DIODES

Diodes Notes ECE 2210 A. Stolp4/8/03,

D D
Diodes are basically 510. Ω check valves. They allow current1 to flow freely
R electrical R 510 .Ω
in one 2/27/07
direction, but not the
other. Check valves require a small forward pressure to open the valve. Similarly, a
12. V 4. V 12. V 4. V
diode requires a small forward voltage (bias) to "turn on". This is called the forward
voltage drop. There are many different types of diodes, but the two that you areDmost 2
likely to see are silicon diodes and light-emitting diodes (LEDs). These two have
forward voltage drops of about 0.7V and 2V respectively.
silicon diode LED
Mechanical check valve Diode

The electrical symbol for a diode looks like an arrow which shows the forward current direction and a small
perpendicular line. The two sides of a diode are called the "anode" and the "cathode"
(these names come from vacuum tubes). Most small diodes come in cylindrical
packages with a band on one end that corresponds to the
small perpendicular line, and shows the polarity, see the picture. Normal diodes are rated by the average forward
current and the peak reverse voltage that they can handle. Diodes with significant current ratings are known as
"rectifier" or "power" diodes. (Rectification is the process of making AC into DC.) Big power diodes come in a variety
of packages designed to be attached to heat sinks. Small diodes are known as "signal" diodes because they're
designed to handle small signals rather than power. be 2n possible states, all of which may have to
be tried until you find the right one. Try to
Diodes are nonlinear parts guess right the first time.
So far in this class we've only worked with linear parts. The diode is
Constant-voltage-drop model This is the
definitely NOT linear, but it can be modeled as linear in its two regions
most common diode model and is the only
of operation. If it's forward biased, it can be replaced by battery of 0.7V
one we'll use in this class. It gives quite
(2V for LEDs) which opposes the current flow. Otherwise it can be
accurate results in most cases.
replaced by an open circuit. These are "models" of the actual diode. If
you're not sure of the diode's state in a circuit, guess. Then replace it id
with the appropriate model and analyze the circuit. If you guessed the
open, then the voltage across the diode model should come out less forward bias

than +0.7V (2V for LEDs). If you guessed the battery, then the current
through the diode model should come out in the direction of the diode's vd
arrow. If your guess doesn't work out right, then you'll have to try the
reverse bias 0.7V
other option. In a circuit with multiple diodes (say "n" diodes), there will
1 Assume the diode is operating in one of the linear regions (make an
educated guess).
2 Analyze circuit with a linear model od the diode.
3 Check to see if the diode was really in the assumed region.
forward bias
4 Repeat if necessary.
50-100s of volts,
unless designed
Actual diode curve to break down
The characteristics of real diodes are actually more complicated than
the constant-voltage-drop model. The forward voltage drop is not quite
constant at any current and the diode "leaks" a little current when the
voltage is in the reverse direction. If the reverse voltage is large Silicon pn
enough, the diode will "breakdown" and let lots of current flow in the junction diode,
reverse direction. A mechanical check valve will show similar reverse bias
the most
common type.
BY: MUHAMMAD AHMAD JAVED

V Z = zener
voltage
DIODES
characteristics. Breakdown does not harm the diode as long as it isn't
overheated.

Zener diodes are special diodes designed to operate in the reverse breakdown region.
Since the reverse breakdown voltage across a diode is very constant for a large range of
current, it can be used as a voltage reference or regulator. Zener diodes are also used for
over-voltage protection. In the forward direction zeners work the same as regular diodes.

ECE 2210 Diode Notes p1

I recommend that you try some of the DC analysis volts ECE 2210 Diode Notes p2

BY: MUHAMMAD AHMAD JAVED

DIODES

V
0.7.V pt

1 tp

0.7.V.t p

t1= Vp

BY: MUHAMMAD AHMAD JAVED

DIODES
V
R2peak=

Sometimes it's
helpful to figure out R1
what the voltage
across the diode
would be if it never R2
conducted (light
dotted line).

