Bipolar transistor Logic Circuits-Surendra ShresthaDepartment of Electronics and Computer EngineeringPulchowk Campus, Institute of Engineering, T.U.2.1Types of Devices2.2Propagation Delays2.3Power Dissipation2.4Fan-IN and Fan-OUT2.5Noise Margin

Department of Electronics and Computer EngineeringPulchowk Campus, Institute of Engineering, T.U.Chip: miniature circuits on the surface of a small piece of semiconductor material using advance photolithographic process

Integrated Circuit (IC):The finished network is so small that need a microscope to see the connection. Integral part of the chip components are transistor, diodes, resistors

Department of Electronics and Computer EngineeringPulchowk Campus, Institute of Engineering, T.U.Saturated family(IC are driven between cut-off and saturation states, e.g. inverter)Bipolar Logic Family:Uni-polar logic familyResistor Transistor Logic (RTL)Direct Coupled Transistor Logic (DCTL)Integrated Injection logic (I2L)Diode Transistor Logic (DTL)Transistor Transistor Logic (TTL)High Threshold Logic (HTL)Schottky TTLEmitter Coupled Logic (ECL)PMOS (p-channel) and NMOS (n-channel)CMOS (Complementary MOS)(transistors are operated between cut-off and active states)Non-saturated family

Department of Electronics and Computer EngineeringPulchowk Campus, Institute of Engineering, T.U.Bipolar Logic Family:Based on the number of transistors (BJT or MOSFET) Small Scale Integration (SSI)- 0 to 10 Medium Scale Integration (MSI)- 10 to 100 Large Scale Integration (LSI)- 100 to 1,000Very Large Scale Integration (VLSI)- 1,000 to 10,000 or thousands gatesSuper-Large Scale Integration (SLSI) - 10,000 and 100,000 transistorsUltra-Large Scale Integration (ULSI) - more than 1 million transistors transistors

Speed of operation (propagation delay, tpLH and tpHL) power dissipation (power consumption under static condition, 0,1; during the switching intervals or dynamic conditions)Current and voltage parameters (High level input and output voltages and low level input and output voltages; IiH, IoH, IiL, IoL)Noise immunity (measure of how much stray noise voltage the device can handle without giving any error at the output level)Fan-out (No. of gates that gate in HIGH output state can feed without voltage dropping by more than the allowable noise margin (NM)H)Department of Electronics and Computer EngineeringPulchowk Campus, Institute of Engineering, T.U.Characterization of Digital ICsThe various characteristics of digital ICs that can be used to compare their performance are:

Department of Electronics and Computer EngineeringPulchowk Campus, Institute of Engineering, T.U.Resistor Transistor Logic (RTL):Resistor-transistor logic gates use Transistors to combine multiple input signals, which also amplify and invert the resulting combined signal. Often an additional transistor is included to re-invert the output signal. This combination provides clean output signals and either inversion or non-inversion as needed.A simple N-input RTL NOR Gate

Department of Electronics and Computer EngineeringPulchowk Campus, Institute of Engineering, T.U.Resistor Transistor Logic (RTL) Advantages:- RTL gates are almost as simple as DL gates, and remain inexpensive.- Using low power supply for each gate.- RTL integrated circuits are sometimes used as inexpensive small-signal amplifiers, or as interface devices between linear and digital circuits.

Limitations:- RTL gates cannot switch at the high speeds used by today's computers- These are not designed for linear operation- low noise margin

Department of Electronics and Computer EngineeringPulchowk Campus, Institute of Engineering, T.U.Resistor Transistor Logic (RTL) Logical NOR Gate

VinAVinBVoLLHLHLHLLHHL

Department of Electronics and Computer EngineeringPulchowk Campus, Institute of Engineering, T.U.Resistor Transistor Logic (RTL) VinL = 0.2 (LOW input to both transistors)

