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Descibes how to use NPN and PNP transistors as switches.
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Using Transistor as a SwitchDecember 23, 2008 by rwb, under Electronics.
Most of microcontrollers work within 5 volt environment and the I/O port can only handle current up to
20mA; therefore if we want to attach the microcontroller’s I/O port to different voltage level circuit or to
drive devices with more than 20mA; we need to use the interface circuit. One of the popular method is to
use the Bipolar Junction Transistor (BJT) or we just called it transistor in this tutorial. I have to make
clear on this BJT type to differentiate among the other types of transistors family such as FET (Field
Effect Transistor), MOSFET (Metal Oxide Semiconductor FET), VMOS (Vertical MOSFET) and UJT
(Uni-Junction Transistor).
A. The Switch
The transistor actually works as a current gainer; any current applied to the base terminal will be
multiplied by the current gain factor of the transistor which known as hFE. Therefore transistor can be
used as amplifier; any small signal (very small current) applied to the base terminal will be amplified by
the factor of hFE and reflected as a collector current on the collector terminal side.
All the transistors have three state of operation:
Off state: in this state there is no base current applied or IB = 0.
On active state: in this state any changes in IB will cause changes in IC as well or IC = IB x
hFE. This type of state is suitable when we use transistor as a signal amplifier because
transistor is said is in the linear state. For example if we have a transistor with gain of 100 and
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1 of 25 3/29/2012 6:17 PM
we increase the IB from 10uA to 100uA; this will cause the IC to swing from 1000uA to
10000uA (1 mA to 10 mA).
On saturate state: in this state any changes in IB will not cause changes in IC anymore (not
linear) or we could say IC is nearly constant. We never use this state to run the transistor as a
signal amplifier (class A amplifier) because the output signal will be clamped when the
transistor is saturate. This is the type of state that we are looking for on this tutorial.
From the picture above we could see the voltage and current condition of transistor on each state; if you
notice when transistor is in off state the voltage across collector and emitter terminal is equal to the
supplied voltage, this is equivalent to the open circuit and when transistor is in saturate state the
collector to emitter voltage is equal or less then 0.2 Volt which is equivalent to the close circuit.
Therefore to use transistor as a switch we have to make transistor OFF which equivalent to the logical
“0” and SATURATE which is equivalent to the logical “1“.
One of the famous diagrams that show the transistor operating state is called the transistor static
characteristic curve as shown on this following picture:
When we operate transistor as the class A common emitter amplifier usually we choose to bias the
transistor (apply voltage on VBE and VCE) in such a way (Q-Point) that IC and VCE (output) will swing to
its maximum or minimum value without any distortion (swing into the saturation or cut-off region) when
the IB (input) swing to its maximum or minimum value; but when we operate the transistor as switch
we intentionally push the transistor into its saturation region to get the lowest possible VCE (i.e. near 0.2
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2 of 25 3/29/2012 6:17 PM
volt) when we need to make the transistor ON (switch ON) and into its cut-off region when we need to
make the transistor OFF (switch OFF).
The above diagram show a typical microcontroller interface circuit using NPN transistor; the RB resistor
is used to control the current on base terminal that make transistor OFF and ON (saturate); while the RC
resistor is the current limiter for the load. if the load operate with the same voltage as the supplied
power (Vcc) you can by pass the RC (not use).
Notice the diode (also known as the clamp diode) in the inductive load circuit is needed to protect the
transistor again the EMF (Electromotive Force) voltage generated by the inductor component when the
transistor is switched on and off rapidly, this voltage is oppose the source voltage. The diode will act as a
short circuit to the high voltage generated by the inductor component, you can use any general purpose
diode with capable on handling minimum 1 A of current such as 1N4001, 1N4002, etc.
On the picture shown above you could see how we connect the transistor as the high active switch
(logical high) also known as low side switch using NPN transistor and the low active switch (logical low)
also known as high side switch using PNP transistor.
Ok let’s calculate each of the RB and RC value on this following NPN transistor circuit:
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3 of 25 3/29/2012 6:17 PM
On the circuit above we are going to use 2N3904 (the cheap general purpose transistor where you could
easily found on your local market) to drive 5 LED from microcontroller port, from the 2N3904 datasheet
we get this following information:
IC max = 200mA (this is maximum value that will make your transistor smoked, in practical application
always use just half of the maximum value mentioned on the datasheet), hFE = 100 to 300, VBE saturate
= 0.65 Volt, VCE saturate = 0.2 Volt
For most transistor in general we can use VBE = 0.7 Volt (should be saturate) and VCE = 0 Volt. Using the
5 volt power supply (VCC) and assuming VLED = 2 Volt, with each of them consuming 15 mA, we could
calculate the RC value using the Ohm’s law as follow:
IC = 5 x 15 mA = 75mA (0.075 A), this current is still far bellow the maximum IC allowed by 2N3904
transistor.