0.7. V .t
t1 = p
R1
Vp .
R1 R2

BY: MUHAMMAD AHMAD JAVED

DIODES
ECE 2210 Diode Notes p2

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DIODES

BY: MUHAMMAD AHMAD JAVED

DIODES
Other Useful Diode Circuits ECE 2210 Diode Notes p4
Simple limiter circuits can be made with diodes.
A common input protection to protect circuit from excessive input voltages
such as static electricity.
The input to the box marked "sensitive circuit" can't get higher than
the positive supply + 0.7V or lower than the negative supply - 0.7V.
Put a fuse in the Vin line and the diodes can make it blow, providing
what's known as "crowbar" protection.
Another example of crowbar protection:
If the input voltage goes above 16 V. the fuse will
blow, protecting the circuitry.
Or, If the input voltage is hooked up backwards the
fuse will blow, protecting the circuitry.
AM detector
v(t) AM modulation
A simple rectifier circuit
Returns the modulation signal

AM And a coupling capacitor can remove the DC

Battery Isolator
Alternator
Like you might find in an RV. One alternator is used to charge
two batteries. When the alternator is not charging, the + Isolator
rectified
batteries, the circuits they are hooked to should be isolated internally to
from one another. If not, then one battery might discharge give a DC
through the second, especially if second is bad. Also, you output -
wouldn't want the accessories in the RV to drain the starting
battery, or your uncle George from South Dakota might never
leave your driveway.
D1
Battery Backup Power
Normally the power supply powers the load through D1.
D2
However, if it fails, the load will remain powered by the battery
through D2. Finally, D3 and R may be added to keep the D3 R
battery charged when the power supply is working. These
sorts of circuits are popular in hospitals.

Diode Logic CircuitsActually, both of the previous circuits are logic

"OR" gate "Flyback" Diode +V

BY: MUHAMMAD AHMAD JAVED

DIODES
circuits as well.Every time the switch
inputsopens the inductor current
"AND" gatecontinues to flow through Inductive V+the diode for a moment. If load
the diode weren't there, then the current would arc across the switch.

inputs
Switch

ECE 2210 Diode Notes p4

ECE 2210 Diode Circuit Examples
conducting not conducting
Basic diode circuit analysis

1 Make an educated guess about each diode's state. or

2 Replace each diode with the appropriate model:

+
3 Redraw and analyze circuit. -

4 Make sure that each diode is actually in the state you assumed: current Vd< 0.7V
Check

Note: 0.7V is for silicon junction diodes & will be different for other types. (2V for LED)
If any of your guesses don't work out right, then you'll have to start over with new guesses. In a circuit with n
diodes there will be 2n possible states, all of which may have to be tried until you find the right one. Try to guess
right the first time.
Ex1

R 1 kΩ . Try reverse-biased,
R 1 kΩ
Try
4. V
VD VD 0.7. V

forward-biased, 4. V 0.7. V
I =
I 0 conducting model non-
conducting model .

4.V > 0.7V

Doesn't work,

I Ddiode must be D .kΩ = 3.3 mA >0D forward biased. 1

The current is in the forward direction,
Ex2 confirming the assumption.

BY: MUHAMMAD AHMAD JAVED

 R 2. kΩ Try forward-biased, conducting model
DIODES R 2. kΩ
4. V Try reverse-biased, non-conducting
VD 0.7. V model
Check the
ID diode voltage
4.V 0.7.V
ID = = 4.7 mA< 0 VD
1. kΩ

ID 0. mA

4.V < 0.7V

Confirms diode
Doesn't work, diode must be reverse biased. is reverse biased

IEx3
did that intentionally to. show the process. I expect that you
R 820 Ω Try reverse-biased, Doesn't work, diode
non-conducting model must be forward
biased.
V = 12.V 4. V = 8 V > 0.7V
12.V 4. V D Try forward-biased, conducting model
VD 0.7. V VR 12.V 4.V V D
12.V 4. V V R = 7.3 V
R 820.Ω
12.V 4. V
Check the
diode current
In each of these examples, my first guess was pretty stupid. VR
I= = 8.9 mA > 0

can make better guess and thus save yourself some work. R Confirms diode
is forward biased
Ex4 R1 1. kΩ R2 1. kΩ ECE 2210 Diode Circuit Examples p1 ECE
2210 Diode Circuit Examples p2
5. V I1
VD
I2 ID
V R2
VD I = 0.7 mA

