dc -dc convertor help.

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Click, go advanced, select manage attachments, click add files and click select files. Find the fie you want to attach, click open, then upload files and click insert attachment.

I'll post some screenshots if you have trouble.

By the way, use PNG or GIF, for schematics, not fuzzy JPG.
 
I designed my circuit by mosfet but,now there is a problem.

In real state when i start fetch much current.so R2 will be so hot.
 
now i achieved to my goal but a little problem!!!!

please help abut R2 in above circuit.

what's your suggestion to improve this circuit ,that work properly without increasing R2 temp.
 
Here's where simulators in the wrong hands can be dangerous.

You don't need R2.

Connect it up like this.

It only worked before because the transistor was short circuiting the relay causing R2 to get very hot.

You'll need to reverse the signal applied to the MOSFET, before turning the MOSFET on, turned the relay off, not turning the MOSFET on, turns the relay on.

EDIT:
I got the diode the wrong way round. I've corrected it for you.

There also needs to be a current limiting resistor in series with the LED.
 

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The diode isn't critical, 1N4001 will do.

There should be two connections for the realy coil, look at the datasheet.

R1 can be any low value resistor 10R to 100R.

The resistor in series with the LED depends on the forward current and voltage.

Why are you using a relay to switch an LED? Why not just power it from the 5.9V supply?
 


I'm using 24V relay beacuase it is my first goal to drive a relay.

secondary,it uses 100-200mA current so i can drive so many components that need medium level current.

can you improve your circuit to 1 wire in output.
LED is only an indicator lamp to show circuit is running.

"Why not just power it from the 5.9..."
it comes from another circuit that has 5.9VDC & 5mA.
 
You need two wires for the relay coil.

The indicator LED is normally connected across the coil.

What are you going to be switching with the relay?
 

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Dear magnatro

I like to use it,but I have only 5mA current.
do you know optocouplers work by 5mA properly?
i read MCT2E datasheet.
what does "pin 6(base)" work??
do you have any driver circuit for thisIC?
 
Dear hero999

i want to use it in industrial field.
our circuit isn't good.
perhaps people want to use this circuit for another objects like instruments ,sensors, etc.
it will be better to have 1 wire for 24VDC and 1 wire for GND.
 
So you need to switch the positive, not the negative, then you need some level shifting which requires two transistors.
 

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jorjo said:
Dear magnatro

I like to use it,but I have only 5mA current.
do you know optocouplers work by 5mA properly?
i read MCT2E datasheet.
what does "pin 6(base)" work??
do you have any driver circuit for thisIC?
Click to expand...


there is no need for a driver IC the mct 2e just only need a current limitter for high current applications..
 

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and if there's only 5mA is current, no need for the resistance in series with the optocoupler input..
the base pin must left unconnected. ok?
 
Why use an opto-coupler?

The relay already provides isolation.

Anyway, the opto-coupler is connected up incorrectly and won't work.

If you're going with an opto-coupler, get rid of the relay and use it to drive the transistor directly, so you have a solid state relay.

See attached but it's only good for up to 800mA and looses about 1V at full load.
 

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Dear maqnatro

It doesn't work by 5mA.
because it need 40mA minimum to run optodiode.
can we work around transistor circuits?
I think,it is better to focus 1 target.
I need only 1 wire in outlet for 24VDC and 1 wire for GND.
 
I labelled both transistors as Tr1, which I've now corrected.

It's not critical.

Tr1 can be any PNP transistor rated to 30V and 300mA or more, e.g. BC327.
Tr2 is any NPN transistor with a voltage rating of at least 30V, e.g. BC548.
R2 needs to be rated to at least 0.4W

The diode is not important, 1N4148, 1N914 or 1N4001 will do.
 
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