On 1 Jul 2004 17:52:06 -0700,
in Msg. said:
No, the "charge pump" in the '4080 series works like most other chips,
and only charges an external capacitor (through an external diode) on
each cycle of the driver IC. So if it's continuously on, the charge
drains away and then a sad end to the story. You need a part with an
internal oscillator and charge-pump switch. There are a few such parts.
What you describe here is what the HIP408x data sheets call
bootstrapping, and indeed it is the main source of energy for the
high-side gate drivers in normal switching apps. Yet the IC seems to
have some more to offer (HIP4081a app note (page 1, 2nd paragraph):
A combination of bootstrap and charge-pumping techniques
is used to power the circuitry which drives the upper halves
of the H-Bridge. The bootstrap technique supplies the high
instantaneous current needed for turning on the power
devices, while the charge pump provides enough current to
maintain bias voltage on the upper driver sections and
MOSFETs.
From page 4 of the same app note:
There are two charge pump circuits in the HIP4081A, one for
each of the two upper logic and driver circuits. Each charge
pump uses a switched capacitor doubler to provide about
30µA to 50µA of gate load current.
Another passage (p. 5)
Before discussing bootstrap circuit design in detail, it is
worth mentioning that it is possible to operate the HIP4081A
without a bootstrap circuit altogether. Even the bootstrap
capacitor, which functions to supply a reservoir of charge for
rapidly turning on the MOSFETs is optional in some cases.
In situations where very slow turn-on of the MOSFETs is tolerable,
one may consider omitting some or all bootstrap
components. Applications such as driving relays or lamp
loads, where the MOSFETs are switched infrequently and
switching losses are low, may provide opportunities for boot
strapless operation. Generally, loads with lots of resistance
and inductance are possible candidates.
Maybe the 408x will noe become even more favorites of yours?
--Daniel