RC = (VCC – VLED) / IC = (5 – 2) / 0.075 = 40 Ohm
Power Dissipation on the RC resistor will be
P = (VCC – VLED) x IC = (5 – 2) x 0.075 = 0.225 Watt
Base on the above calculation we could use the nearest higher value available on the market; which is 47
Ohm, 0.5 watt resistor (for heat dissipation usually we use twice of the watt value calculated).
Assuming the hFE minimum is 100; the minimum current required in the transistor’s base terminal to
drive the LED is:
IC = hFE x IB
IB = IC / hFE = 0.075 / 100 = 0.00075 A (0.75 mA)
This current can easily be supplied by most microcontroller I/O port; which is capable to drive up to 20
mA output current. Again by applying the Ohm’s law we could calculate the RB value as follow:
RB = (VPORT – VBE) / IB
Assuming the minimum average voltage of microcontroller I/O port (VPORT) with logical “1” is about 4.2
volt (the microcontroller is powered by 5 volt supply):
RB = (4.2 – 0.7) / 0.00075 = 4666.66 Ohm
Power dissipation on the RC resistor will be
P = (VPORT – VBE) x IB = (4.2 – 0.7) x 0.00075 = 0.002625 Watt
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Base on the result you could use 4K7 Ohm, 0.25 Watt resistor (this is the common resistor which you
could easily found on the local market i.e. 0.25 watt and 0.5 watt).
Use this RB calculation as your maximum reference value; in the real world most of the transistors hFE is
vary and being measured (tested) with different VCE and IC value not to mention different specification
even though you use the same transistor type. Therefore the real RB value could be lower than 4K7 if
you really want to drive the transistor into its fully saturate mode where the VCE near 0.2 volt.
Now the question is how we determine the exact value? To answer to this question I build this following
testing circuit base on the RC and RB calculated value above using the Atmel AVR ATTiny25
microcontroller to blink the five LED:
Note: the reason I used RC = 3×150 Ohm because at that time I run out the required 47 Ohm resistor,
therefore you could use just single 47 Ohm resistor or if you only have 150 Ohm as I did, you could use
them as I did.
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Bellow is the C Program that I used to test this circuit:
//***************************************************************************
// File Name : trswitch.c
// Version : 1.0
// Description : Transistor as Switch: Simple LED Blinker
// Author : RWB
// Target : Atmel AVR ATTiny25 Microcontroller
// Compiler : AVR-GCC 4.3.0; avr-libc 1.6.2 (WinAVR 20090313)
// IDE : Atmel AVR Studio 4.17
// Programmer : Atmel AVRISPmkII
// Last Updated : 1 November 2009
//***************************************************************************
#include <avr/io.h>
#include <util/delay.h>
int main(void)
{
// Initial I/O
DDRB |= (1<<PB3); // Set PB3 as Output, Others as Input
PORTB &= ~(1<<PB3); // Reset the PB3
for(;;) { // Loop Forever
PORTB |= (1<<PB3); // Port PB3 High
_delay_ms(3000); // Delay 3 Second
PORTB &= ~(1<<PB3); // Port PB3 Low
_delay_ms(1000); // Delay 1 Second
}
return 0; // Standard Return Code
}
/* EOF: trswitch.c */
The program simply blink all the LED by toggling the AVR ATTiny25 microcontroller PB3 output port high
for about 3 second and low for about 1 second and here is the test result when the PB3 port swing to the
logical high:
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As you’ve seen from the result there is about 0.4 volt drop on the collector to emitter (VCE) terminal
instead of 0 Volt as we assume on the above calculation and the DC current gain is about 58 instead of
100 again as we assume on the above calculation. Now you understand there are tremendous different
result between the 2N3904 transistor datasheet and my test circuit, this is because the 2N3904
datasheet is measured using the PWM (Pulse Width Modulation) with period for about 300 us (micro
second) and duty cycle for about 2%, the reason to use this very short pulse period method in the
measurement is because they don’t want to overheat the transistor junction; where this junction heating
will vary the transistor hFE measurement significantly.
On my test circuit above; I used 3 second to make the 2N3904 transistor ON (saturate, VBE = 0.81 Volt,
VCE = 0.4 Volt) and 1second to make it OFF. The other factor that make the test result differ is the
various manufacture specification even though we used the same transistor type. Therefore the answer
to the above question is; there is no exact value for RC and RB; is depend on your application but it save
to use the above method to calculate the RC and RB and then do the circuit prototyping to test your
design, next adjust your RC and RB value accordingly.