Under this condition, both transistors will be cutoff, making the output high. The circuit diagram with circuit models for the cutoff transistors is shown in Figure. The output voltage then is 3.0 Volts. We will call this a no-load condition when there is no load connected to the output.VoH = 3.0 Volts (No-Load)RTL gate with both inputs low. The transistors are replaced by the cutoff model.VinLmax = 0.5 Volts.IinL = 0

Department of Electronics and Computer EngineeringPulchowk Campus, Institute of Engineering, T.U.Resistor Transistor Logic (RTL) INPUT HIGH

In this case we want the transistor to be saturated. Either transistor saturated will cause the output to go low, to 0.2 volts. This case is shown in Figure. We show the circuit with one of the transistors cutoff, although both saturated would produce the same result.We start by determining the minimum input current that will keep the transistor in saturation.RTL gate with one input highthe minimum input voltage that will be guaranteed to be recognized as a "high"

Department of Electronics and Computer EngineeringPulchowk Campus, Institute of Engineering, T.U.Resistor Transistor Logic (RTL) Any input voltage greater than or equal to 0.866 volts will cause the transistor to be saturated, and thus be recognized as a "high". We previously found VinLmax =0.5 volts. Any input voltage between 0.5 and 0.866 is an invalid logic level and the manufacturer assumes no responsibility if it provide an input voltage in that range. For the gate with no load, the relationshipsbetween input and output voltagesare presented graphically in Figure.Noise Margins (High, H and Low, L) NMH and NML.Graphical representation of inputvoltage ranges and output voltages for the unloaded RTL gate.

Department of Electronics and Computer EngineeringPulchowk Campus, Institute of Engineering, T.U.Resistor Transistor Logic (RTL) Terminal specifications for the RTL Gateload requires a minimum of 0.146 mA as input current,

Department of Electronics and Computer EngineeringPulchowk Campus, Institute of Engineering, T.U.Diode Transistor Logic (DTL):By letting diodes perform the logical AND or OR function and then amplifying the result with a transistor, we can avoid some of the limitations of RTL. DTL takes diode logic gates and adds a transistor to the output, in order to provide logic inversion and to restore the signal to full logic levels.

Department of Electronics and Computer EngineeringPulchowk Campus, Institute of Engineering, T.U.the diode AND function on the front end and the transistor NOT at the output end. The extra resistors and diodes areused to maintain appropriate currents, to maintain proper functioning, and to guaranteecertain noise margins.

Digital NAND circuitDiode Transistor Logic (DTL): Diode logic and truth tables

Department of Electronics and Computer EngineeringPulchowk Campus, Institute of Engineering, T.U.If all inputs are high, (+5v), no current will come out of the input diodes at the input and current will flow down through the first 5K resistor and through the diodes D1and D2 toward the base of the transistor. Some current will split off and go down through thelower 5K resistor to ground. However, most of the current will go into the base of the transistor causing it to saturate, pulling the output low, VO = 0.2 Volts. We will show this condition quantitatively shortly. If one or more of the inputs to the gate are held low (0.2 V), then the current down through the 5K resistor will go out the input diode, away from the transistor base. Underthis condition, the transistor will be cutoff and the output will be high with VO = 5 Volts.Diode Transistor Logic (DTL):

Department of Electronics and Computer EngineeringPulchowk Campus, Institute of Engineering, T.U.Diode Transistor Logic (DTL): DTL gate model with input lowFrom this circuit we can calculate all voltages and currents and prove (or disprove) our assumptions about the condition of each element.

The voltage at point P isVP = 0.2 + 0.70 = 0.90 Volts.

ANALYSIS WITH INPUT LOW

Quantitatively, we will start with one or more inputs held low, at 0.2 Volts. From the logic function of the NAND gate, we know that the output is supposed to be high. Therefore, the transistor must be cutoff. To begin with, we will assume the two diodes D1 and D2 in series will also be cutoff. All the current coming down through the 5K resistor must all go out through the input diode, causing it to be on. The circuit with the models indicated is shown in Figure.

