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CHAPTER ELEVEN
Basic Emitter-Coupled Logic
[ ECL ]
Digital Electronics.
© Dr. Anas
Introduction
Ch
11
Analogous to the
analog difference
amplifier
Emitter-Coupled Logic (ECL)
The BJTs in ECL circuits do not operate in saturation
mode, but either in cut-off or forward-active modes
The ECL circuits are the fastest switching time of
commercially digital circuits.

Typical propagation delay times are on the order of
1ns, allowing for clock frequencies up to 1GHz.
2
However, ECL circuits have the highest power
dissipation of all logic families, typically 25mW per
gate.

© Dr. Anas
BJT Current Switch
Ch
11
This figure shows an
ideal BJT current switch
The input is at the
base of QI , and VBB is
a constant reference
voltage
The coupled emitters
are ideally connected
to a constant current
source IEE.
VI  VBB
VI  VBB
3
Basic ECL current switch
VCC
RCI
 
RCR
Non-Inverting
VNINV
VINV
Inverting
QI is OFF
VINV is High
QR is FA
VNINV is Low
QI is FA
VINV is Low
QR is OFF
VNINV is High
VI
QI
QR
VBB
I EE
VEE
 
© Dr. Anas
BJT Current Switch
This figure shows an
early
ECL
implementation
VE  VEE
RE
Ch
11 Outputs are taken at the
collectors of QI and QR.
VO ,1  VINV  VC , I  VCC  I C , I RCI
and
VO , 2  VNINV  VC , R  VCC  I C , R RCR
VI  VBB
VI  VBB
4
Resistor ECL current switch
VCC
I RE 
RCI
Non-Inverting
VNINV
Inverting
VINV  VCC  I C , I RCI
VNINV  VCC
RCR
VINV
VINV  VCC
QI is OFF
VNINV  VCC  I C , R RCR
QI is FA
 
VI
QI
QR
VBB
VE
I RE
RE
VEE
 
© Dr. Anas
Voltage Transfer Characteristics
of ECL Current Switch
Resistor ECL current switch
VOH
VCC
Output High Voltage
For VI<VBB, QI is off,
QR is in the forwardactive mode
Ch
11
RCI
 
RCR
Non-Inverting
VNINV
VINV
Inverting
Proof:
F.A.:
with
QR
VI
QI
5BE
VBB
VE
VE  VBB  VBE , R ECL
and
VBE , I  VI  VE  VI <
 VBBFor
 VBE , R ECLVI<VBB
(Low),
0
VBE , I  VBE ECL
QI is off, i.e.
IC,I=0
V ECL  0.75V
QR
I RE
RE
VEE
 
VINV  VOH  VCC
© Dr. Anas
Voltage Transfer Characteristics
of ECL Current Switch
Threshold Voltage
Resistor ECL current switch
VCC
VTH
For VI=VBB, both QI
and QR are in the
forward-active
mode
(VBE,I=VBE,R)
Ch
11
RCI
and
I RE
2
VINV  VCC
6
Non-Inverting
VNINV
Inverting
QI
assuming
QR
VBB
VE
 F  1
I
 RE RCI
2
VTH  VBB
RCR
VINV
VI
I C ,I  IC ,R 
 
I RE
RE
VEE
 
© Dr. Anas
Voltage Transfer Characteristics
of ECL Current Switch
Input high
Voltages
VIH VIL
and
low
Resistor ECL current switch
VCC
For VI is slightly less
than
VBB,
QI
is
forward-active
mode
Ch
BUT not conducting as
11
heavily as QR.
For
VI is
slightly
greater than VBB, QR is
forward-active
mode
BUT not conducting as
heavily as QI.
Experimentally,
the
transition width is found
to be about VTW=0.1V
and
centered
around
VI=VBB
7
RCI
 
RCR
Non-Inverting
VNINV
VINV
Inverting
VI
VIL  VBB  0.05
QI
QR
VBB
VE
I RE
RE
VEE
&
 
VIH  VBB  0.05
© Dr. Anas
Voltage Transfer Characteristics
of ECL Current Switch
Ch
11
Output low Voltage
Resistor ECL current switch
VCC
VOL
For VI > VBB, QI begins
to conduct.
VE  VI  VBE , I ECL
VINV  VCC  I C , I RCI
As VI increases, VE also
increases, while VB,R=VBB
(fixed)
Thus, raising VI by 0.05V
beyond
VBB,
VBE,R
sufficiently decreases to
cut-off
QR. voltage which
The
input
turns QR off is
resulting in VO=VOL.
8
VI=VIH
RCI
 
RCR
Non-Inverting
VNINV
VINV
Inverting
VI
QI
QR
VBB
VE
I RE
RE
VEE
I C , I  I RE
 
VE  VEE VI  VBE, I ECL  VEE


RE
RE
© Dr. Anas
Voltage Transfer Characteristics
of ECL Current Switch
Ch
11
Output low Voltage
Resistor ECL current switch
VCC
 
