decreases. This can be computed entirely from the $$\\ref{ic}\$$ equation. $$\V_\text{BE}\$$ varies from about $$\-1.8\:\frac{\text{mV}}{^\circ C}\$$ to about $$\-2.4\:\frac{\text{mV}}{^\circ C}\$$. This must be derived by examining all three equations above. Note here that the $$\\ref{isat}\$$ equation overwhelms the implications of the $$\\ref{vt}\$$ equation and reverses the sign of the behavior. For small signal silicon BJTs, the commonly accepted operating voltage for $$\V_\text{BE}\$$ is about $$\700\:\text{mV}\$$. Broadly speaking, this can be assumed to be nearby with a few milliamps of collector current. You can adjust this value up or down based on the estimated magnitude of change for the operating collector current. For small signal BJTs, there is about a 60% increase in $$\I_\text{SAT}\$$ for each 1% change in absolute temperature. There is a small "dynamic resistance" called $$\r_e=\frac{V_T}{I_\text{E}\approx I_\text{C}}\$$ that can be derived directly from the $$\\ref{ic}\$$ equation. In active mode, the value of $$\\beta\$$ can be treated as remarkably "flat" for typically at least 3 orders of magnitude change and often for 5 orders or more. It does vary with temperature (and it varies somewhat with $$\V_\text{CE}\$$ due to the Early Effect.) You can reflect resistances at the emitter backwards to the base by multiplying them by $$\\beta+1\$$. You can reflect resistances at the base forwards to the emitter by dividing them by the same factor, $$\\beta+1\$$. $$\\beta\$$ is a function of temperature and will usually increase with increasing temperature. You cannot compute this variation over temperature with the equations I've provided. Well, the BJT's $$\V_\text{BE}\$$ is too high, so the collector current responds by increasing -- exponentially so. This increase has several immediate effects: (1) The collector voltage drops because the collector load experiences a larger voltage drop; (2) the emitter current increases, as well, and so the emitter voltage rises, too; and (3) the base current rises (assuming $$\\beta\$$ remains unchanged -- but increasing temperature does increase $$\\beta\$$ so this effect complicates this part of the answer, though the KVL equation will still work out close if you can approximate the new value for $$\\beta\$$ at the new temperature), which increases the voltage drop across the Thevenin equivalent of the biasing and therefore lowers the base voltage. In short, the response is that both the $$\V_\text{BE}\$$ and $$\V_\text{CE}\$$ get pinched a bit. And this fact counters (opposes) the original change due to temperature. 3 added 652 characters in body edited Apr 2 '18 at 8:28 jonk 38.1k11 gold badge3131 silver badges8282 bronze badges So either way you look at it, whether it is the AC gain or the DC circuit operating point, providing more room for the emitter voltage is better for thermal stability. You want the following two things to be simultaneously true:\begin{align*} V_\text{E} &\gg V_T\\\\ V_\text{E} &\gg I_\text{B}\cdot R_\text{TH} \end{align*} The second of these two conditions is the reason why it is important that the biasing pair are stiff with respect to the base current. Making them stiffer improves thermal stability. Making them weaker does the opposite. So keep that condition in mind when designing the biasing pair. The negative feedback is pretty simple. A rising BJT temperature decreases the $$\V_\text{BE}\$$ required for the quiescent current. This reduced base-emitter drop allows a slight increase in the base and emitter currents. But these current increases require a slightly increased $$\V_\text{BE}\$$, again. So the currents rise a little and increase the voltage drops across their associated impedances and meanwhile the $$\V_\text{BE}\$$ makes back up some of the temperature-related decrease by a current-demanded increase and these things all conspire to meet in the middle somewhere. The negative feedback is pretty simple. A rising BJT temperature decreases the $$\V_\text{BE}\$$ required for the quiescent current. This reduced base-emitter drop allows a slight increase in the base and emitter currents. But these current increases require a slightly increased $$\V_\text{BE}\$$, again. So the currents rise a little and increase the voltage drops across their associated impedances and meanwhile the $$\V_\text{BE}\$$ makes back up some of the temperature-related decrease by a current-demanded increase and these things all conspire to