Department of Electronics and Computer EngineeringPulchowk Campus, Institute of Engineering, T.U.Diode Transistor Logic (DTL): We need to show that this voltage is low enough that the two series diodes and the transistor will be cutoff. The argument is that if either diode carries current, then both must. Since 0.9 Volts is not enough across the pair to maintain conduction, then neitherconducts. Given that the diodes are off, then the voltage at the base of the transistor is zero, and is also cutoff. We can verify that the input diode is conducting by observing that current I1 isI1 = (5-0.9)/5K = 0.82 mA.

This current leaves through the input diode, hence it is on. The current entering the input terminal isIin = - I1 = -0.82 mAThe negative sign occurs because the input current is defined as going into the terminal.

We can now verify that the output voltage is 5 Volts because the transistor has been shown to be cutoff. Thus,VO = 5.0 Volts

Department of Electronics and Computer EngineeringPulchowk Campus, Institute of Engineering, T.U.Diode Transistor Logic (DTL): One other characteristic of the DTL gate that can be obtained at this point is the range of input voltages that will be recognized as a "low". From the logic function of the NAND gate, this can be translated into the question of how high the input voltage may rise and still keep the transistor cutoff.The transistor will remain cutoff as long as the voltage at the base does not rise above 0.5 volts. At this voltage, there will be current down through the lower 5K resistor to ground. This current must come from the +5 supply down through the upper 5K resistor and the diodes, D1 and D2. Hence the diodes must be conducting. This current will be 0.1 mA. The voltage at point P will beVP = 0.5 + 0.7 + 0.7 = 1.9 VoltsThe current I1 isI1 = (5-1.9)/5K = 0.62 mAThe current going out through the input diode will beIin = -(0.62 - 0.1) = -0.52 mA

Department of Electronics and Computer EngineeringPulchowk Campus, Institute of Engineering, T.U.Diode Transistor Logic (DTL): indicating that the input diode is still conducting. Figure next slide shows the resulting circuit with the circuit models included. The maximum voltage at the input that is guaranteed to be recognized as a low isVinLmax = 1.9 -0.7 =1.2 Volts.DTL circuit model with Vin = VinLmax

Department of Electronics and Computer EngineeringPulchowk Campus, Institute of Engineering, T.U.Diode Transistor Logic (DTL): ANALYSIS WITH ALL INPUTS HIGHWhen all inputs are high, all current down through the upper 5K resistor will go toward the base of the transistor, causing it to saturate. The series diodes will obviously be conducting, and we will show that the input diodes are cutoff. Figure shows the circuit with these models.DTL circuit model with inputs highThe voltage at point P isVP = 0.8 + 0.7 + 0.7 = 2.2 voltsThus, I1 isI1 = (5-2.2)/5K = 0.56 mA

Current going down through the 5K resistor will beI2 = 0.8/5K = 0.16 mAThe base current then isIB = 0.56 - 0.16 = 0.4 mA

Department of Electronics and Computer EngineeringPulchowk Campus, Institute of Engineering, T.U.Diode Transistor Logic (DTL): If the transistor is to be saturated, the maximum collector current isICmax = IB = 30 * 0.4 mA = 12.00 mA

at saturation, the current coming down through the 2.2K collector resistor isI3 = (5-0.2)/2.2K = 2.182 mAthis current is much less than the maximum saturation current and we see that with no load, the transistor will, indeed, be in saturation.

In fact, there is excess capacity in collector saturation current. This excess capacity can be used to sink external load current. This current is called Io or load current. The maximum load current this gate can sink isIoLmax = 12.00 mA - 2.182 mA = 9.818 mANote that this current is entering the terminal of the gate, hence, is positive.

Department of Electronics and Computer EngineeringPulchowk Campus, Institute of Engineering, T.U.Diode Transistor Logic (DTL): CALCULATION OF FANOUTIf several load gates are connected to the output terminal of the gate we are looking at, we need to look at the current output drive capability compared to the input current requirements of the load gates. Because the input current is zero when high, an infinite number of load gates can be driven when high. However, the DTL gate requires current when the input is low. This situation is shown in Figure. IinLmax = - 0.82 mA. Negative meaning that it is coming out of the input terminal. The output of a gate can sink 9.818 mA when it is low. DTL driver gate with N identical DTL load gatesComparison of Vout and Vin voltages for DTL gate.