VOL
RCI
VINV  VCC  I C , I RCI
RCR
Non-Inverting
VNINV
VINV
Inverting
VINV  VCC
9
 VI  VBE, I ECL  VEE 

 RCI 
RE


 V  VBE , I ECL  VEE 

VINV  VCC  RCI  IH
RE


 VOL
VI
QI
QR
VBB
VE
I RE
RE
VEE
 
© Dr. Anas
Voltage Transfer Characteristics
of ECL Current Switch
VTC beyond VIH
Resistor ECL current switch
As VI increases beyond
VIH
VCC
VI  VBE, I ECL  VEE
IC ,I 
Ch
11The output
RE
voltage VO
decreases linearly with VI
RCI
VINV
RCR
 
Non-Inverting
VNINV
Inverting
VI
QI QR
VBB


V

V
ECL

V
 I

BE , I
EE

VINV  VCC  RCI 
RE
VE


QI will eventually saturate
with further increase of VI
RC , I

if:



sat   VBER
sat

V
 VCC  VBC , IIRE
,I E
EE
RE

V

V

S
I
 
VINV  VI  VBC sat 

RVCEE
,I
1
Saturation

RE

voltage
10 This region of saturation is avoided






© Dr. Anas
Voltage Transfer Characteristics
of ECL Current Switch
VO
VOH  VCC
0.1V
Ch
11
VINV
VOL
VNINV
11
QR( DOES NOT SATURATE)
QI(sat)
VS
VI
© Dr. Anas
Voltage Transfer Characteristics
of ECL Current Switch

Example
Calculate the critical values of the VTC of VI for ECL current switch
shown previously assuming:
VEE=0V, VBB=2.6V, VBE(ECL)=0.75V, VBC(sat)=0.6V
VCC=5V,RCI=RCR=RE=1kΩ,
Ch
11

Solution
VOH  VCC  5V VIL  VBB  0.05  2.55V
VIH  VBB  0.05  2.65V
VOL  VCC
R

 VCC  VBC , I sat   VBE , I ECL  VEE  C , I
RE
VS  

R
1  C ,I

RE

12


  3.2V



 VIH  VBE, I ECL  VEE 
  3.10V
 RCI 
RE


VINV VI  VS   VS  VBC , I sat   2.6V
© Dr. Anas
Basic ECL NOR/OR Gate
Adding additional input transistors
with coupled collectors and coupled
emitters to the ECL current switch:
VINV becomes NOR output and
RCI
VNINV becomes OR output.
VNOR
For any high-state input, the
Ch
corresponding
transistor
is
VI 1
11
V
I
2
forward-active and then the
QI 2
corresponding collector current
flows through RCI and

VNOR  VINV  VCC  I C , I RCI Low

VOR  VNINV  VCC High

VNOR  VINV  VCC High
VOR
QI 1
QR
VBB
VE


 
RCR

RE I
RE
If all inputs are low, then all
the corresponding transistors
are cut-off and then
13
VCC
VEE

VOR  VNINV  VCC  I C , R RCR Low

 
© Dr. Anas
MECL I NOR/OR Gate with
Output Buffer
First ECL gate in its
(NOR/OR
form)
for
Motorola
using output
buffers is called MECL I
Ch
11 V
I1
VCC
RCI
RCR
QB , N
VNOR
VNOR
QB ,O
VI 1
VI 2
QI 2
VOR
VI 2
 
QI 1
VE
RD , N
੍ QB,N and QB,O are output buffers and
provide large sourcing current {note:
VC>VB} to the load (larger fan-out).
QR
I RE
VOR
VBB
RD ,O
RE
 
VEE
੍ QB,N and QB,O also provide voltage level shift by VBE(ECL) resulting in
an output voltage swing in a range compatible with the input voltage
swing
14
© Dr. Anas
Ch
11
VTC of MECL I NOR/OR
Gate with Output Buffer
Output
High
Voltage
VOH
When
all inputs
are low; VI‘s are
low, QI‘s are cutQB , N
off
VNOR
VCC
RCI
RCR
QB ,O
VI 1
VI 2
QI 2
VNOR  VCC  RCI I B , BN  VBE , BN FA
QI 1
VCC  VEE  VBE , BN FA
RCI  1   F RD , N
VOH  VCC
15
 VCC  VEE  VBE , BN FA 
  VBE , BN FA
 RCI 

 RCI  1   F RD , N 
QR
VE
RD , N
I B , BN 
 
I RE
VOR
VBB
RD ,O
RE
VEE
 
© Dr. Anas
Ch
11
VTC of MECL I NOR/OR
Gate with Output Buffer
VCC
Input Low and
High Voltages
RCI
VIL VIH
VIL  VBB  0.05
 
