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Although I read the article advising against using TIP120:

http://sensitiveresearch.com/elec/DoNotTIP/index.html

I only have couple of other transistors on my hand:

sc2240
sc2581
sc2837
sc3118
bd137
buk455
c3832
c4747
irf830
mje15030
tip32c
bd240b
tic2260 triac
l7805 scv

Could somebody advise me which one can be used to replace TIP120 in case my circuit would be controlling a solenoid:

http://www.instructables.com/id/Controlling-solenoids-with-arduino/

with a difference that I want to control the solenoid from an 555 timer.

So the 555 circuit would be powered from 12V directly, the solenoid I checked consumes 138mA current when it is turned on. What resistor should be put in front of the base/gate leg of the transistor/mosfet?

Thank you!

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closed as off-topic by Dmitry Grigoryev, laptop2d, uint128_t, Andy aka, brhans Apr 25 '17 at 21:09

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  • \$\begingroup\$ From 12V the tip120 should be fine. Or a FET... irf830? \$\endgroup\$ – George Herold Mar 22 '16 at 13:21
  • \$\begingroup\$ The point was that I don't have TIP120, only those are on the list. \$\endgroup\$ – munin24 Mar 22 '16 at 13:39
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    \$\begingroup\$ I'm not familiar with all those part numbers, but at ~150 mA any npn transistor should be fine. \$\endgroup\$ – George Herold Mar 22 '16 at 13:44
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    \$\begingroup\$ The link you posted about TIP120 is dead. \$\endgroup\$ – Chupacabras Apr 23 '17 at 5:31
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Looking at that list (and someone feel free to correct me) but I'd use the IRF830, firstly the IRF830 is an N-Channel MOSFET not a BJT.

Now, MOSFETs do typically have a resistor on the gate but it doesn't play the same function that a base resistor on a transistor does. On a MOSFET a gate resistor is used to dampen the ringing caused by the gate capacitance and the inductance in the wire which affects the switch on time of the MOSFET. A low ohm resistor uses solves the issue (I typically stick 100R in)

Edit: I though I'd add that it's worth putting a pull down resistor in before the gate resistor just to make sure that the gate isn't floating when it's not being driven. But in the case with a 555 if I remember it's an active-push pull so you shouldn't get that issue

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  • \$\begingroup\$ Hello, thank you I will use this as I see this even have a built in protection diode for connecting inductive loads like the solenoid so I don't have to use an additional external one. \$\endgroup\$ – munin24 Mar 22 '16 at 13:39
  • \$\begingroup\$ You still need a flyback diode across the solenoid. The diode in the MOSFET is only going to conduct in breakdown so you still need a diode across the solenoid \$\endgroup\$ – Doodle Mar 22 '16 at 14:51
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    \$\begingroup\$ @munin24 No, it doesn't have a built-in protection diode. The diode you're seeing wasn't added, it's a parasitic component that arises as a side effect of how a MOSFET is internally constructed. It's not something that was added on purpose. It's called the 'body diode' and it is included in the symbol because you need to be aware of its existence. MOSFET body diodes have extremely slow turn-off times, so you generally want to avoid having it conduct. Definitely follow Doodle's advice on the flyback diode. The body diode something you're protecting, it is not there to protect. \$\endgroup\$ – metacollin Apr 25 '17 at 11:25
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This is meant more as supplementary material, but let me begin by saying I second Doodle's choice of the IRF830, given the choices at hand.

If you were driving it from an arduino, it would be a terrible choice (it would have on the order of 33Ω resistance when turned on with just 5V) but with the qualifier that you're driving it from a 555 powered from 12V, it will fully turn on and should dissipate less power than any other option on the list.

OK. Now, instead of just recommending a part, wouldn't it be better if you could understand when to use a certain type of transistor? Then you can always know what will make a good alternative, make substitutions, and not have to worry about this stuff. It's also easy, it just requires a basic conceptual understanding.

