Recap in last lecture EE2301: Block B Unit 2

p-n Junction The pn junction forms the basis of the semiconductor diode Within the depletion region, no free carriers exist since the holes and electrons at the interface between the p-type and n-type recombine. EE2301: Block C Unit 1

Diode Symbol and Operation Forward-biased Current (Large) Reverse-biased Current (~Zero) iD + - - + Forward Biased: Diode conducts Reverse Biased: Little or no current EE2301: Block C Unit 1

Ideal diode model Circuit assuming that the ideal diode does not conduct Circuit containing ideal diode Circuit assuming that the ideal diode conducts EE2301: Block C Unit 1

Rectification: from AC to DC One common application of diodes is rectification. In rectification, an AC sinusoidal source is converted to a unidirectional output which is further filtered and regulated to give a steady DC output. Supply is AC DC required EE2301: Block C Unit 1

Rectifier with regulator diagram Filter Regulator Bi-directional input Steady DC output Unidirectional output We will look at two types of rectifiers and apply the large signal models in our analysis: 1) Half wave rectifier 2) Full wave rectifier EE2301: Block C Unit 1

Starting from this lecture EE2301: Block B Unit 2

The Transistor A transistor is a 3-terminal semiconductor device (cf Diode is a 2-terminal device) Performs 2 main functions fundamental to electronic circuits: Amplification – magnifying a signal Switching – controlling a large current or voltage across 2 terminals (on/off) 2 major families of transistors: Field Effect Transistors (FETs) – Unit 2 Bipolar Junction Transistors (BJTs) – Unit 3 Unit 8

Transistors as Switches (1) Controlling I or V Controlling I or V I or V applied at 3rd terminal switches the device on/off Unit 8

Transistors as Amplifiers (1) vin vout Controlling I or V Gain is determined by I or V applied at 3rd terminal Unit 8

Unit 8

Block C - Unit 2 Outline MOSFET operation Construction of the MOSFET Basic working principle of the MOSFET Operating modes of the MOSFET MOSFET amplifiers (In this course, the focus will be on their use in amplifier circuits) Biasing the MOSFET Small signal equivalent circuit EE2301: Block C Unit 2

Transistors as Amplifiers Unit 3 BJT Unit 2 FET Whereas a diode is a 2 terminal device, transistors in contrast have 3 terminals. In a diode, the current is controlled by the voltage across the diode. For transistors, the current through the device is controlled by the voltage across the device as well as the voltage or current applied to a 3rd port. EE2301: Block C Unit 2

Classification of FETs FET: Field Effect Transistor MOSFET: Metal Oxide Semiconductor FET JFET: Junction FET EE2301: Block C Unit 2

Enhancement mode MOS n+ n+ p Gate D Source Drain Metal Oxide For NMOS, the source and drain are n-type, while the bulk substrate is p-type. Gate D Source Drain Metal Oxide n+ n+ p Semiconductor G S Bulk (substrate) The above figure shows an n-channel enhancement mode MOS (also known as NMOS) with its respective symbol. It contains three terminals: Source (S), Gate (G) and Drain (D). When in operation, the current flows between the source and drain. The carriers move in the direction from the source to the drain. This current is controlled by the voltage applied to the gate. In the symbol, the arrow points towards the gate for n-channel MOS. EE2301: Block C Unit 2

Enhancement NMOS Operation G D S D - VGD + ID n+ n+ p ID: Channel current G +VGS- Bulk (substrate) S Amount of current in the channel is controlled by the voltage applied at the terminals (VGS and VGD) Device is “switched on” (formation of a conducting channel) by the voltage on the Gate EE2301: Block C Unit 2

Gate Current G D S D Gate IG=0 Oxide n+ n+ p Substrate G IG=0 Bulk (substrate) S Gate current sees a capacitor formed by the gate, oxide (which is insulating), and substrate (conducting) EE2301: Block C Unit 2

NMOS is normally off n+ p Bulk (substrate) G S D p n+ Reverse-biased S VDD VDD +_ D When VGS = 0V, ID = 0A When no voltage is applied to the gate, the enhancement mode NMOS will not conduct between S and D (even for a voltage applied across S and D). This is because, as we can see from the figure on the right: Drain-Bulk n+p junction is strongly reverse-biased Source-Bulk n+p junction is also reverse-biased No undisrupted conductive path between source and drain: no current EE2301: Block C Unit 2