RCR
QB , N
VNOR
QB ,O
VI 1
VI 2
QI 2
VIH  VBB  0.05
QI 1
QR
VE
RD , N
I RE
VBB
RD ,O
RE
VEE
16
VOR
 
© Dr. Anas
VTC of MECL I NOR/OR Gate
with Output Buffer
VCC
Output Low Voltage
VOL
If one input is high;
VI is high, QI is
QB , N
forward-active
Ch
11
VNOR
RCI
VI  VBE, I ECL  VEE
RE
VI 1
VI 2
QI 2
QI 1
I B , BN  I C , I
 VIH  VBE, I ECL  VEE 
  VBE, BN FA
 RCI 
RE


QR
VE
RD , N
 V  VBE, I ECL  VEE 
  VBE, BN FA
VNOR  VCC  RCI  I
R
E


VOL  VCC
17
RCR
QB ,O
VNOR  VCC  RCI I C , I  VBE , BN FA
IC ,I  I E ,I 
 
I RE
VOR
VBB
RD ,O
RE
VEE
 
© Dr. Anas
VNOR
Region
VTC of MECL I NOR/OR
Gate with Output Buffer
Saturation
VCC
As VI increases
beyond VIH
RCI
RCR
QB , N
Ch
11
VNOR
VS  VIH
QB ,O
VI 1
VI 2
QI 2
QI 1
QR
VE
RD , N
R

 VCC  VBC , I sat   VBE , I sat   VEE  C , I
RE
VS  VI  

RC , I
1


RE

18
 






I RE
VOR
VBB
RD ,O
RE
VEE
 
© Dr. Anas
VTC of MECL I NOR/OR
Gate with Output Buffer

Example
Calculate the critical values of the VTC of VI for MECL I circuit
shown previously assuming:
VEE=-5.2V, VBB=-1.175 V, VCC=0V, βF=49
VBC(sat)=0.6V, VBE(FA)=0.75V, VBE(sat)=0.8V
RCI=0.27kΩ, RCR=0.3kΩ , RE=1.24kΩ , and RD,O=RD,N=2kΩ
Ch
11

Solution
 V  VEE  VBE , BN FA 
  VBE , BN FA
VOH  VCC  RCI  CC

 RCI  1   F RD , N 
 0  5.2  0.75 
VOH  0  0.27
  0.75  0.762V
 0.27  50  2 
VIL  VBB  0.05
VIL  1.175  0.05  1.225V
VIH  VBB  0.05
VIH  1.175  0.05  1.125V
19
© Dr. Anas
VTC of MECL I NOR/OR
Gate with Output Buffer

Example
Calculate the critical values of the VTC of VI for MECL I circuit
shown previously assuming:
VEE=-5.2V, VBB=-1.175 V, VCC=0V, βF=49
VBC(sat)=0.6V, VBE(FA)=0.75V, VBE(sat)=0.8V
RCI=0.27kΩ, RCR=0.3kΩ , RE=1.24kΩ , and RD,O=RD,N=2kΩ
Ch
11

Solution
 V  VBE, I ECL  VEE 
  VBE, BN FA
VOL  VCC  RCI  IH
RE


R

 VCC  VBC , I sat   VBE , I sat   VEE  C , I
RE
VS  VI  

R
1  C ,I

RE

20
  1.125  0.75  5.2 
VOL  0  0.27
  0.75  1.474V
1.24








0.27 

 0.6  0.8  5.2

1
.
24
  0.29V
VS  
0
.
27


1


1.24


© Dr. Anas
VTC of MECL I NOR/OR
Gate with Output Buffer

Example
Calculate the critical values of the VTC of VI for MECL I circuit
shown previously assuming:
VEE=-5.2V, VBB=-1.175 V, VCC=0V, βF=49
VBC(sat)=0.6V, VBE(FA)=0.75V, VBE(sat)=0.8V
RCI=0.27kΩ, RCR=0.3kΩ , RE=1.24kΩ , and RD,O=RD,N=2kΩ
Ch
11

Solution
VOH  0.762V
VOL  1.474V
VIL  1.225V
VOH  0.762V
VIH  1.125V
VI
0.1V
VOR
VIL  1.225V
VIH  1.125V
VS  0.29V
21
VOL  1.474V
VO
VNOR
QI(sat)
VS  0.29V
Fan-Out of MECL I NOR/OR
with Output Buffer
When Q Gate
is
© Dr. Anas
I1
cut- off, VC1 is
high
and
therefore VNOR
is
also
high
state
VCC
I
M  OH

