Ferrite Chip Inductors

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Product Description:

1.ferrite chip inductors

2.leaching resistant rerminations due to metal teb electrodes

3.Coils encapsulated in heat-proof

 

Features

1.Ferrite chip inductors

2.leaching resistant rerminations due to metal teb electrodes.

3.Coils encapsulated in heat-proof resin make high accurate

4.dimesions and resostant dimensionsand resostant to mechanical shock or pressure.

5.High resistance to heat and humidity.

 

Applications

Personal computers. Disk drives and comuter peripherals. Telecommunications devices. VCD, DVD and TV circuits,

Test equipment. Electronic control control boards for automobiles.

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Q:what does Inductor do?
An inductor delays a change in current (so to speak). If an inductor is connected to a d.c. source through an open switch; at the instant the switch closes, full votage is applied across the inductor, but current does not flow due to a phenomon referred to as reverse electromotive force or reverse voltage. As a magnetic field builds in the inductor, current begins to flow, with the magnitude of current gradually increasing until the magnetic field builds completely. The inductor then acts as a short circuit. The time delay is a function of circuit resistance multiplied by inductance. At the instant the switch opens, the current in the inductor continues to flow until the field begins to collapse, at which point the current gradually reduces to zero.
Q:An inductor L of Henry is connected across an A.C. source of 220 V, 50Hz?
It depends upon the value of the inductor.
Q:Can I overlap inductor coils?
Capacitors and inductors within the circuit will make the viberations at desired frequency to drift at slash resistance than different (undesired) vibrations. As a consequence reducing the noises in reception. Variable capacitors are exceptional to high-quality tune your stations. However the added circuitry will mean more vigor requirement which defies use of the term 'easy Radio'
Q:A student wants to figure out if an unlabeled item is a capacitor or inductor. He applies a voltage with dife?
inductor. Inductive reactance is proportional to frequency, so it goes up as frequency goes up. Since Current is inversely proportional to reactance, as frequency goes up, current goes down. A cap would be the opposite. .
Q:Determine the energy stored in the inductor?
V L(di/dt), i (1/L)∫V(t)dt ; this is the current in an inductor If V 112 sin(2π87t) ∫112sin(2π87t)dt -0.2 cos(547t); this is the integral of V(t)dt i -13.3 cos(547t). this is the current in the inductor i -13.3 cos(547*0.0076) -13.26 amps at 7.6 milliseconds Energy in an inductor (1/2)(L)(i^2) E (1/2)(0.015)(-13.26)^2 1.32 joules
Q:Physics-- Magnetism, resistance, and inductors?
Once the current has reached its max. then all the power will go to the resistor, so: a) p v^2/r 9.5^2/4.92 b) see a) above c) 0 d) I don't remember these equations, sorry.
Q:1300 Yamaha air inductor?
No.
Q:Can you help me how to tap an inductor of 40 turns? It is tapped at 25%.?
For small air-core inductors there are a couple of methods that I have used. 1. Wind 10 turns and carry the end lead out about 1 inch away from the coil, then carry the wire back next to the 1 inch lead, back to the coil and continue winding for another 30 turns. Scrape off the insulation of the double-wire lead and use that as the tap. 2. Wind 10 turns. Scrape off the insulation at the 10th turn, leaving an opening to solder a wire on the coil. Continue winding 40 turns. Solder a wire onto the scraped off opening for use as the tap. I like the first method best, but both will work. .
Q:physics, current, circuits and inductors?
After the switch is closed, the initial current will only flow through the resistor, since current through an inductor cannot change instantaneously. The steady state current will only flow through the inductor due to lower impedance. To make things easier, we can change the parallel current source and resistor to their thevenin equivalent series voltage source and resistor. Vth I*R Rth R So now the series equations will apply: v.L(t) (IR)exp(-tR/L) L*di/dt i.L(t) I*(1-exp(-tR/L)) Resistor inductor currents are equal when: i.L(t) I*(1-exp(-tR/L)) 0.5*I Solving for time t gives: t -(L/R)ln(0.5)
Q:Intro to Circuit analysis help! Inductor, short and open circuit, and capacitor problem! help please!?
resistors and inductors in series add up R12R1+R2 resistors and inductors in parallel dont: 1/R121/R1 + 1/R2 or R121/[(1/R1) + (1/R2)] or R12R1*R2/(R1+R2) remember this formula, it will be time saver on tests (even if it only for two elements) for capacitors it is the opposite: parallel capacitors add up C12C1+C2 series capacitors dont: C12C1*C2/(C1+C2) a) Inductors - output is open circuit (nothing connected to it). in this case D and B are in, series DBD+B. this is in parallel with A, so ADB AD*B/(AD+B) this ADB is then in series with C so we get ABCD ADB+C using given numbers we get Ldb4+26H La1H Ladb6*1/(6+1)6/7H0.857H Lc3H Labcd3+0.8573.857H now the output is shorted and B and C are in parallel: LbcLb*Lc/(Lb+Lc)2*3/(2+3)6/5H1.2H this is then in series with A so LabcLa+Lbc1H+1.2H2.2H finally this is in parallel with D so LabcdLabc*Ld/(Labc+Ld)2.2*4/(2.2+4) b) capacitors, first output is open: you get Cb in series with Cd. this is then in parallel with Ca. finally Cabd is in sereis with Cc. numerical results will match second part of a. then output X-Y is shorted and you get Cb and Cc in parallel CbcCb+Cc. This is in series with Ca CabcCa*Cbc/(Ca+Cbc) finally this is in parallel with Cd. CabcdCd+Cabc and results must match first part of a.
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1. Manufacturer Overview

Location Guangdong,China (Mainland)
Year Established 2010
Annual Output Value US$10 Million - US$50 Million
Main Markets North America; South America; Eastern Europe; Southeast Asia; Africa; Oceania; Mid East; Eastern Asia; Western Europe
Company Certifications ISO 9001:2000

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