\[ \begin{align*} I_{\rm C}\approx B I_{\rm B} \end{align*} \] for simple operating-point estimates.
The electronics in personal computers, mobile phones, electric toothbrushes, and like all other digital companions, are based on transistor circuits. All logic circuits can be traced back to NAND and NOR gates as building blocks. These logic gates consist of transistors. In the simulation below, the structure of a NAND gate is shown in the current CMOS structure. CMOS here indicates the structure of the circuit and semiconductor structure: Complementary metal-oxide-semiconductor - an oppositely complementary circuit of semiconductors of the metal-oxide-semiconductor structure. The complementary structure is shown by the fact that.
These two different kinds of MOS-transistors and further used kinds shall be explained in this chapter.
A variable resistor can be developed from the diode.
A diode has two terminals and one pn junction, as we have seen in Block11. It can conduct or block depending on the applied voltage.
Therefore, there is only one path on which the current is dependent on the voltage on this path
With this controlled transition resistor (“transfer resistor” or better transistor) this gets a bit more extreenally controlled:
Imagine a water valve.
For a transistor:
A bipolar junction transistor (BJT) has three doped regions and three terminals.
There are two basic BJT types:
Depending on the layer sequence (or “direction of the diodes”), PNP or NPN transistors result, represented by different circuit symbols with three terminals (see figure 1). In both transistor variants, charge carriers are emitted from the emitter terminal (E) toward the collector terminal (C) if a suitable current flows through the base terminal (B). In simplified terms, the negative charge carriers of the n-doped sides could represent a current through an NPN structure if negative charge carriers were also present in the P-doped layer. The current $I_\rm C$ flowing with it in the technical current direction is illustrated in the circuit symbol by the arrow direction at the emitter. In the NPN transistor, the current $I_\rm C$ flows from the collector to the emitter. Since positive charge carriers enable conductivity in the PNP transistor, the technical current direction here points from the emitter to the collector, and the arrow on the emitter points towards the collector. The direction of the arrow is similar to the direction of the diode or the PN junction.
The arrow indicates the technical current direction at the emitter in normal operation.
A BJT can be drawn as two pn junctions, but it is not simply two independent diodes.
The base region is very thin. This is essential: it allows charge carriers injected from the emitter to reach the collector.
Things to try:
For an npn transistor in normal active operation, the base-emitter junction is forward-biased and the collector-base junction is reverse-biased.
\[ \begin{align*} U_{\rm BE} &> 0, & U_{\rm CE} &> 0, & U_{\rm CB} &> 0. \end{align*} \]
With the reference arrows commonly used in this course:
\[ \begin{align*} I_{\rm B}>0, \qquad I_{\rm C}>0, \qquad I_{\rm E}<0. \end{align*} \]
Kirchhoff's current law gives
\[ \begin{align*} I_{\rm B}+I_{\rm C}+I_{\rm E}=0. \end{align*} \]
Using current magnitudes:
\[ \begin{align*} |I_{\rm E}|=I_{\rm B}+I_{\rm C}. \end{align*} \]
| Type | Voltage signs | Current signs with the shown reference arrows |
|---|---|---|
| npn | \(U_{\rm BE}>0\), \(U_{\rm CE}>0\), \(U_{\rm CB}>0\) | \(I_{\rm B}>0\), \(I_{\rm C}>0\), \(I_{\rm E}<0\) |
| pnp | \(U_{\rm BE}<0\), \(U_{\rm CE}<0\), \(U_{\rm CB}<0\) | \(I_{\rm B}<0\), \(I_{\rm C}<0\), \(I_{\rm E}>0\) |
Here, the figures in figure 2 shall be described:
The base-emitter junction behaves approximately like a diode. For a silicon transistor, noticeable base current often starts around
\[ \begin{align*} U_{\rm BE}\approx 0.6\ldots0.7~{\rm V}. \end{align*} \]
When enough base current flows, the thin base region allows many carriers to pass from emitter to collector. Thus a small base current can control a larger collector current.
Because both electrons and holes contribute to the physical operation, the component is called bipolar.