Unfortunately, you will get no such understanding from the article at http://sensitiveresearch.com/elec/DoNotTIP/index.html . It was written by someone who didn't understand what they were doing, chose the wrong part, and rather than admit that they made a mistake, decided to misrepresent the situation as one of old vs. new instead. He isn't incorrect in the details, but the entire argument is based on the false assumption that a transistor is a switch, can only be used as one, and only is used as one.

The 'rant' does little beyond drive home what the rest of us already knew: it's a bad idea to use a darlington in a wildly inappropriate application that would be much better suited to a FET. The only truth that author has found is that MOSFETs work well in situations suited to MOSFETs, and totally different components that are not MOSFETs do, in fact, perform poorly in those situations. In other words, he discovered that two different things are, in fact, different things.

The TIP120 is a darlington transistor. It is a bipolar junction transistor that controls a larger, second bipolar junction transistor. It is strictly a high gain current amplifier.

I can think of nothing less similar to a MOSFET while still being a transistor.

Darlington transistors are used in applications where one needs a current amplifier but with higher gain (A=1000+) or power than one can get with a single BJT.

The word switch appears nowhere in that application. A TIP120 is fantastic for its intended use, which is why it is still made and still useful. It just is not useful as a switch, but that isn't something that wasn't true in the 1970s but is true now. It was never useful as a switch, and the author of that article doesn't really understand transistors or how to pick the correct type for what he is doing, and made the absolute worst possible choice. Then it of course didn't work very well, but he's tried to turn it into 'a thing' when there is nothing there beyond his own mistake.

Just to drive this point home, when one is using a transistor as a switch, you drive it into what is known as its saturation region. This is where they behave the most like a switch. There is also the linear region, in which they behave like a current amplifier in the case of the BJT, or a voltage-controlled resistor in the case of a MOSFET.

A darlington transistor cannot be driven into saturation. This is an inherent property of how they are constructed, and why they have such a large (~2V) voltage drop regardless of how hard you drive the base. You actually can't use a darlington as a switch. You can try, but it can't be made to behave like one, and thus, won't.

So the author of that article tried to use the only type of transistor that is actually impossible to use as a switch, as a switch. And because the thing that wasn't a switch continued to not be a switch even though he tried to use it like it was, that makes all darlingtons some how obselete, even though no one used them like that to begin with. But yes, he is right that they are not as good as MOSFETs as being switches.

Let's get to the actual question though. There are 3 main types of transistors. I'll go in chronological order of their invention (which, unsurprisingly, is not correctly represented in the 'Do not TIP' article.

  1. Field Effect Transistor

    The first transistor ever constructed was a point-contact FET, built at Bell Labs on December 20th, 1947. MOSFETs are, are a type of FET, and FETs all operate via the field effect principal. There is a conductive channel, and a capacitive plate near it known as the gate. By pulling charge carriers off (or forcing them on) the plate with voltage, one can generate an electric field. This field will either remove (deplete) charge carriers from this conductive channel, or add (enhance) carriers to the channel. This causes the resistance of the channel to become extremely high, or very low, often just a few mΩ in the case of modern FETs. This is very much an ohmic change, and it is controlled by the charge on the gate, which is in turn controlled by the voltage across the gate and the source.

    So, when I say they're a voltage-controlled resistor, they very much are. You want a MOSFET any time you want to switch something almost as if you're using a real switch. They also are controlled purely by voltage, and only require a trivial amount of current when actually transitioning from on to off or off to on. The gate can be thought of as a capacitor, more or less. It only consumes current while charging, and release it while discharging. You do not need any sort of resistor on the gate, and adding one can actually prolong the time it takes the FET to turn on, which could potentially damage or destroy the FET in high power applications.

    It would take a fairly massive FET to have a gate capacitance of even 10nF. A 10nF capacitor is used to directly drive the RESET pin of your arduino. For a non-ridiculously powerful FET, you can drive it directly from an IO pin.

    Note: Most FETs require 10V above (or below, if its a P-channel) the voltage on the source. In the case of an n-channel FET with a grounded source, that means 10V at the gate. There are logic level FETs which are optimized to fully turn on from just 5V at the gate - so you would want one of these. Another option is to use a totem pole BJT driver, but that requires an additional 12V rail (to obtain ~10V at least), and its sort of beyond the scope of this question anyway. Just look for a logic level FET with a low enough on resistance that it won't dissipate too much power. I think there are very few FETs that WON'T meet your requirements, as 138mA is not much at all.