Some calculation suggestion is to use the collector to base current ratio of 10 (regardless of the
transistor hFE value) to force the transistor into fully saturate (VCE = 0.2 Volt, as shown on the
datasheet above) by using this following formula:
IB = IC / hFE = IC / 10
This is what I called a “maximum saturate calculation method” (also known as worst-case design
procedure), again as you’ve seen from the real test circuit result above even though we drive the VBE
more than 0.7 volt, we still get the hFE for about 58 and IB for about 0.88 mA which is useful in the
microcontroller application (for more information you could read “Powering Your Microcontroller’s Base
Project” on this blog), therefore for practical application I would suggest; if you want to use this
maximum saturate calculation method to determine the base resistor (RB) value, make sure at least you
double the calculated value. For example to determine the RB on the test circuit above using this
maximum saturate calculation method:
IB = IC / hFE = 0.075 / 10 = 0.0075 A (7.5 mA)
RB = (4.2 – 0.7) / 0.0075 = 466.66 Ohm
By using twice the calculated value you will get 933.32 Ohm, or you could use the 1K Ohm standard
resistor.
In typical rapid switching transistor application actually we don’t drive the transistor into its full saturate
state (i.e. VCE = 0.2 Volt), because when the transistor is fully saturate, it tend to have a longer
switching time (i.e. from ON to OFF to ON again). The VCE = 0.4 volt as shown on the real test circuit
above is already adequate for most switching application, while we could still take advantage of the low
transistor base current (i.e. IB = 0.88 mA). You could see this test circuit on the video at the end of this
article.
B. Driving the Relay
Relay perhaps is one of the oldest electronic components that could be tracked back from the early years
when we first use the electricity in our life. A relay basically is an electrical switch that uses the
electromagnetic solenoid to control a switch contact. Because it use the solenoid (inductive load),
therefore we need to use a diode to protect the transistor against the EMF. The main advantage of using
a relay is that we could “relaying” or pass on the switch effect from a low power side on its solenoid to
the high power side on its metal contact by using the electromagnetic effect, where both of the solenoid
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and contact has its own separate electrical specification.
Now using the same principal we could easily calculate the RC and RB value on this following circuit:
By using 5 Volt power supply and relay with 5 Volt and 60mA operating current:
RC = 0 Ohm (not use, connect relay directly to VCC)
IB = IC / hFE = 0.06 A / 100 = 0.0006 A
RB = (VPORT – VBE) / IB = (4.2 – 0.7) / 0.0006 = 5833.33 Ohm, use 5K6 Ohm resistor
P = (VPORT – VBE) x IB = (4.2 – 0.7) x 0.0006 = 0.0021 watt, use 0.25 Watt resistor
C. Increasing the Collector Current
What if the load current is more than 1 A, let’s say you want to drive a DC motor? Perhaps you will think
to use bigger transistor such as 2N3055 power transistor; unfortunately the big power transistor tends to
have small hFE mostly less then 20, so it’s mean we have to supply bigger base current. We know that
most microcontrollers I/O port can only supply a current up to 20mA, therefore by using this type of
transistor the maximum current that we could achieve in the collector terminal is about 400mA; which is
far bellow our expectation. The solution for this situation is to use what known as Darlington pair
circuit:
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By using the Darlington pair circuit we could combine two transistors; one with high hFE2 factor usually
has a low collector current and the one with high collector current usually has a low hFE1factor. This will
give you a total hFE of hFE1 x hFE2. In the Darlington pair circuits the VBE will be twice the normal
transistor saturated voltage which is about 1.4 Volt. One of the popular ready made Darlington pair
transistors on the market are TIP120 (NPN type) and TIP125 (PNP type) which could handle the collector
current up to 3 A (max 5 A), and has the hFE minimum of 1000.
The TIP120 and TIP125 is called a pair Darlington transistors as they have similar characteristic but have
an opposite type (i.e. NPN and PNP), this Darlington transistor pair is popular used in motor controller
with the H-Bridge circuit. Remember when you use a power transistor to drive a large collector current,
you need to supply the transistor with the adequate heat sink to help cooling the transistor by
dissipating heat through the heat sink surface into the surrounding air.
Using the same principal we’ve learned before, we could easily calculate the RB value of the DC motor
circuit interface bellow:
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By using 5 Volt power supply and DC Motor with 12 Volt and 1 A maximum operating current:
RC = 0 Ohm (not use, connect directly to the 12 Volt power)
IB = IC / hFE = 1 A / 1000 = 0.001 A
RB = (VPORT – VBE) / IB = (4.2 – 1.4) / 0.001 = 2800 Ohm, use 2K7 Ohm resistor
P = (VPORT – VBE) x IB = (4.2 – 1.4) x 0.001 = 0.0028 watt, use 0.25 Watt resistor
D. The Darlington Transistor Array
For more compact version of the Darlington pair transistor you could use the Texas Instrument ULN2803A
which is contain 8 Darlington pair transistors with has build in 2K7 base resistor and clamp diode for
each Darlington pair transistors. This makes this Darlington transistor array suitable for driving the relay
or motor up to 500mA (this is a maximum datasheet value) directly from the microcontroller output.
