Chapter 7 Transistor Logic TTL 74 xx and

Chapter 7 Transistor Logic – TTL (74 xx and 54 xx series chips) Digital Electronics ELE 450 Dr. Idrees Al-kofahi 1

Transistor Logic • DTL was able to improve fan-out compared to RTL. However, that was done at the expense of: – Transient response. – Chip area. • A solution for the problem was proposed in the form of a new logic family which utilizes only transistors and resistors. – BJTs are smaller than diodes. Digital Electronics ELE 450 Dr. Idrees Al-kofahi 2

Basic TTL Inverter VCC RB Vin Digital Electronics ELE 450 RC Vout QI Dr. Idrees Al-kofahi Qo 3

TTL vs. DTL • If we compare the basic DTL and TTL gates, we find that the input and level-shifting diodes of DTL can be combined into the input BJT of TTL. V CC VCC RC RB RB Vout Vin Qo DI DL V in RC V out QI Qo – The advantage is that the BJT requires less silicon area and the propagation delay is improved by a factor of 10. Digital Electronics ELE 450 Dr. Idrees Al-kofahi 4

Calculating the VTC • VOH: – For Vin very low, the base-emitter junction of QI will be forward biased. – The base-collector junction will also be forward biased. – Therefore, QI will be saturated. – The base-emitter voltage of QO is: VBE, O = VCE, I(Sat) + Vin – Therefore, QO will be cut-off – Therefore, Vout = VOH = VCC Digital Electronics ELE 450 Dr. Idrees Al-kofahi 5

Calculating the VTC (Contd. ) • VIL: – As Vin is increased, VB of QO will also increase. Eventually, QO will turn on. – This happens when: Vin = VIL = VBE, O (FA) – VCE, I (Sat) • VOL: – As Vin is increased even more, QO comes closer to Saturation and eventually saturates. – At that point: Vout = VOL = VCE, O (Sat) Digital Electronics ELE 450 Dr. Idrees Al-kofahi 6

Calculating the VTC (Contd. ) • VIH: – The point where QO is just saturating: Vin = VIH = VBE, O (Sat) – VCE, I (Sat) V out V OH = V CC V OL = V CE (Sat) VIL = VBE, O (FA) – VCE, I (Sat) VIn VIH = VBE, O (Sat) – VCE, I (Sat) Digital Electronics ELE 450 Dr. Idrees Al-kofahi 7

What about the currents? • If we look at the currents in the circuit, we find that QI and QO cannot both remain saturated at the same time. • If QI is saturated, a positive IC, I must flow into the collector of QI. • IF QO is saturated, a positive IB, O must flow into the base of QO. • Impossible!!!! • If we look at the voltages, we find that right after QO saturates, the base-emitter junction of QI will become reverse biased while the base-collector junction is still forward biased. – Therefore, QI will turn into Reverse Active mode. • Under reverse active mode, IC, I flows out of the collector of QI. • This current will flow into the base of QO maintaining it in saturation. Digital Electronics ELE 450 Dr. Idrees Al-kofahi 8

The Currents • If both QI and QO are saturated, then both IC, I and IB, O need to be positive. VCC Vout Vin > VIH IB, O QI Sat • RC RB QO Sat IC, I Impossible!! Digital Electronics ELE 450 Dr. Idrees Al-kofahi 9

The Voltages VIH = VBE, O (Sat) – VCE, I (Sat) = ~0. 6 V V CC R R B C Vout VBE, I < ~ 0. 8 Vin > VIH > ~0. 6 V QI cannot remain in Saturation VBC, I (FB) ~ 0. 6 QI QO Sat VBE, O (Sat) ~ 0. 8 VB, I = ~1. 4 V Digital Electronics ELE 450 Dr. Idrees Al-kofahi 10