  2. Bipolar Junction Transistors

    These were invented a couple years after the FET, and operate on a completely different principle. They have nothing in common with FETs, and somewhat arbitrarily were also referred to as 'transistors'. These days, transistor is a blanket term for anything that behaves as a transconductor. A transconductor is something with transconductance (a relationship between input voltage and output current of a device).

    BJTs are constructed like two silicon diodes back to back, but with a shared junction, so pn-np becomes pnp and np-pn becomes npn. You can measure the diode drop from the base (middle semiconductor) to the collector and the emitter. The base behaves just like a diode, and has the same exponential voltage to current curve as a diode. This is why a base resistor is generally needed.

    They are also current controlled, not voltage controlled. And they have current gain, and behave like a true current amplifier, multiplying the current through the base to emitter in the form of current from the collector to emitter. Their behavior is NOT ohmic, and instead when saturated or fully 'on', you get a diode drop across the collector and emitter, rather than a specific ohmic resistance (though, there is of course some resistance too, but it is constant regardless of how turned on or off the device is).

    When using these as switches, you generally just size the base resistor such that it will be driven into saturation (fully 'on' and no longer acting like a current amplifier) at a given voltage. You can find the saturation current in the datasheet. Be aware that no matter what, you will dissipate 0.6V * collector-emitter current worth of heat. In your case, this is 82mW. You can really use just about any BJT in this situation. A MOSFET would be better, though.

    Darlington transistors are arranged as a small BJT used entirely to amplify it's base current, then that current is in turn used to drive the base of a much larger transistor. So it's effectively just adding a current preamplifier to a larger current amplifier, and integrating it all into one convenient package. They are not designed to be switches, nor can they be made to act like switches. At best, they will always have a voltage drop of about 2.1V from the collector to emitter, so in your case, they would dissipate 2.1V*.138A = 290mW. That's not as good, and it won't take much more current before the power dissipation becomes unmanageable. Which is why you don't use darlington transistors as switches.

  3. Insulated Gate Bipolar Transistors

    These are relatively complex compared to the other two types, so I am just going to touch on them briefly. There are situations where the 0.6V diode drop is actually lower than the ohmic voltage drop seen in MOSFETs ('situations' usually meaning very high current, on the order of hundreds of amps and higher), making a BJT superior to a MOSFET when being used as a switch. The problem is that high power BJTs tend to have very low gain, and so require large amounts of base current to control correctly. So there is this usage niche where something that can be controlled a lot like a MOSFET, but behaves like a BJT in terms of the conduction that results, would be useful.

    Enter the IGBT. It's often described as a MOSFET-controlled BJT, but that is not really correct. Unlike a darlington, which is actually just two transistors, an IGBT is not a MOSFET and BJT connected and wrapped into a single package. They're their own, third kind of transistor that is constructed as having an isolated gate like a MOSFET, but sandwiched on top of alternating pnpn doped layers. They're a little weird and actually somewhat complex to drive correctly, but they are also strictly high power devices. You need not worry about them for this application.

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Most moderate to heavy loads these days are switched with MOSFETs. (Until you get to very heavy loads like electric cars, at which point IGBTs enter the picture.)

The reasons MOSFETs are desirable are:

  • Essentially no gate (base) current required
  • Very low voltage between drain and source; less voltage means less loss and heat
  • Abundant, easily sourced
  • Very fast switching time given an appropriate gate driver circuit, again reducing losses
  • Easily sourced, parts available for any application.

To select an appropriate MOSFET, look for these things:

  • A gate threshold voltage appropriate for your application. Many power MOSFETs are intended to operate with around 12V on the gate, but you can find some for 5V, especially for lighter loads.
  • An RDS(on) low enough to keep losses low so cooling the transistor won't be a problem. Calculate the loss with Ohm's law: \$P=IR^2\$, where \$I\$ is the maximum current drawn by your load.
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