Department of Electronics and Computer EngineeringPulchowk Campus, Institute of Engineering, T.U.Diode Transistor Logic (DTL): Terminal Specification for the DTL GATE

VinLmax = 1.2 V VoL = 0.2 V

VinHmin = 1.6 V VoH = 5 (at IoH =0) = 1.6 (at IoH = max)

IinL = - 0.82 mA IoLmax = 9.818 mA

IinH = 0 IoHmax = -1.545 mA (at VoH=VinHmin)

Fanout = 11

Department of Electronics and Computer EngineeringPulchowk Campus, Institute of Engineering, T.U.Other logical functions:DTL AND-OR-INVERT gateDTL OR gateDTL AND gateDTL NOR gate

Department of Electronics and Computer EngineeringPulchowk Campus, Institute of Engineering, T.U.930 Series DTL (ca 1964 A.D.):One of the series diodes is replaced by Q1, providing more base drive for Q2 and improving the fan-out (Nmax= 45)tPLH >> tPHLtPLH = tS+tr /2

Department of Electronics and Computer EngineeringPulchowk Campus, Institute of Engineering, T.U.Transistor Transistor Logic (TTL): The DTL input uses a number of diodes which take up considerable chip area.

In TTL, a single multi-emitter BJT replaces the input diodes, resulting in a more area-efficient design.

- DTL was removed by faster TTL gatesby 1974.Why TTL?

Department of Electronics and Computer EngineeringPulchowk Campus, Institute of Engineering, T.U.Transistor Transistor Logic (TTL): - ALL INPUTS HIGH. Q1 is reverse active. Q3 is saturated. VOL = VCES

- ANY INPUT LOW. Q1 is saturated. Q3 is cut off. VOH = VCCBasic TTL NAND gateThe diode D2 from DTL circuit can be replaced by a transistor whose collector is pulled up to the power supply; transistor Q2 in Figure. The p-n junction of D2 is replaced by the BE junction of Q2 and with the current gain of the transistor, the current going into the base of Q3 is greatly increased, increasing the fanout.Figure 1Figure 2

Department of Electronics and Computer EngineeringPulchowk Campus, Institute of Engineering, T.U.Transistor Transistor Logic (TTL): The input diodes and D1 are replaced by the multi-emitter NPN transistor, Q1, and represented by the drawing in right Figure. Later on, we will make additional modifications to this circuit to improve its performance further.

The analysis of this circuit follows very much the same path as the analysis of the DTL gate. For the most part, we will consider the input transistor, Q1, to act just like two diodes. The transistor Q2, however, will operate in all three regions. The treatment of the output voltages and currents will be treated the same as the DTL gate and Q3 will either be cutoff or saturated, corresponding to an output high and an output low, respectively.

Department of Electronics and Computer EngineeringPulchowk Campus, Institute of Engineering, T.U.ANALYSIS WITH ONE OR MORE INPUTS LOWTTL circuit model with one input lowFigure 3Figure 4TTL circuit model to determine VinLmax.

Department of Electronics and Computer EngineeringPulchowk Campus, Institute of Engineering, T.U.ANALYSIS WITH ONE OR MORE INPUTS LOW With an input low, Q3 should be cutoff. We will assume Q2 is cutoff and then check our assumption. If Q2 is cutoff, then there can be no current coming out of the collector of Q1, hence its base-collector junction can be modeled as an open circuit. The base-emitter junction of Q1 will be conducting. The circuit with these models substituted for the transistors is shown in Figure 3. Note the similarity to the DTL circuit under the same conditions. The two unused inputs are assumed to be high, and are thus, modeled as open. From this case, we can see that VoH = 5 volts with no load, and