Conducting channel in NMOS iD Bulk (substrate) G S D p n+ +- VDD VGG D VDD +_ G VGG +_ S A key point on MOS devices is that for the device to conduct (or turn on), there must be a conducting channel between the source and drain. For NMOS, this channel must be n-type. For this n-channel to form, VGS must be larger than VT, which is known as the threshold voltage. The value of VT depends on the design and properties of the device. For conduction to occur, VGS > VT EE2301: Block C Unit 2

Formation of channel in NMOS VGG (+ve) Initially when VGS < VT: The majority carriers in p-type silicon are holes (positively charged) Holes are repelled away from the surface at the silicon-oxide interface The interface becomes depleted of majority carriers No conduction occurs between S and D + + + + + + + + + + + + _ _ _ _ _ _ _ _ _ _ _ _ _ _ p-type substrate When VGS > VT: The minority carriers in p-type silicon are electrons (negatively charged) Electrons are attracted towards the surface at the silicon-oxide interface A narrow conducting n-type channel near this surface forms Electrons from the source can now flow towards the drain resulting in a a drain current ID Increasing VGS (while keeping VDS fixed) increases the concentration of carriers, thereby increasing conduction EE2301: Block C Unit 2

p-Channel MOS (PMOS) p The PMOS operates on the same principle as the NMOS only that the charge carrier types are reversed. As a consequence of this swap: It is p-type conducting channel that is formed in an n-type bulk It requires a negative threshold voltage in order to form the channel EE2301: Block C Unit 2

Enhancement vs Depletion mode Enhancement mode Only when VGS is greater than VT will a conducting channel form at the surface of the bulk as free charge carriers are created, otherwise device remains off. As such, we say the device is normally off. Depletion mode In contrast, depletion mode MOS devices are designed with a built-in conducting channel. Hence even with no gate voltage, the MOS will still conduct between source and drain. The device is turned off by applying a gate voltage so as to deplete this channel. In this course we will focus on the enhancement mode MOS, which we shall see has three regions of operation unlike the diode which has only 2 (either on or off). EE2301: Block C Unit 2

Different Operation Regions of a MOSFET Device Con’t in next lecture EE2301: Block B Unit 2

Cutoff mode ID Cutoff Region VGS and VGD both less than VT The first mode we shall consider is when the device is off. This is known as Cutoff. As previously pointed out, this occurs when the threshold voltage has not been reached. D S G Cutoff Region VGS < VT , VGD < VT As a result of this, ID = 0A ID Cutoff Region VGS and VGD both less than VT Channel is off at both source and drain No conduction and no current flow between S & D EE2301: Block C Unit 2

Triode mode ID D Triode/Ohmic Region VGS > VT, VGD > VT As mentioned previously, when the gate voltage exceeds the threshold the MOS can conduct and therefore turns on. When it turns on, it can operate in one of two possible regions. One of these 2 modes is known as the triode or ohmic region. D S G ID Triode/Ohmic Region VGS > VT, VGD > VT The drain current is given by: ID = K [2(VGS – VT)VDS – VDS2] K is known as the conductance parameter, and is defined as: W: channel width L: channel length µ: mobility of carriers Cox: Oxide capacitance Triode/Ohmic Region Both VGS and VGD are greater than VT Channel is on at source and also on at drain Current is dependent of both VDS and VGS EE2301: Block C Unit 2

MOS in Triode mode n+ p G S D VGS > VT VGD > VT Bulk (substrate) G S D p n+ Near both the drain and source: Voltage difference > VT Conducting channel is formed on both sides In fact everywhere under the gate oxide is larger than VT so there is a continuous channel between the drain and source. VGS > VT VGD > VT +_ VDD Increasing VGS thus increases the number of carriers, resulting in a larger drain current Increase VDS increases the voltage drop across the channel, resulting in a larger drain current ID is dependent therefore on both VDS and VGS (Refer to equation on previous slide) EE2301: Block C Unit 2

Saturation mode - VGD + ID +VGS- D Saturation Region While the MOS is still on, if we continue to increase VDS without increasing VGS, there will come a point when the voltage near the drain becomes less than the threshold (VGD < VT). This is known as the saturation region. D S G - VGD + Saturation Region VGS > VT, VGD < VT The drain current is given by: iD = K (VGS – VT)2 ID Saturation Region VGS > VT and VGD < VT Channel is on at source but off at drain Current is nearly independent of VDS, depending only on VGS One might expect the MOS to operate like in cutoff under these bias conditions but it does not. +VGS- EE2301: Block C Unit 2

