Q: Is it necessary to control the power supply's static electricity?

A. Since the power supply contains static-sensitive electronic components (semiconductors&#44 etc.)&#44 it is necessary to control static electricity. Related Information AC-DC Power Supplies Product Information Power Supplies Tech Notes Contact Information Inquiries on products&#44 sales and technical matters c:faq cid:1433021523330 def:faq_detail_D locale:en -->

Question 1

Why is the primary side fuse of the power supply in the live side (L phase)?

Accepted Answer

A. The primary side fuse is in the live side (L phase).
Fuses on live sides can melt when there are short circuits between L-FG or N-FG inside power supplies.
The primary side fuse is in the live side (L phase) because there is the possibility of electric shock with fuses not melting when there are short circuits between L-FG or N-FG if the fuse is in the neutral side (N phase).

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Question 2

Please explain the specs of the vibration test.

Accepted Answer

A. Standard number: JIS60068-2-6
Standard name: Sinusoidal vibration test method
The conditions in “instruments used for large power-generating equipment and general industry” for sweep endurance tests have been extracted with the assumption that there are vibrations during transportation.

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Question 3

Please explain the difference between the rating voltage on the nameplate seals and specifications.

Accepted Answer

A. The input ranges shown on power supplies (nameplates) are the figures used when applications for safety standards are submitted so approval of the safety standards is already granted at the time of use within these ranges.
Furthermore, the specifications shown on documents such as catalogs are the input ranges of the operational assurance that we offer. Therefore, assurance for use within the specifications is given for use exceeding the ranges shown on the power supplies (nameplates) although the safety standards are excluded.

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Question 4

Please explain the heat dissipation methods for power modules / onboard power supplies.

Accepted Answer

A. They can be classified into two major types depending on the structure of the products.

(1) The type where the heat generated from mounted components is directly dissipated in the surrounding air
The heat generated from components mounted on products with exposed mounting boards is directly dissipated in the surrounding air.
The electrical power that can be outputted with this type of product can be increased when it is forcedly cooled with external fans (CC-E series have cases to cover their mounting boards but are classified as this type because they have intake ports for surrounding air and mold resin is not used). The operable temperatures are specified in the ambient temperatures with the type above.
Refer to the specifications, output deratings, and instruction manuals of the products for details.

(2) The type where heatsinks are attached for the use of heat conduction and dissipation in the surrounding air
Aluminum baseplate-type products (with heat-generating components mounted on aluminum plates) require the attaching of appropriate heatsink on their baseplates.
With this type, the heat generated from mounted components placed on the baseplates undergoes heat conduction (conduction cooling) in the heatsink through the baseplates and is efficiently dissipated in the surrounded air with the heatsink. This type is different from the one described above in that the operable temperatures are specified in the baseplate temperatures. It is adopted by products that have relatively large capacity and high levels of heat generation. It allows electrical power that can be outputted to be designed freely through the shape of the heatsink and forced cooling of the heatsink with an external fan (please contact us for details).

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Question 5

Is it alright to clean the products?

Accepted Answer

A. Washing only the terminal pins by brush with IPA (isopropyl alcohol) is recommended, with the exception of some products (cleaning solutions that enter the products may cause deterioration of the components used).
Refer to the instruction manuals of the respective products for details.

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Question 6

Is it a problem if the load currents of each output are not equal in ±2 output products with insulated onboard power supplies?

Accepted Answer

A. Please note that the output voltages can fluctuate significantly. The output voltages can be stabilized by controlling the voltages between +Vout and -Vout in circuits such as the one shown in the Figure below as±2 output products are compact.

Figure 1

The output voltages fluctuate with the current values of each output because each output voltage does not undergo stabilized controlling individually. We recommend use combined with a single-output converter when load currents of each output are unequal and precise output voltages are required.

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Question 7

What do the terms such as EHFP and HFP at the end of the product nameplates mean?

Accepted Answer

A. They are identification codes that we use.

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Question 8

Please explain power loss of power supplies.

Accepted Answer

A. The power loss of power supplies is the difference between input active power and output power and is calculated with the formulas below.
Power loss= input active power - output power
Input active power can be calculated with the output power ÷ efficiency, so
power loss= (output power ÷ efficiency) - output power.

(example) When calculating the power loss in HWS1500-48 with the maximum output power when the input voltage Vin = 200VAC and the ambient temperature Ta = 25°C：
The power loss is as shown below because the efficiency is 90% of the specifications with the maximum output power (1536W).
Power loss = (1536W ÷ 0.9) - 1536W = 171W

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Question 9

Please explain how to interpret the output power when the set voltage is used after being raised above the rated output voltage.

Accepted Answer

A. The maximum output power is calculated with the formulas below.
Maximum output power = rated output voltage × maximum output current
Please note that there are specifications on the maximum output power when the set voltage is raised above the rated output voltage and that the levels of the maximum output currents that can be used will be lowered.