I IH
22
 
VCC
QI1
 1 VNOR
I IH
VI 1
QI 1 QR V
BB
 

I IHM
 
VE
RD , N
VEE
RE
VEE
QB ,O
I OH
RE

QIM
RCR
VOH QB , N
 

RCI
 
RCI

RCI
Ch
11The
maximum
fan-out of an ECL
is
therefore
dependent on the
output high state
of the driving gate
VCC
M-Fan-Out
M- Load gates
I RE
VOR
RD ,O
RE
VEE
 
© Dr. Anas
Ch
11
Fan-Out of MECL I NOR/OR
Gate with Output Buffer
VCC

I IH
VOH  VBE, I ECL  VEE


I IH  I B, I 
1   F RE
I OH
I OH  I E , BN 
I OH
VOH  VEE
RD, N
VOH QB , N
QI1

RCI
 1 VNOR
I IH
VI 1
QI 1
I OH
RE

VCC  VBE, BN FA  VOH VOH  VEE QIM


RCI
RD, N
RE
1   F 
VE
RD , N
VEE
 

I IHM
 
 
RCI
 
VEE
23
 

RCI
VCC
VCC
M-Fan-Out
M- Load gates
I RE
RE
VEE
 
© Dr. Anas
Fan-Out of MECL I NOR/OR
Gate with Output Buffer
Ch
11


Example
Calculate the maximum fan-out
in the last example, assuming

RCI
that the load gates have
reduced VOH of the driving gate
 
from -0.76V to -0.79V:
V
Solution
 0.79  0.75  5.2
 0.059mA
50 1.24
I OH  1   F 
I
OH
24
VCC  VBE , BN FA  VOH
RCI

RCI
 
VOH QB , N
QI1
 1 VNOR
I IH
V  VEE
 OH
RRDE , N
I OH
2
VE
RD , N
VEE
 

I IHM
I
 M
  OH  88
VEE
0  0.75  0.79  0.79  5.2

I IH
 50 

 5.2mA
0.27
VI 1
QI 1
RE

QIM
 
RCI
CC
VOH  VBE, I ECL  VEE


I IH  I B, I 
1   F RE
 
I IH
VCC
VCC
I RE
RE
VEE
 
© Dr. Anas
Power-Dissipation in MECL I
circuits
VCC
Output high current supplied
(ICC(H))+(IEE(H))+(IBB(H))
For High output,Input is low
I CC OH   I C , BN  I RCI  I RCR  I C , BO
Ch
V  VBE , BN FA  VOH
11
1   F 
I CC OH   CC
RCI

I EE OH  
RCR
V V  
 OL EE  F
RD ,O  1   F
QB ,O
VI 1



QI 1 QR V
BB
VNOR
VOH  VEE VOL  VEE VBB  VBE , R ECL  VEE


RD , N
RD ,O
RE
I BB OH  
VBB  VBE, R ECL  VEE
1   F RE
Very small compared to IEE(OH)
25
RCR
VOH QB , N
RCI
VCC  VBE , BO FA  VOL
 
VE
RD , N
I RE
RD ,O
RE
VEE
 
VOR
VOL
© Dr. Anas
Ch
11
TTL Power-Dissipation in
MECL I circuits
VCC
Output low current supplied
(ICC(L))+(IEE(L))
For Low output,Input is high
RCI
I CC OL   I C , BN  I RCI  I RCR  I C , BO
I CC OL  
VCC  VBE , BN FA  VOL
RCI

I EE OL  
V V
 OL EE
RD , N
VCC  VBE , BO FA  VOH
RCR
VOL  VEE VOH  VEE


RD , N
RD ,O
 F

1 F
RCR
VOL QB , N



VNOR
1   F 
QB ,O
VI 1
QI 1 QR V
BB
VIH  VBE , I ECL  VEE
 I OL   I EE OH  
PEE avg   PCC avg   VEE  EE

2


 I OL   I CC OH  
 VCC  CC

26
2


VOR
VOH
VE
RD , N
RE
 
RD ,O
I RE
Zero if VCC=0
RE
VEE
 
© Dr. Anas
Power-Dissipation in MECL I
circuits
Ch
11

Example
Calculate the dissipated power in the driver gate for the last
example

Solution
I EE OH  
VOH  VEE VOL  VEE VBB  VBE , R ECL  VEE


RD , N
RD ,O
RE
 0.762  5.2  1.474  5.2  1.175  0.75  5.2


2
2
1.24
 6.723mA
I EE OH  
I EE OL  
I EE OL  
27
VOL  VEE VOH  VEE VIH  VBE , I ECL  VEE


RD , N
RD ,O
RE
 1.474  5.2  0.762  5.2  1.125  0.75  5.2


2
2
1.24
 6.76mA
 6.76  6.723 
PEE avg   5.2
  35.06mW
2


© Dr. Anas
Ch
11
28
HW #9:Solve Problems: 11.1-3, 11.08-10, and
11.13-19