5. V
=V
0.7.V R2
Assume diode conducts: Analyze

V I
R2 2 R2 2

BY: MUHAMMAD AHMAD JAVED

DIODES
V
R1

V R1 VD V R1 = 4.3 V I1 R1 I 1 = 4.3 mA

We assumed conducting (assuming a voltage), so check the current.

ID I1 I2 I D = 3.6 mA > 0, so assumption was correct

Assume diode conducts

VD 2. V = V R2
R1 1 kΩ
Analyze
5. V I1 V R2
V R2 VD I2 I 2 = 2 mA
R2 1. kΩ R2

I2 ID V R1 5. V VD V R1 = 3 V I1 I 1 = 3 mA
R1

Ex5 Now with an LED

.

V
R1

We assumed conducting (assuming a voltage), so check the current.

ID I1 I2 I D = 1 mA > 0, so assumption was correct,

but the current is probably
too small to create
noticeable light

VD
Ex6 Regular diode, but smaller R1 Assume diode conducts 0.7.V= V R2

R1 1 kΩ R2 100 Ω . . V R2
I = 7 mA
I1 Analyze
5. V
Assume diode does not conduct
I2 ID

BY: MUHAMMAD AHMAD JAVED

 DIODES
R1 1. kΩ

5. V I1
R2 100. Ω
0. mA
I2 ID

V I V VD
R2 2 R 2 2 R1 V R1 VD V R1 = 4.3 V I 1 R1 I 1 = 4.3 mA
5.V
We assumed conducting (assuming a voltage), so check the current.

ID I1 I2 I D = 2.7 mA < 0, so assumption was WRONG !

ID 0.mA

5.V
R1 R2
Analyze I1 I2 I1
We assumed not conducting (assuming a current), so check the voltage.

V R2 I 2.R 2 V R2 = 0.455 V < 0.7V, so assumption was correct

Actually, this final check isn't necessary, since first assumption didn't work, so this one had to.

ECE 2210 Diode Circuit Examples p2

Ex7ECE 2210 Diode Circuit Examples p3
R1 1 kΩ .

I1
R2 300. Ω You can safety say that diode D1 doesn't conduct without
VS 5. V rechecking later because no supply is even trying to make current
I2 I D2
flow through that diode the right
way. V D2
R3 150. Ω
I D1 0.7.V V D3
I3 I D3
Assume both D2 and D3 conduct.
0.7.V
Analyze V R1 V S V D2 V D2

V R1 = 3.6 V
V
R1 I 1 R 1 I 1 = 3.6 mA

V
D2 I 2 R 2 I 2 = 2.333 mA

V
D3

BY: MUHAMMAD AHMAD JAVED

DIODES
I3 R3 I 3 = 4.667 mA

We assumed D1 & D2 conduct (assumed a voltage), so check currents.

I
I D2I12 I D2 = 1.267 mA > 0, so assumption OK
I1
I D3I 3 I D3 = 1.067 mA
______________ < 0, so assumption wrong

Assume D2 conducts and D3 doesn't.

V
0.7.V AnalyzeI 2 RD22 I 2 = 2.333 mA
V D2

I D3 .
S D2
0 mA V V
R1 R3
I1 I 1 = 3.739
mA I D2 I1

V
Assumed D2 conducts, so check D2 current. I2 I D2 = 1.406
mA > 0, so assumption OK

Assumed D3 doesn't conduct, so check D3 voltage. R3 I 1.R 3 V R3 = 0.561 V < 0.7V, so OK

Once you find one case that works, you don't have to try any others.

