• Intellectualization HG-SR202BJ Motor For Semiconductor Manufacturing System 1
  • Intellectualization HG-SR202BJ Motor For Semiconductor Manufacturing System 2
  • Intellectualization HG-SR202BJ Motor For Semiconductor Manufacturing System 3
  • Intellectualization HG-SR202BJ Motor For Semiconductor Manufacturing System 4
  • Intellectualization HG-SR202BJ Motor For Semiconductor Manufacturing System 5
Intellectualization HG-SR202BJ Motor For Semiconductor Manufacturing

Intellectualization HG-SR202BJ Motor For Semiconductor Manufacturing

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Loading Port:
Shanghai
Payment Terms:
TT OR LC
Min Order Qty:
1 kg
Supply Capability:
2000 kg/month

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Motor series: low inertia, medium power. Rated output power: 30.0kw. Rated
speed: 1500r / min with brake or not: No. Shaft end: Standard (straight shaft).
Voltage: 400V type.
Protection level: IP65 (p67) hg-sr202bj basic knowledge.
Features: low inertia, from low speed to high speed, three modes can be
selected, suitable for different application ranges. As a standard production
port, hg-sr202bj is installed with 30kW or higher dry flange and feet
Application examples: basic knowledge of injection molding machine,
semiconductor manufacturing installation, large transmission machinery hg-
sr202bj. High resolution encoder 131072p / rev (17 bits).
High resolution encoder is included to ensure excellent performance and
stability at low speed.
All motors are the same size as the previous products and the wiring is
compatible. Motor series: low inertia, low power. Rated output power: 0.75KW.
Rated speed: 3000r / min. with brake or not: attached.
Shaft end: keyway (slotted position).
Protection level: IP55, excluding the shaft threading part and the basic
knowledge of connector hg-sr202bj. Features: low inertia, suitable for high
frequency operation.
Application examples: inserter, assembly machinery. Motor series: medium
inertia, medium capacity. Use serial pair / incremental encoder. Rated output
power: 3.0kw. Rated speed: 1000rpm. With or without band brake: No. Shaft end:
straight shaft.
The stator structure of AC servo motor is basically similar to that of
capacitor split phase single-phase asynchronous motor hg-sr202bj (type
selection data). The stator is equipped with two windings whose positions are
90 ° different from each other. One is the excitation winding RF, which is
always connected to the AC voltage UF, and the other is the control winding L,
which is connected to the control signal voltage UC.
Therefore, AC servo motor is also called two servo motors. Motor series: medium
inertia, medium power. Rated output power: 3.5kw rated speed: 2000r / min.
With brake or not: hg-sr202bj (model selection data) is attached. Shaft end:
Standard (straight shaft). Protection level: IP65 (IP67).
Features: there are three modes of medium inertia from low speed to high speed,
which are suitable for different application ranges. Application examples:
transmission machinery, robot, X-Y worktable.
High resolution encoder 131072p / rev (17 bit) hg-sr202bj (selection data).
High resolution encoder is included to ensure excellent performance and
stability at low speed.
All motors are the same size as the previous products and the wiring is
compatible. Motor series: low inertia, medium power. Rated output power: 6.0kw.
Rated speed: 1000r / min. with brake or not: attached. Shaft end: Standard
(straight shaft). Protection level: IP65 (IP67)
Features: low inertia, from low speed to high speed, three modes can be
selected, suitable for different application ranges hg-sr202bj selection
manual. As a standard production port, 30kW or higher power is suitable for dry
process blue plate (type) installation and foot installation. Basic knowledge
of hg-sr202bj. Application examples: injection molding machine, semiconductor
manufacturing device, large conveying machinery. High resolution encoder
131072p / rev (17 bits).
High resolution encoder is included to ensure excellent performance and
stability at low speed. All motors are the same size as the previous products
and the wiring is compatible.

