Hi Guys,

I have a Circuits as follows:

8.4v Power Source
Voltage regulator / Adjuster (needs a minimum of 4.5 A)
Output

What I need is a switch to be able to turn the circuit on and off.
Previously I've been using fairly large push buttons to make the circuit due to their current carrying ability. But these are very large and to be frank don't look nice at all.

So I was researching and came across transistors. If my understanding is correct I can use a small tactile switch to switch the the high current circuit.
My issue is that I cant seem to find out how to select the correct transistor.

Is there an alternative that can be used or would a transistor be the best thing?
Which transistor should I use? Any pointers on where to go to find out?
Any help is really greatly appreciated.

 

Without a schematic we are only guessing, or, your words are not sufficient for us to come up with an excellent answer.

 

ok, i will try to upload an image here with more details

 

Go Advanced, Manage Attachments, browse, open, upload.
.png is preferred.

 

Ok, Here is the Circuit as it is.

The SWADJ3 is a DC-DC Converter. Its a 25w regulator that can handle 4.5 A.
What I need to replace is the Fire Switch at the top.

Thanks

 

I'd use a MOSFET, such as a IRF540N (only because I have a bunch of them). The 8.4 voltage is a bit low for that ordinary MOSFET, so another choice - a logic level fet - might be wise.

Use a small switch to apply voltage to the gate pin. Connect the source to ground, and the drain pin to the ground for the IC. When the gate goes high, the MOSFET completes the path to ground for the IC, which needs to be wired to the 8.4V. Which raises a question about the outputs when the chip is "hot" but has no ground.

 

Here's an idea.

 

I'd use a MOSFET, such as a IRF540N (only because I have a bunch of them). The 8.4 voltage is a bit low for that ordinary MOSFET, so another choice - a logic level fet - might be wise.

Use a small switch to apply voltage to the gate pin. Connect the source to ground, and the drain pin to the ground for the IC. When the gate goes high, the MOSFET completes the path to ground for the IC, which needs to be wired to the 8.4V. Which raises a question about the outputs when the chip is "hot" but has no ground.

Hmmm.... Ill try to read up a bit on MOSFETs. Thanks

Here's an idea.

Oops. The top terminal of the switch is supposed to be connected to the + of the battery.

Sorry for my ignorance but I dont quite understand that image

 

What part of one transistor and one resistor are you having trouble with?

 

Here's an idea.

You have the source and drain reversed on your schematic. For a P-MOSFET, the source must go the the plus voltage.

 

What part of one transistor and one resistor are you having trouble with?

Im really sorry i should have explained that i dont really have a clue when im looking at a schematic.

I see where the resistor is, Is the transistor where the 3 dots are? and whats the item to the top right of the resistor?

Many thanks once again/.

 

That's the MOSFET, a special kind of transistor.

Just to be clear, the schematic offered is very similar to what I described but cures the concern I mentioned earlier. It switches the high side, making a path to + instead of completing a path to ground. Switching the high side requires a P channel MOSFET instead of the "normal" N type.

I think the "dots" you asked about are just connection terminals to the rest of your circuit.

 

That item on the top right of the resistor is a transistor as demonstrated an pages 1, 3, and 4 of the datasheet.

You have to actually look at the datasheet to understand how the transistor is used, or even what the symbol for a transistor is.

 

The 3 dots are part of what represents a transistor ...

Mmmm...pretty sure he meant "dots" are the little circles at the node points. Hard to know.

 

The three "dots" or circles represent the terminals on a single single-pole double-throw (SPDT) switch. The thing on the upper right is the transistor, a MOSFET.

 

Here's an idea.

Ive been searching for the 2SJ464 locally (within UK) and havent found anywhere that has it, the closest one I could find locally (without ordering) is the IRF540A & looking through all the specsheets it seems to have a slightly higher rating then the 2SJ464 although all voltages are marked in + instead of - with the 2SJ464. Whats the benefit of having the mosfet attached to the negative side of the circuit instead of positive where I currently have my switch placed? Also if I bridge the gate and ground with a 100 ohm resistor would that make it have the same effect as a momentary push switch rather then push-to make/push-to-break?

Also I did some research and your image is all very clear now, sorry for my ignorance.

So would it be ok to use the IRF540A instead or is there something i missed in the specsheets?

 

Thanks for all the info guys, the thread didnt update till I put in a new post

 

So would it be ok to use the IRF540A instead or is there something i missed in the specsheets?

One big thing, maybe two: 1. You need a P-channel MOSFET, and that one is an N-channel. 2. You really should use a logic level MOSFET since your gate voltage is less than the 10V or so needed to fully switch a regular MOSFET.

Just about any p-channel logic MOSFET you can find locally will be fine, since your current load is relatively small. Unfortunately "logic level" is not usually labeled as such and you have to read the datasheets. Find the chart of current versus gate voltage, and look for a device that is fully on at your voltage.

 

The crux of the biscuit is, "what transistor can do this swtching". I pointed out a transistor that can do the job and where to buy it. Though I am not so good at MOSFETs that I got the drawing right, the principle holds. You should go to the top of the page and click on, "USER CP" to enter what country you are in. That makes life easier for the helpers on this site because we have people all over the planet asking questions here.

The advantage to having the transistor on the negative side is that N type MOSFETs are easier to find...as you already discovered.

Turning this into a momentary circuit can be configured after you choose which transistor you will use. Adding a 100 ohm resistor to ground after a 10,000 ohm resistor will only defeat the 10,000 ohm resistor and the transistor will stay stuck in one condition or the other.

I tried to fix the drawing.

 

Just about any p-channel logic MOSFET you can find locally will be fine, since your current load is relatively small. Unfortunately "logic level" is not usually labeled as such and you have to read the datasheets. Find the chart of current versus gate voltage, and look for a device that is fully on at your voltage.

Hey Wayne. Looking through the chart for the IFR540 it states the following:
Gate-Source Voltage +/- 20V

but also says later on:

Gate Threshold Voltage 2-4V

I thought that meant that the gate needed at least 2-4v to fully open. Am i mistaken and its actually 20V?

Heres a link to the Datasheet:
http://www.fairchildsemi.com/ds/IR/IRF540A.pdf

The advantage to having the transistor on the negative side is that N type MOSFETs are easier to find...as you already discovered.

Turning this into a momentary circuit can be configured after you choose which transistor you will use. Adding a 100 ohm resistor to ground after a 10,000 ohm resistor will only defeat the 10,000 ohm resistor and the transistor will stay stuck in one condition or the other.

I tried to fix the drawing.

Many thanks again #12 Ive got a circuit working on my bread boards and it only works with a 10k resistor between the gate and source with nothing prior to the button.
Attached is a picture.
My only concern now is that it may not fully open if i mis-understood the datasheet as wayne pointed out above: