Technical data

60 APPLICATIONS INFORMATION
10 Most frequently asked questions
about using dc power products
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1) How do I put the power supply in the constant current mode?
The power supply cannot be “put” into the constant current
mode. The output settings of the power supply combined with
the ohmic value of the particular load determine whether or
not the power supply is in constant current.
ie: The power supply inherently resides in the constant voltage
mode. If the output voltage were set to 24 volts and a 6 load
were placed across the output terminals, Ohm’s Law would
require that 4amps would flow (24V/6). This presumes that
the constant current setting of the power supply were set to a
value greater than 4 amps; lets say 5amps. Now, if the 6load
were replaced by a 2 load, Ohm’s Law would suggest that
12 amps (24V/2) would flow. However, the power supply
is set to go into constant current at 5 amps. Therefore, the
actual output voltage would be 10 volts (2 x 5A). The power
supply will now remain in constant current for values of
load = 0 R 4.8. Once the ohmic value of the load
becomes greater than 4.8 (24V/5A), the power supply will
again revert to constant voltage operation at the value of 24 volts.
2) I have 208 vac, 3 φ phase power; can it be used to operate a
product requiring 208 V single phase?
Yes, see below.
3) Why are the required Watts and VA so different?
Watts is a scalar quantity which is frequently used to measure
system efficiency. It is the energy supplied by the utility
company over a given period of time and is commonly referred
to as power. Except for heavy industrial users, the utility
company only bills users for the watts consumed. Watts are
directly convertible into mechanical work or BTUs (British
Thermal Units) of heat. Wasted power is paid for a second time
in terms of additional loading on the user’s air-conditioning
system. Mathematically, it is a scalar quantity resulting from
the vector product of two vector quantities (volts and amps).
It is NOT the simple algebraic product of the rms volts times
rms current.
VA on the other hand IS the scalar quantity resulting from
multiplying the magnitudes (rms) of the vector quantities
(volts and amps). This resulting quantity will never be smaller
than the watts demanded by an instrument. Uninformed users
incorrectly use VA to assess the device’s over-all efficiency and
power demands. VA is most frequently and correctly used by
electricians to determine proper ac mains conductor gage and
circuit breaker sizing.
4) How much cooling do I need for my power supply?
Users frequently rack power supplies into an enclosure
to supply power to some remotely located external load.
Under these conditions, to properly determine the cooling
requirements, the systems integrator needs thermal data from
the manufacturer for the specific enclosure in question. This
data is generally in the form of a curve which relates the rise
of the enclosure’s internal air temperature to the amount of
power (or BTU’s) dissipated within the enclosure.
The difference between the maximum power demanded by
the external load, and the ac power demanded by the power
supply to support the load’s needs, is the power dumped into
the internal air of the enclosure. Using this number and data
for the enclosure, the internal rise can be determined. The
internal rise added to the external ambient temperature will
determine the temperature of the environment for the power
supply. This must be within the ratings of the product or
premature failure will occur.
A valuable conversion factor between Watts and BTU’s is
listed below:
1 BTU/Hr = 0.293 Watt
5) Can Agilent power supplies sink current?
Yes! Sinking, or downprogramming, is the ability of a power
supply to pull current into the positive power terminal. Sinking
is necessary to discharge the power supply’s own output
capacitor, or the capacitors that are part of an external load.
Sinking is particularly important, for example, in printed
circuit board test systems. The relays in test board systems
typically must be switched only when the power supplies have
discharged to zero volts, to avoid arcing and burn-out of the
relay contacts. Sinking allows the power supply outputs to go
to zero quickly, thus providing faster test times, an important
factor for reducing overall test cost.
The value of the sink current is fixed and is not programma-
ble, with the exception of the 6630 series, where sink current is
set to the same value that is programmed for source current.
In general, sinking is provided to improve a power supply’s
transition time from a higher to a lower constant voltage
operating level, and is not intended to be a steady-state
operating condition.
Series Current Sinking Capability
6620 Multiple Output 110% of source current rating
6620 Precision Output 110% of source current rating
6630 100 Watt 110% of source current rating
6030 Autorangers 50 W/actual output voltage in volts
or actual output voltage volts/0.05
ohms, whichever is less
6640 200 Watt 25% of source current rating
6650 500 Watt 20% of source current rating
6670 2000 Watt 50 W/actual output voltage in volts
or actual output voltage in
volts/0.05 ohms, whichever is less
6680 5000 Watt 50 W/actual output voltage in volts
or actual output voltage in
volts/0.05 ohms, whichever is less
6) I want to put a microswitch on the safety cover over my UUT so
that lifting the cover will program my ATE power supplies to
zero volts and protect the operator from harm. Do HP power
supplies have this capability?
Yes, all of the GPIB programmable supplies in the 6030, 6640,
6650, 6670 and 6680 series have this capability built-in at no
extra cost. It’s called “Remote Inhibit” (RI). RI is available as
an option at extra cost on the 6620 and 6630 series. A contact
closure or TTL low signal programs the output of the supply
to zero volts. The power supply can also be programmed to
generate a service request (SRQ) via the GPIB in the event
that RI is pulled low.
Note: Connections are made
from phase to phase.
Instrument requires an
ac input voltage in the
window 191-250 Vac
Earth
(Safety Ground)
NL
120 Volts (Typ) 208 Volts (Typ)
Rear of Instrument
C Phase
A Phase
B Phase
Assume V (nominal) = 208 Vac
208V @ 8% low = 191V
208V @ 12% high = 233V
(Instrument operates between 191 & 250 Vac)
instrument will not operate on a 120 Vac line
657xA/667xA Connection to a 3-phase Line
A
B
C
D
E
F
G
H
I
K
App.
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