Q Q O I R. Sat A. Q Q O I S Sa at t Q O Q I F. Sa A. t VOH = VCC Q O Vout Q I C u Sa tof t f The VTC VOL = VCE, O (Sat) VIn VIL = VBE, O (FA) – VCE, I (Sat) VIH = VBE, O (Sat) – VCE, I (Sat) Digital Electronics ELE 450 Dr. Idrees Al-kofahi 11

Do we need a pull-down resistor? • For DTL, a pull-down resistor was added to quickly discharge the charge built into the base of the saturated QO when it is switching to cut-off. – When QO is saturated, there is about 0. 8 V worth of charge built up in the base. – When QO turns off, this charge will VCC flow primarily through RD down to ground. RC RB – The current through RD will be: Vin (using typical values) Digital Electronics ELE 450 Vout Qo DI Dr. Idrees Al-kofahi DL RD 12

What about TTL? • • When Vin is switched from high to low, QO is still in saturation. – So, VC, I = VBE, O (Sat) VE, I is connected to the output of a previous TTL gate. – VE, I = VOL = VCE, O’ (Sat) – The base emitter voltage of QI will be VBE (FA) – Therefore, • VB, I = VCE, O’ (Sat) + VBE, I (FA) • VBC, I = VCE, O’ (Sat) + VBE, I (FA) – VBE, O (Sat) VCC – This is not enough to saturate QI. Therefore, it operates in FA mode. – The resulting collector current is IC, I = bf IB, I RB – Using typical values, IC, I = 102. 5 m. A – Comparing to DTL, the TTL current is 600 times larger. Vin QI RC Vouy QO – So, a TTL gate can switch 600 times faster than DTL without a pull-down resistor. Digital Electronics ELE 450 Dr. Idrees Al-kofahi 13

The TTL NAND Gate • For DTL, a NAND gate was built as shown below on the left. – The same thing can be implemented in TTL by combining the input diodes and the level shifting diode into multiple transistors. – Or, the input transistors can be combined into a “multi-emitter” BJT as shown in the figure on the right. VCC RC RB Vout VA Qo DA VB RB RC DL VA VC RB VCC DB Vout VA VB RC Vout Qo VC Qo VB DC VC Digital Electronics ELE 450 Dr. Idrees Al-kofahi 14

The Multi-Emitter BJT B E 3 E 2 E 1 E 3 E 1 B C C Circuit Symbol • E 2 Physical Structure All three emitters share the same base and collector. – The only difference is that instead of having one IE, we now have IE. – So, for a multi-emitter BJT, the basic current relationship becomes: IE = IE 1 + IE 2 + IE 3 = IC + IB Digital Electronics ELE 450 Dr. Idrees Al-kofahi 15

The TTL NAND Operation • If any input is low: – The corresponding B-E junction will be forward biased. – This allows a large base current to flow in RB and makes QI saturate. VCC – QO will turn off and the output will be high. • RB If all inputs are high: VA VB – – All B-E junctions are reverse biased. V The B-C junction is forward biased. QI will operate in reverse active mode. A large current will flow into the base of QO sending it into saturation. – The output voltage will be low. Digital Electronics ELE 450 RC Vout Qo C Dr. Idrees Al-kofahi 16

TTL with Totem-Pole Output VCC Element RC RB RCP QP VA VB QI QS DL Vout DCA DCB QO RD Digital Electronics ELE 450 Purpose QI Multi-emitter input BJT. RB Limits IIL QS Drive splitter, base driving current to QO RC Together with QS provide logic inversion. QO Output inverting BJT, active pull-down. DL Level-shifting diode. RD Discharge path for QO QP Active pull-up RCP Part of active pull-up DCA, DCB Input clamping diodes to limit the negative swing of the inputs. Dr. Idrees Al-kofahi 17

TTL with Totem-Pole Output (Contd. ) • The combination of RCP and QP provide active pull-up. – This increases the amount of sourcing current available for turning the load gates on when the output is changing from low to high. • QS acts as an emitter follower increasing the amount of current going into the base of QO ensuring that it will saturate. – It also provides logic inversion to make sure that QP and QO are not on at the same time. • Diode DL is also used to ensure that the transistors do not operate at the same time. Digital Electronics ELE 450 Dr. Idrees Al-kofahi 18