To increase the output current up to 1 A (2 x 500mA, remember this is a maximum datasheet value, for
practical application use just half or 2 x 250 mA) you could simply use two Darlington transistor array
connected in parallel, the following is the sample circuit for driving two DC motors using the ULN2803A
Darlington transistor array:
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Thanks to the build in internal 2K7 base resistor and the two clamp diode, you don’t need any external
component when using ULN2803A to drive the DC motor from your microcontroller port. The Darlington
transistor array ULN2803A could be used to drive up to 50 volt voltage load.
E. Isolating your Circuit
Sometimes we need to isolate our microcontroller circuit from the interface circuit especially in the
environment that generating a lot of noise which could disturb our microcontroller operation. When we
use a relay from the above example, the driver ground is still directly connected to the microcontroller
circuit, so there is a change the noises will interfere the microcontroller circuit.
To completely isolate the circuit we could use the optocouplers (also called optoissolator) circuit, this
circuit will completely isolate your microcontroller from the interface circuit:
The popular optocouplers circuit available on the market is 4N35 which has the hFE of 500 (in the
optocouplers terminology this is also known as the transistor static forward current transfer ratio, Texas
Instrument SOES021C, measured with infrared LED current = 0) and maximum collector current of
100mA.
Differ from the ordinary transistor in the optocouplers we don’t use the transistor base terminal for
driving the collector current; instead we use the internal infrared LED to transfer the infrared LED light
intensity to the phototransitor; based on this infrared LED light intensity the phototransistor will be
turned ON or OFF; giving more current to drive this infrared LED will effect more current to flow on the
phototransistor collector; This effect is known as the current transfer ratio (CTR). The 100% CTR means
that all the current flow on the infrared LED will be transferred 100% to the phototransistor collector.
Therefore by driving the internal infrared LED with 15 mA (in the optocouplers terminology this is also
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known as the input diode static forward current), we could assure that the phototransistor will be in the
saturate state (ON), because the minimum current to make the phototransitor on is about 10 mA. The
following circuit is use optocoupler to interfacing the relay:
By using 5 Volt power supply and relay with 5 Volt and 60mA operating current:
RC = 0 Ohm (not use, connect relay directly to 5 Volt)
Idiode = 15 mA (0.015 A), VLED = 2 Volt
RB = (VPORT – VLED) / IB = (4.2 – 2) / 0.015 = 146.66 Ohm, use 150 Ohm resistor
P = (VPORT – VLED) x IB = (4.2 – 2) x 0.015 = 0.033 watt, use 0.25 Watt resistor
If you need to drive more current you could use the Darlington pair circuit above or you could use the
high gain Darlington optocopuler such as 4N45 (CTR minimum about 350 %).
F. Controlling your DC motor direction
Using just one transistor to control the DC motor as the above example; we only can turn the DC motor
in one direction if we want to change the direction than we also have to change the DC motor voltage
polarity. The other way to work around this condition is to use the relay to switch the DC motor’s voltage
polarity, but using this method means the DC motor will always ON and we can not control the DC motor
speed using digital signal or known as the PWM (Pulse Width Modulation).
The best and popular way to solve this issue is to use the H-bridge circuit:
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When we apply current (IB1) to the TR1 and TR2 transistors, IB2=0 to the TR3 and TR4 transistors, then
TR1 and TR4 transistors will be turned ON, TR2 and TR3 will be turned OFF; this will cause the current to
start flow through TR1 transistor, passing the DC motor and going into the TR4 transistor (blue color).
When we apply current (IB2) to the TR3 and TR4 transistors, IB1=0 to the TR1 and TR2 transistors, then
the TR3 and TR2 transistors will be ON while TR1 and TR4 transistors will be turned OFF; this will cause
the current to flow through TR3, passing the DC motor in reverse polarity and going into the TR2
transistor (red color). By not applying current to both IB1 and IB2 all the transistors will be turned OFF.
