SaM - Classical Band Gap Based Temperature Sensor (3 Pins)

Remeber:

To understand this circuit you need to rememebr:

• How current mirror is made.
• The BJT formula (in active region).

Also let’s see how many BJT in parallel behave, this is the circuit:
The current $I_C$, will be the sum of all the currents of each parallel BJT.
we consisder them equal and define the currents of each parallel BJT as "$I_{C_2}$", so we have:$I_C=N\cdot I_{C_2}$This is like having a more conducting device, where the current is $N$ times the current of the single the basic device, so if we use the BJT formula, we can say:$I_C=N{I}_{S_2}{e}^{\frac{q{V}_G}{KT}}$

This is the strucuture of a “classical band gap based temperature sensor”:

• We have a differential structure with two identical resistances, and two different BJT.
• The 2nd BJT is the strucutre we have seen before with $N$ parallel BJT.
• The bases of the BJT (1) and (2) are at the same voltage, and the circuit here is concluded with a feedback loop.
• This is an amplifier which has a very high gain.
  ⇒ Therefore these two voltages, $V_{-}$ and $V_{+}$ are approximately at the same value.
• So (1) and (2) have also the the same collector voltage.
• Since the two resistances $R_C$ are nominally identical we have the same current in these two branches here ($I_C$).
• Here’s a circuit. (Circuit at rest $I_S=100\text{fA}$, BjTs NOT in active region):
  
• Her’s another circuit (Go to one of the BjT in the circuit >> Double click it >> “Edit Model” >> Change the value of the “Saturation Current”, and see how the output voltage change.)
  Here’s it with each BjT at $I_S=100\text{pA}$, and they are in active region:
  

Because you know that both BJT are in the active region ⇒ we can neglect the base current, so $I_C\simeq I_E$. So now we can relate the $I_C$ current to this voltage $\Delta V_{BE}$:$\Delta V_{BE}=V_{B{E}_1}-V_{B{E}_2}=I_C{R}_T$And since we know that a BJT in active region has this formula:$V_{BE}=\frac{kT}{q}\ln \left(\frac{I_C}{I_S}\right)$And the property of logaritms $\ln (a)-\ln (b)=\ln (\frac{a}{b})$, we can say:$\Delta V_{BE}=\frac{KT}{q}\ln \left(\frac{I_C}{I_{S_1}}\frac{I_{S_2}}{I_C}\right)$We simplify:$V_{out}=\Delta V_{BE}=\frac{KT}{q}\ln \left(\frac{I_{S_2}}{I_{S_1}}\right)$Note that the ratio between $I_{S_1}$ and $I_{S_2}$ is equal to $\frac{K_2}{K_1}$, we can also say that:$V_{out}=\Delta V_{BE}=\frac{KT}{q}\ln \left(\frac{K_2}{K_1}\right)$

We know that:

• $V_{out}=\Delta V_{BE}=\frac{KT}{q}\ln \left(\frac{I_{S_2}}{I_{S_1}}\right)$
• $I_C=N{I}_{C_2}=N{I}_{S_2}{e}^{\frac{q{V}_G}{KT}}$
• $I_C=I_{C_1}=I_{S_1}{e}^{\frac{q{V}_G}{KT}}$
• $I_C=\frac{\Delta V_{BE}}{R_T}$
• $V_{out}=I_C{R}_T$

So:$\frac{I_{S_2}}{I_{S_1}}=\frac{1}{N}$And we can finally calculate the function of this sensor:IMPORTANTE $V_{out}=R_T\frac{kT}{q{R}_1}\ln (N)$It has a linear dependance on temperature.

The usual values for the sensitivity for this kind of devices are in the range:IMPORTANTE $\alpha \in \left[2\frac{\text{mV}}{°\text{C}}÷10\frac{\text{mV}}{°\text{C}}\right]$

So this device that we have here, we have discussed is a device with three pins:

• In the theory we have found no offset, but in reality there could be one.
• So this is a 3 pin device, but there also exists other circuits based on the same philosophy and the same topology or similar topology which instead have 2 pins.

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Memory Card

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This is the structure that can be used also to obtain regulated voltages where the temperature dependence is compensated (or eliminated).

• We have a differential structure with two identical resistances, and two different BJT.
• The 2nd BJT can be seen more clearly (how it is strucutered) on the right.
  it’s like having many BJT in parallel.
  ⇒ This means that the current here $I_C$, will be the sum of all these currents, of each parallel BJT.
  This is like having a more conducting device, where the current is $N$ times the current of the single the basic device.
• The bases of the BJT (1) and (2) are at the same voltage, and the circuit here is concluded with a feedback loop.
• This is an amplifier which has a very high gain.
  ⇒ Therefore these two voltages, $V_{-}$ and $V_{+}$ are approximately at the same value.
• So (1) and (2) have also the the same collector voltage.
• Since the two resistances $R_C$ are nominally identical we have the same current in these two branches here ($I_C$).

Because you know that both BJT are in the active region ⇒ we can neglect the base current, so $I_C\simeq I_E$. So now we can relate the $I_C$ current to this voltage $\Delta V_{BE}$:

And so we have the output voltage here:

• We use the property of logarithms: $\ln (a)-\ln (b)=\ln (\frac{a}{b})$
• So we have: $\Delta V_{BE}=\frac{KT}{q}\ln \left(\frac{I_C}{I_{S_1}}\frac{I_{S_2}}{I_C}\right)$
• We can simplify $I_C$ and obtain:$V_{out}=\Delta V_{BE}=\frac{KT}{q}\ln \left(\frac{I_{S_2}}{I_{S_1}}\right)$
• Where there ratio between $I_{S_1}$ and $I_{S_2}$ is equal to $\frac{K_2}{K_1}$.
• So finally: $V_{out}=\Delta V_{BE}=\frac{KT}{q}\ln \left(\frac{K_2}{K_1}\right)$
• We have supposed that the ratio $\frac{K_2}{K_1}=10$ but this depend on the structure, since we have that:$I_C=I_S{e}^{\frac{q{V}_G}{KT}}$Then we wil have that since both $I_C$ are equal and $I_{C_2}$ is the sum of $N$ identical collector currents (since the 2nd BJT is the parallel of many BJT):
  • $I_C=I_{C_1}=I_{S_1}{e}^{\frac{q{V}_G}{KT}}$
  • $I_C=N{I}_{C_2}=N{I}_{S_2}{e}^{\frac{q{V}_G}{KT}}$
  • So finally we have: $\frac{I_{S_2}}{I_{S_1}}=\frac{1}{N}$There must be an error somewhere becouse in this case the output comes out as negative with respect to what there is written in the notes, since:
    $\ln \left(\frac{1}{N}\right)=-\ln (N)$
    This is not a big problem, but it is different.
• So in this particular example we have supposed $N=10$, so the BJT (2) is composed of 10 stacked BJT.
• NOTE: If we choose to stack only 1 BJT, so $N=1$, we have that $\ln (1)=0$ so we need at least $2$ BJT in parallel in position (2), to have a reading.

• So now $V_{out}$ has a linear realtionship with temperature, really good, this means that is also has a fixed sensitivity $\alpha$.
• NOTE: We have supposed a $V_{ref}$ independent on temperature.

So this device that we have here, we have discussed is a device with three pins.

• In the theory we have found no offset, but in reality there could be one.
• So this is a 3 pin device, but there also exists other circuits based on the same philosophy and the same topology or similar topology which instead have 2 pins.