Power Electronics	
No.4: Step-down Chopper			Takeshi Furuhashi	Furuhashi_at_cse.nagoya-u.ac.jp		
1Basic idea for high efficiency	
SW ON 	SW

iovSWSW OFF 	SWvSWvSW = 0
PSW = vSW io=0ioio = 0
PSW = vSW io=0

Loss by SW is zero.	

2Output voltage control by switching	
voSW

iovSW
EvE

VE
RL

votTonTSW(a) Switching circuit	 TSW(b) Waveform of output voltage	 

Ton
1 Ton
vo =VE dt =VE∫TSW 0TSW
TSW:	 Switching period	 

3

Transistor → used as a switching device	

iCiCCiB

vCE

B
E

CB

E

CTransistor	 
(left: 2SC2120(NPN-type)right: 2SA950(PNP-type))

BvBECiB

E(a) NPN-type transistor	 

vCEB
vBE

E

(b) PNP-type transistor	 Voltage and current polarities definition	 

4

Operating region of the transistor in an A-class amplifier	
IC [mA]

IC
IB
VCE
VBE

Collector current	 

Common emitter	VCE [V]

Amplifier using an NPN-type transistor	 Collector-Emitter voltage	 5

Operating region of a transistor used as a switch	IC [mA]

IC
IB
VCE
VBE

Collector current	 

Common emitter	

Switch ON	
Switch OFF	

VCE [V]

Amplifier using an NPN-type transistor	 Collector-Emitter voltage	 
6

Drive transistor scheme	
Tr 2SA950
voERB510

RLvoVEtTonTSW

TSWTransistor
driving circuit 	 (a) Switching circuit	 (b) Waveform of output voltage	 

7

PWM (Pulse Width Modulation) control method	If vcom ≥ vtri , then Tr is turned on.
If vcom < vtri , then Tr is turned off.

vcom
vtrit

ON OFFON

OFFON	t8

PWM control method	 PWM (Pulse Width Modulation) control method	

vcom

vtri vcom
Vtpvtritt

ON OFFONOFFON	OFF ONOFFONOFF	tt
9

PWM control method	 Drive transistor scheme (Tr:ON)	
vtri vcom
Vtpt

6V	

Tr	
2SA950	vEBio-iB

RB	 R v ≈ 6 [V]L	 ovPWMTransistordriving circuit	 

vPWM = 0 [V]	tvot
ON OFFONOFFON	

10

The emitter-base circuit of a PNPtype transistor such as 2SA950 canbe modeled by a diode.	 

When vPWM = 0 [V]	

E	-iB
6V	

VEvEB= 0.7 [V]B	RB	

VE − vEB− iB =RB
6 − 0.7=≈ xx [mA]RB

Equivalent circuit between emitter-base circuits	 11

Drive transistor scheme (Tr:OFF)	
vtri vcom
Vtpt

6V	

Tr	
2SA950	vEB
io = 0 [A]
-iB

RB	 R

L	 vo = 0 [V]vPWM
Transistor
driving circuit 	 vPWM = 6 [V]	tvot
ON OFFONOFFON	

12

When vPWM = 6 [V]	

E	

-iB
6V	

vEB= 0 [V]
B	

RB	

VE
vPWM
6 [V]	VE − vBE − vPWM
− iB =RB
6−0−6== 0 [mA]RBiB = 0 [mA]

Equivalent circuit between emitter-base circuits	 13

STEP 3. Circuit construction practice (Step-down chopper)
Construct the step-down chopper shown below. Design the base resistance oftransistor Tr, so that the base current IB < -6 [mA] when Tr is ON.
Tr 2SA950

IB < -6 [mA]	

R2510

RB6V
LED1vo

The forward bias voltage ofLED (LED1) is approximately
1.7[V].	

LED2RS50
R1510
VR12kΩ

PIC12F615
PIC16F615
PIC12F615
A2 A1 A0vcom（A1 A0 = 01）	

vPWM
CS1µF

A1 A0 = 01: Mode for very
slow PWM	
A2 = Voltage command input	
14
P1A = 0.5HzPWM output	

Step 3. Problem 1	 	 
The ratio between on-period Ton
and switching period Tsw is called
the “duty ratio,” δ. Derive thefollowing when the triangularvoltage vtri , command voltage
vcom , and output voltage ofchopper vo are given in the rightfigure. The peak value of thetriangular voltage is Vtp. Theoutput voltage of the chopper is
VE when Tr is ON:	 (a) Relation between δ and vcom.	 
(b) Relation between the averageof output voltage Vo and vcom	 	 vtrivcom
Vtpt

voVEt
Ton
TSW

Ton
δ=TSW

δ: duty ratio	 Ton: on-period of transistor 	 
TSW: switching period	 Step 3. Problem 2
The figure below shows a series circuit of resistor R and reactor L. Assume that switches SW1and SW2 are OFF.
At t = 0, switch SW1 is turned on, at t = L/R, switch SW1 is turned off, and switch SW2 is turned
on. Then, at t = 2L/R, switch SW1 is turned on, and switch SW2 is turned off. At t = 3L/R,switch SW1 is turned off, and switch SW2 is turned on. This switching is repeated (i.e., at t =(2n-1)L/R (n = 1, 2, …), switch SW1 is turned off, and switch SW2 is turned on, and at t = 2nL/R (n = 1, 2, …), switch SW1 is turned on, and switch SW2 is turned off.
Draw the waveform of current i at t ≥ 0.	 

SW1VE
6V

SW2i
R
10ΩvR

L
400µHvLvs