The VTC of the Basic TTL Gate • VOH: – For a low Vin, IB, I will be large. – However, IC, I will only be the leakage current flowing out of the base of QS. • Therefore, IC, I << IB, I and QI is saturated. – The voltage at the base of QS is VB, S = Vin + VCE, I (Sat) – This is not enough to turn QS on, QS is cut-off. VCC VA VB – IE, S = 0 IB, O = 0 QO is also cut-off. – VB, P = VCC – Therefore, VB, S Sat IC, I IE, S DCA Vout = VOH = VCC – VBE, P (FA) – VD, L (ON) Digital Electronics ELE 450 VB, P IB, I QI Dr. Idrees Al-kofahi RCP RC RB QS off FA QP DL ON Vout IB, O DCB QO off RD 19

The VTC of the Basic TTL Gate • VIL: – As Vin is increased, so will VB, S. – This will continue until VBE, S = VBE (FA). • At that point QS will be at Edge Of Conduction. VCC – Vin at this point is: VIL = VBE, S (FA) – VCE, I (Sat) RC RB RCP FA VA VB QP QI VB, S Sat IE, S DCA QS E. O. C DL ON Vout IB, O DCB QO off RD Digital Electronics ELE 450 Dr. Idrees Al-kofahi 20

The VTC of the Basic TTL Gate • VIB: – As Vin is increased even more, QS goes into forward active mode. – IE, S is no longer 0. But IB, O is still 0. VCC – The current will go through RD. This creates a voltage difference across RD. I RB RC RC – As Vin rises, so will the voltage across RD. IB, P – Eventually, this will be enough to put QO IC, S at edge of conduction. QI VA – The input voltage needed for that is: QS VIB = VBE, O (FA) + VBE, S (FA) – VCE, I (Sat) • What about QP and DL? VB FA IE, S DCA – As IE, S 0, IC, S cannot be 0. – Most of IRC will go to IC, S and IB, P will approach 0. – Therefore, QP and DL will start to go into cutoff mode. Digital Electronics ELE 450 Sat Dr. Idrees Al-kofahi RCP Off QP DL Off Vout IB, O DCB QO E. O. C IRD RD 21

The VTC of the Basic TTL Gate • VOB: VCC – At that point, Vout = VCC – IRC * RC – VBE, O (FA) – VD, L (ON) RC RB RCP Off VA VB QP QI VB, S Sat FA IE, S DCA QS DL Off Vout IB, O DCB QO E. O. C IRD RD Therefore, Digital Electronics ELE 450 Dr. Idrees Al-kofahi 22

The VTC of the Basic TTL Gate • VIH: – As Vin is increases still more, QS and QO will both saturate. – The input voltage needed for that is: VIH = VBE, O (Sat) + VBE, S (Sat) – VCE, I (Sat) V CC • VOL: – At that point: Vout = VOL = VCE, O (Sat) RCP RC RB VB, P VA VB QI VB, S Sat FA IE, S DCA QS FA QP DL ON Vout IB, O DCB QO E. O. C IRD Digital Electronics ELE 450 Dr. Idrees Al-kofahi RD 23

Will QI Switch to Reverse Active? • Yes. – For QI to switch to reverse active, we need – VBE, I < 0. 7 – For Vin > VIH – VB, I = VBC, I (FB) + VBE, S (Sat) + VBE, O (Sat) • For typical values, VB, I = 0. 6 + 0. 8 +0. 8 = 2. 2 V – Therefore, QI will switch to RA mode when Vin > 1. 5 V Digital Electronics ELE 450 Dr. Idrees Al-kofahi 24

