SAR Series Fully Automatic A.C.Voltage Regulator

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Shanghai
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TT OR LC
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10000pcs pc/month

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Product Description:

1.Application
These special products of SAR series Voltage Regulator have advanced equipment, abundant technology, reliable quality and distinguished credit. SAR series are designed at JS3717-84 request, these products conform to fully automatic control of integrated circuit. They are of quick speed, good reliable of sensitive action, convenient use and assure the stale resuit. They are suitable for families, schools, enterprises, hotels and diets etc., where need a stable civil power. They can make the illumination lamp, TV set, air-conditioner, refrigerator, computer SAR-2000VA and copy machine etc. work at a normal condition and have a long using time.

2. Technical specification

FrequencyRegulating TimeTemperature RiseWithstand VoltageLow Voltage ProtectionOver Voltage Protection
50-60HzLess Than 0.5<60oCAccord With Ministry Issued StandardOutput 100V/160VOutput 270V/260V
ModelRangeInput Civil Power Voltage(V)Output Precision of Stabling Voltage

Current160-260220V±5%
0.5KVA To 5KVAOver-Low Voltage130-270220V±8%
Special Over-Low Voltage100-270220V±10%
High Precision of Stabling Voltage160-270220V±3%


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Q:800KVA transformer itself is the loss of how much?
Are 800kVA transformer, the model is different, loss of access great. You find out the model to say, there are standards can be found.
Q:What is the meaning of the transformer without excitation?
No excitation refers to the transformer should be off the state of the power regulator, and his relative to the excitation voltage regulator should be on-load regulation.
Q:Taiwan, Taiwan, the concept of transformer
Transformer refers to equipment; Taiwan change is installed in a place of the transformer, is specifically installed, running a transformer; Taiwan area refers to the regional range of variable power supply. Metering point is for the two sides to determine the installation of electric energy meter position, in general, the measurement point should be set at the demarcation point of property rights.
Q:630 kilovolt transformer secondary side of the current is how much
630kVA that his rated capacity, it is ideal to say that with the load of 630kW (but actually can not bring so much) Rated current calculation formula: rated current = capacity / rated voltage / root number 3 (for 3-phase transformer) 630 / 0.4 / 1.732 = 909.3 A Pure hand to play, hope to adopt ~
Q:Transformer turns around the number and how much V is calculated
3, seeking wire diameter What is the current required to output 8 volts? Here I assume 2 am. Transformer output capacity = 8 ╳ 2 = 16 volt Transformer input capacity = transformer output capacity / 0.8 = 20 VA Primary coil current I1 = 20/220 = 0.09A Wire diameter d = 0.8√I Primary coil wire diameter d1 = 0.8√I1 = 0.8√0.09 = 0.24 mm Secondary coil wire diameter d2 = 0.8√I2 = 0.8√2 = 1.13 mm The voltage after the bridge rectifier capacitor is 1.4 times the secondary voltage of the original transformer. Transformer (Transformer) is the use of electromagnetic induction principle to change the AC voltage of the device, the main components are primary coil, secondary coil and core (core). The main functions are: voltage conversion, current conversion, impedance conversion, isolation, voltage regulator (magnetic saturation transformer) and so on. According to the purpose can be divided into: power transformers and special transformers (electric furnace change, rectifier, frequency test transformer, voltage regulator, mine transformer, audio transformers, IF transformers, high-frequency transformers, impact transformers, instrument transformers, electronic transformers , Reactors, transformers, etc.).
Q:50kva transformer price
2, Transformer (Transformer) is the use of electromagnetic induction to change the principle of AC voltage device, the main components are primary coil, secondary coil and core (core). The main functions are: voltage conversion, current conversion, impedance conversion, isolation, voltage regulator (magnetic saturation transformer) and so on. According to the purpose can be divided into: power transformers and special transformers (electric furnace change, rectifier, frequency test transformer, voltage regulator, mine transformer, audio transformers, IF transformers, high-frequency transformers, impact transformers, instrument transformers, electronic transformers , Reactors, transformers, etc.). Circuit symbols commonly used as the beginning of the number. Example: T01, T201 and so on.
