Datasheet

TC7660
DS21465C-page 6  2002-2011 Microchip Technology Inc.
4.0 DETAILED DESCRIPTION
4.1 Theory of Operation
The TC7660 charge pump converter inverts the voltage
applied to the V
+
pin. The conversion consists of a two-
phase operation (Figure 4-1). During the first phase,
switches S
2
and S
4
are open and switches S
1
and S
3
are closed. C
1
charges to the voltage applied to the V
+
pin, with the load current being supplied from C
2
. Dur-
ing the second phase, switches S
2
and S
4
are closed
and switches S
1
and S
3
are open. Charge is trans-
ferred from C
1
to C
2
, with the load current being
supplied from C
1
.
FIGURE 4-1: Ideal Switched Capacitor
Inverter.
In this manner, the TC7660 performs a voltage inver-
sion, but does not provide regulation. The average out-
put voltage will drop in a linear manner with respect to
load current. The equivalent circuit of the charge pump
inverter can be modeled as an ideal voltage source in
series with a resistor, as shown in Figure 4-2.
FIGURE 4-2: Switched Capacitor Inverter
Equivalent Circuit Model.
The value of the series resistor (R
OUT
) is a function of
the switching frequency, capacitance and equivalent
series resistance (ESR) of C
1
and C
2
and the on-resis-
tance of switches S
1
, S
2
, S
3
and S
4
. A close
approximation for R
OUT
is given in the following
equation:
EQUATION
4.2 Switched Capacitor Inverter
Power Losses
The overall power loss of a switched capacitor inverter
is affected by four factors:
1. Losses from power consumed by the internal
oscillator, switch drive, etc. These losses will
vary with input voltage, temperature and
oscillator frequency.
2. Conduction losses in the non-ideal switches.
3. Losses due to the non-ideal nature of the
external capacitors.
4. Losses that occur during charge transfer from
C
1
to C
2
when a voltage difference between the
capacitors exists.
Figure 4-3 depicts the non-ideal elements associated
with the switched capacitor inverter power loss.
FIGURE 4-3: Non-Ideal Switched
Capacitor Inverter.
The power loss is calculated using the following
equation:
EQUATION
V
+
GND
S
3
S
1
S
2
S
4
C
2
V
OUT
= -V
IN
C
1
+
+
-
+
R
OUT
V
OUT
V
+
R
OUT
1
f
PUMP
C1
----------------------------- 8R
SW
4ESR
C1
ESR
C2
++ +=
R
SW
on-resistance of the switches=
ESR
C1
equivalent series resistance of C
1
=
ESR
C2
equivalent series resistance of C
2
=
f
PUMP
f
OSC
2
-----------=
Where:
LOAD
C
1
C
2
R
SW
S
1
I
DD
ESR
C1
V
+
+
-
R
SW
S
2
R
SW
S
3
R
SW
S
4
ESR
C2
I
OUT
++
P
LOSS
I
OUT
2
R
OUT
 I
DD
V
+
+=