(example) When HWS100A-5 is set at 5.5V and used:
The maximum output current that can be used = maximum output power ÷ set voltage
=100W÷5.5V = 18.2A

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Question 10

Please explain how to interpret the output power when the set voltage is used after being lowered below the rated output voltage.

Accepted Answer

A. The maximum output current that can be used increases if the set voltage is lowered when “maximum output power = rated output voltage × maximum output current” is used, but please use at the specified values of the maximum output current or below because there is the danger of damage due to factors such as the generation of heat in the components inside the power supply.

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Question 11

Please describe the power consumption (loss).

Accepted Answer

A. The power consumption (loss) is the difference between input power and output power and is calculated with the formulas below.

Pd: Power consumption (W)
Pin: Input power (W）　Pin = Vin × Iin
Pout: Output power (W）　Pout = Vout × Iout
η: Efficiency
Vin: Input voltage (V)
Iin: Input current (A)
Vout: Output voltage (V)
Iout: Output current (A)

Example 1) The efficiency is 92.5% of the specifications with the maximum output (30W) when the input voltage of the CC30-4805SFP-E is 48V and the ambient temperature is 25°C.
The power consumption (loss) in this case is Pd = (30 ÷ 0.925) – 30 = 2.43W with the formula above.

Example 2) The following figures were produced when the DC-DC converter was operated and the input power voltage and output power current were measured.
Input voltage (Vin) = 47.996V
Input current (Iin) = 0.348A
Output voltage (Vout) = 5.014V
Output current (Iout) = 2.998A
The power consumption (loss) in this case is Pd = (47.996 × 0.348) – (5.014 × 2.998) = 1.67W with the formula above.

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Question 12

How much output power (current) can actually be used?

Accepted Answer

A. The specifications stipulate the maximum output current and maximum output power. It is necessary to use the products under conditions where none of these figures are exceeded so that it can be operated safely.
Please note that they will be as shown below particularly when using an output voltage variable function.

(1) When the output voltage is equal to the rated output voltage (initial settings)
　　The maximum output power = the maximum output current × the rated output voltage so there is no problem if it is confirmed that one of them is not exceeded.

(2) When setting the output voltage higher than the rated output voltage
　　The maximum output power < the maximum output current × the set output voltage so it is limited with the maximum output power (the output current must be decreased).

(3) When setting the output voltage lower than the rated output voltage
　　The maximum output power > the maximum output current × the set output voltage so it is limited with the maximum output current (the output power must be decreased).

Output derating by operating temperature is also specified for each product. This must also be taken into consideration along with the other factors.

Example) The maximum output current is 2A and the maximum output power is 10W under the specifications if the output voltage of the CC10-4805SF-E is adjusted to 6V with an ambient temperature of 85°C and convection cooling is used, but the output current is decreased so that it does not exceed the maximum output power when the output voltage is set higher than the rated output voltage.
It is necessary to make a further decrease down to 40% or below during use with an ambient temperature of 85°C with output derating.
Therefore, the maximum output current is as follows:
Maximum output current = maximum output power ÷ set output voltage × 0.4 = 10 ÷ 6 × 0.4 = 0.66A

Figure 1

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Question 13

Are there specifications for input surge currents (inrush currents) with DC input power modules / on-board power supplies?

Accepted Answer

A. There are regularly no stipulations in the specifications.
Please refer to the typical waveforms listed in the data for the models of the respective products.
Input surge currents (inrush currents) are charging currents that flow into capacitors that are connected between input lines when the power is turned on. These currents also flow when input voltage is instantly applied with a device such as a switch in DC input power modules / on-board products. However, the values of the input surge currents (inrush currents) change according to the capacity of input capacitors that are additionally connected outside or impedance components of wiring patterns.

Figure 1

*The input surge current (inrush current) is the total value of the charging current that flows into the capacitor inside the product ① and the charging current that flows into the input capacitor Cin that connects to the exterior ②.

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Question 14

How do I measure the output ripple and noise?

Accepted Answer

A. *Applicable product
Power module type and on-board type power supplies

1. Measurement method prescribed by TDK-Lambda
We measure output ripple and noise based on JEITA RC-9131 for AC input power supply and JEITA RC-9141 for DC input power supply.

Put the attachment (50Ω resistance and 4700pF ceramic capacitor) to the 1.5m length 50Ω coaxial cable, and solder the coaxial cable to both ends of the output ceramic capacitor (C15). Also, set the frequency band of the oscilloscope to 100MHz for usage. The method used to measure the output ripple and noise differs depending on the power supply series. For further details, please refer to the instruction manual for each series.

2. When using a 10:1 probe without using a coaxial cable
If you use a 10:1 probe that is used for normal waveform measurement without using a coaxial cable or attachment, the probe loop becomes long and susceptible to the impact of common mode noise. As a result, a great amount of output ripple and noise will be observed.

