PBP Series Shielded SMD Power Inductor

Ref Price:

Loading Port:: China Main Port

Payment Terms:: TT or LC

Min Order Qty:: 1000 Pieces pc

Supply Capability:: 20,000 Pieces per Day pc/month

Inquire Now

Add to My Favorites

OKorder Service Pledge

Quality Product, Order Online Tracking, Timely Delivery

OKorder Financial Service

Credit Rating, Credit Services, Credit Purchasing

You Might Also Like

Muti-functional Inverter with Battery Charger 1500 Watt 12V 220V

Frequency Inverter

0.4KW~800KW CMAX Control AC Motor Inverter Frequency

CE Approved CMAX Inverter Frequency

AC/DC Adapters 8W USB SAA Certificate

AC/DC Adapters 100MA-800MA 8W USB BS, UK Certificate

AC/DC Adapters with Line UL, PES Certificate

AC/DC Adapters 900MA-1000MA 12W USB CE GS Certificate

1. pbp series shielded smd power Inductor

2. Rated current:0.5-10A

3. Inductance0.5~6000uH

4. Quality assured

5. competitive price

smallest possible size and high performance they are with high energy storage

Feartures:

The Surface Mount Inductors are designed for the smallest possible size and high performance,

They are with high energy storage and very low resistance making them the ideal inductors for

DC-DC conversion in the following application.

Q: Help! I am having trouble trying to compute the inductor current changes in a parallel R (1Kohms), and L (1mH). They are being driven by a current source of 1A. How do I verify that the current flows in the inductor changes by 63% of its initial value in one time constant.Also, How to,1. Analytical derivation of voltage and current across L2. How the specific values of R and L affect the transient behavior of voltage and current across the inductor3. Compare the analytical solution with the numeric results (I used LT SPICE to successfully obtain the voltage and current waveform across R and L)any help on any of the questions, particularly, the 63% question and Q1 and Q2 would be greatly appreciated. Thanks.: This problem is easy to solve using LaPlace transforms For the circuit indicated; The time constan, tc L/R In s domain; Solving for inductor current, Convert R and I to thevenin equivalent, a voltage source I*R in series with R Ohms. IL(s) (I*R)/[s*(R + s*L)] Transforming to the time domain; IL(t) I(1 - exp[(R/L)*t] IL(t) I(1 - exp[(1/tc)*t] at one time constant t tc t L/R Then, IL(t) I{1 - exp[1]} IL(t) I(1 - 0.367879) IL(t) I(0.632121) IL reaches 63.2% of maximum in 1 time constant Solving for volatge across the inductor, VL(s) I*R/(s + R/L) then; VL(t) I*R*exp[(R/L)*t] VL(t) I*R*exp[t/(tc)]

Q: A 4 mH inductor is connected to an AC voltage source of 141 V rms. If the rms current in the circuit is 0.85 A, what is the frequency of the source?: reactance X V/I 141/.85 165.88 ohms X 2(pi)fL f X/2(pi)L 165.88/2x3.14x4x10^ --3 6603 (check the calculation)

Q: All the current from the wire before it will go through the inductor because there is no resistance, right?: If you're working with DC, and inductor is a short circuit. If you're working with AC, the inductor has an impedance and is not a short. The higher the frequency the higher the equivalent impedance. Z jwL (where w 2*pi*f)

Q: An RLC circuit with a capacitance of 0.010 uF is found to resonant with a frequency of 1.30×104 Hz. What is the value of the inductor? Also what are the units of the inductor. Thanks: X ind2TTfL , X cap1/(2TTfC), at resonance,X ind Xcap,2TTfL1/(2TTfC), 4TT^2f^2LC1, L1/(4TT^2 f^2 C)1/[4TT^2 (13000)^2 (0.01 x 10^-6)] L0.014988 henry14.98 mili-henry God bless you.

Q: If a series RC, RL, or resistor-inductor-capacitor (RLC) circuit was given to you in a black box (with access only to the two input terminals), how could you determine the component values?: w 2pi*f 2pi*60 377 rad/s Z R + j(wL - a million/wC) the place j sqrt(-a million) (in case you do no longer comprehend this line, pass directly to the subsequent) |Z| sqrt[R^2 + (wL - a million/wC)^2] |Z| sqrt{4.84e4 + [377*27e-6 - a million/(377*16e-6)]^2} sqrt[4.84e4 + (a million.02e-2 + a million.66e2)^2] 275.6 ohms (word thet the inductor seems especially very like purely a chew of cord with L0) Then rms i one hundred twenty/275.6 0.435A answer

Q: A capacitor C charged to voltage V is discharged into an inductor L. What is the voltage on C at the instant when its stored energy and inductor's energy are equal?: Short answer: energy in capacitor goes as square of voltage across it, and since there are not dissipative losses (ideal inductor and capacitor), there will be half the energy in the capacitor and therefore equal energy in the inductor when the capacitor voltage is V / sqrt(2) Long answer: Energy in inductor is (1/2)Li^2, energy in capacitor is (1/2)Cv^2, and i C dv/dt, so energy in both components is the same when L i^2 L C^2 (dv/dt)^2 C v^2 LC (dv/dt)^2 v^2 But we also know that the voltage as a function of time will be v V cos( t / sqrt(LC) ) dv/dt -V / sqrt(LC) sin( t / sqrt(LC) ) Substitute this expression for dv/dt into above energy balance to get LC (V^2 / LC) sin^2( t / sqrt(LC) ) V^2 cos^2( t / sqrt(LC) ) sin( t / sqrt(LC) ) +/- cos( t / sqrt(LC) ) The first point of equality is when t (Pi/4)sqrt(LC), and substituting that into expression for v gives: v V cos(Pi/4) V / sqrt(2)

Q: For an ideal conductor in an ac circuit explain why the voltage across the inductor must be zero when the current is at the maximum: I would answer as i know: From equation V L (di/dt), it tells us something. The nature phenomenon of an ideal or any practical inductor is that it will protect current not to change rapidly.For example, if the current at first condition is 5 Amp. but suddenly you increase it to 6 Amp. the inductor will act aginst the change so it will increase by some amount of time not suddenly increase.The time depend on many factors-induction,etc. At the maximum point of waveform,the current is quite constant.So,it's not nessessary for the inductor to act against the current.The Vol. so would Zero. Does this help?

Q: An ac circuit has a 20 ?F capacitor, an 80 mH inductor, and a 60 Ω resistor in series. If the voltage source has an angular frequency of 1000 rad/s, what is the phase constant for the circuit?: First calculate the reactance of the inductor and capacitor XL ωL 1000*80x10^-3 80Ω XC 1/ωC 1/(1000*20x10^-6) 50Ω So the phase constant φ arctan((XL - XC)/R) arctan((80-50)/60) 26.6o

Q: How would you go about fixing (if possible) an electrical inductor on a comp. monitor?: Please!!! do not try to do this your self with out the proper equipment and training. CRT monitors can hold charge for up to 2 weeks so if not properly grounded or protected you could be SERIOUSLY injured.

Our products have highest quality and competitive prices.Our well-equipped facilities and excellent quality control throughout all stages of production enable us to guarantee total customer satisfaction. As a result of our high quality products and outstanding customer service, we have gained a global sales network.

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

2. Manufacturer Certificates

a) Certification Name
Range
Reference
Validity Period

3. Manufacturer Capability

a) Trade Capacity
Nearest Port
Export Percentage	41% - 50%
No.of Employees in Trade Department
Language Spoken:
b) Factory Information
Factory Size:
No. of Production Lines
Contract Manufacturing	OEM Service Offered Design Service Offered Buyer Label Offered
Product Price Range