Again by applying the Ohm’s law we could easily calculate the RB1 and RB2 on this following circuit
(Updated! Thanks for the nice discussion and correction from the All About Circuits Forum discussion
here, in order for this circuit to work you have to put a resistor on each of the TIP 120 Darlington
transistors base terminal):
The above H-Bridge circuit use 5 Volt supply and DC motor with 5 Volt and 1 A maximum operating
current rating; assuming the TIP120 Darlington transistor hFE is 1000, the RB1 and RB2 resistors could
be calculated as follow:
IB = IC / hFE = 1 A / 1000 = 0.001 A, for each of the transistor base current
RB1a,b = (VPORT – VBE) / IB = (4.2 – 1.4) / 0.001 = 2800 Ohm, use 2K2 Ohm resistor
RB2a,b = (VPORT – VBE) / IB = (4.2 – 1.4) / 0.001 = 2800 Ohm, use 2K2 Ohm resistor
P = (VPORT – VBE) x IB = (4.2 – 1.4) x 0.001 = 0.0028 watt, use 0.25 Watt resistor for RB1 and RB2
To test the TIP 120 Darlington transistors H-Bridge circuit above I used this following circuit using Atmel
AVR ATTiny13 microcontroller as shown on this following picture:
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Bellow is the C Program that I used to test this circuit:
//***************************************************************************
// File Name : trhbridge.c
// Version : 1.0
// Description. : Transistor as Switch: Simple All TIP120 H-Bridge
// Author : RWB
// Target : Atmel AVR ATTiny13 Microcontroller
// Compiler : AVR-GCC 4.3.2; avr-libc 1.6.2 (WinAVR 20090313)
// IDE : Atmel AVR Studio 4.17
// Programmer : Atmel AVRISPmkII
// Last Updated : 18 June 2010
//***************************************************************************
#include <avr/io.h>
#include <util/delay.h>
int main(void)
{
// Initial I/O
DDRB |= (1<<PB3)|(1<<PB4); // Set PB3,PB4 as Output, Others as Input
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PORTB &= ~(1<<PB3); // Reset PB3 (OFF)
PORTB &= ~(1<<PB3); // Reset PB4 (OFF)
for(;;) { // Loop Forever
PORTB |= (1<<PB3); // Turn ON PB3
_delay_ms(3000); // Delay 3 Second
PORTB &= ~(1<<PB3); // Turn OFF PB3
_delay_ms(2000); // Delay 2 Second
PORTB |= (1<<PB4); // Turn ON PB4
_delay_ms(3000); // Delay 3 Second
PORTB &= ~(1<<PB4); // Turn OFF PB4
_delay_ms(2000); // Delay 2 Second
}
return 0; // Standard Return Code
}
/* EOF: trhbridge.c */
One of the advantage using all NPN transistors in the H bridge circuit is the NPN transistor tends to have
faster turn on time comparing to the PNP transistor, beside by using the same transistor type we could
have similar transistor characteristic in the circuit. You could read more example of using all NPN
transistor H-Bridge in “H-Bridge Microchip PIC Microcontroller PWM Motor Controller” on this blog.
Actually most of the modern H-Bridge circuit design for higher voltage (e.g. more than 9 volt) is rarely
use the BJT anymore; instead we use the MOSFET because MOSFET is more efficient on higher voltage
(i.e. less power dissipation) compare to the ordinary BJT. The other advantage of using MOSFET is that it
has very high input impedance, therefore we could easily connect parallel a couple of the same MOSFET
to achieve the higher current output and at the same time we could decrease the output resistance of
the MOSFET (Rds), which mean we could get more lower power dissipation as shown on this following
picture:
The “The Line Follower Robot with Texas Instruments 16-Bit MSP430G2231 Microcontroller” article is a
good example of how we use the N-Channel MOSFET to control the DC motor.
Driving the Stepper Motor
One type of the brushless electric motor that is designed specifically for digital signal input is called the
stepper motor. The stepper motor usually is used when we need to control the precise rotation movement
and speed with the open loop control. These advantages make the stepper motor is widely found in many
applications such as printers, scanners, disk drives, automotives, CNC machines, toys, and many more.
Today the most common used stepper motor types are Unipolar Stepper Motor and Bipolar Stepper Motor.
The unipolar stepper motor usually has two windings with a center tap on each of windings, therefore the
current could move from the center tap either to the left winding or to the right winding. Usually the
unipolar stepper motor comes with 5 or 6 terminal leads. On the other hands the bipolar stepper motor
actually is similar to the unipolar type but without the center tap.
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Therefore the unipolar stepper motor has advantage of more simple driving circuit over the bipolar
stepper motor but has a torque less than the bipolar motor for the same size. The following circuit uses
four BC639 transistors to drive the unipollar stepper using the Atmel AVR ATTiny13 microcontroller to
provide the required stepping signal to the unipolar stepper motor:
From the schematic above you could see that each transistor is connected to half windings of the
unipolar stepper motor phase. You need to experiment with your own stepper motor to get the right
windings connection. Assuming maximum 100 mA unipolar stepper motor current on 5 volt supply, and
using minimum BC 639 transistor hFE of 40, we could calculate the RB (the base resistors) as follow:
IB = IC / hFE = 100 mA / 40 = 0.0025 A
RB = (VPORT – VBE) / IB = (4.2 – 0.7) / 0.0025 = 1400 Ohm, use 1K5 Ohm resistor
P = (VPORT – VBE) x IB = (4.2 – 0.7) x 0.0025 = 0.0086 watt, use 0.25 watt resistor
Two clamp diodes on each transistor are required because the winding has a center tap. Therefore when
one end of winding is high (Vcc) the other end is low (GND) the lower diode will bypass the back EMF
(Electromotive Force) voltage that appear on BC 639 transistor collector and emitter terminals.
