Hello,
i need a load switch to arduino, which is by default open. I tried to draw some circuits, can anybody tell me, if it will or will not work?
I am not sure wit the resistor values. Or if this is the right way how to do it. I would prefer the scheme on the right side, when there will be Atmega failure it will just let the load to work...
The load will be connected or unconnected for 10-20 hod.
Thank you for any comments.
Jan
The load will be connected or unconnected for 10-20 hod.
What is a hod?
Normally you put the load in the collector of a transistor. As you have drawn it here you have an emitter follower so the load will never see a voltage greater than 0.7V of the base voltage, so that would be 4.3V for the one on the left and 11.3V for the one on the right.
You could move the load or you might have to go for a high sided driver. That is using a PNP transistor instead of a NPN one.
What sort of current are you switching?
Oh... silly mistakes...
This should be better?
hod meand hour... sorry i wrote czech abbrv. The load will be switched on for 10-20 hours, 1-3 Amp max.
Thank you
Jan
Yes that's better.
However for a 3A load you will have to watch the power dissipation. The Vsat for this transistor (voltage across collector / emitter when fully on) is 2V. That means the transistor will be burning 3 * 2 = 6W. So each transistor will need a hefty heat sink to stop it overheating.
If you use a power FET these burn much less power because the voltage drop is so much smaller when they are on.
Which do you recomend?
I want the circuit on the right side, i founf IRFZ44 which is quite cheap. But the resistor values shuld be much higher.
R3 like 100kOhm
R4 like 10kOhm
R5 1kOhm
?
Thank you.
Jan
That should be OK, just replace T2 with a FET and keep T1 as a transistor. There is no need to change the resistors.
The values of resistors were just guesses...
The power dissipation should be now RI^2, so now it should be under 0,1W. What size of current will open this FET transistor? And at what type of current will it close?
Thank you for your advices.
Jan
What size of current will open this FET transistor?
FETs work on voltage not on current, that's why the values of the resistors in the base circuit don't matter much.
So look at the data sheet for the exact turn on / turn off voltage, but I think I remember it takes 10V to fully turn it on, that's why the circuit on the right is best to use.
So uf tehre is not going hig current througth FET, the R3 is useless, isn't it?
Yes but it's not doing any harm.
i need a load switch to arduino, which is by default open.
You say "by default open", but the circuits you propose are for a normally closed switch. That is your load will be powered unless the Arduino is actively pulling the base of the TIP121/BC547 to low.
Also your designs have a series base resistor equivalent of 11k which would limit the load to less than 3A. With a gain of 1000 there is no need however to use a transistor to switch the TIP121 as the Arduino would be capable of doing so directly even for a 3A load.
For a normally closed switch (single transistor) you would connect Arduino directly to the base of the TIP121 and use a pullup/base resistor to 5V (not to 12V as in your T1 design). To switch "On" - you configure the Arduino pin as input with no pullup (this is power on default) and to switch "Off" - you set the pin to output and digital low.
For a normally open switch (single transistor) you connect a base resistor in between Arduino and the TIP121 base. Set the digital pin to output and digital high to switch power on.
To fully saturate the TIP121 you need 12mA base current at 3A and this calls for a base resistor of about 350 Ohm.
As Mike points out however - at 3A the TIP121 will drop 2V (6W) and this would require a good heatsink - so in the end a power FET may altogether be a better choice. Either way you also need to take precautions (add a 3A reverse current diode across the load terminals) if you will be switching an inductive load.