Zener Diodes
Zener diodes are special diodes designed to operate in the reverse breakdown region. Since
+
the reverse breakdown voltage across the diode is very constant for a large range of current,
it can be used as a voltage reference or regulator. Diodes are not harmed by operating in V Z= zener this
region as long as their power rating isn't exceeded. In the forward direction zeners work _ voltage the
same as regular diodes.
Now there are three possible regions of
operation:
Same basic diode circuit analysis
1 Make an educated guess about each diode's state.
2 Replace each diode with the appropriate model:
3 Redraw and analyze circuit.

4 Make sure that each diode is actually in the state you assumed: V D= 0.7V 0.7V > V D> VZ V
Z

ECE 2210 Diode Circuit Examples p3

BY: MUHAMMAD AHMAD JAVED

DIODES
Zener Diode Circuit Examples I1 ECE 2210 Diode Circuit Examples p4
Ex1 Typical shunt regulator circuit:
VZ 4.5.V R L 500. Ω
I1
IL
R1 250. Ω Assume conducting in breakdown region
V S 10.V
IL
VD VZ
VZ
IL I L = 9 mA
R L 500.Ω RL
VS VZ
VZ 4.5.V I1 I 1 = 22 mA

R1

Assumed a conducting region, so check the current to see

if the current flows in the direction shown.

ID I1 IL I D = 13 mA > 0, so assumption OK

Ex2 What if RL is smaller? RL 150.Ω

VZ
VD Assume conducting in breakdown region
VZ IL R L I L = 30
mA I I
R1
Circuit "falls out of regulation"
V V
AssumeSnot conducting
Z
I1 I 1 = 22 mA
VS
ID I1 1 L I D = 8 mA< 0, so assumption is WRONG !
R1 R

IL

+
I L= I 1 I 1 = 25
VD
mA R1 R
_
L
Assumed a non-conducting region, so check the voltage to see if it's in the right range.

BY: MUHAMMAD AHMAD JAVED

DIODES
L
VD R .V S V D = 3.75 V < V Z = 4.5 V
L so this assumption is OK

Ex3 What if VS is smaller instead of RL? V S 6.V RL 500.

Ω

VD V Z Assume conducting in breakdown region

VZ IL RL I L= 9
mA I

V V
S Z
I1 R1 I 1 = 6mA ID 1 IL I D = 3 mA < Circuit "falls out of regulation"0, so
assumption is WRONG !

VS
Assume not conducting I L= I 1 I1=8
mA
R1 RL

Assumed a non-conducting region, so check RL

.
the voltage to see if it's in the right range. VD R1 R VS VD=4
V < V Z = 4.5 V
L so this assumption is OK

ECE 2210 Diode Circuit Examples p4

ECE 2210 Diode Circuit Examples p5
Exam-type Diode Circuit Examples
On an exam, I usually tell you what assumptions to make about the diodes, then you can show that you know how to
analyze the circuit and test those assumptions. Since everyone starts with the same assumptions, everyone should
do the same work.
In the circuit shown, use the constant-voltage-drop model for the silicon diode.
a) Assume that diode D1 does NOT conduct. Assume that D1 D2
diode D2 does conduct.
I R1
Find VR2, VR1, IR1, & ID2, based on these assumptions. I D2
Stick with these assumptions even if your answers come
V2
out absurd. Hint: think in nodal voltages.2.V R1 50. Ω
1.8. V
V V R2
V R2 = _________ 1 R2 260. Ω

BY: MUHAMMAD AHMAD JAVED

DIODES
V R1 = _________

I R1= __________

I D2= __________

1.3. V

I R1
V R2 = 1.3 V I
D2
V R1 = 0.5 V V2 2. V
R1 50. Ω
V1 1.8. V
I R1 = 10 mA V
R2
R 260. Ω
2
I R2 = 5 mA

I D2 = 5 mA

0.7.V
Solution to a)

V R2 V2 0.7.V

V I V
V R1 V1 V R2 R1 R1 R 1
I
R2 R2 R 2

I I I
D2 R2 R1

b) Based on your numbers above, does it look like the assumption about D1 was correct? yes
no
How do you know? (Specifically show a value which is or is not within a correct range.) (circle one)

yes V D1= V R1 = 0.5 V < 0.7V

c) Based on your numbers above, does it look like the assumption about D2 was correct? yes no
How do you know? (circle one)

no I D2 = 5 mA < 0

d) Based on your answers to b) and c), which (if any) of the following was not correctly calculated in part a.

