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參數(shù)資料
| 型號(hào)： | TCM680EOA |
| 廠商： | TelCom Semiconductor, Inc. |
| 英文描述： | +5V TO 【10V VOLTAGE CONVERTER |
| 中文描述： | 5V至【10V的電壓轉(zhuǎn)換器 |
| 文件頁(yè)數(shù)： | 4/7頁(yè) |
| 文件大小： | 96K |
| 代理商： | TCM680EOA |
4-16
TELCOM SEMICONDUCTOR, INC.
TCM680
+5V TO
±
10V VOLTAGE CONVERTER
EFFICIENCY CONSIDERATIONS
Theoretically a charge pump can approach 100% effi-
ciency under the following conditions:
The charge Pump switches have virtually no offset
and extremely low on resistance
Minimal power is consumed by the drive circuitry
The impedances of the reservoir and pump capaci-
tors are negligible
For the TCM680, efficiency is as shown below:
Efficiency V
+
= V
DD
/(2V
IN
)
V
DD
= 2V
IN
– V
+
V
+
DROP
DROP
= (I
+
OUT
)(R
+
OUT
)
Efficiency V
–
= V
SS
/(– 2V
IN
)
V
SS
= 2V
IN
– V
–
V
–
DROP
DROP
= (I
–
OUT
)(R
–
OUT
)
Power Loss
= (V
+
DROP
)(I
+
OUT
) + (V
–
DROP
)(I
–
OUT
)
There will be a substantial voltage difference between
OUT
– V
IN
) and V
IN
for the positive pump and between
OUT
and V
–UT
if the impedances of the pump capacitors C
1
and C
2
are high with respect to the output loads.
Larger values of reservoir capacitors C
3
and C
4
will
reduce output ripple. Larger values of both pump and
reservoir capacitors improve the efficiency. See "Capacitor
Selection" in Applications Section.
(V
+
V
+
APPLICATIONS
Positive and negative Converter
The most common application of the TCM680 is as a
dual charge pump voltage converter which provides positive
and negative outputs of two times a positive input voltage.
The simple circuit of Figure 6 performs this same function
using the TCM680 and external capacitors, C
1
, C
2
, C
3
and C
4.
Figure 6. Positive and Negative Converter
C
1
C
1
C
2
C
3
22
μ
F
C
2
V
IN
V
IN
V
OUT
V
OUT
V
OUT
GND
GND
TCM680
22
μ
F
22
μ
F
C
4
22
μ
F
8
7
6
5
4
3
2
1
V
OUT
C
2
C
1
+
–
+
–
+
Capacitor Selection
The TCM680 requires only 4 external capacitors for
operation. These can be inexpensive polarized aluminum
electrolytic types. For the circuit in Figure 6 the output
characteristics are largely determined by the external
capacitors. An expression for R
OUT
can be derived as shown
below:
R
+
OUT
= 4(R
SW1
+ R
SW2
+ ESR
C1
+ R
SW3
+ R
SW4
+ ESR
C2
)
+4(R
SW1
+ R
SW2
+ ESR
C1
+ R
SW3
+ R
SW4
+ ESR
C2
)
+1/(f
PUMP
x C1) + 1/(f
PUMP
x C2) + ESR
C4
R
–
OUT
=
4(R
SW1
+ R
SW2
+ ESR
C1
+ R
SW3
+ R
SW4
+ ESR
C2
)
+4(R
SW1
+ R
SW2
+ ESR
C1
+ R
SW3
+ R
SW4
+ ESR
C2
)
+1/(f
PUMP
x C1) + 1/(f
PUMP
x C2) + ESR
C3
Assuming all switch resistances are approximately
equal...
R
+
OUT
= 32R
SW
+ 8ESR
C1
+ 8ESR
C2
+ ESR
C4
+1/(f
PUMP
x C1) + 1/(f
PUMP
x C2)
R
–
OUT
= 32R
SW
+ 8ESR
C1
+ 8ESR
C2
+ ESR
C3
+1/(f
PUMP
x C1) + 1/(f
PUMP
x C2)
R
OUT
is typically 140
at +25
°
C with V
IN
= +5V and C1
and C2 as 4.7
μ
F low ESR capacitors. The fixed term
(32R
SW
) is about 130
. It can be seen easily that increasing
or decreasing values of C1 and C2 will affect efficiency by
changing R
OUT
. However, be careful about ESR. This term
can quickly become dominant with large electrolytic capaci-
tors. Table 1 shows R
OUT
for various values of C1 and C2
(assume 0.5
ESR). C1 and C4 must be rated at 6VDC or
greater while C2 and C3 must be rated at 12VDC or greater.
Output voltage ripple is affected by C3 and C4. Typically
the larger the value of C3 and C4 the less the ripple for a
given load current. The formula for V
RIPPLE(p-p)
is given
below:
V
+RIPPLE(p-p)
= {1/[2(f
PUMP
/3) x C4] + 2(ESR
C4
)}(I
+
V
–RIPPLE(p-p)
= {1/[2(f
PUMP
/3) x C3] + 2(ESR
C3
)}(I
–
OUT
)
OUT
)
For a 10
μ
F (0.5
ESR) capacitor for C3, C4,
f
PUMP
= 21kHz and I
OUT
= 10mA the peak-to-peak ripple
voltage at the output will be less than 100mV. In most
applications (I
OUT
< = 10mA) 10-20
μ
F output capacitors and
1-5
μ
F pump capacitors will suffice. Table 2 shows V
RIPPLE
for different values of C3 and C4 (assume 1
ESR).
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