Hi everyone. I'm looking for a transistor that will switch a 12vdc supply voltage on and off using a 2 to 2.5vdc trigger. I can't seem to find one that will do this as there are thousands to choose from. Anyone have any ideas of what switching transistor might accomplish this? Much thanks to anyone who can help.

 

Does the trigger signal just go from 2V to 2.5V or from 0V to 2-2.5V?
If it's the former, then you could use a comparator such as an LM339 to detect the change and use that to drive a transistor.
If it's the latter, then you can use just about any transistor directly of the proper voltage and current rating.
What is maximum current that you want to switch?
Can the switch be either high-side or low-side?

 

Does the trigger signal just go from 2V to 2.5V or from 0V to 2-2.5V?
If it's the former, then you could use a comparator such as an LM339 to detect the change and use that to drive a transistor.
If it's the latter, then you can use just about any transistor directly of the proper voltage and current rating.
What is maximum current that you want to switch?
Can the switch be either high-side or low-side?

Hi crutscow....thanks for the quick response. The trigger voltage would go from 0vdc to 2vdc or just above that. I would like to switch on and off a 12vdc 5amp supply voltage for some LEDs. Not sure If id need high or low side. Anything to switch that voltage on and off with that trigger voltage will work. Thanks again for your help!!

 

What details do you have about the power supply that are not secret? Power level, control input, etc.

ak

 

What details do you have about the power supply that are not secret? Power level, control input, etc.

ak

Hi AK.....Nothing secret about it, basically just a 12vdc 5A "Wall Wort" power supply thats hot all the time and the transistor doing the switching with that trigger voltage. Thanks!

 

does your solution HAVE TO BE a transistor. Relays that operate on 2 volt coil voltage are available. A quick search revealed this one:http://www.mouser.com/ProductDetail...GAEpiMZZMtGt%2bn33CgIP49XJpVNs/lYXCzQ5rU67w0=

 

Hi Kermit...Yes preferably a transistor. The switching can be really fast at 250 triggers a minute, and I'm not sure if a mechanical relay could keep up with that depending on the response time..... I researched some solid state relays and actually bought one to test with but was not able to trigger the relay with the 2v. After this I thought that a transistor should be able to do it. Thanks for your replies!

 

By wall wart, I assume that this supply does not have an enable input, so you're asking about switching the full 12 V / 5 A on and off with a power device? If so, a P-channel power MOSFET driven by an N-channel signal MOSFET like 2N7002 should do what you have described so far. What is the load? Can it produce inductive kicks? Any rise time or fall time requirements?

ak

 

Hi AK....It looks like the 2V7002 might do it looking at the specs of it. I'm going to order a few and do some testing. I'm also looking at a few converters to up the trigger voltage 2VDC to 5VDC so I have more options to work with. Thanks so much for your replies and knowledge on the subject. I will post my results once I get the testing done.

 

The 2N7002 will not switch the power supply output directly. It is a voltage level translator between the 2.0 V control signal and the 12 V power supply output. It must be combined with a power MOSFET.

ak

 

2V @ 13.5uA switches 12V@ 1/2A to a grounded load. What is not to like?

View attachment 87265

 

12V @ 13.5uA switches 12V@ 1/2A to a grounded load. What is not to like?

View attachment 87265

First datasheet I find says 115mA continuous and the op requested 5A. I guess it depends on what the actual load is, but the 2N7002 isn't up to switching a full 5 amps!

 

First datasheet I find says 115mA continuous and the op requested 5A. I guess it depends on what the actual load is, but the 2N7002 isn't up to switching a full 5 amps!

My schematic/sim shows an IRF7204 with is about a 5A PFET. There are 100s of PFETs avail that would work in the this circuit without modification...

 

Copy that AK. Thanks. I will ad a mosfet to the circuit.

Hi Mike, Thanks for the schematic. I'm going to build it and see what happens.

Thank you for your help everyone!!!

 

My schematic/sim shows an IRF7204 with is about a 5A PFET. There are 100s of PFETs avail that would work in the this circuit without modification...

Oops, sorry. Read post without looking at schematic and misunderstood you - thought you were saying the little one would work by itself. I see now what you've done. Sorry for the mix up.

 

My schematic/sim shows an IRF7204 with is about a 5A PFET. There are 100s of PFETs avail that would work in the this circuit without modification...

R4 is too big and the Vce may not get into the saturation status very well, R3 maybe a little big and then it can't make the Vds of Pfet get into the saturation very well, and the Rds will reach up some more, so it can't reach to 5A, please see the datasheet.

If the Pfet to switching the 12V/5A power, at least using 3 times of 5A as 15A, the better is 5 times of 5A as 25A.
Calculation for IRF7204:
Vds = 5A*0.06Ω = 0.3V.
W=V*I=0.3V*5A=1.5W
1.5W*5=7.5W (for practical application)
IRF7204 only has 2.5W, so it is not enough, even it can be damaged.

 

R4 is too big and the Vce may not get into the saturation status very well, R3 maybe a little big and then it can't make the Vds of Pfet get into the saturation very well, and the Rds will reach up some more, so it can't reach to 5A, please see the datasheet.

There is absolutely no requirement that Q1 be driven to saturation. The min Hfe for the 2n3904 is 40 for low currents, so the 13.5uA into the base will sink 540uA, which produces about -8V of gate drive to the PFET; more than enough to turn on the PFET!

If the Pfet to switching the 12V/5A power, at least using 3 times of 5A as 15A, the better is 5 times of 5A as 25A.
Calculation for IRF7204:

Vds = 5A*0.06Ω = 0.3V.
W=V*I=0.3V*5A=1.5W
1.5W*5=7.5W (for practical application)
IRF7204 only has 2.5W, so it is not enough, even it can be damaged.

I got it into my head that the load requirement is 1/2A as it shows on the schematic "your 500mA load"; not 5A. Just use a TO220 PFET instead of the surface mount one. The gate drive is more than good enough...

 

There is absolutely no requirement that Q1 be driven to saturation. The min Hfe for the 2n3904 is 40 for low currents, so the 13.5uA into the base will sink 540uA, which produces about -8V of gate drive to the PFET; more than enough to turn on the PFET!

I knew you will said so, before I come here, I didn't care much about the saturation status of bjt, almost concentrated in the practical values of Vce through measuring, how low is it, does the Vce lower than 0.2V yet.

Just for the Vgs of IRF7204, when Vgs=-10V, Id=-5.3A, so I think when Vgs=-8V, Id can't reach to -5A.

I got it into my head that the load requirement is 1/2A as it shows on the schematic "your 500mA load"; not 5A. Just use a TO220 PFET instead of the surface mount one. The gate drive is more than good enough...

I knew you used a different current to do test from your circuit, but I still have to pointed out, that's not for you, it is for other members who didn't know much about the mosfet as you.

 

i think this will do the job. please see attachment.
relay must be 6V on the triggering site and 12V 5A on the power side.
If relay can't triggered use a diode on 7805 between ground pin and ground. This will add 0,7V on the 7805 output.
Free wheel diode is a must on the relay trigger side.
as you have the 2-2,5 V on the transistor gate you have 12V on the load.
If 2V signal is a pulse and you need the power circuit to stay on then is a different circuit.

Hope it helps.

 

djb:
Are you trying to burn the bjt?