V V I I
R2 R1 R1 D2
(circle any number of answers)

Circle all in this case

BY: MUHAMMAD AHMAD JAVED
DIODES
ECE 2210 Diode Circuit Examples p5
ECE 2210 Diode Circuit Examples p6
Assume that diode D1 is conducting and that diode D2 is not conducting.
a) Find VR1, IR1, IR3, ID1, VR2 based on these assumptions.
Do not recalculate if you find the assumptions are wrong. D1
R1 200. Ω
V R1
V R1 = __________ I D1 I R1
3.V
I R1 = __________ V in

I R3 = __________ D2
R2 100. Ω
V R2
I D1 = __________ I R3
I R2

V R2 = __________
R3 400. Ω
Solution:
V R1 0.7. V
D1
V R1 R1 200. Ω
I R1 V R1
R1
I D1 I R1
V in 0.7. V 3.V
I R3
R2 R3 I R1 = 3.5 mA I R3 = 4.6 mA

I D1 I R3 I R1 D2
R2 100. Ω
V R2
I R2 I R3
I R3
I R2
V R2 I R2. R 2
R 400. Ω

V in
I D1 = 1.1mA

V R2 = 0.46 V

3
(circle one)
b) Was the assumption about D1 correct? yes no

How do you know? (Specifically show a value which is or is not within a correct range.)

yes I R2 = 4.6 mA> 0

c) Was the assumption about D2 correct? yes no

How do you know? (circle one)

BY: MUHAMMAD AHMAD JAVED

DIODES
yes VD2 = V R2 = 0.46 V < 0.7V

d) Based on your answers to b) and c), which (if any) of the following was not correctly calculated in part a.

V I I I V
R1 R1 R3 D2 R2
(circle any number of answers)

Circle none in this case

ECE 2210 Diode Circuit Examples p6

ECE 2210 Diode Circuit Examples p7
A voltage waveform (dotted line) is VZ 4. V
applied to draw the output waveformvo) the circuit shown.
R1 10. Ω v
( Accurately you expect to
see. and voltage
levels. Label important times o
vo 0
zener
If diode doesn't conduct: v in
R2 20. Ω
Positive half

. 0.7.V
Diode conducts at: 0.7 V input at time: .10.ms = 0.7 ms
10.V
Maximum:
R1 0.7. V
R2
vo ( 10. V 0.7. V ) . v o = 6.2 V
R1 R2
10. V
R2

Negative half

BY: MUHAMMAD AHMAD JAVED

DIODES

ECE 2210 Diode Circuit Examples p7

ECE 2210 Diode Circuit Examples p8
A voltage waveform (dotted line) is applied to the circuit R1 60. Ω
shown. vo
Accurately draw the output waveform (vo) you expect to see.
Label important times and voltage levels.
LED R2 40. Ω

v
in

If diode doesn't conduct:

R 2 v o =.v in 1 2
R R

BY: MUHAMMAD AHMAD JAVED

DIODES
2
R .10.V = 4 V

R1 R2

R1 R 2.2.V
v Diode begins to
When: in v in = 5 V at: 5.ms
conduct
R2
When diode conducts:

ECE 2210 Diode Circuit Examples p8

Diode Physics (The simple version) EE 2210 A.Stolp 3/22/01,
rev, 2/25/16
FYI Only, You don't need to know this
Silicon atoms
Silicon atoms each have 4 valence electrons (electrons in their outermost shell). That leaves 4
spaces in the outer shell of 8. This makes silicon a very reactive chemical, like carbon, which
has the same valence configuration.

BY: MUHAMMAD AHMAD JAVED

DIODES
Silicon crystals
Each atom covalently bonds with four neighboring atoms to
form a tetrahedral crystal, which we'll represent in 2D.

