PBH Series SMD Power Inductor

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Product Description:

1. pbh series smd power inductor

2. Rated current: 1-10A  

3. Inductance0.5~6000uH

4. ROHS

5. competitive price

 

Features:

1.SMD Power Inductor

2.Magnetic shieled surface mount inductor with high current rabing low D.C resistance

3.Excellent terminal strength

4.Packed in embossed carrier tape and can beused by automatic mounting machine.

5.Various hogh power inductors are superior to be high saturation for suiface mounting.

 

Applications:

Power supplu for VTR,OA equipment Digital camera, LCD television set notebook PC,

portable communication equip,ents, DC/DC converters, etc.

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Q:How do you get resistance and henry from inductor?
L 2 henry and R 1/3 ohm Since the switch was in the A position for a long time before it was switched to the B position at t 0, there is no current in the inductor at t 0+ At t 0+, 1 amp V/1Ω + (1/L)(∫ Vdt) The voltage across the inductor will be V(t) 1(e^ -(t/τ) for 0+ ≤ t 2 seconds where τ L/1Ω L since the resistor is 1Ω The current through the inductor will be i (t) (1 - e^-(t/τ) amps from 0+ ≤ t 2 seconds where τ L/1Ω L since the resistor is 1Ω At t 2- seconds i (2-) (1 - e^-(2/L) amps (1 - e^-(1)) since t/τ 2/L 1 at t 2-, L must be 2 Henry At t 2 seconds the switch is thrown back to A. The inductor current will be i( t2+) (1 - e^-1)(e^-(t/τ) where τ L/R 2/R i(6) (1 - e^-1)(e^-(6/τ) (1 - e^-1)(e^-(1) so 6/τ 1 6/(2/R) 3R R 1/3 ohm
Q:What is the current through the inductor?
At t0, switch closed and a potential came across the inductor. So it is like a Alternating current for a Moment. That alternating current makes inductor to behave like in AC circuit. That affects reduce continuously after t+ time because of dc voltage. There will a opposing electromagnetic field in the inductor due to sudden voltage. that will stop current to flow through the inductor
Q:if an inductor has a parasitic resistance does it effect the resonance frequency, if so how?
No, it doesn't affect the resonant frequency but it does effect the circuit Q or quality factor. Q will effect the sharpness of the bandpass of circuits using the inductor and the losses(to heat) in the inductor. In general, the lower the resistance, the sharper the bandpass. To elaborate on the other poster's answer: For SERIES LC circuits( L and C in series) , the resonant frequency is not dependent on series resistance of L. For PARALLEL LC circuits (L and C in parallel) there is a small effect due to the series resistance of L. However, in practical circuits this effect is swamped by the tolerance in the components. By the same vein, capacitors exhibit leakage that can be modeled as a parallel resistance that will effect resonance. Inductors have parasitic capacitance that affects resonance. These effects are usually ignored by most EE's unless the currents or frequencies are very high since the component tolerances of 10%-50% or greater will swamp these effects. So, the answer they want on the exam depends on whether this is a high school, university or graduate school class. Talk to your teacher.
Q:An RLC circuit contains a 4.100e-6 F capacitor in series with an inductor. What should the value of the induct?
Capacitive reactance, Xc 1/WC Inductive reactance, XL WL At resonace the capacitive and inductive reactance are equal. Xc XL Therefore, 1/W x C WL L 1/(W^2 x C) L 1/(63 x 63 x 4.1 x 10^-6) L 61.45 H. Wow! that's a big inductor. Are you sure W is correct?
Q:Inductor time constraints question?
Please convert your 10 mH to 10000H and do your calculation. The time constant is L/R 10 sec. The time constant gives you 63.2 percent of current. See the formula I (E/R)*(1 - Є^-Rt/L). Є^-Rt/L always gives you a positive result whatever the R,t,L is. So (1 - Є^-Rt/L) always less than one. That is why the maximum value of current theoretically never been reach unless t is very long to make Є^-Rt/L minimum.
Q:A current in a 100uH inductor is known to be i(L) 20te^-5t for t>0. Can you show me how you got the answer?
I don't have time to check the work, but here is the general idea. i(l) integral [v(t) dt], in this case, for t0 i 20te^-5t is given A) e(t) voltage across the inductor Ldi/et L20[ e^-5t(dt/dt) + t d(e^-5t)dt] L20[e^-5t + t(-5e^-6t)] (20 e^-5t)(1 - 5te^-t) B) Power e(.1) times i(.1) C) Absorbing, since e(.1) is still positive D) E (1/2) Li(.1)^2
Q:Wiring inductor fan? help?
It all comes down to wattage and what the wiring in your walls is and how much wattage is being consumed at the outlets. If your room wiring is 12 gauge that is usually made to handle a 20 amp fuse so 120V * 20A 2400 watts of capacity So if that circuit is good for 2400 watts, I would unplug anything else that is plugged into your strip to keep the wattage there available for your big fan. Buy a Kill-a-watt meter from Home Depot or Walmart. They are very handy little meters and you can plug your power strip right into the Kill-a-watt meter and see directly how much wattage is used as you plug in loads to your power strip. The cord gauge being 18 is pretty light. But usually they have a built in fuse that will blow if you exceed the wattage of the power strip for a minute or so. Try to plug into the wall recepatacle directly for the big fan as the wiring is probably 12 gauge and will take the load a lot better. Look for a input label on your 500 cfm fan. Check the amperage draw at 120V. Usually fans aren't real bad once they are up to running speed.
Q:At a given instant the current through an inductor is 50.0mA and is increasing at the rate of 115 mA/s?
Energy in an inductor 0.5 L I^2 0.5 (60 x10^-3)(50 x10^-3)^2 7.5 x10^-5 J 7.5 x10^-4 J 0.5 (60 x10^-3) I^2 I SQRT [ 7.5 x10^-4 J / 0.5 (60 x10^-3) ] 1.58 x10^-1 Amps 158mA increase needed 158mA - 50mA 108mA time 108mA /115mA/s .939 sec
Q:What would be the voltage across an inductor just after the battery is disconnected?
In theory infinite. In practice there will always be some resistance/reactance in parallel with the inductor that will limit the voltage in accordance to ohms law.
Q:Inductor Testing, anyone?
TESTING INDUCTORS An inductor is a device consisting of one or more windings of wire with or without a magnetic core. Frequent causes of inductor coil failures are shorted turns, open turns, and changes in inductor value. Small power supply transformers are similar in construction to inductors and can be tested with the capacitor-inductor analyzer shown in figure 4-30. Inductors may be tested in the circuit, but the circuit impedance will have some effect on the readings. It is recommended that you remove the inductor or transformer from the circuit before you perform any tests.
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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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