Q:The difference between transformer power kva and kw
KVA is apparent power, kW is active power. inspecting power: S = 1.732UI = 1.732 x 0.38 x 1000 ≈ 658 (kVA) Active power: P = 1.732UI cosφ = 1.732 × 0.38 × 100 × 0.8 ≈ 526 (kW) AC power, power divided into three kinds of power, active power P, reactive power Q and apparent power S, at any time these three power always exist at the same time.       The cosine of the phase difference (Φ) between the voltage and the current is called the power factor, denoted by the symbol cosΦ, where the power factor is the ratio of the active power to the apparent power, that is, cosΦ = P / S       Three power and power factors cosΦ is a right-angle power triangular relationship: two right-angled edges are active power P, reactive power Q, and oblique edge is apparent power S.       S² = P² + Q² S = √ (P² + Q²)       Apparent power S = 1.732UI       Active power P = 1.732UIcosΦ       Reactive power Q = 1.732UIsinΦ          DC = P = UI = U² / R = I²R
Q:Transformer if the boost and buck current will change?
You here is actually a rated power problem, the transformer rated power is determined by the core size and winding diameter. Actual current X The actual output voltage can not be greater than the rated power otherwise there is a possibility of damage, which is the principle. If you are a constant load, the actual current will increase at the time of boost, and the actual current will decrease when it is lowered.   Disengaged the actual use of the transformer, the transformer as a boost when the input current must be greater than the output side of the current, the transformer as a buck when the input current must be less than the output side of the current.
Q:10KV distribution transformer how to calculate the tap voltage
10kV distribution transformer tap voltage can be Ue ± Ue X% (X is the percentage of each range of adjustment). General distribution transformer tap adjustment range of Ue ± 2 × 2.5%; tap for 1 - 5, the tap voltage is calculated as follows: 1 ----- 10+ (10 x (2 x 2.5%)) = 10.5 kV 2 ----- = 10 + (10 x 2.5%) = 10.25 kV 3 ----- 10kV rated voltage, 4 ----- = 10- (10 x 2.5%) = 9.75 kV 5 ------ 10- (10 x (2 x 2.5%)) = 9.5 kV Transformer tap on the high side, according to the needs of the low side to adjust the high pressure side of the tap position. If the low voltage side of the low pressure should be raised to 4 or 5, and vice versa to 2 or 1 up. Because the adjustment tap is used to change the transformer ratio. In the case of a transformer with a rated voltage of 10 / 0.4kV, its change ratio K = U1 / U2 = 10 / 0.4 = 25; when the secondary voltage is low, if the split is set to 4, the position K4 = U1 / U2 = 9.75 / 0.4 = 24.375, the primary voltage is still 10kV secondary voltage = 10 / 24.375 = 410V. Increased secondary voltage. In fact the adjustment of the split is adjusted by the number of turns of the coil, the change in the ratio and the number of turns of the secondary coil is constant. To reduce the secondary voltage, increase the number of turns of the coil.
Q:What is the transformer capacity ratio? Capacity than 100%.
Transformer capacity is the power of the transformer, the capacity ratio is the power ratio, such as 1000KA transformer and 200KVA transformer is not suitable for parallel use. Capacity ratio = 5: 1 greater than 3: 1 Capacity than 100% -------- is the meaning of 1: 1 (such as 500KVA and 500KVA, 200KVA and 200KVA and so on. Transformer side by side conditions: ① to participate in parallel operation of the transformer must be the same wiring group. ② the primary voltage of each transformer should be equal, the secondary voltage are equal. Otherwise the secondary side of the circulation caused by overload, heat, affecting the load, and increase the power loss, reduce efficiency; ③ the transformer impedance voltage (short circuit voltage) percentage should be equal, or with load after the load distribution unreasonable. Because the large capacity of the transformer short circuit voltage percentage, small capacity of the transformer short circuit voltage percentage is small, and the load distribution and short circuit voltage percentage is inversely proportional to this will cause large transformer distribution load is small, the equipment is not fully utilized; and small transformer distribution of the load Large, easy to overload, limiting the parallel operation of the transformer with load operation.

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