This following is the C code is used for testing the circuit above:
//***************************************************************************
// File Name : upstepper.c
// Version : 1.0
// Description. : Transistor as Switch: Simple Unipolar Stepper
// Motor Driver - Full Step Method
// Author : RWB
// Target : ATTiny13
Using Transistor as a Switch | ermicroblog http://www.ermicro.com/blog/?p=423
16 of 25 3/29/2012 6:17 PM
// Compiler : AVR-GCC 4.3.2; avr-libc 1.6.2 (WinAVR 20090313)
// IDE : Atmel AVR Studio 4.17
// Programmer : Atmel AVRISPmkII
// Last Updated : 03 Nov 2010
//***************************************************************************
#include <avr/io.h>
#include <util/delay.h>
#include <avr/interrupt.h>
// Unipolar Stepper Motor CW/CCW Stepping Sequence
#define MAX_STEP 4
unsigned char cwstep_seq[MAX_STEP]= {0b00000110,
0b00000011,
0b00001001,
0b00001100};
unsigned char ccwstep_seq[MAX_STEP]= {0b00001100,
0b00001001,
0b00000011,
0b00000110};
volatile unsigned char step_index;
volatile unsigned int ovftimes;
volatile unsigned char status;
ISR(TIM0_OVF_vect)
{
static unsigned int count=1;
count++;
if (count >= ovftimes) {
cli(); // Disable Interrupt
// Stepping Output
if (status)
PORTB = ccwstep_seq[step_index++];
else
PORTB = cwstep_seq[step_index++];
if (step_index >= MAX_STEP)
step_index=0;
count=0; // Reset Count
TCNT0=0; // Start counter from 0
sei(); // Enable Interrupt
}
}
int main(void)
{
// Initial I/O
DDRB = 0b00001111; // Set PB0, PB1, PB2, and PB3 as Output, Others as Input
PORTB = 0b00000000; // Reset PORTB Output
// Set ADCSRA Register on ATTiny13
ADCSRA = (1<<ADEN) | (1<<ADPS2) | (1<<ADPS1);
ADCSRB = 0b00000000;
// Set ADMUX to PB4 (ADC2)
ADMUX=0b00000010;
// Disable Digital Input on PB4 (ADC2)
DIDR0=0b00001000;
// Initial TIMER0
TCCR0A=0b00000000; // Timer/Counter 0 Normal Operation
TCCR0B=(1<<CS01); // Use prescaller: Clk/8 with 9.6 MHz Internal Clock
TCNT0=0; // Start counter from 0
TIMSK0=(1<<TOIE0); // Enable Counter Overflow Interrupt
step_index=0;
ovftimes=10;
status=0; // 0 - CW, 1 - CCW
sei(); // Enable Interrupt
for(;;) { // Loop Forever
// Start conversion by setting ADSC on ADCSRA Register
ADCSRA |= (1<<ADSC);
// wait until convertion complete ADSC=0 -> Complete
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while (ADCSRA & (1<<ADSC));
// Get the ADC Result
ovftimes = ADCW;
if (ovftimes > 800)
status^=0x01; // Toggle the Direction
_delay_ms(50);
}
return 0; // Standard Return Code
}
/* EOF: upstepper.c */
The method to rotate this unipolar stepper motor rotor is known as the full step mode method; in full
step mode we always excite two windings at the same time, with the right current sequence we could
rotate the stepper motor in 4 repeatable steps. Reversing the step sequences will make the stepper
motor to turn into opposite direction. In this tutorial I used NMB-PM20S-020 permanent magnet motor
where the step required is shown on this following diagram:
You could easily adapt the step sequence to your own unipolar stepper motor by changing both the
cwstep_seq (clockwise rotation) and ccwstep_seq (counter clockwise rotation) array variables data in
the program.
Using the AVR ATTiny13 microcontroller TIMER0 interrupt we could easily supply the required output
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steps to the stepper motor. I used the ADC (Analog to Digital Conversion) to control the stepper motor
step sequence delay as well as to change the rotation direction by adjusting the 10K trimport. For more
information about AVR ADC and TIMER0 you could read these blog’s articles “Analog to Digital Converter
AVR C Programming” and “Working with AVR microcontroller Communication Port Project“.
You could also replace the BC639 transistor with the Darlington pair transistor array such as ULN2803A
from Texas Instrument mention above. Using this Darlington pair transistor array make the unipolar
stepper circuit become simpler (less components) because it has the required clamp diode on each
Darlington transistor pair and you could take advantage of higher current gainer provided by the
Darlington pair transistors.
To drive the bipolar stepper motor; each of the two windings will require the H-Bridge circuit similar to
the H-Bridge circuit for driving the DC motor mention above. Therefore we need at least 8 transistors to
drive the bipolar stepper motor (4 transistors on each windings). By forwarding and reversing the current
flow on each winding we could achieve the required steps sequence to drive the bipolar motor.
