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I'm wondering if there is a transistor which, when a voltage is applied to the base / gate, opens the circuit instead of closing it. If there is not, how would such a configuration be created?

EDIT: Okay, here's some clarification. I have a circuit set up like this:

Motor-------------Nmos-------------Gnd
                   |
                   |
            voltage from IC

Currently, I am supplying a voltage to the gate of the nmos from my arduino and controlling the motor that way. I want to replace the nmos with something which will perform oppositely. As the voltage increases on it, the current between motor and gnd should decrease.

I hope that helps clarify.

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Are you asking exclusively for a one component solution or for a circuit too? –  alexan_e Dec 21 '13 at 9:01
    
Standard NPN. Connect base to a reference voltage - say 0.8V. Apply drive to emitter. As Vemitter rises transistor will turn off. Main issue is that drive input must sink load current when transistor is on. –  Russell McMahon Dec 21 '13 at 14:20
    
You say the current should decrease, does this mean that you are not looking for an ON/OFF switch but a circuit that gradually limits the current? Where does the voltage from IC come from , is a a variable DC (like filtered PWM)? –  alexan_e Dec 21 '13 at 17:11
    
The voltage is a PWM signal. And I'm looking for a circuit which acts just like a normal transistor but opposite, so a high voltage is off, and a low voltage is on, and a voltage in between is between current. –  PitaJ Dec 21 '13 at 17:37
    
"I want to replace the nmos with something which will perform oppositely." Why not change the program in your microcontroller? –  jippie Dec 21 '13 at 17:49
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5 Answers 5

up vote 6 down vote accepted

Here is a circuit that is on by default and turns off when you apply voltage to the input.

enter image description here

When the input is floating or grounded Q2 is off so R2 works as a pullup resistor that turns Q1 on.

When there is a positive voltage applied to the input then Q2 turns on and sinks current trough R2, that creates a voltage drop across R2 that drops the voltage applied to the base of Q1 and turns it off.

It can also be done with mosfets with a similar logic.

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A depletion MOSFET does what you want. An enhancement MOSFET is the more common type that needs a gate-to-source voltage to start conducting.

The delpletion mosfet characteristics can be seen in the following graph enter image description here

And the comparison between a depletion and enhancement N mosfet in the following graph

enter image description here

Images taken from An introduction to depletion-mode MOSFETs

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Does that require that I use a negative voltage? –  PitaJ Dec 21 '13 at 7:48
    
Negative voltage is a relative concept. It depends on your circuit. –  jippie Dec 21 '13 at 8:07
    
A depletion type MOSFET is not what is being asked here. A depletion type nMOS is simply an nMOS with a zero, or a negative threshold voltage, allowing conduction in a lower voltage value compared to an ordinary nMOS. Instead of opening the circuit, a depletion type nMOS will enter saturation at a lower VG voltage, closing the circuit at a lower voltage value. –  Cem Dec 21 '13 at 12:47
    
@Cem I think the question is sufficiently vaguely worded that this could easily be considered a possible answer. –  Phil Frost Dec 21 '13 at 13:01
    
@Cem wait...are you saying that the channel of an N-channel, depletion mode MOSFET will conduct as the gate becomes sufficiently negative with respect to the source? I think that is the opposite of true, as the second graph here demonstrates. –  Phil Frost Dec 21 '13 at 13:08
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A fairly generic solution is to use an analogue switch. It is a device that behaves very much like a relay in that it has an input that can switch on or off internal transistors. You can get these devices in SPDT format. You haven't stated what current or voltage you need to switch so it may be a problem using this device but it's worth consideration.

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Many ways to do this. A few ideas:

schematic

simulate this circuit – Schematic created using CircuitLab

The third example might require some explanation if you are not familiar with JFETs: normally the channel of J1 is open, but if you apply a sufficiently negative (relative to ground) voltage to the gate, it will "pinch off". A depletion-mode MOSFET would work as J1 also, but that's a rather esoteric part. I can't say I've ever encountered a JFET used in this application either, since JFETs are usually more expensive, not present in my parts drawer in great quantities, can't usually switch large currents, etc.

The first example works great if your load voltage is the same as your logic voltage. Not always the case.

The middle example simply demonstrates inverting the logic before applying it to the gate of M2. Here the load voltage can be anything within the capabilities of M2, shown here as 24VDC.

And of course, M1 and M2 could be replaced by their BJT equivalents, with the addition of an appropriate resistor to limit base current.

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I am pretty surprised why no one gave the simplest answer, a PMOS or a PNP BJT will do exactly what you have asked.

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Are you sure? It seems too simple to be true. –  PitaJ Dec 21 '13 at 17:05
    
Imagine a PNP with the emitter connected to 12v and the collector connected to the motor, how exactly will you turn off the transistor with a 5v control voltage? The problem with the P mosfet is the same, when the source is connected to 12v it can't be turned off with 5v to the gate. –  alexan_e Dec 21 '13 at 17:14
    
Yes, but imagine a pMOS in a digital NAND gate, in an IC. If its input is VDD, then the pMOS goes into cutoff mode, and the corresponding nMOS in the nMOS network goes into saturation, yielding a logic 0 in the output. When I answered the question, there was no detail about the application area of the device(controlling a motor from a microcontroller), therefore I logically answered with the usage in my area of expertise. –  Cem Dec 21 '13 at 17:39
    
And it is quite true, PitaJ, but not in your application, as you would need a voltage higher than the source voltage in order to put the pMOS into cutoff. –  Cem Dec 21 '13 at 17:41
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