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DMS Accessories
50mV and 100mV Base-mounted DC Shunts
3. Lead Extensions: The shunt’s Kelvin outputs (i.e., their 50mV or 100mV
output signal) should not exceed 24 inches in length (61cm), assuming
The following formulas can be used to calculate the maximum pulsed-
current ( I puls) for a given shunt:
22AWG (0.5mm2) wire is used. If longer leads are necessary, the
additional voltage drops in the leads must be taken into consideration
I puls =
P puls/ R , and P puls = P a/ K 1, and K 1 =
D
P e = P a x [1-( T e-25°C)/100°C], and I e =
= 924A
0.0000625 Ω
as they will degrade accuracy and stability.However,MPS’s DCA5 series
dc?ammeters feature a calibration potentiometer that can be used to
compensate for some additional losses when long leads are employed;
contact MPS for calibration information.
4. Continuous-Operation Derating: If continuous operation is required,
a shunt must only be allowed to carry 2/3 of its maximum nameplate
amperage. The 2/3 derating factor provides an adequate safety margin
for convection cooled shunts operating in an ambient temperature of
+25°C. The following two sections contain additional derating that is
required when shunts will be exposed to higher operating temperatures
or pulsed (intermittent) currents.
5. Temperature Derating: If the shunt is to be operated at ambient
temperatures above +25°C, the maximum continuous current must
be further derated from the 2/3 value previously noted.
To ?nd the maximum permissible continuous current (Ie) at an elevated
temperature (Te), one must ?rst calculate the maximum allowable
power dissipation (Pe) at Te, using the following formulas:
P e
R
Where
I e = Maximum permissible continuous current at elevated
temperature T e
Where
I puls = Maximum pulsed current
P puls = Maximum pulsed power
P a = 0.667 x shunt’s rated power at 25°C ambient
D = Ratio of ON TIME to ON TIME + OFF TIME
(period not to exceed 5 minutes)
K 1 = Pulsed-operation rating factor (the square root of D)
R = Shunt’s resistance (see selection guide for values)
Example: An 800 Amp 50mV shunt will be operated for 15 seconds
each minute. What is the maximum current this shunt can carry
during the 15-second ON time?
D = 15/60 = 0.25 and K 1 = 0.25 = 0.5
P a = 0.667 (800A x 0.050V) = 26.7 Watts
P puls = P a/ K 1= 26.7W/0.5 = 53.4 Watts
53.4W
I pulsed =
0-50mV
OUTPUT
P e = Maximum power dissipation at the elevated temperature Te
R = Shunt’s resistance (see selection guide for values)
–
+
P a = 0.667 x shunt’s rated power at 25°C ambient
T e = Elevated temperature
–
LOAD
+
+
DC SUPPLY
–
Example: How much current can a 150 Amp 50mV shunt safely carry
at an ambient temperature of +100°C?
T e = +100°C
P a = 0.667(150A x 0.050V) = 5.0 Watts
P e = 5.0W[1-(100-25)/100] = 5.0W x (1-0.75) = 1.25W
R = 0.00033 Ohms
P e
Lastly, use the formula I e = R to ?nd the maximum derated
continuous-current at T e = +100°C:
50mV SHUNT
LOAD CURRENT
Figure 2. Typical Low-Side Shunt Connections
0-50mV
OUTPUT
POWER
GROUND
(0Vdc)
I e = 0.00033 Ω = 61.5A
1.25W
6. Intermittent (Pulsed) Operation: Shunts that will not be exposed to
–
+
+
DC SUPPLY
–
continuous operation at 2/3 rated current can be operated at levels
close to, or?even above, their nameplate ratings for portions of short
periods of times (5 minutes maximum) at 25°C ambient.
50mV SHUNT
LOAD CURRENT
POWER
GROUND
+
LOAD
–
(0Vdc)
Figure 3. Typical High-Side Shunt Connections
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MPM_DMS Accessories.C01 Page 2 of 4
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