L6223 STMicroelectronics, L6223 Datasheet - Page 19

• Manufacturer Part Number: L6223
• Description: DMOS PROGRAMMABLE HIGH SPEED UNIPOLAR STEPPER MOTOR DRIVER
• Manufacturer: STMicroelectronics

2) Back EMF (BEMF) equal to 80% of its peak
during the phase change and equal to 50% of its
peak during the chopping period.
3) Constant slope of the current during t
and for power calculation during the phase
change (See t
4) Current imbalance supposed to be zero.
5) Current ripple during the chopping neglegible.
As was previously stated, the current chopping is
obtained by means of one PWM Loop that con-
trols the charge time t
windings, A and B for example in fig.22.
This time starts each clock pulse and stops when
Q5 is switched OFF because of the condition:
V
A factor 2 is required because the single sensing
resistor R
ing through each of the two energized windings
(A of MA; B of MB).
This configuration can produce an imbalance be-
tween the two peak currents because at the
phase change the BEMF of one winding (MA) can
be out of phase with respect to the BEMF of the
other one (MB); in addition, an imbalance may
also occur at the phase change when the Power
Supply Voltage selected is too low and/or when
one motor is driven with too large Lm/Rm ratio.
Nevertheless in most of the applications the dissi-
pated power is not increased and there is no sig-
nificant change in torque.
During t
phase A (seg. Fig. 24), is defined by V
V
where R
in which R
phase A and R
Q1 and Q5: R
Figure 24.
At the end of t
and jumps to I
ductance becomes four times L
pling) that is the inductance of the phase A alone.
The recirculation time t
V
since R
The current through Q1 is shown in Fig. 26: the
current ripple is on lp and I
respectively. It can be obtained the Duty Cycle:
DC = V
since 2V
The slow decay allows a small current ripple as
earlier It is considered equal to zero. The current
through the phases A and B can be seen in Fig.
27 where the InA and InB signals (see Fig. 22)
are shown as well.
These two signals are 90 out of phase with each
other and they are 180 out of phase with the cor-
responding inputs of the IC. In A and In B are not
ref
ON
OFF
= 2 R
V
2BEMF + I
OFF
DSONQ1
S
tot
ON
ON
S
S
- R
I
= R
p
/ (2V
is crossed by the peak current I
t
m
.
ON
the current Ip, flowing through the
tot
1
ON
S
m
p
is the winding resistance of the
Ip - BEMF
DSON tot
= R
in Fig. 27).
/2 (see Fig. 25) since the total in-
= V
ON
+ R
and BEMF are not shown on the
, the current starts its slow decay
P
DSONOQ2
OFF
+ V
m
(R
+ R
m
ON
OFF
OFF
t
is the sum of the R
OFF
+ R
DSON tot
of the inductance of the
)
is defined by:
P
.
DSONQ1
/2 during t
m
)
(perfect cou-
ON
ON
and t
ON
DSON
p
, t
flow-
OFF
OFF
of
shown in the Figure.
During the time Tp the motor goes through four
steps and the rotation speed V
be given by:
V
By considering what was stated above, the follow-
ing can be applied:
1) Dissipated power by the 4 sink power DMOS
2) Dissipated power by Q5 (PdH).
The chopping produces little power dissipation.
It’s value can be approximated by:
3) Pdch
The sum of 1) + 2) + 3) gives the dissipated
power of the output stage. To obtain the total
amount of dissipated power it’s necessary to in-
clude the power dissipation produced by the qui-
escent currents I
(from the Logical circuits):
P
considering I
P
Example
Supply Voltage
Logic Voltage
Peak current (per phase)I
Motorresistance
Motorinductance
Rotation speed
Peak of the BEMF
Max ambient temperature T
Max junction temperature T
From the Electrical Characteristics of the L6223
(Typical value):
Internal Reference Voltage V
Sink DMOS R
Source DMOS R
Power Supply Current
Logic Supply Current
From Fig. 3 (see pag. 6) the following is obtained:
The DMOS ON-Resistances become (worst case):
where the phase change duration is:
rot
do
tot
T
PdL
PdH 4R
(Q1 to Q4).
1
1.65 at Tj = 125 C.
= PdL + PdH + Pdch + Pdo
= 4/T
= V
S
R
p
I
L
4 R
S
.
8 10
m
tot
+ V
S
DSONQ5
DSONQ1
log
DSON
constat versus V
T
SS
-3
P
e
DSON
S
V
I
SS
(from the power stage) and I
1
s
I
,
I
I
2
p
p
2
p
V
V
R
L
V
BEMF = 1 Vp
R
R
V
DC
m
rot
S
SS
m
DSON
DSON
s
I
I
T
3
S
SS
p
= 6mH
= 36V
= 9
1
= 500 step/sec (const)
= 5V
= 0.7A
= 4 mA
1.6 BEMF R
= 20 mA
amb
j
ref
T
T
L = 1.2
H = 0.7
= 125 C
2
1
p
S
T
= 0.5V
2
R
. Finally:
rot
= 50 C
p
4
3
tot
(step/sec) can
I
T
at T
p
4DC
1
amb
Worst
1
Tj = 25 C
Case
tot
at
L6223
= 50 C
2
I
DC
P
19/33
SS