IinL = -I1 = -(5-0.9)/4K = -1.025 mA

Department of Electronics and Computer EngineeringPulchowk Campus, Institute of Engineering, T.U.ANALYSIS WITH ONE OR MORE INPUTS LOW We turn now to finding VInLmax. We will use the criterion that Vin will be considered as a low as long as Q3 is kept cutoff. If the base voltage for Q3 can be raised to 0.5 Volts without turning it on, then there will be 0.5 mA current in the 1K resistor. This current can only come from Q2, which means it must be conducting. Even assuming all this 0.5 mA comes through the collector of Q2, the voltage drop across the 1.4 K resistor will be 0.7 Volts, not enough to cause the transistor to saturate. Thus, the active model for Q2 is appropriate as shown in Figure 4.If we assume that =30, the base current in Q2 is

Department of Electronics and Computer EngineeringPulchowk Campus, Institute of Engineering, T.U.Because this current is coming out of the collector of Q1, the base- collector junction of Q1 is on, and is modeled as a diode in Figure 4.The voltage at B1, the base of Q1, isVB1 = 0.5 + 0.7 + 0.7 = 1.9 Volts

The current coming down through the 4 K resistor, I1, isI1= (5.0-1.9) / 4K = 0.775mA

This is considerably more than is going into the base of Q2, therefore, the input BE junction of Q1 will also still be conducting. The maximum voltage at the input isVinLmax = 1.9 - 0.7 = 1.2 VoltsANALYSIS WITH ONE OR MORE INPUTS LOW

Department of Electronics and Computer EngineeringPulchowk Campus, Institute of Engineering, T.U.The circuit model for the TTL gate with all inputs high is shown in Figure 5. Both Q2 and Q3 are modeled as saturated, an assumption that must be verified. With the inputs high, Q1 is modeled as two diodes with the B-E diodes cutoff, and B-C diode conducting.The voltage at the base of Q1 isVB1 = 0.8 + 0.8 + 0.7 = 2.3 Volts.CALCULATIONS WITH INPUT HIGHFigure 5TTL gate circuit model with all inputs highThe current down through the 4 K resistor, I1 is

I1 = (5.0-2.3) / 4K = 0.675mA

Department of Electronics and Computer EngineeringPulchowk Campus, Institute of Engineering, T.U.All this current goes into the base of Q2.IB2 = 0.675 mA

If Q2 is saturated, voltage at its collector terminal isVC2 = 0.8 + 0.2 = 1.0 VoltsAnd the collector current isIC2 = I2 = (5.0 1.0) / 1.4K = 2.857mA

Clearly, if = 30, B2 > IC2 , and, therefore, Q2 is saturated.

The current coming out of the emitter of Q2 is the sum of the base and collector currents. Part of this current will go down through the 1 K resistor to ground and the rest will enter the base of Q3.IB3 = IB2 + IC2 - I3 = 0.675 + 2.857 - 0.8 = 2.732 mACALCULATIONS WITH INPUT HIGH

Department of Electronics and Computer EngineeringPulchowk Campus, Institute of Engineering, T.U.The maximum collector current that Q3 can carry and still be in saturation is IB3 = 81.96 mA, assuming =30. The maximum current the gate can sink when the output is lowIoLmax = ICsatmax - I4 = 81.96 - 1.2 = 80.76 mA

Now let's turn our attention back to the input and determine VinHmin and IinH . We will define the input voltage to be high as long as no current goes out the input terminal.

Thus, all we have to do is keep the input voltage high enough so that the B-E p-n junction of Q1 does not turn on. Thus,VinHmin = 2.3 - 0.6 = 1.7 VoltsCALCULATIONS WITH INPUT HIGH

Department of Electronics and Computer EngineeringPulchowk Campus, Institute of Engineering, T.U.As the designers of TTL gates became more sophisticated, they developed modifications which would provide special characteristics. The original series of TTL was designated as 74XX, where the XX is replaced by logic function ( 00 is a quadruple 2-input NAND, 04 is a hex inverter, etc.) The 74LXX series is a low power family. 74HXX is a high speed family. 74SXX is a family based on Schottky diodes and transistors. 74LSXX is a family of low power Schottky. A 54xXX is also provided as a companion family to the 74xXX families. The 54... families are identical to the 74... families, except for operating temperature range and tolerance on power supply voltage.TTL FAMILIES

Emitter Coupled Logic (ECL)

ECL gates use differential amplifier configurations at the input stage. A bias configuration supplies a constant voltage at the midrange of the low and high logic levels to the differential amplifier, so that the appropriate logical function of the input voltages will control the amplifier and the base of the output transistor. The propagation time for this arrangement can be less than a nanosecond, making it for many years the fastest logic family.