meet in the middle somewhere. So either way you look at it, whether it is the AC gain or the DC circuit operating point, providing more room for the emitter voltage is better for thermal stability. You want the following two things to be simultaneously true:\begin{align*} V_\text{E} &\gg V_T\\\\ V_\text{E} &\gg I_\text{B}\cdot R_\text{TH} \end{align*} The second of these two conditions is the reason why it is important that the biasing pair are stiff with respect to the base current. Making them stiffer improves thermal stability. Making them weaker does the opposite. So keep that condition in mind when designing the biasing pair. The negative feedback is pretty simple. A rising BJT temperature decreases the $$\V_\text{BE}\$$ required for the quiescent current. This reduced base-emitter drop allows a slight increase in the base and emitter currents. But these current increases require a slightly increased $$\V_\text{BE}\$$, again. So the currents rise a little and increase the voltage drops across their associated impedances and meanwhile the $$\V_\text{BE}\$$ makes back up some of the temperature-related decrease by a current-demanded increase and these things all conspire to meet in the middle somewhere. 2 edited body edited Apr 2 '18 at 8:12 jonk 38.1k11 gold badge3131 silver badges8282 bronze badges $$\V_\text{BE}\$$ varies by about $$\60\:\text{mV}\$$ for each 10-fold change in $$\I_\text{C}\$$. It increases when the collector current increases and decreases when the collector current decreases. This can be computed entirely from the $$\\ref{ic}\$$ equation. $$\V_\text{BE}\$$ varies from about $$\-1.8\:\frac{\text{mV}}{^\circ C}\$$ to about $$\-2.4\:\frac{\text{mV}}{^\circ C}\$$. This must be derived by examining all three equations above. Note here that the $$\\ref{isat}\$$ equation overwhelms the implications of the $$\\ref{vt}\$$ equation and reverses the sign of the behavior. For small signal silicon BJTs, the commonly accepted operating voltage for $$\V_\text{BE}\$$ is about $$\700\:\text{mV}\$$. Broadly speaking, this can be assumed to be nearby with a few milliamps of collector current. You can adjust this value up or down based on the estimated magnitude of change for the operating collector current. For small signal BJTs, there is about a 63%60% increase in $$\I_\text{SAT}\$$ for each 1% change in absolute temperature. There is a small "dynamic resistance" called $$\r_e=\frac{V_T}{I_\text{E}\approx I_\text{C}}\$$ that can be derived directly from the $$\\ref{ic}\$$ equation. In active mode, the value of $$\\beta\$$ can be treated as remarkably "flat" for typically at least 3 orders of magnitude change and often for 5 orders or more. It does vary with temperature (and it varies somewhat with $$\V_\text{CE}\$$ due to the Early Effect.) You can reflect resistances at the emitter backwards to the base by multiplying them by $$\\beta+1\$$. You can reflect resistances at the base forwards to the emitter by dividing them by the same factor, $$\\beta+1\$$. You wanted to know how the emitter resistor helps to counter variations with temperature. It's not too difficult to follow why. Well, the BJT's $$\V_\text{BE}\$$ is too high, so the collector current responds by increasing -- exponentially so. This increase has twoseveral immediate effects: (1) The collector voltage drops because the collector load experiences a larger voltage drop; (2) the emitter current increases, as well, and so the emitter voltage rises, too; and (3) the base current rises, which increases the voltage drop across the Thevenin equivalent of the biasing and therefore lowers the base voltage. In short, the response is that both the $$\V_\text{BE}\$$ and $$\V_\text{CE}\$$ get pinched a bit. And this fact counters (opposes) the original change due to temperature. The AC voltage gain, as I'm sure you know, is about: $$\V_\text{BE}\$$ varies by about $$\60\:\text{mV}\$$ for each 10-fold change in $$\I_\text{C}\$$. It increases when the collector current increases and decreases when the collector current decreases. This can be computed entirely from the $$\\ref{ic}\$$ equation. $$\V_\text{BE}\$$ varies from about $$\-1.8\:\frac{\text{mV}}{^\circ C}\$$ to about $$\-2.4\:\frac{\text{mV}}{^\circ C}\$$. This must be derived by examining all three equations above. Note here that the $$\\ref{isat}\$$ equation overwhelms the implications of the $$\\ref{vt}\$$ equation and reverses the sign of the behavior. For small signal silicon