The current in $T_1$ will be equal to $I_{\text {OUT }}$, so that

$$
V_{B E 1}=V_T \ln \frac{100 \mu \mathrm{~A}}{10^{-14} \mathrm{~A}}=598 \mathrm{mV}
$$

technologies. A corresponding circuit utilizing npn transistors can be used in $n$-substrate CMOS technologies. The feedback circuit formed by $M_2, M_3, M_4$, and $M_5$ forces the current in transistor $Q_1$ to be the same as in resistor $R$. Assuming matched devices, $V_{G S 2}=V_{G S 3}$ and thus

$$
I_{\mathrm{OUT}}=\frac{V_{B E 1}}{R}
$$

An alternate source for the voltage reference is the thermal voltage $V_T$. The difference in junction potential between two junctions operated at different current densities can be shown to be proportional to $V_T$. This voltage difference must be converted to a current to provide the bias current. For the Widlar source that shows that the voltage across the resistor $R_2$ is

$$
I_{\mathrm{OUT}} R_2=V_T \ln \frac{I_{\mathrm{IN}}}{I_{\mathrm{OUT}}} \frac{I_{S 2}}{I_{S 1}}
$$


Thus if the ratio of the input to the output current is held constant, the voltage across $R_2$ is indeed proportional to $V_T$. This fact is utilized in the self-biased circuit. Here $Q_3$ and $Q_4$ are selected to have equal areas. Therefore, if we assume that $\beta_F \rightarrow \infty$ and $V_A \rightarrow \infty$, the current mirror formed by $Q_3$ and $Q_4$ forces the collector current of $Q_1$ to equal that of $Q_2$. Figure also shows that $Q_2$ has two emitters and $Q_1$ has one emitter, which indicates that the emitter area of $Q_2$ is twice that of $Q_1$ in this example. This selection is made to force the gain around the positive feedback loop at point $B$ to be more than unity so that point $B$ is an unstable operating point and a stable nonzero operating point exists at point $A$. As in other self-biased circuits, a start-up circuit is required to make sure that enough current flows to force operation at point $A$ in practice. Under these conditions, $I_{S 2}=2 I_{S 1}$ and the voltage across $R_2$ is

$$
I_{\mathrm{OUT}} R_2=V_T \ln \frac{I_{\mathrm{IN}}}{I_{\mathrm{OUT}}} \frac{I_{S 2}}{I_{S 1}}=V_T \ln 2
$$


Therefore, the output current is

$$
I_{\mathrm{OUT}}=\frac{V_T}{R_2} \ln 2
$$

Thus,

$$
R=\frac{598 \mathrm{mV}}{0.1 \mathrm{~mA}}=5.98 \mathrm{k} \Omega
$$

$$
T C_F \simeq \frac{-2 \mathrm{mV} /{ }^{\circ} \mathrm{C}}{598 \mathrm{mV}}-1.5 \times 10^{-3} \simeq-3.3 \times 10^{-3}-1.5 \times 10^{-3}
$$

and thus

$$
T C_F \simeq-4.8 \times 10^{-3} /{ }^{\circ} \mathrm{C}=-4800 \mathrm{ppm} /{ }^{\circ} \mathrm{C}
$$


The term ppm is an abbreviation for parts per million and implies a multiplier of $10^{-6}$.
For the threshold-referenced circuit,

$$
I_{\mathrm{OUT}}=\frac{V_{G S 1}}{R} \simeq \frac{V_t}{R}
$$


gives

$$
T C_F=\frac{1}{I_{\mathrm{OUT}}} \frac{\partial I_{\mathrm{OUT}}}{\partial T} \simeq \frac{1}{V_t} \frac{\partial V_t}{\partial T}-\frac{1}{R} \frac{\partial R}{\partial T}
$$


Since the threshold voltage of an MOS transistor and the base-emitter voltage of a bipolar transistor both change at about $-2 \mathrm{mV} /{ }^{\circ} \mathrm{C}$, the temperature dependence of the threshold-referenced current source  is about the same as the $V_{B E}$-referenced current source.
$V_{B E}$-referenced bias circuits are also used in CMOS technology. An example is shown in Fig. 4.40, where the $p n p$ transistor is the parasitic device inherent in $p$-substrate CMOS The temperature variation of the output current can be calculated as follows. 

$$
\begin{aligned}
\frac{\partial I_{\mathrm{OUT}}}{\partial T} & =(\ln 2) \frac{R_2 \frac{\partial V_T}{\partial T}-V_T \frac{\partial R_2}{\partial T}}{R_2^2} \\
& =\frac{V_T}{R_2}(\ln 2)\left(\frac{1}{V_T} \frac{\partial V_T}{\partial T}-\frac{1}{R_2} \frac{\partial R_2}{\partial T}\right)
\end{aligned}
$$

$$
T C_F=\frac{1}{I_{\mathrm{OUT}}} \frac{\partial I_{\mathrm{OUT}}}{\partial T}=\frac{1}{V_T} \frac{\partial V_T}{\partial T}-\frac{1}{R_2} \frac{\partial R_2}{\partial T}
$$


This circuit produces much smaller temperature coefficient of the output current than the $V_{B E}$ reference because the fractional sensitivities of both $V_T$ and that of a diffused resistor $R_2$ are positive and tend to cancel. We have chosen a transistor area ratio of two to one as an example. In practice, this ratio is often chosen to minimize the total area required for the transistors and for resistor $R_2$.