Example • Calculate the VTC using typical values: – VOH = VCC – VBE, P (FA) – VD, L (ON) = 5 – 0. 7 = 3. 6 V – VIL = VBE, S (FA) – VCE, I (Sat) = 0. 7 – 0. 2 = 0. 5 V – – – VIB = VBE, O (FA) + VBE, S (FA) – VCE, I (Sat) = 0. 7 + 0. 7 – 0. 2 = 1. 2 V – VOL = VCE, O (Sat) = 0. 2 V – VIH = VBE, O (Sat) + VBE, S (Sat) – VCE, I (Sat) = 0. 8 + 0. 8 – 0. 2 = 1. 4 V Digital Electronics ELE 450 Dr. Idrees Al-kofahi 25

The VTC Region 1 1 VOH = 3. 6 V 2 2 VOB = 2. 5 V 3 3 4 5 4 VOL = 0. 2 V VIH = 1. 4 V VIB = 1. 2 V 5 VIL = 0. 5 V Digital Electronics ELE 450 Dr. Idrees Al-kofahi Element State QI Sat QS Off QP & DL FA, On QO Off QI Sat QS FA QP & DL Off, Off QO FA QI Sat QS Sat QP & DL Off, Off QO Sat QI RA QS Sat QP & DL Off, Off QO Sat 26

TTL Fan-out VCC R’CP RB Q’P RC VCC RCP RB QP D’L QI QS DL QI Q’O DCA RC QS QO DCA RD RD Digital Electronics ELE 450 Dr. Idrees Al-kofahi 27

IIL • The low input comes from the saturated Q’O of a previous similar gate. – Therefore, Vin = VCE, O’ (Sat) – QI will be saturated – VB, S will be VCE (Sat) + VCE (Sat) – QS is off. – IIL = IE, I – IE, I = IC, I + IB, I – IC, I = 0. VCC R’CP RC RB IB, I Q’P D’L Sat Q’O VB, S QI IIL Sat IC, I QS Off DCA RD Digital Electronics ELE 450 Dr. Idrees Al-kofahi 28

IOL • The low output comes from QO being saturated and both QP and DL being off. – IOL = IC, O – IC, O = sb. FIB, O VCC • Max fan-out, when s = 1 – IB, O = IE, S – IRD RB RC RCP Off QP – IE, S = IC, S + IB, S QI RA QS Sat Off DL IOL DCA QO Sat – IB, S = IC, I – QI is R. A. , therefore, RD • IC, I = (1 + b. R) IB, I Digital Electronics ELE 450 Dr. Idrees Al-kofahi 29

Example • Calculate output-low fan-out using typical values. IB, S = IC, I = (1 + 0. 1). 675 m =. 743 m. A IE, S =. 743 m + 2. 5 m = 3. 243 m. A IB, O = 3. 24 m – 0. 8 m = 2. 44 m. A IOL = IC, O = 1 X 25 X 2. 44 m = 61 m. A Digital Electronics ELE 450 Dr. Idrees Al-kofahi 30

IIH • The high input comes from QP and DL of the driving gate both being on. – This makes QI RA, and QS and QO both saturated. – We can determine VB, I = VBE, O (Sat) + VBE, S (Sat) + VBC, I (RA) VCC R’CP FA IB, I Q’P ON D’L – Since QI is RA, • IIH = IE, I = b. R IB, I Q’O RB RC QP VB, I IIH QI RA RCP QS Sat DCA DL QO Sat RD Digital Electronics ELE 450 Dr. Idrees Al-kofahi 31

IOH • Since the QI’s of the load gates must be kept in RA mode, – Vout > VB, I – VBE, I (FA) VCC RC RCP FA RB QP QS OFF DL ON IOH • QO Therefore, – VB, P > VB, I – VBE, I (FA) – VD, L (ON) – VBE, P (FA) • VB, I QI RA Off IOH = IE, P = (1 + b. F) IB, P Digital Electronics ELE 450 Dr. Idrees Al-kofahi 32