NMOS in Saturation n+ p Bulk (substrate) G S D VDD VGG +- VDD VGG The channel in saturation mode is typically represented as tapered or wedged (as shown in the figure). Large currents still conduct via the channel due to the large voltage drop across drain and source, but changes little with increasing VDS. Pinch-off (going from triode to saturation) The channel near the drain begins for fall below the threshold condition when VGD = VT In terms of the drain-source voltage, this corresponds to: VDS(pinch-off) = VGS - VT EE2301: Block C Unit 2

Summary of operating modes Cut-Off: VGS < VT, VGD <VT Linear/Triode/Ohmic: VGS > VT, VGD> VT Saturation: VGS > VT, VGD < VT For n-channel enhancement MOS EE2301: Block C Unit 2

Operating Regions BREAKDOWN SATURATION CUTOFF When VGS - VT = VDS ID = K VDS2 TRIODE/OHMIC SATURATION CUTOFF EE2301: Block C Unit 2

Operating state example Determine the operating state of the MOSFET shown in the circuit for the given values of VDD and VGG if the ammeter and voltmeter shown read the following values: VGG = 1V; VDD = 10V; ID = 0mA; RD = 100Ω VGG = 4V; VDD = 10V; ID = 72mA; RD = 100Ω VGG = 6V; VDD = 10V; ID = 270mA; RD = 26Ω For the MOSFET in the circuit, VT = 2V; K = 18mA/V2 a. Drain current is zero, which implies that the MOS is in the cutoff state. We should also check that VGS and VGD are both less than VT. VGS = 1V VGD = - 9V EE2301: Block C Unit 2

Operating state example b. VGS = 4V (hence larger than VT); we now still need to find VGD to decide whether the MOS is in the triode or saturation region. VDS = VDD – IDRD = 2.8V VGD = VGS – VDS = 1.2V (less than VT) MOS is in saturation c. VGS = 6V (hence larger than VT); we now still need to find VGD to decide whether the MOS is in the triode or saturation region. VDS = VDD – IDRD = 3V VGD = VGS – VDS = 3V (larger than VT) MOS is in the ohmic region EE2301: Block C Unit 2

MOSFET Q-point As mentioned at the start, in this course our focus is on the use of transistors for amplifiers. A key takeaway to remember is that when building an amplifier, we use the saturation region. Why this is so will become clear in the next few slides. For this to be realized, the MOS amplifier circuit has to be designed to ensure the MOS is working at the operating point (which corresponds to a fixed value of ID, VGS and VDS). This is called the Q-point. We say that the circuit needs to be properly biased to give us the desired Q-point. EE2301: Block C Unit 2

Determining Q-point (Graphical) The I-V characteristic of the MOS describes the full range of values that the MOS can take. However, the MOS works only at one value of ID, VDS and VGS when put in a circuit. It is the circuit that imposes this constraint on the MOS. Hence to find the Q-point we also need to consider the characteristics of the circuit as well: this is defined by the load line. Given: VGG = 2.4V; VDD = 10V; RD = 100Ω The Q-point is found at the intersection of the load line and the I-V curve. Load line: Apply KVL around the right side mesh, VDD = VDS + IDRD Since we want to draw this on the I-V curve, we should express this as ID against VDS: ID = VDD/RD – VDS/RD EE2301: Block C Unit 2

MOSFET amplifiers But how do we achieve the function of an amplifier from the concept of the Q-point? We can see this graphically. Initially, VGS = 2.4V, while ID = 52mA and VDS = 4.75V Now if VGS was dropped slightly by 0.2V, then VDS ≈ 6.4V (+1.65V) Now if VGS was instead increased from 2.4V by 0.2V, VDS 2.8V (-1.95V) VGG forms the input and VD is the output We see that for a peak to peak change in the input by 0.4V, we see a peak to peak change in the output by 3.6V  Output is amplified! EE2301: Block C Unit 2

Find Q-point by calculation We have seen how to determine the Q-point of the MOS graphically. This example shows how we can do the same but now using calculation. Given: VGG = 2.4V; VDD = 10V; RD = 100Ω; VT = 1.4V; K = 52mA/V2 Since the source is connected to ground, it becomes obvious that VGS = 2.4V Now we still need to find VDS and ID (2 equations). To do so we can apply KVL around the right-hand mesh. To get another equation, we use the formula to find ID when MOS is in saturation (this is a lot simpler than that for triode state). Applying KVL to right-hand mesh: VDD = VDS + IDRD 10 = VDS + 0.1*ID Next, we use the drain current formula for saturation mode: ID = K(VGS – VT)2 ID = 52*(2.4 – 1.4)2 = 52mA From this, we can then find VDS VDS = 4.8V Is it saturation? Check: VGD = -2.4V (less than VT) YES!! EE2301: Block C Unit 2

