Product Description:
1. pbp series shielded smd power Inductor
2. Rated current:0.510A
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
DCDC conversion in the following application.
 Q:A series RLC circuit is connected across a 10kHz source. The 1.2 H inductor and 1000 ohm resistor are fixed?
 At resonance the inductor and capacitor will be a short. C1/w^2 L where w2 pi f.
 Q:What is value of the inductor?
 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 milihenry God bless you.
 Q:Resistor and inductor in parallel?
 A) VrmsIrms*R therefore Irms Vrms/ R Vrms Vmax (sqrt2) Irms (100)/(50) 2 A B) Since we are talking about a inductor, it will lead by 90 degrees
 Q:A 12.6 V battery is in series with a 30.0 mH inductor and 0.150 ohm resistor connected through a switch. When?
 Let V_s the voltage of the battery 12.6 V Let i the current through the series circuit Let R the resistance of the resistor 0.150 Ω Let L the inductance of the inductor 0.03 H Let V_r the voltage across the resistor (i)R Let V_l the voltage across the inductor L(di/dt) The source voltage must equal the sum of the voltages across the components: V_s V_r + V_l 12.6 V (i)R + L(di/dt) di/dt + (i)(R/L) (12.6 V)/L The integrating factor for this is e^{∫ (R/L)dt} e^{(R/L)t} e^{(R/L)t}di/dt + e^{(R/L)t}(i)(R/L) e^{(R/L)t}(12.6 V)/L The left side integrates as the reverse of the product rule and the right side integrates with the reciprocal of the coefficient with a constant, C: e^{(R/L)t}(i) e^{(R/L)t}(12.6 V)/R + C Multiply both sides by e^{(R/L)t}: (i) (12.6 V)/R + Ce^{(R/L)t} i (12.6 V)/0.150 Ω + Ce^{(R/L)t} i 84 A + Ce^{(R/L)t} We find the value of C by knowing that i 0 at t 0 0 84 A + Ce^0 C  84 A i (84 A)(1  e^{(R/L)t}) To find the time constant set (R/L)t 1: t L/R 0.03/0.150 0.2 s One time constant means that (R/L)t 1 i (84 A)(1  e^1) ≈ 53.1 A The current is 0 at t 0 so V_r R(0) 0 The current is 53.1 A at t 0.2 s so V_r (0.150 Ω)(53.1 A) ≈ 7.97 V The charge rate is di/dt and we have an equation involving that: di/dt + (i)(R/L) (12.6 V)/L Solve for di/dt: di/dt (12.6 V)/L  (i)(R/L) di/dt 12.6 V/0.03 H  (53.1 A)(0.150 Ω/0.03 H) di/dt 154.5 A/s
 Q:Help with capacitors and inductors?
 At t0, the capacitor acts like a short circuit. Therefore, all the current will flow through it. Then, at t 0+, the inductor current will be zero. Current source: L[2]2/s ; Resistor: L[1]1 ; Inductor: L[1]s ; Capacitor: L[1]1/s The current flowing through the inductor (2/s) * [(s+1) * (1/s)]/(s +1 + 1/s) 2/[s * (s^2 + s + 1)] (A/s) + [(B*s + C)/(s^2 + s + 1)] A*s^2 + A*s + A + B*s^2 + C*s (A+B)*s^2 + (A+C)*s +A 2 A 2 ; B 2 ; C 2 The current flowing through the inductor (2/s) + [(2*s  2)/(s^2 + s + 1)] (2/s) + 2*[(s + 0.5)/((s + 0.5)^2 + sqrt(0.75)^2)] + [1/sqrt(0.75)]*[sqrt(0.75) / ( (s + 0.5)^2 + sqrt(0.75)^2) )] The current flowing through the inductor i(t) i(t) 2  2*exp(0.5*t)*cos[sqrt(0.75)*t] – [1/sqrt(0.75)]*exp(0.5*t) * sin[sqrt(0.75)*t] i(0+) 2 2 + 0 0 A ; Note that, cos0 1, sin0 0, exp(0) 1
 Q:What is the maximum magnitude current across inductor in LC Circuit?
 The energy move back and forth between the inductor and capacitor. Each component will store all of the energy of an instant. You can figure out Imax and Vmax by using E.5L I^2 or .5 C V^2 and letting E1 J. Just for clairification Imax and Emax are 180 degees out of phase. The angular frequency is w1/sqrt(L C). The period is 2 pi/w.
 Q:why the winding of inductor for crystal radio cannot being overlapping to each other.?
 They can be overlapping if you do not use the traditional slider on the coil to tune stations. The slider needs to be able to adjust the length of the coil to change frequency and if the coils are overlapping you would limit its range of tuning. In some crystal radio configurations a variable capacitor is used for selecting the frequency and these usually have overlapping windings. Most crystal radio enthusiasts agree that the sensitivity and selectivity is better with a single wound air core coil.
 Q:240 V at 50 Hz applied to an inductor yields a current of 4A calcuate the maximum stored energy during a cycle?
 Assuming the current is 4 amps RMS the impedance is j60 ohms, so the inductance is .191 H. The peak current is 1.414*47.07 amps The peak store energy is .5 L Ip^24.77 joules. There is 0 average power consumed by a lossless inductor.
 Q:AM Radio Reciever?
 Basically the inductor is a coil. When connected to a battery or some other power source in parallel it discharges, or shorts. On the other hand a capacitor stores electricity. In a tuning circuit, these two are connected parallel. So you can imagine the outcome. when power is supplied, capacitor stores the electricity while the stored electricity quickly gets discharged via the inductor. again the same process. This process happens again and again and the outcome is a signal of a fixed frequency. This frequency depends on the capacitor value and the inductance. So the frequency can be varied when one or both of these are varied. Thats where the tuning capacitor comes in. Hence now we can make a variable frequency. In the radio, the antena absorbs all signals from the space that it can absorb, so various frequencies are absorbed. We get a signal with all frequencies in it. Then a selected frequency is absorbed to a transistor. This selected frequency depends on the frequency I explained at first. After going through the transistor we get the desired radio frequency which we listen through the earphones. Hence by varieng the frequency I described first you get different chanels.
 Q:Using resistors instead of inductors in an audio amp crossover circuit?
 Best to get building instructions (radio shack used to have books on this). I changed my home theater to all 4 ohms to get the full spectrum as I like at moderate listening levels. 8 ohm speakers have to be driven hard to sound right in my opinion. Using a crossover I salvaged from a passive subwoofer (with a dual magnet speaker so I got two separate crossovers, they were 8 ohm design) by using 4 ohm 10 subs I got a crossover of 115 Hz. that's about where my 6 1/2 mid woofers go down to without crossover, and I found some really cheep tweeters from China that I double up on because you have to take in mind high freqs don't travel well but these tweeter were high enough that they don't double up with the mid woofers which would create too brilliant a sound. Car speakers at 4 ohms work great but watch for overheating the reciever/amp. My new amp has no problem handling 4 ohm. When using two tweeters keep them vertically in line for clear separation. I added cooling fan (computer case fan) in old amp.
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1. Manufacturer Overview 
Location 
Guangdong,China (Mainland) 
Year Established 
2010 
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US$10 Million  US$50 Million 
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