1. Transistors Appendix

2. Transistors are scalable electronic switches, made from doped silicon
Transistors are scalable electronic switches, made from doped silicon. Silicon forms a crystal, and have no free electrons at low temperature (around 0K). At higher temperatures (300K), thermal energy is sufficient to release some electrons from their covalent bonds. The availability of conducting electrons is, however, limited (more quantitative) because the band-gap energy is high relative to a good conductor (copper).

3. Remove electrons from valence shell, holes created
Dopant :
Phosphorus
Boron
Means :
Add electrons to
conduction band
Remove electrons from valence shell, holes created
Creates :
N-type material
P-type material N P
N- and P-type doped silicon are the components necessary to build switching devices, like diodes and transistors. Dopant atoms can integrate with silicon’s crystal lattice, and create additional holes or conducting electrons. Because phosphorus has five valance electrons, an additional weakly bound electron is present when it integrates. Thermal energy frees this electron, producing a large number of conducing electrons in this material, known as N-type. Tri-valent Boron creates a hole upon integration, producing P-type material.

4. Establishes current equilibrium
Diffusion
Sets up a field
Establishes current equilibrium N P N P N P
Diffusion
Diffusion Si Si Si Si Si Si
Field P Si P+ Si P+ Si Si B Si B- Si B- Si Si Si Si Si Si + -
The diode device architecture fuses an N and a P-junction together. The NP junction with no voltage applied establishes an equilibrium such that the force of diffusion, which draws electrons into the P-type material, is opposed by the force of the electric field, which draws electrons back into the N-type material. This “equilibrium” electric field represents a potential difference around 0.76V (proven, p. 18 of Electronic Circuit Design and Analysis).

5. N-type
P-type
Relative concentration of majority carrier in N-type material: electrons Si Si P Si Si B
Diffusion “force” Si Si
Relative concentration of minority carrier in P-type material: electrons
The N-region contains a higher concentration electrons than the P-region. Electrons diffuse from the N to P-type material.

6. Electric field (oriented from positive to negative), ~0.75V
Electric force (on electrons) due to equilibrium field
N-type
P-type
Electrons Si Si P+ Si Si B-
Diffusion “force” Si Si
Electrons
Diffusion establishes a charge separation, which sets up an electric field. The field exerts a force on the electrons opposite the direction of diffusion.

7. Electrons : equilibrium
Electric field (oriented from positive to negative) , ~0.75V
Electric force (on electrons)
N-type
P-type
Diffusion “force”
Electrons : equilibrium
Diffusion continues and the charge separation continues to grow in magnitude, until the force of the electric field due to the charge separation equally opposes the force of diffusion. At this point, there is no net current flow across the junction, and the electron concentration at either side of the junction reaches an equilibrium value.

8. Reverse Bias
Applied
Field N
Net Field > Equilibrium P Si Si
Force P+ Si Si B- Si Si
Equilibrium
Field, ~0.75V + -
In reverse bias voltage, the fields are oriented in the same direction, and the magnitude of the electric field in the space charge region increases above the equilibrium value. This holds back electrons, and no current flows. With forward bias, the net result is that the eclectic field at the junction is lower than the equilibrium value, and electrons diffuse.

9. Applied electric field from reverse-bias voltage polarity
Electric field (oriented from positive to negative)
Net electric force due to fields
N-type
P-type
Diffusion “force”
Electrons drawn from P-type material near junction
Electrons
Reverse-bias voltage increases net field at the junction, opposing diffusion. Charges are drawn away from the junction by the field.

10. Forward Bias
Applied
Field N
Net Field < Equilibrium P Si Si
Net diffusion P+ Si Si B- Si Si
Equilibrium
Field - +
In reverse bias voltage, the fields are oriented in the same direction, and the magnitude of the electric field in the space charge region increases above the equilibrium value. This holds back electrons, and no current flows. With forward bias, the net result is that the eclectic field at the junction is lower than the equilibrium value, and electrons diffuse.