Example • Calculate output-high fan-out using typical values. – VB, I = VBE, O (Sat) + VBE, S (Sat) + VBC, I (RA) = 0. 8 + 0. 7 = 2. 3 V – IIH = IE, I = b. R IB, I = 0. 1 X 0. 675 m = 0. 0675 m. A – VB, P > VB, I – VBE, I (FA) – VD, L (ON) – VBE, P (FA) – VB, P > 2. 3 – 0. 7 = 0. 2 V – IOH = IE, P = (1 + b. F) IB, P = (1 + 25) X 3 m = 78 m. A Digital Electronics ELE 450 Dr. Idrees Al-kofahi 33

TTL Power Dissipation • Output Low State: VCC – ICC (OL) = IRCP + IRC + IRB RB IRC RC IRCP Off QP – QP is cut off IRCP = 0. QI RA QS Sat DCA – QS and QO are Sat. Off DL QO Sat RD – QI is RA, therefore Digital Electronics ELE 450 Dr. Idrees Al-kofahi 34

TTL Power Dissipation • Output High State: VCC – ICC (OH) = IRCP + IRC + IRB RB IRC RC IRCP FA QP – If we assume no loads IRCP = 0, IRC = 0. QI Sat QS Off DCA – QI is saturated and this gate is driven by a similar gate, therefore Digital Electronics ELE 450 Dr. Idrees Al-kofahi ON DL QO Off RD 35

Example • Calculate the average power dissipation of the TTL gate using typical values – ICC (OL) = IRC + IRB – ICC (OH) = IRB Digital Electronics ELE 450 Dr. Idrees Al-kofahi 36

Example (Contd. ) ICC (OL) = 2. 5 m + 0. 675 m = 3. 175 m. A ICC (OH) = 1 m. A • Should there be loads connected, ICC (OH) will increase and so will the power dissipation. Digital Electronics ELE 450 Dr. Idrees Al-kofahi 37

Open Collector TTL • • • If we remove the active pull-up section, we end up with a gate that can be used for connecting to a common bus. VCC For a low output, QO RC RB saturates and pulls the bus line low. VA QS QI For a high output, VB an external pull-up Vout resistor is added to DCA DCB QO the bus line. RD Digital Electronics ELE 450 Dr. Idrees Al-kofahi 38

Open Collector NAND gates • • If any NAND gate produces a low, the whole line is drawn low. The NAND gates cannot produce a logic high. – They will basically produce a high impedance state. – The pull-up resistor will pull the line up to VCC Digital Electronics ELE 450 Dr. Idrees Al-kofahi 39

Typical TTL Values (74 xx) • • VOH = 3. 6 V VIL = 0. 5 V VOB = 2. 5 V VIB = 1. 2 V VOL = 0. 2 V VIH = 1. 4 V IIL = 1 m. A IOL = 100 m. A VCC 4 k. W 1. 6 k. W 120 W QP Vin QI QS DCA DL QO – Max N = 100 1 k. W • Ave. PDisp = 10 m. W • Propagation Delay = 10 n. S • PDP = 100 Digital Electronics ELE 450 Dr. Idrees Al-kofahi 40 Vout

Low Power TTL – LTTL (74 Lxx) • To reduce the power, we need to reduce ICC. – The easiest way is to increase the sizes of the resistors. – Ave PDisp = 0. 9 m. W • The price is a reduction in Fan-out and switching speed. VCC 40 k. W 20 k. W 500 W QP Vin QI QS DCA DL QO 12 k. W – N = 50 Digital Electronics ELE 450 Dr. Idrees Al-kofahi 41 Vout

High Speed TTL – HTTL (74 Hxx) • To increase the transition speed, we need to increase the currents. Darlington Pair VCC – The easiest way is to reduce the sizes of the resistors. • • In addition, a combination known as the Darlington Pair replaces QP to increase the amount of current that can be supplied to the load when the output is switching from low to high. 2. 8 k. W 58 W 760 W QP QP 2 Vin QI QS 4 k. W Vout DCA Of course, the price is increased power dissipation. QO 470 W – Ave. PDisp = 20 m. W Digital Electronics ELE 450 Dr. Idrees Al-kofahi 42