Tetrahedral crystal
2-dimensional representation
In the pure, "intrinsic" crystal, practically all the electrons are used in bonds and all the spaces are filled, which
leaves almost no electrons free to move and thus no way to make current flow.
By the effects of heat, light and/or large electric fields, a few electrons do break free of the bonds and become
"free" carriers. That is, they're free to move about crystal and "carry" an electrical current.

hole electron
carrier movement more carrier movement
Interestingly, the space that was vacated by the electron also acts like a carrier. This pseudo-carrier is called a
"hole" and it acts like a positively charged carrier.
Unless there's a lot of heat or light, the intrinsic silicon is still a very bad conductor.
Silicon is considered a semiconductor.
Doping
p-typen-type
Some atoms, like boron and aluminum Some atoms, like arsenic and naturally
have 3 valence electrons in phosphorus naturally have 5 valence their outer
shells.electrons in their outer shells.
If you replace some of the silicon atoms in a crystal If you replace some of the silicon atoms in a
with boron there won't be quite enough electrons to fill crystal with arsenic the 5th electron doesn't fit into
the crystalline bond structure and unfilled spaces will the crystalline bond structure and is therefore free
BY: MUHAMMAD AHMAD JAVED
DIODES
act just like free holes. This "doped" silicon crystal is to roam about and be a carrier. This "doped"
now called an p-type semiconductor. The p refers to silicon crystal is now called an n-type
the "extra" "positive" carriers. semiconductor. The n refers to the "extra"
negative carriers.
free electron

free hole

Diode Physics (The simple version)

Diode Physics (The simple version) p.2
p-type n-type
It turns out that the free carriers are the most important things in the semiconductor crystals, so we can
simplify the drawings to show only these free carriers.

When a p-type semiconductor is created next to an n-type, some of the free electrons from the n side
will cross over and fill some of the free holes on the p side. This makes the p side negatively charged
and leaves the n side positively charged. When the voltage across the junction reaches about 0.7 V
the electrons find it too difficult to move against the charge
A region near the junction is now depleted of

carriers and (surprise) is called the depletion

region. an external voltage across the diode in the

\_______/
depletion region
Reverse bias "positive" holes move negative electrons move toward the
This pn junction is now a diode. If you place
negative voltage toward the positive voltage
reverse bias direction, the depletion region gets

bigger and no current flows.

BY: MUHAMMAD AHMAD JAVED

DIODES
This reverse bias region can be used as a heat + or light sensor since the only current flow should be due to a
few carriers produced by these effects.
The reverse biased diode can also be used as a

voltage variable capacitor since it is essentially With reverse bias the depletion region gets bigger
an insulator (the depletion region) sandwiched
between two conducting regions.
Forward bias
If you place an external voltage across the _ diode in the forward bias direction, the +
depletion region shrinks until your external voltage reaches about 0.7V. After that the diode conducts
freely..
Diode Physics (The simple version) p.2
With forward bias the depletion region gets
smaller and eventually (at about 0.7V) conducts
freely.

Diode
Folder:____ Name:_______________________ ECE 2210 hw # DO1 Due: Fri, 11/6/20
Fill in the blanks in the following circuits. For some of the simple calculations, you may simply write down A.Stolp rev b
the answer without showing work.
Assume the diodes are silicon with a 0.7V forward voltage drop:
R 330. Ω V
1.R= _________ Assume the LEDs have a 2V forward voltage drop:
2. R 330. Ω V
4. V

4. V

R= _________

I = _________
BY: MUHAMMAD AHMAD JAVED
DIODES
I = _________ V D2= _________

Note: In problems 5 and 6 you'll have to make some assumptions about which diode(s) is/are conducting.
Work the problem with those assumptions and see if you arrive at impossible answers. If so, change your
assumptions and try
again.
5. V D1= _________ V D2= _________ There are four possible assumptions.
1. Neither diode conducts.
2. Only D1 conducts.
D1 D2 3. Only D2 conducts.
IR
4. Both diodes conduct.
R 220. Ω
NOTE: You don't have to try all four possibilities. As soon as
10. V 8. V
you find one that works, that's the answer. So make your
best guess first.