11. Applied electric field from forward-bias voltage polarity
Electric field (oriented from positive to negative)
Net electric force due to fields
N-type
P-type
Diffusion “force” exceeds the net electric force, electrons diffuse across junction, where electron concentration is above equilibrium
Electrons diffuse from junction into the bulk
Electrons
Forward-biased voltage reduces net field at the junction, reducing the width of the depletion region. The force of diffusion dominates.

12. Device schematic
Transfer function I
Two-state:
Non-linear response
at “on” threshold
Vin
Input: voltage
V forward bias
V reverse bias
Output: current
Current
No current
Voltage input
Sensitive:
Only 0.7V “on” threshold
I = Is*[ e^(Vin/Vt) -1 ]
Is = Reverse bias current, 5*10^(-14)A
Vt = Thermal voltage = 0.026
The output is highly responsive to increases in the forward-biased input, above the “cut in” voltage threshold ~0.7V. At zero or reverse bias input voltage, the output current is the very small saturation value, around 5*10^(-14)A. Because the forward bias voltage is included in the exponent, the exponential terms increases with respect to the input voltage. At the cut in voltage ~0.7V, the input voltage driven exponential term begins to dominate the very small reverse bias saturation current term, resulting in the exponential behavior. Physically, this means that diffusion is unrestricted once the forward bias voltage establishes a field that fully counters the built-in equilibrium field.

13. V=IR
I=Vs/R
Energy balance:
I= -(1/R)*Vd+Vs/R R
Diode Vd
Voltage
Source
Vd = Vs Vd
KVL (energy balance):
Vs=IR+Vd
I= -(1/R)*Vd+Vs/R
The diode current and voltage are given by the intersection between the circuit load line and the diode performance curve. This intersection, or Q-point, gives the DC voltage and current for the forward-biased diode in the circuit: this intersection identifies the feasible diode operating conditions that also satisfy the energy balance of the circuit.

14. Polarity forward to reverse bias
Voltage drop
0.7V
Switch voltage
Polarity forward to reverse bias 5V I
4.3V,
I=4.3/R R R
0V,
I=0 + -
Supply : 5V
Time + -
Zero output
A diode circuit can make a switch. If the output is defined as the voltage drop across the resistance, then output is zero in the reverse bias condition (red). This is because the diode is reverse biased in this condition, resulting in no circuit current and no resistor voltage drop.

15. Supply : 5V R I
Vi (1)
V (o)
Vi (2)
A diode circuit can make an AND gate. If both inputs are high (5V), then there is no potential difference across either diode (no voltage difference between input and supply). Neither diode is forward biased, no current flows. Since there is no current in the circuit, there is no voltage drop across the resistance and output voltage is 5V (high). If either input is low, then the diode is forward biased and the voltage drop across the diode will be ~0.7V. Thus, ~4.3V will drop across the resistance, resulting in a 0.7V (low) output.

16. Vi (1)
V (o)
Vi (2) R I
A diode circuit can make an OR gate. If either input is high (5V), the corresponding diode is forward biased with a 0.7V drop. The output voltage is 4.3V, and the resistance determines the output current (4.3V/R).

17. N P
Net diffusion P+ Si
Equilibrium
Field
Input Voltage - +
Resistance, R
10V
Supply Voltage, V
Voltage
Out = Always High
A diode in the circuit will not switch the output because the applied field from the supply voltage will over-ride the applied field from the input voltage. The diode is forward biased with respect the supply voltage , so current will always flow through the circuit.

18. Current through first diode controlled by input voltage
Second diode is reverse biased with respect to supply, so no current
Input Voltage - +
Supply Voltage, V - +
Resistance, R
10V
Voltage
Out = Always Low
Adding a second diode establishes a three-terminal device in which the voltage across the first diode is set by the input and unaffected by the supply. The second diode is reverse biased with respect to the supply voltage, which shields the first diode. However, no current will flow through the outer circuit because the second diode is reverse biased.

