Ok guys I need some help building a circuit. This is going to be used to activate launch controll on a car engine management unit. Basically I have a 5v pin and when this pin gets grounded is when the launch control system is triggered. The easiest solution would be to just connect the pin to the clutch switch right? Well this works but switch is backwards from what I need it to be. When the clutch pedal is up the pin gets grounded. I need the pin to get grounded when the pedal is pushed down.

I had a few 2n3906 transisters and tried to build something that would fix this problem but everything failed. I have spent several hours working on this and cant figure out anything. Any help will be appreciated. I tried to explain everything but if you need anything clear up just let me know. Thanks guys.
-Jason

 

A transistor can be used as an inverter: http://hyperphysics.phy-astr.gsu.edu/hbase/Electronic/buffer.html#c4

 

You need to use an NPN transistor.
2N3906 transistors are PNP.
2N3904 transistors are NPN.
2N2222 transistors are also NPN, and are much more rugged than 2N3904 transistors.

 

ok thanks for the replys guys. I will look into it some more and see if I can get it worked out.

 

Hey guys, ok I have 2 circuit ideas that seem like they will work ok. This one is the simplest but Im not 100% it will work. Wanted to get some opinions before I build it.

The plan was that when the switch is open there is enough current to turn the transistor on and when the switch is closed most of the current is drained so that the transistor will turn off. Let me know if you have any suggestions.

http://i189.photobucket.com/albums/z173/SilvrEclipse/LaunchControl.gif?t=1220039046

 

You can add photos to your posts by clicking on the "Go Advanced" button.

.PNG types usually work out the best; they don't get blurry like .jpg or .gif types.

When the transistor is biased ON, the diode will get full current through it, perhaps 70mA based on the transistor having a gain (hFE) of 160, with 0.44mA flowing through it's base.

The diode will likely have somewhere betwee 0.7v-1.1v dropped across it, depending upon the specific model. The remaining voltage will be dropped across the transistor, so roughly 2.9v to 3.3v, perhaps 231mW of power dissipated by the transistor.

Replace the diode with a 510 Ohm resistor.

 

I just wanted to add a link so that it wouldn't take up a lot of space on the page. I just threw the LED on the circuit just so I would know if it was working correctly. So the 4v source will get grounded only when the switch is in the off position and wont be grounded when the switch is on right?

 

OK, I didn't know it was an LED. The symbol you used was for a standard diode. You should always use a current limiter in series with LEDs.

So the 4v source will get grounded only when the switch is in the off position and wont be grounded when the switch is on right?

Correct.

 

There is such a small amount of current going into the circuit from the 4v source I didn't think it was really needed. But I could add one just in case.

Thanks for all the help guys. Ill try to get this build tomorrow or Sunday and maybe have it wired up this weekend also so Ill let you know how it works.

 

You don't really need two sources, you know.

See the attached schematics and the simulated oscilloscope outputs.

The 2.2k resistor allows more current to the base of the 2N3904 (about 1.8mA)

The 1k collector resistor may be too high for your load; I don't really know what it is.

You can always use the output of the NPN's collector to drive a PNP transistor to source current.

PNP transistor connections to this circuit:
1) Emitter to +4v.
2) Collector to load.
3) Connect a 1k resistor to the PNP transistor's base, and the other end of that same 1k resistor to the Vout connector.

 

Well the reason I used 2 sources is because the 4v source is a trigger. When that source gets grounded it turns on an output function on an ECU. So this would mean I couldn't use that one source to drive the whole circuit correct?

 

Right you are. I didn't mean to confuse you - you do in fact need another source.

 

ok sounds good man. Thanks for all the help guys. I will let yall know if it works out for me.

 

Thanks again for all the help guys. The circuit worked perfect.

 

Could someone please explain this in really simple terms for me please?

The way I understand the above circuit is that inorder for the transistor to switch there must be a Vbe of (atleast?)~0.7V (so a 0.7v difference between the base and the emitter).

Currently the switch is open so the 5V supply is feeding the transistor? By pressing the switch we're drawing the current towards the ground on the left hand side therefore creating the right amount of Vbe to activate the transistor and cause the 4V output to drop to 0v?


And that is why the two sources needed?

Well the reason I used 2 sources is because the 4v source is a trigger. When that source gets grounded it turns on an output function on an ECU.

I understand that the 4V source is a trigger. When you say "the source gets grounded" you mean when a value of 0V is present the ECU ouput is activated (so it being active low) ?

 

I understand that the 4V source is a trigger. When you say "the source gets grounded" you mean when a value of 0V is present the ECU ouput is activated (so it being active low) ?

This is correct. The the switch is open the transistor gets enough voltage to turn on but when the switch is closed the voltaged is drained faster than it can build up so the transistor turns off. It works something like that.

 

The way I understand the above circuit is that inorder for the transistor to switch there must be a Vbe of (atleast?)~0.7V (so a 0.7v difference between the base and the emitter).

That's right. You don't get any significant current flow through the base-emitter junction until the voltage of the base is roughly 0.6v-0.7v above the emitter voltage when you're talking about a typical silicon-based transistor. There are still some germanium transistors around that have a much lower Vbe before they start conducting, but their use is limited, and they are expensive compared to silicon-based transistors.

Currently the switch is open so the 5V supply is feeding the transistor? By pressing the switch we're drawing the current towards the ground on the left hand side therefore creating the right amount of Vbe to activate the transistor and cause the 4V output to drop to 0v?

When the switch is open, current can flow through the base-emitter junction because there is a current path to the 5v supply via the 10k resistor. This keeps the transistor biased ON, or conducting, so that current can flow from the 4v source through the LED (which looks like a regular diode in the 1st schematic) and down throught the collector to the emitter and ground.

When the switch is depressed, the base of the transistor is connected directly to ground. There is still current flow through the 10k resistor, but it is going directly to ground instead of through the transistor's base. The transistor turns OFF, and since there is no current flow through the collector, the LED turns off.

And that is why the two sources needed?

No, it's because his "4v supply" was actually a 4v signal output; he needed to ground it to activate the device.

 

Thanks for the explanation