Can use a darlington (Tip110) as a switch ?

[panalog9]

Hi -

Is it possible to use a darlington transistor (Tip110) as a switch ? I want to switch a fairly high-current load at 12V using a microcontroller output pin at 5V. It's a 12V car siren.

I don't actually have a TIP110 here right now, or I would try it.

Darlingtons are usually used for higher currents, right ? Would I still get the full 12V across Vce ?

TIA

-pana

 

[flatulent]

You will have two problems which may not be that bad for you.

The first is that the Vce sat will be more like 0.9 V than 0.2 V which will increase the heating and reduce the power to your siren.

The other is slower on-off switching times which in your case is not a problem.

 

[alphi]

first,how amp current across tip110?
you can use MCU pin to control tip110,but have something to note!
Pls ref:

 

[panalog9]

OK, got one to test with. Strange results ... On the breadboard I have it as this :

tip110 collector to +12V (+ of battery)
tip110 emitter to plug A
tip110 base to 1k resistor with switch to +5V
other side of resistor to GND
plug B to GND (-ve of battery)

Put voltmeter across GND/-ve and emitter. It shows 0V when tip110 is off, and 5V when it's turned on. No difference if the base resistor is changed.

Why doesn't it show the full 12V across the emitter and GND ? I don't have the damn siren yet, so I can't check the output from that. I believe it's about 0.8A at 12V btw.

-pana

 

[flatulent]

OK, got one to test with. Strange results ... On the breadboard I have it as this :

tip110 collector to +12V (+ of battery)
tip110 emitter to plug A
tip110 base to 1k resistor with switch to +5V
other side of resistor to GND
plug B to GND (-ve of battery)

Put voltmeter across GND/-ve and emitter. It shows 0V when tip110 is off, and 5V when it's turned on. No difference if the base resistor is changed.

Why doesn't it show the full 12V across the emitter and GND ? I don't have the damn siren yet, so I can't check the output from that. I believe it's about 0.8A at 12V btw.

-pana

You are using the transistor as an emitter follower. You need to either put the base up to 12V or use the transistor on the ground side of the load. That is ground the emitter and put the load between the 12V and the collector of the transistor.

 

[panalog9]

Ah, OK. I thougt that the darlington could do it. Best to replace it with a P-ch FET then I think.

-pana

 

[panalog9]

Anything special to keep in mind with this ? I'm assuming to run the micro IO pin to the gate via a 1k resistor, Source to +12V and Drain to siren +input. Siren -input to GND.

-pana

 

[flatulent]

You are going to have to make the gate 12V to turn it off and about 6 V to turn it on. One easy way for this is to use a npn transistor emitter grounded and collector to two resistors in series to 12V. Scale them so the mid point goes 12V and 6V. Put the PFET gate to this point. Drive the npn with a series base resistor to your logic.

 

[panalog9]

Dang, I'm out of room on the pcb High-side switching is nicer since you only need to send the switched +12V wire out to the siren. The GND lead on the siren can go right to the body of the car.

If I go with an N-ch fet to switch the GND lead, I'll have to send both +12V and GND wires out to the siren. Not THAT big a problem, since the +12V can come from something close to the siren.

Another option is to use an N-ch fet to switch the gate on the P-ch fet. That's only one small TO-92 footprint, which I t h i n k I can squeeze in.

-pana

 

[Rope]

Usually in the cars the loads are directly connected to the +12V and are switched on by grounding the other side -why dont you use this TIP110 NPN darlington in the way proposed by Flatulent connecting the E to GND and wirinng the C to the siren.You can yse a short cable between the other input of the siren and +12V so you wont need extra wires. I think it is the best way.
PS If you want to reduce the power over the switching element you can use a very low Rdson FET switching transistor with on resistance lower then 0.1 ohm and Vgson about 2.5 V. You can find such part on an old broken computer mothher board there are some of these FETs in the CPU power supply converter.

 

[panalog9]

Hmm - I thought it was usually the other way. The + side of the device was switched, and the - side was connected directly to the chassis.

That way you only needed one wire to the device. If you switch the - side, you need a wire to the +12V of the battery, and 2nd wire to the switch for the - side.

Sticking with the tip110 is the least parts count, and smallest board area. A P-ch fet will require another switching device (npn, n-ch fet) to actually switch it, as per flatulent.

The tip110 only needs to carry up to max about 1A for a short while, 5 seconds or so.

 

[Rope]

I dont think so for example all the switches in the doors are connectet direct to shassis and switch the lamps when the door opens, the same is with the brake lights switch and the manual brake indikator switch etc.
You can use bipolar transistors from Zettex in sot223 case 1 npn and one PNP switch if you dont have enough space the PNP will be in saturation and the voltage across it will be not more then 0.1V and Zettex have tranzistors in SOT223 which can handle up to 3A, so the current isnt problem. For example FZT790A-50V/3A/Hfe=300-800