Q: Do any adults play that game online anymore? I am 19, big fan of transformers since I was a kid watched the G1 series, Transformers prime, seen the michael bayfail movies and played all the games. I love TFFOC its a gream game, with great unique multiplayer and its in cybertron! I love it and can't wait for the next game as that will be on Earth!.I have been playing this game for quite a while now and everytime I put my headset/mic on and play a match online I always come across kids that are like 13 years old and get friend requests from them lol.Should I stop playing this game?
No. All the adults are playing The Last of Us multiplayer.
Q: Parallel can provide the total capacity is not the sum of the capacity of the two transformers, or how to calculate? What is the principle of allocation of load load, that is, how to allocate two transformer load capacity?
After the parallel capacity can be provided is the sum of the capacity of the two transformers ---- If the two transformers capacity is the same, after the parallel capacity can provide the total capacity of the two transformers are the sum of capacity. If a 1000kva transformer and a 800kva transformer in parallel, the parallel can provide the total capacity of 1600kva. The distribution of load load and what principles, that is, how to allocate two transformer load capacity --- transformer parallel to meet the three conditions: 1, two transformer transformer ratio equal; 2, two impedance voltage equal; 3, two Transformer capacity should be as close as possible, should not exceed 3: 1. To meet these three conditions in parallel, the load is evenly distributed to two transformers.
Q: How to test E-I Transformer having 10 terminals (pins) 5 on each side ? thanks
If a transformer has shorted turns, every winding will show drastically reduced inductance than is required for the circuit to operate. For example, the inductance of an audio transformer has to be at least L Z/(2piF), where F is the lowest frequency, audio ~20Hz and Z is the circuit impedance. So an 8 Ohm speaker winding should have 60mH or more inductance. The resistance of the winding should be less than 10% of the operating impedance. The maximum operating frequency of a transformer is something less than the self resonance and the minimum load is that which provides critical damping of the self resonance. So large transformers just work better since they have more inductance and less resistance With a signal generator, scope and some series resistors, you can find the inductance etc.
Q: According to your measurements when we have a step-up or a step-down transformer?
A transformer is a device that transfers electrical energy from one circuit to another through inductively coupled conductors—the transformer's coils. A varying current in the first or primary winding creates a varying magnetic flux in the transformer's core and thus a varying magnetic field through the secondary winding. This varying magnetic field induces a varying electromotive force (EMF) or voltage in the secondary winding. This effect is called mutual induction. If a load is connected to the secondary, an electric current will flow in the secondary winding and electrical energy will be transferred from the primary circuit through the transformer to the load. In an ideal transformer, the induced voltage in the secondary winding (Vs) is in proportion to the primary voltage (Vp), and is given by the ratio of the number of turns in the secondary (Ns) to the number of turns in the primary (Np) as follows:- Ns/Np Vs/Vp ---------------------(i) By appropriate selection of the ratio of turns, a transformer thus allows an alternating current (AC) voltage to be stepped up by making Ns greater than Np, or stepped down by making Ns less than Np.
Q: I am completing an essay on how to improve the performance of a transformer. Please could someone advise me on the problems incurred in transformers and ways that have been designed to reduce these problems. Thanks in advance
To improve the performance of a transformer, you have to:- [1] Reduce the copper losses, and [2] reduce the iron losses. Copper losses can be reduced by using copper wire of the highest purity; and by using copper wire that has the thinnest possible layer of insulating material on it. These measures, taken together, reduce the length of the wire and therefore its resistance. Iron losses are reduced by using magnetically soft iron in the laminations. This is the type of iron that has a tall and narrow hysteresis loop. Using copper wire with thin insulation (as mentioned above), keeps the windings close to the iron and so reduces magnetic leakage. A third measure is to use very thin laminations and to insulate them with very thin material as this will reduce eddy currents. It is very expensive to build a 99% efficient 250 MVA transformer. The losses in a 94% efficient version, are often cheaper to manage than the extra cost of making the 99% version.
