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Topic: Preventing shorts and power failures (Read 2 times) previous topic - next topic

lost_and_confused

Ok, here is what I think may be what you need.

Required:
1 POT/variable resistor.
1 pushbutton swith.
1 power NPN transistor.
1 power PNP transistor.

I can't attatch a picture because just now I can't get the drawing thing working.

Basically:

Power supply
Emitter of PNP transistor.
Base of PNP connected to pushbutton switch.
Other side of pushbutton swith connected to Collector of PNP transistor.
Collector of PNP connected to Pot (stationary part) and Collector of NPN transistor.
Base of NPN transistor connected to to the moving part of the pot.  More on this later.
The other "stationary part" of the pot is connected to the Emitter of the NPN transistor.
The Emitter is also connected to the "LOAD".

What is going on - as far as I can work out:
At normal everything is not conducting.
The POT needs adjusting so with the load connected it turns  on the transistor.
To test, you will need to keep the press button pressed.
At one extreme of the POT it should turn off because the base current would be too low to turn on the transistor if there was a short.  This is where the "Load" comes into the equation.
As it isn't a short, there will be a voltage drop over it and so with a bit of tweaking, the pot will set the trip voltage to turn off the NPN transistor.  So from there let's look at the PNP.  It is turned off because of the base being not connected to anything and so there it  would not be allowing any current.  When the button is pressed, it takes the base low and it turns on.  In doing so, the base current starts to flow and the NPN turns on.  The load is active.  If there is a short, the base current on the NON stops and so the NPN turns off.  Doing so the PNP is turned off because its base it taken high.

Hope that works.


pgmartin

#11
Oct 27, 2012, 03:37 am Last Edit: Oct 29, 2012, 05:11 pm by pgmartin Reason: 1
Dear lost_and_confused,
I tried to understand your idea, but wasn´t able do it and, to get a working circuit.
But your help drove me ina a way out of SCRs and into simple transistor solutions. 

So I kept searching and found this circuit:


This one has passed a simulation, now I´ll have to build and test it.

lost_and_confused

H again.

Ok, I am glad you found what you needed.

But just for the record, here is my idea.

You will have to read my earlier post to get an idea of how I think it should work.

If you use bigger transistors (higher power) then you can use more current.   Ofcourse the big trade off is the power used by R1.

I don't know if it IS needed, and may not actually be needed.


pgmartin

#13
Dec 04, 2012, 04:34 pm Last Edit: Dec 04, 2012, 05:01 pm by pgmartin Reason: 1
I built the circuit from my last post (Or you can find here http://320volt.com/12v-ve-5v-600ma-irf9530-mosfetli-akim-sinirlama-devresi/). Worked like a Charm. Added a couple of LEDs to monitor if it was in normal mode or in overcurrent protection (One at the Gate and one at the Drain of T2, plus 1K resistors in series).

I understad the basic working principle of this circuit: when the voltage drop in R1 is bigger tha 0.7V, T1 starts sto conduct and shuts T2 off. Thats the easy part.

What I do not get is the role of the two capacitors. Any ideas?

C1 is quite big, and and of the drawbacks of this solutions is that it has quite a big charge after plugging it off from the power source.

MarkT


PG,

Couple of things:

As far as I know, SCR's aren't "Conducting at startup".  They need a signal on their gate to start them conducting.

Also, I didn't know SCR's do current limiting.

They are ON or OFF.

I have bashed a couple of thoughts out but haven't tried them as similar to the origina circuit.
Few components and SHOULD work.

I can try to make a JPG of the schematic if you are interested.  But it is really difficult to make drawings with the programs I have at hand.



SCRs or thyristors turn on when when there is current from the gate OR when the rate of change of voltage is high enough (due to internal
capacitance).  Simply connecting the supply might trigger conduction due to the sudden increase in voltage.

And yes you are right they don't do current limiting other than via melting!
[ I won't respond to messages, use the forum please ]

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