Recap in last lecture EE2301: Block B Unit 2

Summary of operating modes Cut-Off: VGS < VT, VGD <VT Linear/Triode/Ohmic: VGS > VT, VGD> VT Saturation: VGS > VT, VGD < VT For n-channel enhancement MOS EE2301: Block C Unit 2

Operating Regions CUTOFF BREAKDOWN TRIODE/OHMIC SATURATION CUTOFF EE2301: Block C Unit 2

Biasing aMOSFET (Q-point) A change in VGS is AMPLIFIED by 9 times in VDS!! VGS=2.2V, VDS=2.8V VGS=2.4V, VDS=4.75V Vpp =0.4V VGS=2.6V, VDS=6.4V Vpp = 3.6V EE2301: Block C Unit 2

Let’s con’t EE2301: Block B Unit 2

Self-biasing MOS amplifier The previous circuit requires a voltage source to bias the gate. One disadvantage is this takes up space. The self-biasing MOS amplifier takes care of this problem. Determine VGS, VDS, and ID for the following MOS amplifier, given that: R1 = R2 = 1MΩ; RD = 6kΩ; RS = 6kΩ; VDD = 10V; VT = 1V; K = 0.5mA/V2 Begin by first finding the gate voltage: Next we assume the MOS is in saturation and then check to see if this assumption (VGS>VT and VGD<VT) is correct (formula for saturation is a lot simpler than for triode state, thus easier to work with): ID = K(VGS – VT)2 ; where VGS = VG - IDRS ID = 0.5(5 – 6ID – 1)2 18ID2 -25ID + 8 = 0  ID = 0.89mA or 0.5mA EE2301: Block C Unit 2

Self-biasing MOS amplifier Only one of these solutions is valid for our case: thus we need to check each one in turn to see which of these makes physical sense. But what do we use ID to check for? A hint is that we still have not found the value of VDS up till this point. We have assumed that it is a value that corresponds to the saturation mode. We can substitute ID into the right-hand mesh to find VDS since we can see that: VDD = VDS + ID(RD + RS) First we try ID = 0.89mA: VDS = 10 – 0.89*12 = -0.68V (REJECTED: VDS should be positive) Now we try ID = 0.5mA: VDS = 10 – 0.5*12 = 4V (ACCEPTED: this makes sense) Finally, now that we know VDS, we can then use this information to verify if our starting assumption (that the MOS is in saturation) was true. To do this, we find VGD: VGD = VG – (VDS + IDRS) = 5 – (4 + 0.5*6) = -2V <VT ; VGS = VG-IDRS = 5 – (0.5*6) = 2V > VT This indicates the MOS is in saturation, thereby confirming our starting assumption. EE2301: Block C Unit 2

Example (Prob 11.9) The NMOS transistor shown in figure has VT=1.5V and K=0.4mA/V2. If VG is a pulse with 0 to 5V, find the voltage levels of the pulse signal at the drain output. Assume the NMOS is in saturation region when VG=5V Since VT = 1.5 V, with vG = 0 V, vGS < VT, the transistor is cut off. Therefore, vD = 5 V. When vG = 5 V, and assuming that the transistor is in the saturation region: iD = k (vGS - VT)2 = 0.4 (5 - 1.5)2 = 4.9 mA. Therefore, vD = 5 - 4.9´1 = 0.1 V. EE2301: Block C Unit 2

Small signal equivalent circuit The MOS is inherently a non-linear device (basic circuit theory is based on linear problems) Within a small range, we see from the I-V characteristic that we can approximate its behavior as linear This allows us to analyze a MOS amplifier using the I-V graph We can instead define the linear characteristics of the MOS for a given Q-point We can do so using a small signal equivalent circuit Now we can analyze the MOS using the linear circuit theory we have learnt so far EE2301: Block C Unit 2

Small signal equivalent circuit Linearizes the characteristics of a MOSFET (which is otherwise nonlinear) for a given operating point Only valid for a particular operating point Only valid for small varying signals rd id = gmvgs + vds/ro Mathematics will not covered in this course EE2301: Block C Unit 2

End of Block C Unit 2 EE2301: Block B Unit 2