Go Down

Topic: Confused about DataSheet, need help find a cure to a problem. (Read 11718 times) previous topic - next topic

raschemmel

I've seen that too so I would have to say you're both right. Mike is probably right about not doing it without the resistor and you are right that it is done sometimes with the resistor.

crullier

Thank you for all the responses. Lot of information to digest but I am getting there.
In reality as I mentioned I don't need the electromagnet actuated for a very long period of time plus I was able to find a 600 mA version which I need to mearure in my tool box last night.

I have been toying with the idea of using a 555 timer to hold the electromagnet actuated for a brief moment and then release.


Here is a question related to the transistor, I see many board members prefer to use MoFets. Why is this? What is the advantage; It seems that the work the same?

I am also noticing a theme among components, come work with current (transistors) and some work with voltages (555 timer trigger) is this correct?


Grumpy_Mike

Quote
you can easily parallel two or more bjts as long as they are identical and you put a low ohmage hi pow resistor per collector (like 1ohm 1\2W in your case)
Yes you can easily parallel them and they will easily melt.

Quote
I see many board members prefer to use MoFets. Why is this?
Because FETs can switch more current and dissipate less power (not so hot) than transistors. Above 0.5 to 1A you are better off with a FET.

@screwpilot - given this and some recent "advice" you have given, do you actually have any real experience in electronics? You certainly have no real experience in paralleling up transistors.

crullier

Quote
Because FETs can switch more current and dissipate less power (not so hot) than transistors. Above 0.5 to 1A you are better off with a FET.
Mike so maybe I should try a FET. I found this at radioshack today IRF510.




BTW, are the radioshack components "generic" or copies of the brand name ones?

PS: Two transistors in a darlington arrangement are considered to be in series correct?

Grumpy_Mike

Quote
I found this at radioshack today IRF510.
This is not a logic level FET, it requires a gate signal of 10V to turn it on fully. You need a logic level FET, one that only requires 5V to turn it on fully.

Quote
BTW, are the radioshack components "generic" or copies of the brand name ones?
They are real components, but packaged and priced to a ridiculous extent.

Quote
Two transistors in a darlington arrangement are considered to be in series correct?
No they are considered to be in a Darlington arrangement.

screwpilot

PS: Two transistors in a darlington arrangement are considered to be in series correct?
no they are in "cascade"


crullier

Quote
This is not a logic level FET, it requires a gate signal of 10V to turn it on fully. You need a logic level FET, one that only requires 5V to turn it on fully.
Ah, (urgh) how did I miss this. I wish the info on the package was a bit more clear. Or I should say I wish they had noted that on the package and not Vgs 2-4V.

BTW, 

Quote
They are real components, but packaged and priced to a ridiculous extent.
By this I am taking that you are saying they are way overpriced?


Grumpy_Mike

Yes way over priced, many times more than a top end distribuitor. But then you have to pay for a shop and the people in it.

TomGeorge

Hi, if that's an irf510 in the bag, take it back because the specs on the back are not for irf510, even radio shacks web site says that Vgs is +-20V, not 2 to 4V.

Tom...... :)

Everything runs on smoke, let the smoke out, it stops running....

crullier

#25
Jan 28, 2015, 06:54 pm Last Edit: Jan 28, 2015, 09:30 pm by crullier Reason: Been reading all day about mosfets and I need to correct myself.
I think part of the challenge is also my inexperience.. 


I want to add that after some research I realized that Rds(on) is what you guys are looking at when you say the the FET I bought need 10V to be on. However my statement about Vds(th) still applies, to start conducting there needs to be a 2-4V differential to "start" to turn on the mosfet.


Quote
But for instance I do not see that listed anywhere on the data sheet other then as a Test condition for "Drain-Source On-State Resistance".
I am quoting myself: You guys are also looking at Rds(on).



Now with a logic level mosfet all the above goes out the window since it turn on with either 3.3V or 5v.



crullier

Hey guys, I found a good resource that explains things very well for newbs like me,

http://blog.oscarliang.net/how-to-use-mosfet-beginner-tutorial/

the only parts I dont understand which I am hoping you guys can explain is:


Quote
Where to put the load to a MOSFET? Source or Drain?

Because load  has resistance, which is basically a resitor. For N-channel MOSFET the reason we usually put the load at the Drain side is because of the Source is usually connected to GND.

If load is connected at the source side, the Vgs will needs to be higher in order to switch the MOSFET, or there will be insufficient current flow between source and drain than expected.
If I draw a schematic of either scenario, they both seem the same to me. Vcc stops at the gate...


Grumpy_Mike

Can't read that link because it is not formatted correctly for my iPad now the forum software has changed. But the reason is that with the load in the source the voltage on the source can never be more than the voltage on the gate. It is called a source follower.

Paulcet

Maybe you can work it out for yourself:

See image, and suppose we could somehow turn on the mosfet with 12V (this is over the required level, right?) and it would have Rds of .01Ω (Disregard the part #, I'm just pulling these numbers out of the ether!)

What is the voltage from A to Gnd?
What is the voltage from B to Gnd?
What is the voltage from C to B?  (This is Vgs)

Go Up