In the active region, a small base current controls a larger collector current. The base current flows over a diode between base and emitter (depict as arrot in the symbol).
\[ \begin{align*} \boxed{ I_{\rm C}\approx B I_{\rm B} } \end{align*} \]
The factor \(B\) is the DC current gain:
\[ \begin{align*} B=\frac{I_{\rm C}}{I_{\rm B}}. \end{align*} \]
For small changes around an operating point, the small-signal current gain is
\[ \begin{align*} h_{\rm FE} = \beta = \frac{\Delta I_{\rm C}}{\Delta I_{\rm B}} = \frac{i_{\rm C}}{i_{\rm B}} \approx B. \end{align*} \]
\[ \begin{align*} {\color{blue}{I_{\rm B}}} &:\text{ small control current}, \\ {\color{green}{I_{\rm C}}} &:\text{ larger controlled current}, \\ {\color{red}{U_{\rm BE}I_{\rm B}}} &:\text{ input control power}. \end{align*} \]
The useful first approximation is
\[ \begin{align*} {\color{green}{I_{\rm C}}} \approx B\,{\color{blue}{I_{\rm B}}}. \end{align*} \]
For robust switch design, one usually does not rely on the typical value of \(B\). This is treated in Block 14.
A transistor is nonlinear. Therefore we describe it by characteristic curves.
Important BJT characteristics are:
Important parameters around an operating point are
\[ \begin{align*} r_{\rm BE} &= \frac{\Delta U_{\rm BE}}{\Delta I_{\rm B}} = \frac{u_{\rm BE}}{i_{\rm B}}, \\[4pt] r_{\rm CE} &= \frac{\Delta U_{\rm CE}}{\Delta I_{\rm C}} = \frac{u_{\rm CE}}{i_{\rm C}}. \end{align*} \]
A BJT operates mainly in three different regions.
| Region | Approximate condition | Electrical behavior | Typical use |
|---|---|---|---|
| cutoff | \(I_{\rm B}\approx 0\) | \(I_{\rm C}\approx 0\), transistor blocks | open switch |
| active region | \(I_{\rm C}\approx B I_{\rm B}\) | collector current controlled by base current | analog amplifier |
| saturation | \(I_{\rm B}\) large enough, \(U_{\rm CE}\) small | transistor conducts strongly | closed switch |
For a BJT used as a switch:
In saturation, the collector-emitter voltage is small, often approximated by
\[ \begin{align*} U_{\rm CE,sat}\approx 0.1\ldots 0.3~{\rm V}. \end{align*} \]
The conduction loss of a saturated BJT switch is approximately
\[ \begin{align*} P_{\rm on,BJT} \approx U_{\rm CE,sat}I_{\rm C}. \end{align*} \]
The control loss at the base is approximately
\[ \begin{align*} P_{\rm ctrl,BJT} \approx U_{\rm BE}I_{\rm B}. \end{align*} \]
Real transistor switching is not instantaneous.
Typical time intervals are:
During switching, both current and voltage can be significant at the same time.
Therefore switching losses occur during turn-on and turn-off.
\[ \begin{align*} p(t)=u_{\rm CE}(t) \cdot i_{\rm C}(t). \end{align*} \]
A BJT in saturation stores charge carriers in the base region. When switching off, this stored charge must be removed first. This contributes to the storage time \(t_{\rm s}\).
A field-effect transistor controls current by an electric field.
For a MOSFET:
The name MOSFET means:
\[ \begin{align*} \text{Metal-Oxide-Semiconductor Field-Effect Transistor}. \end{align*} \]
For a BJT:
\[ \begin{align*} {\color{blue}{I_{\rm B}}} \quad \text{controls} \quad {\color{green}{I_{\rm C}}}. \end{align*} \]
For a MOSFET:
\[ \begin{align*} {\color{blue}{U_{\rm GS}}} \quad \text{controls} \quad {\color{green}{I_{\rm D}}}. \end{align*} \]
In stationary operation, the ideal gate current is approximately
\[ \begin{align*} I_{\rm G}\approx 0. \end{align*} \]
The gate is separated from the semiconductor by a thin oxide layer. Therefore the gate behaves approximately like one plate of a capacitor.
For an n-channel enhancement MOSFET:
The threshold voltage \(U_{\rm GS(th)}\) indicates when a small channel just begins to form.
\(U_{\rm GS(th)}\) is not the voltage for a fully switched-on MOSFET.
For low conduction loss, use the gate voltage at which the datasheet specifies $R_{\rm DS(on)}$.
In a discrete power MOSFET, the bulk is often internally connected to the source. Together with the drain region, this creates an internal pn junction. This is the body diode.
Things to try:
The drain current $I_{\rm D}$ depends on
A MOSFET has several operating regions.
Their names can be confusing because they are not identical to the BJT names.
| Region | Approximate condition | Electrical behavior | Typical use |
|---|---|---|---|
| cutoff | \(U_{\rm GS}<U_{\rm GS(th)}\) | no useful channel, \(I_{\rm D}\approx 0\) | open switch |
| linear / ohmic region | \(U_{\rm GS}\) high, \(U_{\rm DS}\) small | behaves like a controlled resistor | closed switch |
| MOSFET saturation region | \(U_{\rm GS}\) high, \(U_{\rm DS}\) larger | current mainly controlled by \(U_{\rm GS}\) | analog operation, current-source-like behavior |
A fully switched-on MOSFET is usually operated in the linear or ohmic region.
When a MOSFET is fully switched on, the drain-source path behaves approximately like a small resistance:
\[ \begin{align*} U_{\rm DS}\approx R_{\rm DS(on)} \cdot I_{\rm D}. \end{align*} \]
The conduction loss is
\[ \begin{align*} \boxed{ P_{\rm on,MOS} = R_{\rm DS(on)} \cdot I_{\rm D}^2 } \end{align*} \]
MOSFETs can be classified by channel type and by whether they are normally off or normally on.
| Channel type | Enhancement type / self-blocking | Depletion type / self-conducting |
|---|---|---|
| n-channel | off at \(U_{\rm GS}=0\), on for sufficiently positive \(U_{\rm GS}\) | on at \(U_{\rm GS}=0\), can be reduced by negative \(U_{\rm GS}\) |
| p-channel | off at \(U_{\rm GS}=0\), on for sufficiently negative \(U_{\rm GS}\) | on at \(U_{\rm GS}=0\), can be reduced by positive \(U_{\rm GS}\) |
In many mechatronic power circuits, the most common device is the n-channel enhancement MOSFET.
| Property | BJT | MOSFET |
|---|---|---|
| control quantity | base current \(I_{\rm B}\) | gate-source voltage \(U_{\rm GS}\) |
| stationary control current | required | approximately zero |
| stationary control loss | \(P_{\rm ctrl}\approx U_{\rm BE}I_{\rm B}\) | very small, but gate must be charged and discharged during switching |
| on-state loss | \(P_{\rm on}\approx U_{\rm CE,sat}I_{\rm C}\) | \(P_{\rm on}=R_{\rm DS(on)}I_{\rm D}^2\) |
| switching behavior | storage charge can slow turn-off | often faster, but gate capacitance matters |
| typical risk | current gain \(B\) varies strongly | gate oxide sensitive to overvoltage and ESD |
In many modern digital and power-electronic circuits, MOSFETs are preferred as switches because they need almost no stationary gate current and can have very small \(R_{\rm DS(on)}\).
However, the gate is capacitive. Fast switching requires charging and discharging this capacitance quickly. This is one reason why gate drivers are needed in power stages.
An npn transistor is operated in normal active operation. The reference arrows are chosen as in the core content:
For normal npn operation:
\[ \begin{align*} U_{\rm BE}>0, \qquad U_{\rm CE}>0, \qquad U_{\rm CB}>0. \end{align*} \]
1. State the meaning of the terminals \(B\), \(C\), and \(E\).
Use the terminal names of the bipolar junction transistor:
The names are not arbitrary. They describe the physical function of the regions.
2. State the signs of \(I_{\rm B}\), \(I_{\rm C}\), and \(I_{\rm E}\) in normal npn operation.
For the reference arrows used here, base current and collector current enter the transistor.
The emitter current physically leaves the transistor. Since \(I_{\rm E}\) is also defined as entering the transistor, its sign becomes negative.
3. Calculate \(I_{\rm E}\) for
\[ \begin{align*} I_{\rm B}=50~\mu{\rm A}, \qquad I_{\rm C}=5.0~{\rm mA}. \end{align*} \]
Apply Kirchhoff's current law at the transistor:
\[ \begin{align*} I_{\rm B}+I_{\rm C}+I_{\rm E}=0. \end{align*} \]
Solve for \(I_{\rm E}\):
\[ \begin{align*} I_{\rm E}=-(I_{\rm B}+I_{\rm C}). \end{align*} \]
Insert both currents in the same unit:
\[ \begin{align*} I_{\rm B}=50~\mu{\rm A}=0.050~{\rm mA}. \end{align*} \]
The emitter current magnitude is
\[ \begin{align*} |I_{\rm E}|=5.05~{\rm mA}. \end{align*} \]
An npn transistor is operated in the active region. The base current is
\[ \begin{align*} I_{\rm B}=35~\mu{\rm A}. \end{align*} \]
The DC current gain is approximated by
\[ \begin{align*} B=150. \end{align*} \]
1. Calculate the collector current \(I_{\rm C}\).
In the active region, use the current-gain approximation
\[ \begin{align*} I_{\rm C}\approx B I_{\rm B}. \end{align*} \]
Insert the given current gain and base current.
2. Calculate the emitter current magnitude \(|I_{\rm E}|\).
Using magnitudes,
\[ \begin{align*} |I_{\rm E}|=I_{\rm B}+I_{\rm C}. \end{align*} \]
Convert \(I_{\rm B}\) to \({\rm mA}\):
\[ \begin{align*} 35~\mu{\rm A}=0.035~{\rm mA}. \end{align*} \]
3. Explain why the value of \(I_{\rm C}\) is only an estimate.
The formula
\[ \begin{align*} I_{\rm C}\approx B I_{\rm B} \end{align*} \]
uses a simplified transistor model.
The current gain \(B\) is not an exact constant. It depends on transistor type, collector current, temperature, and manufacturing variation.
For robust circuit design, especially for switches, one should not rely only on a typical value of \(B\).
An npn transistor is used with a resistive load. The supply voltage is
\[ \begin{align*} U_{\rm dc}=12~{\rm V}. \end{align*} \]
The load resistance is
\[ \begin{align*} R_{\rm L}=680~\Omega. \end{align*} \]
The base current is
\[ \begin{align*} I_{\rm B}=0.15~{\rm mA}. \end{align*} \]
Assume a current gain
\[ \begin{align*} B=100. \end{align*} \]
1. Calculate the collector current predicted by the active-region formula.
If the transistor is in the active region, use
\[ \begin{align*} I_{\rm C,active}\approx B I_{\rm B}. \end{align*} \]
This is the collector current the transistor would try to set.
2. Calculate the maximum load current set by the resistor.
The load circuit limits the current. As a first estimate, neglect \(U_{\rm CE,sat}\) and use Ohm's law:
\[ \begin{align*} I_{\rm L,max}\approx \frac{U_{\rm dc}}{R_{\rm L}}. \end{align*} \]
3. Decide whether the transistor is likely in the active region or in saturation.
Compare the current predicted by the active-region formula with the maximum current allowed by the load circuit:
\[ \begin{align*} I_{\rm C,active} \quad \text{versus} \quad I_{\rm L,max}. \end{align*} \]
If \(I_{\rm C,active}\) is smaller than the load limit, the transistor can still be in the active region. If \(I_{\rm C,active}\) is larger than the load limit, the transistor is driven into saturation.
The transistor is not forced into saturation by this base current. It is likely still in the active region.
4. What base current would approximately be required to reach the load-current limit in the active-region model?
Set the active-region collector current equal to the maximum load current:
\[ \begin{align*} I_{\rm L,max}=B I_{\rm B,limit}. \end{align*} \]
Solve for the base current:
\[ \begin{align*} I_{\rm B,limit}=\frac{I_{\rm L,max}}{B}. \end{align*} \]
So a base current above about
\[ \begin{align*} I_{\rm B}\approx 0.18~{\rm mA} \end{align*} \]
would begin to drive the transistor toward saturation.
An n-channel enhancement MOSFET is used as a switch. The datasheet gives
\[ \begin{align*} U_{\rm GS(th)}=2.0\ldots4.0~{\rm V}. \end{align*} \]
The datasheet also gives
\[ \begin{align*} R_{\rm DS(on)}=80~{\rm m}\Omega \end{align*} \]
but only for
\[ \begin{align*} U_{\rm GS}=10~{\rm V}. \end{align*} \]
The load current is
\[ \begin{align*} I_{\rm D}=3.0~{\rm A}. \end{align*} \]
1. Explain why \(U_{\rm GS(th)}\) is not the correct value for switching the MOSFET fully on.
The threshold voltage only describes the beginning of channel formation. At \(U_{\rm GS(th)}\), the datasheet usually defines only a very small drain current.
For low-loss switching, the MOSFET must have a small \(R_{\rm DS(on)}\). Therefore use the gate voltage for which \(R_{\rm DS(on)}\) is specified.
It does not mean: the MOSFET is a good low-resistance switch.
For this datasheet value, low-loss operation is specified at
\[ \begin{align*} U_{\rm GS}=10~{\rm V}. \end{align*} \]
2. Calculate the drain-source voltage \(U_{\rm DS}\) when the MOSFET is fully switched on at \(U_{\rm GS}=10~{\rm V}\).
For a fully switched-on MOSFET, use the resistor approximation:
\[ \begin{align*} U_{\rm DS}\approx R_{\rm DS(on)}I_{\rm D}. \end{align*} \]
Convert the resistance:
\[ \begin{align*} 80~{\rm m}\Omega=0.080~\Omega. \end{align*} \]
3. Calculate the conduction loss \(P_{\rm on,MOS}\).
The conduction loss of a switched-on MOSFET is
\[ \begin{align*} P_{\rm on,MOS}=R_{\rm DS(on)}I_{\rm D}^2. \end{align*} \]
Alternatively, you can use
\[ \begin{align*} P_{\rm on,MOS}=U_{\rm DS}I_{\rm D}. \end{align*} \]
A mechatronic actuator draws the load current
\[ \begin{align*} I_{\rm L}=1.2~{\rm A}. \end{align*} \]
Two possible switching transistors are compared.
For the BJT, assume
\[ \begin{align*} U_{\rm CE,sat}=0.25~{\rm V}, \qquad U_{\rm BE}=0.70~{\rm V}, \qquad I_{\rm B}=15~{\rm mA}. \end{align*} \]
For the MOSFET, assume
\[ \begin{align*} R_{\rm DS(on)}=120~{\rm m}\Omega \end{align*} \]
at the available gate voltage.
1. Calculate the BJT conduction loss.
For a saturated BJT switch, the conduction loss is approximated by
\[ \begin{align*} P_{\rm on,BJT}\approx U_{\rm CE,sat}I_{\rm C}. \end{align*} \]
Here the collector current is the load current:
\[ \begin{align*} I_{\rm C}\approx I_{\rm L}. \end{align*} \]
2. Calculate the stationary BJT control loss at the base.
The base-emitter path behaves approximately like a forward-biased diode.
The stationary control loss is
\[ \begin{align*} P_{\rm ctrl,BJT}\approx U_{\rm BE}I_{\rm B}. \end{align*} \]
3. Calculate the MOSFET conduction loss.
For a fully switched-on MOSFET, use
\[ \begin{align*} P_{\rm on,MOS}=R_{\rm DS(on)}I_{\rm D}^2. \end{align*} \]
Here the drain current is the load current:
\[ \begin{align*} I_{\rm D}\approx I_{\rm L}. \end{align*} \]
Convert
\[ \begin{align*} 120~{\rm m}\Omega=0.120~\Omega. \end{align*} \]
4. Which component has the lower on-state loss in this operating point?
Compare the calculated conduction losses:
\[ \begin{align*} P_{\rm on,BJT} \quad \text{and} \quad P_{\rm on,MOS}. \end{align*} \]
The lower value means lower heat generation in the switched-on state.
For this operating point, the MOSFET has the lower conduction loss.
5. State one important disadvantage or risk of each device.
Use the qualitative comparison:
Both devices must be checked for power dissipation and temperature.
A discrete n-channel MOSFET is used as a low-side switch. The source is connected to ground, the drain is connected to a load, and the load is connected to \(+24~{\rm V}\).
The MOSFET is off:
\[ \begin{align*} U_{\rm GS}=0. \end{align*} \]
1. Can the MOSFET channel conduct a useful load current?
For an n-channel enhancement MOSFET:
Therefore the drain-source path through the channel is off.
\[ \begin{align*} U_{\rm GS}=0 \end{align*} \]
an n-channel enhancement MOSFET is normally off, so the channel does not conduct a useful load current.
2. Why is a real MOSFET still not an ideal bidirectional open switch?
A discrete power MOSFET usually contains a body diode. This diode is caused by the internal pn junction between body and drain/source regions.
Even when the MOSFET channel is off, the body diode can conduct for one drain-source polarity.
Therefore it can block well only in one polarity. In the opposite polarity, the body diode may become forward-biased and conduct.
So a single MOSFET is not an ideal bidirectional open switch.
3. Why is the body diode important for motor-driver circuits?
Motors and coils are inductive loads. Their current cannot change instantly.
When a transistor switches off, the current may continue through diodes or body diodes. This influences voltage spikes, current paths, losses, and braking behavior.
This can be useful, but it also affects the behavior of H-bridges and motor drivers. The designer must know when the body diode conducts and whether its current and power limits are sufficient.
Functional Principle of a Transistor