19. N P N
An NPN junction produces the effect of two opposing diodes in the circuit.

20. Forward Bias
Reverse Bias
Applied field
from input voltage
Applied field from supply voltage N P N
Net diffusion
Force -
Equilibrium
Field
Equilibrium
Field - +
Supply Voltage, V
Base current
Input Voltage - +
Resistance, R
10V
Voltage
Out = Always Low
No current flows in the outer circuit because all electrons entering the P-type region exit through P-region lead, and because no current can flow across the reverse biased PN junction Electrons diffuse from N to P-type material, recombine with holes, and exit through the conductor in the P-type material.

21. N P N
Make base thin
Net diffusion
Force
The electron concentration across the P region varies from high at the forward biased junction through which electrons are passing to low at the reverse biased junction, at which no current is flowing. Transistors are designed to allow diffusion of electrons across this concentration gradient, from the emitter across the base and into the collector, thus completing the outer circuit. This is done by making the base thin. The thin base leads to a sharper concentration gradient, and reduces the likelihood of recombination.

22. N
Emitter P
Base N
Collector
Net diffusion
Force -
Equilibrium
Field - +
Supply Voltage, V Ib
Input Voltage - +
Resistance, R
10V
Voltage
Out = Contollable
In a bi-polar junction NPN transistor, electrons are injected into and diffuse through the base to the collector, completing the outer circuit. Many of the electrons will not recombine with holes in the base (P-region) for two reasons. First, the emitter is heavily doped and the base is lightly doped. Second, the P-region is thin. Because there is a concentration gradient across the region, electrons diffuse towards the reverse-biased base-collector junction. The electrons will be captured by the strong electric field at this junction, and will flow into the collector.

23. Device
Transfer Function
Energy balance:
Ib= -(1/Rb)*Vbe+Vbase/Rb
Intersection gives Ib and Vbe Rb B + E
Load line
I (base)
Vbe
Vbase - Ib
The transistor can be de-coupled into two parts, first one being the base-emitter, which functions like a diode. Diode performance can be determined (along with the base current) by the load line intersection with the diode performance curve.

24. Energy balance:
Vsource = Vt +IcRc C Rc
Vout Rb B Vt + E +
I (base)
Vbe
Vbase - -
Vsource Ib
The current exiting the collector, Ic, is determined by the voltage across base-emitter junction (the input) only. This is because electron injection to the base from the emitter is the limiting factor on the current through the circuit, and base voltage control the degree of electron injection. As a result, Ic is independent of the reverse-bias voltage polarity across the BC junction: if, for example, Rc decreases, the voltage drop across the transistor, Vt, will increase (energy balance). The voltage drop across the BE junction is fixed at 0.7V, for a forward biased diode. Thus, the reverse bias voltage across the BC junction will have to increase. However, this has no affect on the output current. The collector current is related to the base current by a factor B, or gain, which is between 50 – 200 for transistors. The above design is a common emitter (emitter is the common connection) bi-polar junction transistor.

25. Vt must be > Vbe, else there is no reverse bias
Output
Out (Ic)
Various inputs (Vbe) C B On E
In (Vbe) Vt Vt
Off
Vt must be > Vbe, else there is no reverse bias
across the base-collector
Energy Balance:
–(1/Rc)Vt+Vsource/Rc=Ic
Two elements are necessary for the shared emitter, bi-polar junction NPN transistor. First, there must be reverse bias voltage polarity across the base-emitter junction to capture diffusing electrons in the base. This is represented on the x-axis. The output is weakly dependent upon the degree of reverse bias, reflected by weak slope of each line. But the output will drop off rapidly as if the voltage across the transistor is too low to maintain reverse bias across BE junction. Second, there must be forward-biased input voltage. As the input voltage increases (represented by each curve), the output current increases. The lower line represents zero input voltage, and output is zero, as expected.