3. Measurement method which uses a 10:1 probe and reduces output and ripple noise
If you have to use a 10:1 probe, you can reduce the output and ripple noise by removing the clip and the GND wire, and then measuring with a smaller loop. However, the output and ripple noise will be larger than the measurement method specified by TDK Lambda. Therefore, please be careful when attempting to measure the correct value.

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Question 15

Please provide information on support for CCC standards?

Accepted Answer

A. The power supplies of TDK-Lambda are built-in power supply for general equipment. CCC standards are not required.
Please contact TDK-Lambda to receive safety standard materials for each product.

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Question 16

The EMI characteristics of the power supply differ from the published data.

Accepted Answer

A. The published data are the characteristics in the TDK-Lambda measurement environment.
For information the measurement environment and conditions, see the section entitled "Measurement Method" in the model data on the technical data page of each product on the product information website (TDK Product Center).
EMI characteristics may change if the measurement environment is different or if the power supply is installed in equipment.

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Question 17

Are there any instructions concerning the wire to be used?

Accepted Answer

A. There are no instructions specifying the maker or type. Comply with the recommendation concerning the wire maker.

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Question 18

Some products are listed as "not recommended for new designs". Why is this?

Accepted Answer

A.Since the product has been on the market for several years, we may issue a notice of discontinuation in the future depending on the availability of parts and other factors.
If you wish to continue purchasing this product, please contact our sales office.

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Question 19

Is it necessary to control the power supply's static electricity?

Accepted Answer

A.Since the power supply contains static-sensitive electronic components (semiconductors, etc.), it is necessary to control static electricity.

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Question 20

Is there a startup time limit?

Accepted Answer

A.The startup time from the time of power input to the time when the output voltage rises is not usually specified in the specifications.
The Evaluation Data is available for each model, and the startup time can be read as typical data from the output voltage rise characteristics. The ON/OFF CNT function is also available.
The startup time by ON/OFF CNT for power supplies with ON/OFF CNT function is also available in the Evaluation Data. The ON/OFF CNT output voltage rise characteristics of the Evaluation Data can be referenced as representative data.

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Question 21

What altitude does the power supply support?

Accepted Answer

A.Safety standard reports are published for each product, and the corresponding altitude is listed there.
Regarding altitudes, the creepage and clearance distances within the product are designed based on the required values up to the altitude specified in the safety standard.
However, we have not actually evaluated the product for use in environmental conditions where the altitude is high and the air pressure is low.
If the product is used in such an environment, possible problems may arise, including reduced heat dissipation performance and ease of discharge.
Please ensure that everything is okay with your device.

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Question 22

How do you calculate the power supply's generated heat?

Accepted Answer

A.The amount of heat generated in a power supply can be expressed by its internal loss as follows:

Amount of heat generated: 1(W x s) = 1(J) = 4.2(cal)

The calorific value of the power supply can be obtained by the following equation using the internal loss of the power supply and time.
・Heat generation (J) = Internal loss of power supply (W) x Time (s)
　　　　　　　　　 = [Effective power of input (W) – Output power (W)] x Time (s)
　　　　　　　　　 = {[Output voltage (V) x Output current (A) / Efficiency] - [Output voltage (V) x Output current (A)]} x Time (s)

Calculate the amount of heat generated per second based on the following operating conditions (example).
・Output voltage　24 V
・Output current　10 A
・Efficiency　　　 80%
・Time　　　　　 1 second

Heat generation (J) = {[Output voltage x Output current / Efficiency] - [Output voltage x Output current]} x Time (s)
　　　　　　　　 = {[24(V) x 10(A) / 80%] - [24(V) x 10(A)]} x 1(s)
　　　　　　　　 = [300(W) - 240(W)] x 1(s)
　　　　　　　　 =60 (W x s)
　　　　　　　　 =60 (J)
　　　　　　　　 =252 (cal)

For the power supply's efficiency characteristics, refer to the static characteristics "Efficiency vs. Output current" in the Evaluation Data of the technical data for each product.

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Question 23

Can the input-output polarity of DC-DC converters be reversed?

Accepted Answer

A. Polarity reversal outputs can be made with input-output isolation type products. Please refer to the Figures below to see how connections can be made.

Figure 1　For positive voltage inputs → negative voltage outputs

Figure 2　For negative voltage inputs → positive voltage outputs

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Question 24

Is it OK to use the isolated DC-DC converter as non-insulated? What is the connection method in this case?

Accepted Answer

A. Yes, it is OK. For information on the connection method, please refer to the example connection application given below.

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Question 25

I don't want to stop the system&#44 so I would like to operate it redundantly (backup). Is this possible? Also&#44 what should I be cautious of when using remote sensing during redundant operation?