I 1 = _________ I 2 = _________

ECE 2210 homework # DO1 p.1

6. I T = _________ V D1= _________

V D2 = _________
V D1
I D1 I D2 = _________
I R1

I R1 = _________
12.V
R1 150. Ω
R2 820. Ω I R3 = _________
I R2
R3 1. kΩ

I R2= _________

7.
I T = _________ V D1= _________

BY: MUHAMMAD AHMAD JAVED

 12. V R1
R1 150. Ω
R2 820. Ω
I R2 I R3 = _________
DIODES
R 1. kΩ
I R2 = _________ 3

8. V R= _________ 9. R = _________

R = _________
14. V
ID 15. mA
6. V ID 20. mA

10. I R1 30. mA
R 1 = _________

R1

R2 300. Ω R 3 = _________

I D2 20. mA
12. V

ECE 2210 homework # DO1 p.2

11. V R= _________
R = _________ P R = _________
(power)
ID 50. mA
18. V

VZ 12. V

P D = _________

12. I R = _________

R 120. Ω P R = _________

I L= _________
18. V
ID
RL 300. Ω
VZ 12. V

I D = _________ P D = _________

BY: MUHAMMAD AHMAD JAVED

DIODES
13. I R = _________

R 120.Ω P R = _________

I L= _________
18.V
ID
R L 200. Ω
VZ 12. V

I D = _________ P D = _________

Warning: If ID turns out negative, it is

actually 0 and you must recalculate
everything else.

You will need more paper for the next two problems, add a sheet or two.
14. Assume that diode D1 does conduct. Assume that diode D2 does NOT conduct.

a) Find VR1, IR1, IR3, ID1, VR2 based on these assumptions.

Stick with these assumptions even if your answers come
out absurd. D1
R1 50. Ω
V R1 = ? I R1 = ? I R3 = ? I D1 = ? V R1
I D1
I R1
V R2 = ?
4. V

V
in
D2
R2 150. Ω
V R2
I R3
I R2

R 400. Ω

ECE 2210 homework # DO1 p.3 3

b) Was the assumption about D1 correct? yes or no ECE 2210
homework # DO1 p.4
How do you know? (Specifically show a value which is or is not within a correct range.)
c) Was the assumption about D2 correct? yes or no How do you know?

BY: MUHAMMAD AHMAD JAVED

DIODES
D1
Assume that diode D3 does NOT conduct.
LED R1 54. Ω
I D2
I D1
VS 5. V
I R2
D2

R3 30. Ω
R2 100. Ω
V D3

D3

R4 20. Ω
15. Assume that diodes D1 and D2 DO conduct.

a) Find IR2, ID2, ID1, & VD3 based on these assumptions.

Stick with these assumptions even if your answers
come out absurd.

I R2 = ? I D2 = ? I D1 = ? V D3 = ?

b) Based on the numbers above, was the assumption about D1 correct? yes no How do
you know? (Show a value & range.)
c) Was the assumption about D2 correct? yes no d) Was the assumption about D3

correct? yes no How do you know? (Show a value & range.) How do you know?

(Show a value & range.) e) Based on your answers to parts b), c) & e):

i) The real I R2 < I R2 calculated in part a. ii) The real I R2 = I R2 calculated in part a.

iii) The real I R2 > I R2 calculated in part a. You do not need to justify your answer.
Answers
0.7 .V .
DI10.mA I D .mA
1 VD VR 3.3 V 2.0 V D4.V V R0.V
0.7 V I .
BY: MUHAMMAD AHMAD JAVED
0.mA
D1 1
.
R2 D2
D1
DIODES
. . . . .
3. VD . VR 7.3 V I 14.3 mA 4. 0 mA V D2 8V V D1 0V VR 0V
. .
5. V 0.7.V V D21.3 V I42.3 mA I2
. . I . .
6. I D2 0 mA V D10.7 V I13.8 mA R1 =I R3 9.83 mA V2.17 V I D1 =I T 23.6.mA
. . I . .
7. V 0.7.V V D20.7 V I R1 0 mA R2 13.8 mA =I D1 I R3 11.3 mA =I D2 IT 25.1.mA 8.
VR R 267.Ω 9. R 500.Ω