Q: I need to get a transformer that produces 12 watts. Problem is that I burned out the previous 2 I had with my hot wire cutter. I have a couple of 9 watt and 3 watt transformers. Is there anyway I can combine them together so that I can produce the desired 12 watts. I was thinking about rectifying to DC. Then putting the power sources in parallel to power my hot wire cutter. Is this a good idea or is there another way to do it.
Get a transformer that is capable of more than the 12 watts your wire cutter needs. It would be more practical. I don't know what current your wire needs, nor the operating voltage. If one assumed 12V., at 12W the current required would be 1 amp. The operating voltage is the critical factor. Overvoltage will be too hot, undervoltage too cool, assuming no form of temperature control. Divide the 12W by the operating voltage to show the current requirement. Get a transformer that exceeds that figure. AC is fine for such a device. You cannot parallel unlike transformers.
Q: A 2200/220V, single phase transformer has the following resistance and leakage reactance referred to primary: R5.6 ohms, X22 ohms1. calculate V2, I2 and cos(2) when the primary is supplied with 2200V, 9A at a lagging power factor of 0.8. Neglect Io.
you may get the comparable voltage out of one output as in case you had related all 3. the different 2 might have decrease voltages for the period of them. case in point, say the transformer replace into regularly a 240 to 480 3-section delta. in case you have been to place 240 for the period of legs A and B, you may get 480 for the period of A and B of the output. i've got self assurance the B/C and C/A outputs might have 240 each and each, because of the fact the unconnected C section might seem as a midsection-faucet to the A and B legs
Q: Here's photos of the transformer in question. My dad pulled it put of some electronic he can't remember what and has had it sitting on the garage. Recently I've been working with high voltage and he pulled this out for me to work with just having trouble figuring it out
First of all, never assume anything! There are some standard colour codes for transformer wiring, but not all transformers conform to the standards. In practice, a manufacturer can use any colour wire for any purpose, leaving you and me and a lot of other people wondering for all time, which wire is which? The measured DC resistance does not tell you anything about the AC voltage rating of any winding. With most power transformers, if there are two solid-black wires, they are the primary. If it is tapped, or there are two primaries (such as in dual-voltage units), there will be three or more primary wires (and possibly as many as 8!). I think the side with 115V on it is probably -- mind you, this is not for certain -- nothing more than the side on which that number happened to be stamped. So of course you want to know what to do next. Considering the number of wires and the ways in which they are connected, it would seem that an experimental method will be required. The 2006 ARRL Amateur Radio Handbook (and possibly other editions) tells how to figure out what's what with transformers. I strongly recommend using the procedures described therein.
Q: Transformer short-circuit impedance is big good, or small good? The same capacity, voltage ... solution
Transformer impedance depends on the size of many factors: 1 circuit breaker breaking capacity. Transformer below the power distribution board in a variety of electrical equipment has parameters, you should calculate the short-circuit current according to the impedance of the transformer to verify that these electrical equipment can meet the requirements, the same capacity and variable ratio of the transformer, the impedance of small short-circuit current, Large short-circuit current is small; 2 transformer connected to the large-capacity motor starting voltage. If the transformer load capacity close to the transformer capacity of 1/4 of the motor, should pay attention to check its starting voltage to meet the motor terminal voltage above 70% of the rated voltage. Impedance of the transformer itself is also a large pressure drop, more difficult to meet the requirements; small impedance transformer for the start of the pressure drop better performance; because you do not know how much the transformer capacity, it can not be the absolute conclusion, if the transformer capacity below 1000KVA , The above parameters in 4% and 6% or less, the choice of 4% better
Q: explain the role of both types of transformers in getting electric energy from an electric power plant (coal-fired, natural gas or nuclear) to our homes.
The voltage is transformed up to thousands of volts and low current for minimum heat loss in the wires. Then down to low voltage at high current to the neighbor hood.

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