And by supplying the correct steps sequence logic to IN1, IN2, IN3, IN4, IN5, IN6, IN7, and IN8
input from the microcontroller output port we could make this bipolar stepper motor to rotate. The
opposite direction (counter clockwise) rotation could be achieved by reversing the steps sequence (i.e. 4,
3, 2, and 1).
Using Transistor as switch Testing Circuit Video
1. This following video show you of how to drive a transistor which connected with 5 red LED using the
Atmel AVR ATTiny25 microcontroller.
Using Transistor as a Switch | ermicroblog http://www.ermicro.com/blog/?p=423
19 of 25 3/29/2012 6:17 PM
2. The TIP120 H-Bridge Testing Circuit video using Atmel ATTiny13 Microcontroller:
3. The Unipolar Stepper Motor Testing Circuit video using Atmel ATTiny13 Microcontroller:
Using Transistor as a Switch | ermicroblog http://www.ermicro.com/blog/?p=423
20 of 25 3/29/2012 6:17 PM
02.09.09 #1
02.09.09 #2
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16 Responses to “Using Transistor as a Switch”
Comment by slowjoe.
I’ve had a go at making the final H-bridge circuit shown here
using TIP120 darlington pairs and had a bit of trouble. If I
split the connections from the microcontroller after the
resistors RB1 and RB2 (as shown in the circuit diagram) then
it doesn’t seem to work, however if I split the signal before
the resistors and use 2 resistors for each of RB1 and RB2 then
it does work. I’m not sure I understand why yet maybe
somebody can explain.
Comment by rwb.
The original TIP120 H-Bridge schematic has been changed;
now I used resistor on each of the TIP120 base terminal as
Stepper motor parts
Replacement Motors For HMC/VMC/CNC Machines, Express Deliverywww.GlobalMachineParts.com/Motors
Using Transistor as a Switch | ermicroblog http://www.ermicro.com/blog/?p=423
21 of 25 3/29/2012 6:17 PM
23.01.10 #3
20.07.10 #4
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21.07.10 #6
you did. Thank you
Comment by mandomoose.
Thankyou for this great post. I was wondering about using
the transistor as a switch with my avr.
Keep having fun
Comment by kansairobot.
Thanks for the great blog
I am sorry if this sounds like a total newbie question but in
your pics (in which you dont use a optocoupler) you connect
both the 12V circuit and the micro ground to a common
ground.
My question is how to implement the 12V and 5V part? I
mean let’s say I am using common batteries (8 batteries=
12V) I connect the motor to the 12V part but where do I
connect the 5V pin of the micro to??
I dont know if my question make sense sorry
Kansai
Comment by rwb.
You should have two separate DC power sources (e.g. 3 AA
batteries for 4.5 volt and 8 AA batteries for 12 volt); the first
one is the 5 volt or 4.5 volt which is used to power the
microcontroller circuit and the second one is the 12 volt which
is used to power the Darlington transistor and the DC motor.
In order to make the darlington transistor work (ON) we have
to provide adequate voltage between the base and the
emitter terminal; this voltage is provided by the
microcontroller I/O port (powered by 5 Volt source), that is
why we have to connect these two voltage sources on the
same common ground.
Comment by kansairobot.
thank you very much for your reply.
For the H-bridge part i was thinking of using 2N2222′s (since
my motor only needs around 280mA).
or use Toshiba TA7291S bridge circuit.
These circuits are made of transistors it seems but how can
you see if they generate enough current C-E (as we did for
transistors in this tutorial)? I cant seem to understand their
datasheet.
Sorry for all the questions but I am learning a lot with your
tutorials. Thanks always for these resources…
kansai
Using Transistor as a Switch | ermicroblog http://www.ermicro.com/blog/?p=423
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08.05.11 #10
Comment by rwb.
When you choose the transistor (e.g. 2N2222A); from
datasheet Ic max = 800 mA, remember this is the maximum
value, usually in real application we only use just half of its
maximum capacity which is 400 mA. When you measure the
DC motor make sure you also take into the consideration the
DC motor stall current (i.e. motor on heavy load, where its
almost stop) not reach the 400mA limit; unless you only use
the DC motor as a free running DC motor (without or have a
very small load).
When using the Toshiba TA7291S bridge all you need is to
supply the correct standard logic voltage to the IN1 and IN2
input pins; the integrated circuit inside TA7291S will make
sure you get the saturate transistors condition on its output
(average 0.9 volt). The input current on the IN1 and IN2 pins
is very low (about 3 to 10 uA with Vin = 3.5 volt).
Comment by ajoyz124.
Very illustrative, basic and simple facts required to work with
micro controls.
Thanks.
Comment by topx666.
