lecture 2: diodes - university of arizona · lecture 2: diodes •in the first lecture, we...
TRANSCRIPT
Lecture 2: Diodes• In the first lecture, we discussed linear circuit components
– resistors, capacitors, and inductors– we found that we could build some electronic circuits using only
these devices• However, much of electronics (in particular digital electronics)
depends upon non-linear devices• The first such device we’ll consider is the diode:
– An ideal diode is a perfect conductor in one direction, and aperfect resistor in the other
– The current vs. voltage curve looks like:
V
I
Real Diodes• How do we make a device that behaves at least
approximately like an ideal diode?• We use a material such as silicon• Remembering our freshman chemistry, Si has four valence
electrons– meaning each Si atom can bond easily with four other atoms
• We’d really need to do some quantum mechanics tounderstand how the electrons behave in this lattice (orcrystal) of Si
• But we can do some hand-waving and make someplausibility arguments to explain the features that areimportant to us
• In the lattice as shown, all of the electrons are “locked in”to their bound pairs– They won’t move even if a voltage is applied across the
crystal– Pure Si is a good insulator!
Doping• Let’s say we were to replace some of the Si atoms in our
lattice with arsenic (As). This is called doping the silicon– As has 5 valence electrons– So, after it forms bonds with the four Si atoms around it,
there’s an extra electron hanging around:
– Since the extra electron is not strongly bonded anywhere, it’sfree to flow when voltage is applied
• Doping that results in free electrons is called n-type
• Rather than As, we might dope with something likeGallium (Ga)– i.e., something with 3 valence electrons
• In this case electrons again can move, as they try to “fillin” the empty bond around the Ga atom
• Mathematically, this is equivalent to the motion ofpositively-charged “holes” in the distribution of electrons
• This is called p-type doping• Either n- or p-doped Si is a reasonably good conductor
– semiconductor• But it’s still linear
The pn junction• What happens if we join an n-doped piece of Si with a p-
doped piece?• The exact instant after they’re joined we have:
– the “free” charge carriers are shown – both halves areinitially electrically neutral
• The free electrons on the n side see a set of holes to fill inon the right, so they start to diffuse across the boundary
• As they do so, the n side becomes positively charged andthe p side negatively charged– and these charges tend to attract electrons back toward the n
side
p n- -- -- - -
--
- -+++ +
+
+ +
+
+
+
• The system very quickly reaches an equilibrium that lookslike the following:
• Note that there is a net charge on each side of the depletedregion (shown in red)– means that there is an intrinsic voltage drop across the
junction• But no current is flowing (since we’re at equilibrium)
p n--- -- - -
--
- -++ ++
++ +
+
+
+ ++-
Depleted region(no free charge
carriers)- +
pn junction with external voltage• Now we apply an external voltage across our pn junction• Let’s make it positive on the p side. Then initially we
have:
• This acts to prevent electrons moving from the n side tothe p side– and holes from the p side to the n side
• The depleted region becomes smaller
p n- ---
- - -
--
- -++
++
++ +
+
+
+ + +-
- +
• Note that the depleted region is an insulator, since it has nofree charge carriers– As the depleted region becomes very thin, the junction
becomes a very good conductor!• Now let’s apply the opposite voltage:
– This makes it easier for electrons to migrate from the n sideto the p side
– Depleted region becomes larger:
p n--- -- - -
--
- -++ ++
++ +
+
+
+ ++-+ -
• And that’s all that happens!– The system reaches a new equilibrium– No current flows across the crystal
• Our pn junction behaves kind of like an ideal diode!– But not exactly
• Even if the external voltage (or bias) is applied in theproper direction, no current flows until the depleted regionbecomes thin enough– Voltage needed to do this is called the diode drop
• If enough voltage is applied in the reverse bias direction,current can be forced through the junction– This is called the breakdown voltage, and it’s generally not
good for the diode to be exposed to this
Zener diodes• Unlike normal diodes, Zeners are designed to operate in
the breakdown region– From the previous page, we see that the current vs. voltage
curve is even steeper in the breakdown region than in theforward-bias region
• This means that Zeners are nearly an ideal voltageregulator:
• Vout will never exceed the breakdown voltage of the Zenerdiode, no matter how big Vin is– should really put a resistor in the circuit to limit current
Vin Vout
Other diode applications• One simple way to take advantage of a real diode’s
properties is:
• Vout is Vin minus the diode drop (typically < 1 V)– so it’s about 5V– which is just right to power many common digital
electronics components
Diode protection• Look at the following circuit fragment:
• What happens when the switch is opened?– The inductor attempts to keep the current flowing– If the switch is a transistor (we’ll learn about those next
week), that might destroy the switch• Can protect the switch by using a diode to provide an
alternate current path:
I
Rectification• Digital electronics requires a constant input voltage to
supply power• But the voltage that comes out of our wall sockets is AC• Producing a DC Vout from an AC Vin is called rectification• It’s clear that diodes can help with this:
• This does the job, but only for the part of the cycle wherethe AC voltage is positive– this is called a “half-wave” recifier– it means that half of the input power isn’t used
Vin Vout
Bridge rectifier• To obtain full-wave rectification, we can do something like
this:
– current can always flows in the same direction through theresistor, regardless of whether the input voltage is positive ornegative
• Output still isn’t DC voltage, though:
Vin
Vout
• To fix this, we can pass the output through a low-passfilter:
• How much does the output voltage “ripple”?– meaning how big are the variations?
Vin
Vout
out
out1
QV
C
dV dQ I
dt C dt C
=
= =
• Cycles as follows:– As Vin increases from 0 to its maximum Vp current flows
through diodes, charging up the capacitor– When Vin falls below Vp, current flows from capacitor
• diodes won’t let this current go backwards, so it must gothrough the load on the circuit, which we’ll represent with aresistance RL
– This current is the “I” on the previous page• So we have:
out
out
dV I
dt C
IV t
C
=
! = !
• Δt is about 1/2f, and I = V/RL, so:
• By choosing RLC >> f, the ripple can be made negligible– Note that this answer does not depend on the series resistor
in the filter circuit – that resistor is actually left out in realpower supplies
out
out
2
1
2
L
L
VV
R Cf
V
V R Cf
! =
!=
Light-emitting diodes• As current flows though an LED, electrons “drop” into
holes inside the depleted region– Energy released in the form of photons– In “normal” diodes these photons are in the infrared
• LEDs are designed such that visible photons are emitted– Commonly used as visual indicator that current is flowing
(e.g., “Power On” light)– The materials science needed to create LED’s is far from
trivial – blue LEDs have only become widely available inthe last ten years or so
Silicon microstrip detectors• For a particle detector, we often want to measure precisely
the position at which a charged particle goes through– Si pn junctions can be useful for this as well:
– p doped strips can be placed every 50µm or so– gives a position resolution of about 10µm
• Most particle physics experiments built in the last 15 yearsor so use this technology