R&S
®
ZVA / R&S
®
ZVB / R&S
®
ZVT GUI Reference
Trace Menu
Operating Manual 1145.1084.12 – 30 135
device (e.g. an amplifier) to the supplied DC power P
DC
. The added RF power can be expressed as the
difference between the power of the outgoing wave b
2
at the output of the DUT and the power of the
incident wave a
1
at the input of the DUT; hence:
Positive PAE values indicate a gain in the RF power, negative values an attenuation. The PAE is always
smaller than 1.
CALCulate<Ch>:PARameter:MEASure "<Trace_Name>", "PAE21" |
"PAE12" |...
Create new trace and select name and measurement parameter:
CALCulate<Ch>:PARameter:SDEFine "<Trace_Name>", "PAE21" |
"PAE12" |...
DC Power Measurement
The power P
DC
supplied to the DUT can be measured using either one of the DC inputs DC MEAS 10V
(for large voltages), DC MEAS 1V (for small voltages) or both inputs. The DC Power panel in the PAE
dialog suggests different models involving different test setups and approximations. The models are
selected by means of the radio buttons in the DC Power panel; they depend on the properties of the DC
power supply (constant current I
DC
or constant voltage U
DC
) and an optional precision resistor R used to
measure the DC current. The values I
DC
, U
DC
, and R determine the Constants c and k. These constants
must be entered in the DC Power panel, using the appropriate physical units, before a particular model
can be activated.
I
DC
= const., R = 0 —> P
DC
= c * U (DC Meas –10 V...+10 V) Assume that the DC power supply
provides a constant current IDC and that the voltage applied to the DUT is measured via DC
MEAS –10V...+10 V (R = 0). PDC = IDC * U(DC MEAS –10V...+10 V), hence the Constant c must
be set equal to IDC. The unit field shows the SI unit of a current (W/V). The input field for k is
disabled (k is not used).
U
DC
= const., R << R
DUT
, P
R
= 0 —> P
DC
= c * U (DC MEAS –1 V...+1 V) Assume that the DC power
supply provides a constant voltage UDC and that the current through the DUT is measured by
means of a precision resistor R connected in series and DC INP 2. If R << RDUT the power
consumption of the resistor can be neglected so that PDC = UDC / R * U(DC MEAS –1 V...+1 V),
hence the Constant c must be set equal to UDC / R. The unit field shows the SI unit of a current
(W/V). The input field for k is disabled (k is not used).
R << R
DUT
, P
R
= 0 —> P
DC
= k * U (DC MEAS –10 V...+10 V) * U (DC MEAS –1 V...+1 V) Assume
that the DC power supply provides an arbitrary (not necessarily constant) voltage UDC and
current IDC. The current through the DUT is measured by means of a precision resistor R
connected in series and DC INP 2. The voltage applied to the DUT and R is measured via DC
MEAS –10 V...+10 V. If R << RDUT the power consumption of the resistor can be neglected so
that PDC = 1 / R * U(DC MEAS –10 V...+10 V) * U(DC MEAS –1 V...+1 V), hence the Constant k
must be set equal to 1 / R. The unit field shows the SI unit of an inverse resistance (W/V2). The
input field for c is disabled (c is not used).
U
DC
= const., P
R
= U (DC INP2)
2
/R —> P
DC
= c * U (DC MEAS –1 V...+1 V) + k * U (DC MEAS –1
V...+1 V)^2 Assume that the DC power supply provides a constant voltage UDC and that the
current through the DUT is measured by means of a precision resistor R connected in series and
DC INP 2. PDC = UDUT / R * U(DC MEAS –1 V...+1 V) = (UDC – UR) / R * U(DC MEAS –1
V...+1 V) = UDC / R * U(DC MEAS –1 V...+1 V) – 1 / R * U(DC MEAS –1 V...+1 V)2, hence the
Constant c must be set equal to UDC / R and k must be set equal to 1/R. The unit fields for c and
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