Lhank you Mr. Besinga for creating this tutorial. please let me
introduce my self. my name Taufiq Sunar. I am a student of
Electronics and Instrumentation in Gadjah Mada University,
Indonesia. Currently I’m trying to make a switch using
MOSFET with input from the AVR microcontroller PWM. I’m
planning to use MOSFET series IRF740, IRF9530, IRF9540, or
IRFZ44. Whether working principle and the calculation of
MOSFET is similar to BJT transistor? Thank you very much for
your answer!
Comment by rwb.
Nice to know you Taufiq. Bipolar Junction Transistor (BJT) is
different compared to Metal Oxide Semiconductor Field Effect
Transistor (MOSFET), therefore all the calculation for BJT
could not be applied to MOSFET.
BJT will amplify the input current (IB) by the current gain
factor (hFE) on it’s output (IC) as follow:
hFE = IC / IB
On the other hand MOSFET will amplify the input voltage
(VGS) by the transconductance gain factor (gfs) on it’s output
(IDS) as follow:
gfs = (change in IDS / change in VGS)
Where VDS: gate to source voltage and IDS: drain to source
current.
As you notice there is no IG (gate current) on the MOSFET
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gain factor formula; because of the issulation oxide on the
MOSFET gate terminal design, therefore there is no current
flow on the MOSFET Gate terminal (MOSFET is also known as
a very high input impedance transistor).
The IDS (drain to source current) will start to increase when
the VGS voltage is above the VGS threshold voltage. Any small
VGS voltage change applied to the MOSFET Gate input
terminal will be amplified by the factor of gfs and reflected as
a Drain current on the MOSFET Drain terminal side.
Therefore by supplying VGS much greater than VGS threshold
voltage, we could easily push the MOSFET into its saturate
region. This make using the MOSFET is more simple
compared to BJT as all you need is to connect the MOSFET
gate terminal directly to the AVR microcontroller output port.
The AVR microcontroller output port voltage (high logic) could
easily drive the MOSFET into its saturate region, but make
sure you always read the MOSFET datasheet especially the
VDS, IDS, gfs, and VGS threshold when using different type of
MOSFET.
I hope this answer will clarify the differences between BJT
and MOSFET
Comment by topx666.
Thank you very much for the explanation Mr. Besinga.
Actually, I want to make a high-speed switch from AVR
microcontroller PWM. PWM output is squarewave. Whether
after connected into the MOSFET, its output also a
squarewave? Because I’ve tried using a IRF740 MOSFET, and
8-bit fast pwm. But the results was not squarewave and its
currents and voltages are very small. The PWM input was
connected by 4k7 resistor to its gate. And its source supplied
by 12V with 10k resistor.
Comment by rwb.
Yes the output should be a square wave when the input is a
square wave. You could read more about AVR Fast PWM on
“Working with Atmel AVR Microcontroller Basic Pulse Width
Modulation (PWM) Peripheral” article.
Actually you don’t need to use 4K7 resistor (current limiter)
in series with the MOSFET Gate terminal, because there is no
“Gate Current”, therefore you don’t have to “reduce” the
current as it in BJT. If you want to use resistor use the voltage
divider circuit (2 resistors) instead of one resistor. As long as
the VGS > VGS threshold and VGS > VDS, the MOSFET will
turn ON.
Comment by drogge.
I’ve been trying to make a H-Bridge circuit using 4 TIP120′s
as per the above diagram and everything works fine if the
motor voltage is around 5V. However if I raise the motor
voltage to 9 volts I only see about 3.5 volts at the motor. The
problem seems to be that the TIP120 can switch the motor to
ground but it can’t switch the full 9 volts when it’s base is 5
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volts. In other words the circuit labled Using TIP120
Darlington Transistor for Driving Motor works but if I move the
transistor between V and the motor it doesn’t work. It looks
like I need to use something like TIP125′s for the top
transistors in the H-Bridge.
Comment by rwb.
Yes, the all TIP120 H-Bridge circuit only work when the base
and collector has the same power supply voltage. Because the
top TIP120 (NPN) transistor is in common collector
configuration and in order to turn ON the transistor, we need
to forward bias the VBE, therefore if the base voltage is less
then the collector voltage it will not turn ON. When you use
the TIP125 (PNP) for the top transistor then all the transistors
will have the same common emitter configuration.
I will do some correction and add example on the article
above, thanks for your comment.
Comment by akhb.
Hi there, This is a very informative post. Your hard work is
very appreciated by everyone who luckily lands here.
I receive an active low error signal for a chip. The signal and
Vcc on board is +5V. I want to light a blue LED if the signal is
low. So I guess we use PNP transistor. I was looking at digital
transistors. (they have biased resistors pre-built-in). Can you
suggest something like
PDTA143E where the resistors and their ratio is all correct for
+5V power supply?
Thank you in advance.
Comment by rwb.
You could use any PNP potential divider type of digital
transistor as long as you choose a suitable Ic max for your
need.
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