ECL family include the fact that the large current requirement is approximately constant, and does not depend significantly on the state of the circuit. This means that ECL circuits generate relatively little power noise, unlike many other logic types which typically draw far more current when switching than quiescent, for which power noise can become problematic. In an ALU - where a lot of switching occurs - ECL can draw lower mean current than CMOS. Emitter Coupled Logic (ECL)

The ECL circuits available on the open market usually operated with negative power supplies (-5.2 volts), and logic levels incompatible with other families (required additional interface circuits). The fact that the high and low logic levels are relatively close meant that ECL suffers from small noise margins, which can be troublesome.

The drawbacks associated with ECL have meant that it has been used mainly when high performance is a vital requirement.

Other families (particularly advanced CMOS variants) have replaced ECL in many applications, even mainframe computers.

Uses:- Older high-end mainframe computers, such as the Enterprise System/9000 members of IBM's ESA/390 computer family, used ECL; current IBM mainframes use CMOS.

The equivalent of emitter-coupled logic made out of FETs is called source-coupled FET logic (SCFL).Emitter Coupled Logic (ECL)

Integrated Injection Logic (IIL, I2L, or I2L) is a class of digital circuits built with multiple collector bipolar junction transistors (BJT). When introduced it had speed comparable to TTL yet was almost as low power as CMOS, making it ideal for use in VLSI (and larger) integrated circuits.

The logic voltage levels are very close (High: 0.7V, Low: 0.2V), I2L has high noise immunity because it operates by current instead of voltage.

I2L, sometimes also called Merged Transistor Logic (MTL), is relatively new in concept as well as realization. This is quite attractive for LSI/VLSI realization.Integrated Injection Logic (I2L):

OperationThe heart of an I2L circuit is the common emitter open collector inverter. An inverter consists of an NPN transistor with the emitter connected to ground and the base biased with a forward current. The input is supplied to the base as either a current sink (low logic level) or as a high-z floating condition (high logic level). The output of an inverter is at the collector. Likewise, it is either a current sink (low logic level) or a high-z floating condition (high logic level).To understand how the inverter operates, it is necessary to understand the current flow. If the bias current is shunted to ground (low logic level), the transistor turns off and the collector floats (high logic level). If the bias current is not shunted to ground because the input is high-z (high logic level), the bias current flows through the transistor to the emitter, switching on the transistor, and allowing the collector to sink current (low logic level). Because the output of the inverter can sink current but cannot source current, it is safe to connect the outputs of multiple inverters together to form a wired AND gate. When the outputs of two inverters are wired together, the result is a two-input NOR gate because the configuration (NOT A) AND (NOT B) is equivalent to NOT (A OR B). This logical relationship is known as De Morgan's Theorem.Integrated Injection Logic (I2L)

Integrated Injection Logic (I2L)

Important features are:

The circuit uses BJTs, both pnp and npn. Therefore operating speed is quite high.The circuit is similar to RTL, except that there are no base resistors used. This minimizes the circuit area and simplifies the circuit lay-out.All the gate transistors are operated in the inverted mode. Not only does this facilitate the use of low voltage supply for the circuit.The circuit requires no isolation unlike the previous digital ICs. This facilitates higher packing density and also reduces the fabrication cost, as fewer masks (e.g. only four) are required for the complete process.Parasitics in the circuit are greatly reduced. This also improves the operating speed of the IC.Integrated Injection Logic (I2L)