BJTs, the commonly accepted operating voltage for $$\V_\text{BE}\$$ is about $$\700\:\text{mV}\$$. Broadly speaking, this can be assumed to be nearby with a few milliamps of collector current. You can adjust this value up or down based on the estimated magnitude of change for the operating collector current. For small signal BJTs, there is about a 63% increase in $$\I_\text{SAT}\$$ for each 1% change in absolute temperature. There is a small "dynamic resistance" called $$\r_e=\frac{V_T}{I_\text{E}\approx I_\text{C}}\$$ that can be derived directly from the $$\\ref{ic}\$$ equation. In active mode, the value of $$\\beta\$$ can be treated as remarkably "flat" for typically at least 3 orders of magnitude change and often for 5 orders or more. It does vary with temperature (and it varies somewhat with $$\V_\text{CE}\$$ due to the Early Effect.) You can reflect resistances at the emitter backwards to the base by multiplying them by $$\\beta+1\$$. You can reflect resistances at the base forwards to the emitter by dividing them by the same factor, $$\\beta+1\$$. You wanted to know how the emitter resistor helps to counter variations with temperature. It's not too difficult to follow why. Well, the BJT's $$\V_\text{BE}\$$ is too high, so the collector current responds by increasing -- exponentially so. This increase has two immediate effects: (1) The collector voltage drops because the collector load experiences a larger voltage drop; (2) the emitter current increases, as well, and so the emitter voltage rises, too; and (3) the base current rises, which increases the voltage drop across the Thevenin equivalent of the biasing and therefore lowers the base voltage. In short, the response is that both the $$\V_\text{BE}\$$ and $$\V_\text{CE}\$$ get pinched a bit. And this fact counters (opposes) the original change due to temperature. The AC voltage gain, as I'm sure you know, is about $$\V_\text{BE}\$$ varies by about $$\60\:\text{mV}\$$ for each 10-fold change in $$\I_\text{C}\$$. It increases when the collector current increases and decreases when the collector current decreases. This can be computed entirely from the $$\\ref{ic}\$$ equation. $$\V_\text{BE}\$$ varies from about $$\-1.8\:\frac{\text{mV}}{^\circ C}\$$ to about $$\-2.4\:\frac{\text{mV}}{^\circ C}\$$. This must be derived by examining all three equations above. Note here that the $$\\ref{isat}\$$ equation overwhelms the implications of the $$\\ref{vt}\$$ equation and reverses the sign of the behavior. For small signal silicon BJTs, the commonly accepted operating voltage for $$\V_\text{BE}\$$ is about $$\700\:\text{mV}\$$. Broadly speaking, this can be assumed to be nearby with a few milliamps of collector current. You can adjust this value up or down based on the estimated magnitude of change for the operating collector current. For small signal BJTs, there is about a 60% increase in $$\I_\text{SAT}\$$ for each 1% change in absolute temperature. There is a small "dynamic resistance" called $$\r_e=\frac{V_T}{I_\text{E}\approx I_\text{C}}\$$ that can be derived directly from the $$\\ref{ic}\$$ equation. In active mode, the value of $$\\beta\$$ can be treated as remarkably "flat" for typically at least 3 orders of magnitude change and often for 5 orders or more. It does vary with temperature (and it varies somewhat with $$\V_\text{CE}\$$ due to the Early Effect.) You can reflect resistances at the emitter backwards to the base by multiplying them by $$\\beta+1\$$. You can reflect resistances at the base forwards to the emitter by dividing them by the same factor, $$\\beta+1\$$. You wanted to know how the emitter resistor helps to counter variations with temperature. Well, the BJT's $$\V_\text{BE}\$$ is too high, so the collector current responds by increasing -- exponentially so. This increase has several immediate effects: (1) The collector voltage drops because the collector load experiences a larger voltage drop; (2) the emitter current increases, as well, and so the emitter voltage rises, too; and (3) the base current rises, which increases the voltage drop across the Thevenin equivalent of the biasing and therefore lowers the base voltage. In short, the response is that both the $$\V_\text{BE}\$$ and $$\V_\text{CE}\$$ get pinched a bit. And this fact counters (opposes) the original change due to temperature. The AC voltage gain, as I'm sure you know, is about: 1 answered Apr 2 '18 at 7:22 jonk 38.1k11 gold badge3131 silver badges8282 bronze badges