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[Reply #15 on:]

Have you tried http://www.farnell.com/distributors/philippines.htm and http://philippines.rs-online.com/web/?

[Reply #16 on:]

Have you tried http://www.farnell.com/distributors/philippines.htm and http://philippines.rs-online.com/web/?

thanks. i will review that link.

by the way an out of topic question since you look like an intelligence person.

can i charge panasonic sla 6v battery (Panasonic LC-RB064P) by my 7.5v 500mA ac to dc adaptor?
take note that the battery is 4ah and my charger has only 500mA....

[Reply #17 on:]

Your probably referring to this H bridge diagram, http://s.eeweb.com/members/von_wong/answers/1335122304-h-bridge.gif

Or possibly a simple H bridge chip, http://www.elektroda.pl/rtvforum/files-rtvforum/ckt_mtr_1826.jpg

i notice in your first link. does it use npn transistors?

because i think you can only use npn for switching device by putting the load before pin 2(collectors pin) ... can you put it after the emitter pin?.

[Reply #18 on:]

The schematic in the first link does use NPN transistors. I've used that design for years, and it does work.

[Reply #19 on:]

can i charge panasonic sla 6v battery (Panasonic LC-RB064P) by my 7.5v 500mA ac to dc adaptor?
take note that the battery is 4ah and my charger has only 500mA....

4ah is a measure of capacity, not current. If your 7.5v adaptor has a regulated output, then you can charge your 6V SLA battery through a silicon diode (1N4001 or similar) and small series resistor (I suggest 10 ohms 3W if the battery is really flat, or about 2 ohms if it is still reading 6V). The higher the resistor value, the longer it will take to charge. If it has an unregulated output, you can still charge your battery through a 10 ohm 3W resistor, but you should stop charging it when the voltage reaches the value specified in the datasheet (probably about 6.8V).

[Reply #20 on:]

OP PLEASE STOP CROSS-POSTING.
This wastes time.

[Reply #21 on:]

If your following the first link I gave you then, tip 41, should be fine. But looking at the data sheet, its made for high voltage applications. Kinda overkill for your project. You may just want to get some 2n3904 or 2n3906 NPN transistors from RadioShack, or simply a H-bridge DC Driver IC. 

dude! i came from the store from the nearest city, i it is not available. they give me numbers of transistors and i picked 2n551 since its the nearest number from what you have given.

the question is do the 2n5551 can handle the 7v? i wried its data shit but i dont understand what it means

[Reply #22 on:]

The 2n5551 is fine, the maximum voltage it can handle is 160 volts, so 7 volts should be nothing to it.

[Reply #23 on:]

The 2n5551 is fine, the maximum voltage it can handle is 160 volts, so 7 volts should be nothing to it.

wow, what did you say? 160vots for that little tiny transistor? absolutely incredible offer.

[Reply #24 on:]

the led or the load has no problem switching if its in the collectors pin but
when the led is in the emitter, the led will not light. no switching is done.

as well as the hbridge. it seems impossible to make it work. whats wrong with it? connections of pins are correct.

[Reply #25 on:]

Double check your wiring, and make sure only one side of the H bridge is on at a time.

[Reply #26 on:]

Double check your wiring, and make sure only one side of the H bridge is on at a time.

this half of the hbridge didnt also work. as the first transistor doesn't triggered since the load is in the emitter.

[Reply #27 on:]

I don't think 3.5 volts is enough to drive that circuit. Try 6 or 9 volts, and 5v at the base.

[Reply #28 on:]

That circuit will not work as it stands and may damage the lower transistor because there is no resistor to limit the base current. Put a 100 ohm resistor in series with the base of the upper transistor, and a 470 ohm resistor in series with the base of the lower transistor. Increase both the power supply voltage and the drive voltage to 5V. That will give you about 3V or 3.5V across the load.

[Reply #29 on:]

That circuit will not work as it stands and may damage the lower transistor because there is no resistor to limit the base current. Put a 100 ohm resistor in series with the base of the upper transistor, and a 470 ohm resistor in series with the base of the lower transistor. Increase both the power supply voltage and the drive voltage to 5V. That will give you about 3V or 3.5V across the load.

is this what your trying to express?