Accepted Answer

A. Redundant (backup) operation is possible.

Please match the diode to the output side and then connect.
If there is a remote sensing terminal, please be sure to perform sensing on the anode side of the diode.
If sensing is performed on the cathode side of the diode, the power supply operation may become unstable.

Please use the following guidelines when selecting a diode:
(1) The repetitive peak reverse voltage is more than twice the rated output voltage of the power supply
(2) The average output current is more than the rated output current of the power supply
(3) Diode with forward voltage that is as low as possible (Example: Schottky diode)
Furthermore, if the diode is used alone, the element may be damaged by heat. Therefore, be careful when using a radiator, etc., for cooling and confirm that the element temperature is within the range specified by the manufacturer.
*Please use the backflow prevention module RP-60-20 etc. which is a product of TDK-Lambda.

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Question 26

What should I be cautious of when connecting multiple units in series?

Accepted Answer

A. When connecting multiple units in series, high voltage is applied between the power supply output (secondary side) and the FG. Therefore, it is necessary to consider isolation between the power supply output and the FG.
Based on consideration of customer equipment, basic isolation is required between the power supply output and the FG.
Furthermore, if there is an SELV (ES1 in 62368-1) part in customer equipment, reinforced isolation is required between the SELV (ES1 in 62368-1) and the power source output. Also, reinforced/double isolation may be necessary for the output line.
Additionally, if the output voltage exceeds 60V DC, which is the upper limit of the SELV (ES1 for 62368-1) target range, it is also necessary to prevent the electric shock which will result when touching the human body, etc.
Also, the output voltage in series operation and the withstand voltage for the required isolation change depending on the conditions of customer equipment.
Please contact TDK-Lambda for details.
TDK-Lambda products are designed with SELV (ES1 of 62368-1) and the target range is 60V DC or less.
If the power supply outputs are connected in series that does not exceed SELV (ES1 in 62368-1), the safety standards for a "single power unit" take effect.

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Question 27

During capacitive load connection&#44 a surge current that exceeds the rating flows on the output side. Can I use it?

Accepted Answer

A. If the surge current flows for only a few milliseconds and has a low frequency (for example, a few times per day), it does not affect the component temperature and use is possible.
Please avoid use in an environment where repeated surge currents flow may cause the temperature of the parts to rise.
When selecting the power supply, please consider the drooping characteristics of overcurrent protection and the starting characteristics of the load side. 　

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Question 28

Can I connect products with different capacities in a series?

Accepted Answer

A. It is possible to connect products with different capacities in series.
However, please use a load current within the range of the model with the smallest maximum output current.

Example) Operation of a 63V output voltage stack series consisting of (1) 150W-48V output model and (2) 50W-15V output model
(1) 150W-48V output model: Output voltage = 48V, Output current = 3.3A (100%)
(2) 50W-15V output model: Output voltage = 15V, Output current = 3.5A (100%)

The maximum output current for the (1) 150W-48V output model and the (2) 50W-15V output model are 3.3A and 3.5A respectively, so the upper limit of the load current is 3.3A, which is the smaller of the two models (150W-48V output model).

TDK-Lambda products are designed with SELV (ES1 in 62368-1: Safety extra low voltage circuit), and the target is the circuit of DC 60V or less.
If the power supply outputs are connected in series that do not exceed SELV (ES1 in 62368-1), the safety standard acquired as "single power supply" is valid.
If the safety voltage (peak value 42.4V, DC 60V) is exceeded, the obtained safety standard will be invalid.
In this case, there are other points to note, so please contact TDK-Lambda.

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Question 29

I would like to select a breaker. What are the different times for inrush current flow?

Accepted Answer

A. The inrush current varies depending on the model.
For the inrush current waveform, see the section on "Input Surge Current (Inrush Current) Characteristics" in the model data on the technical data page of each product on the product information website (TDK Product Center).

Depending on the breaking characteristics of the breaker, this rush current may cause a break (trip).
When selecting a breaker, make sure that the inrush current is outside the breaker operating range based on the breaker breaking characteristics.
In addition, please perform the evaluation with the device.

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Question 30

There seem to be several types of overcurrent protection characteristics. What are the differences?

Accepted Answer

A.There are three main types: Constant-current voltage droop, F-shaped droop, and Intermittent.

Constant-Current Voltage Droop Type
When the output current reaches the overcurrent detection value, the voltage will drop while maintaining the current.
If the output current drops below the overcurrent detection value, the output voltage will automatically recover.

Foldback Drooping Type
When the output current exceeds the overcurrent detection value, both output voltage and current will drop.
Output may not automatically recover in this mode, so input must be turned on again in such a case.

Intermittent Type
When the output current reaches the overcurrent detection value, both the output voltage and current will periodically repeat pulse operation.
If the output current drops below the overcurrent detection value, the output voltage will automatically recover.

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