. . .
10. R 1R 3 150 Ω 11. V R ID 50 mA R120 Ω 0.3.W0.6.W
. .
12. I LI R 50 mA I D10 mA 0.3.W0.12.W
. .
13. I DI L =I R 56.3 mA VP RP D 0W
I .
14. a) V R1 0.7.V I R1 14.mA I R3 D1V R2 0.9 V b) no I D1 = 8 mA < 0

.
c) no V D2= V R2 = 0.9 V > 0.7V _/ b) yes I D126 mA
. . .
> 0 15. a) I R230 mA I D2 4 mA I D126 mA V D30.92.V c)
no I D24.mA < 0

.
ECE 2210 homework # DO1 p.4 d) no V D30.92 V > 0.7V e) ii)
Folder:____ Name:_______________________
Assume the diodes are silicon
with a 0.7V forward voltage drop:

1. The input voltage to the circuit below is shown at right v

in (dotted line). Show the output voltage across the
(volts) resistor. Make it accurate and label the
important voltages and times. You can draw your
answer right on my drawing, that's why the input is
shown as a dotted line.

R vo
v in

=?

BY: MUHAMMAD AHMAD JAVED

DIODES
2. The voltage waveform shown (dotted line) is applied to the circuit. Accurately R1 10. Ω
draw the output voltage you expect to see across the 20 Ω resistor.
Label the important voltages and times.

R2 20. Ω

v
in

R1 25. Ω

ECE 2210 homework # DO2, p2

3. The voltage waveform shown below is applied to the circuit shown. Accurately draw the output voltage (vo)
you expect to see. The characteristic curve for the 3-V silicon zener diode is also shown.
. v R2
Label important times and voltage levels.15 Ω in

BY: MUHAMMAD AHMAD JAVED

DIODES

ECE 2210 homework # DO2, p2

ECE 2210 homework # DO2, p3 R1 40. Ω
vo
4. A voltage waveform (dotted line) is applied to the circuits shown.
Accurately draw the output waveform (vo) you expect to see.
D2
Label important times and voltage levels. v in LED

D1
R2 23. Ω

BY: MUHAMMAD AHMAD JAVED

DIODES

ECE 2210 homework # DO2, p3

ECE 2210 homework # DO2, p4
5. A voltage waveform (dotted line) is applied to the circuits shown.
Accurately draw the output waveform (vo) you expect to see.

Label important times and voltage levels. V DC 2. V

R 100. Ω

BY: MUHAMMAD AHMAD JAVED

DIODES

Answers
1 Straight lines between the following points: (0ms,0V), (0.7ms,0V), (2ms,1.3V), (3.3ms,0V), (8.7ms,0V), then
ramps up as between 0.7ms & 2ms.
2. Straight lines between the following points: (0ms,0V), (1ms,0V), (10ms,4.2V), (10ms,0V), (21ms,0V), then ramps up as
between 0.7ms & 10ms.
3. Straight lines between the following points: (0ms,0V), (6ms,3V), (16ms,4.875V), (16ms,0V), (17.4ms,-0.7V), (32ms,-
3.438V), (32ms,0V), (38ms,3V), then ramps up as between 6ms & 16ms.
4. Straight lines between the following points: (0ms, 0), (2.86ms, 2V), (10ms, 2V), (10ms, -3V), (19ms, -0.7V), (22.86ms, 2V),
flat at 2
5. Straight lines between the following points: (0ms,1.3V), (0.2ms,1.3V), (0.4ms,3.3V), (0.4ms,1.3V), (1ms,1.3V) .

ECE 2210 homework # DO2, p4

BY: MUHAMMAD AHMAD JAVED