Download FETs You know how to make a NOT gate and a NAND gate using

FETs
1. You know how to make a NOT gate and a NAND gate using CMOS technology.
Now design an NOR gate. A NOR gate’s output is low if any of its two inputs is
high. In any other case it is high. Tip: a two input NOR gate can be built using
only four transistors.
Start by understanding how the NAND circuit works. For simplicity we say that
when the voltage on input or on output is 5V we have a logical state “1”. When
we have 0V then we call it a local “0”.
Our gates are made using N-MOS (arrow pointing in) and P-MOS transistors
(arrow pointing out). N-MOS are on (conducting) when there is a voltage on its
gate. P-MOS are on when there is no voltage on its gate. This kind of structure is
called complementary MOS (CMOS) technology. It is a base for most of electronic
circuits nowadays.
Our inputs can be either logical 1 or 0. In case of NAND gate we can four different
input combinations: 00 (A=0V, B=0V), 01 (A=0V, B=5V), 10 and 11.
A
0
0
1
1
B
0
1
0
1
NAND AND OR NOR
1
0
0
1
1
0
1
0
1
0
1
0
0
1
1
0
We need to design a NOR gate. Compare NOR output with A and B inputs. You
can see that when output is 0 either A or B input is one. We can realize this
behaviour by connecting two N-MOS transistor in parallel between output and
ground. We only need to take care of AB=00 case. The output has to be 1 so we
need to connect the output to one. We use this by connecting two P-MOS
transistors in series.
Can you also design a three input NOR gate?
Three input device would need three transistors in series/in parallel in one
structure.
2. Use Farnell, RS, Digikey or any other supplier database to find a good MOSFET
that will turn on and off a load. When the load is connected it draws 1A of
current. The MOSFET is connected to an Arduino development board. Arduino
can provide a 0/5V control signal.
Be sure that you choose the transistor that can handle the 42V when the
transistor is off.
We need a MOSFET that will be easily solderable so you should look for a
through-hole part.
We don’t want a heatsink so you need to be sure that you will not overheat the
transistor junction. You need to find out how much power your chosen MOSFET
can safely dissipate without the heatsink.
There are 7312 MOSFETs on Farnell (Semiconductors – discrete → Transistors –
MOSFETs). To find a right one you can limit you search only to devices that fulfil
all conditions:
1. VDS > 1.5·V so we need a device with VDS at least 55 or 60V. 1.5 is a safety
margin. It can be lower if we are confident there will be no over-voltages
in the circuit. It can be higher if you want a more reliable design.
2. Case style: TO-220. There are many more but this is a very common one.
3. n-channel MOSFET.
4. Vth < 5V so that we can turn the MOSFET on.
You can use filter on Farnell site to remove devices that don’t fulfil these criteria.
We are left with 357 transistors. Because we do not want to dissipate power we
want to have as low RDSon resistance as possible. You can sort the table to view
the lowest resistance on top.
The best transistor has RDSon = 1.9mΩ but it costs 41dkk.
The worst transistor has RDSon = 9Ω. At this resistance we will loose 9W of power.
TO-220 can dissipate safely around one watt of power. Therefore we can
additionally limit our search to transistors with RDSon < 1Ω.
Now we can choose from the remaining transistors:
The best one in terms of losses is FDP025N06 for 41dkk with RDSon = 1.9mΩ
The cheapest one is STP14NF12 for 4.7dkk with RDSon = 160mΩ
We can also choose any transistor in between. The choice is always a
compromise between performance and price.
3. In some situations we need to limit the speed of MOSFET switching. How would
you do it? Is it possible to e.g. slow down turn-on without affecting turn-on?
You can add an additional resistor to the gate of the MOSFET to slow down the
switching time. This will increase the losses but will decrease di/dt which may
be beneficial in many cases. You can use additional diodes to control turn-on and
turn-off separately.
4. An IXTN36N50 MOSFET is used in a converter as an active rectifier switch. This
means MOSFET’s body diode is used to do the rectification. Whenever the diode
is conducting MOSFET can be turned on so that current can flow through the
channel instead of body diode. Look at the electrical characteristics of MOSFET’s
structure and its body diode and see if you can figure out whether current will
flow through the MOSFET’s channel or though its body diode at 5A, 15A and 25A.
MOSFET and its internal diode have following characteristics:
You can see the voltage scale on MOSFET characteristics is 10× bigger. We can
overlay these two characteristics on a single graph: diode black, MOSFET: red
As you can see voltage drop on MOSFET channel is lower than diode’s voltage
drop for I < 8A. Above 8A diode’s voltage drop is lower so it will conduct instead
of MOSFET channel.
Remember: this analysis is valid only for reverse conduction (current going into
source)!
5. Driving a power MOSFET very often requires an additional gate driver circuit.
This is because a microprocessor can only give 0/5V or 0/3.3V signals are not
good enough to achieve low RDSon values in the MOSFET. This circuit is typically
connected to an external 15 or 20V power supply. It can also supply very high
gate currents (few amps) to speed up MOSFET switching.
A very common problem in power electronics is supply of power to the high-side
gate driver (driving QH). Imagine the DC bus voltage is 600V. When QL is off the
emitter of the top transistor is emitter of the top transistor is at 600V potential.
When QL is on the emitter of QH is at 0V. This means that the power supply of the
Qh gate driver has to be isolated from the main circuit.
Dbootstrap
QH
QL
There is a workaround called a bootstrapping (red circuit). An additional diode is
added. Explain how this works and what are the limitations of this method.
When we start the converter both MOSFETS are off. We have power to the lowside gate driver so we can control low side MOSFET (QL). If we turn it on the
current will flow from 15V supply through bootstrap diode, through capacitor
supplying high-side gate driver (charging it) and through QL. High-side capacitor
will be charged every time QL is on.
Therefore to have control over QH we need to turn on QL first.
Bootstrap diode has to block full DC-link voltage when QH is on.
6. In the circuit below is it possible to replace BJT transistor with a MOSFET?
We can replace BJT transistor with MOSFET because they operate similarly.
However MOSFET’s gate–source is a capacitor and opamps don’t like driving
capacitive loads. Very often MOSFET driven with opamp will oscillate. So it is
possible, however, there may be stability issues.