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Darlingtons vs. MOSFETs?

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Andy1845c

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I need to switch a 400mA load with a PIC. Would it be better to use a logic level MOSFET or a Darlington? Are there alot of advantages one way or the other or is it more a matter of preferance in this case?
 
Andy1845c said:
I need to switch a 400mA load with a PIC. Would it be better to use a logic level MOSFET or a Darlington? Are there alot of advantages one way or the other or is it more a matter of preferance in this case?
A darlington has a voltage loss of 0.8V to 1.5V. A half-decent Mosfet has an extremely low voltage loss.
 
Thanks Audio!

I don't know if i want to waste that much voltage on this project.

Sorry to be a pain, but when reading MOSFET datasheets, what value do I look at to know how many volts it takes on the gate to turn in ON? I have not used a MOSFET yet, so I don't know much about them.
 
Do a parametric search for logic level FETs. They are rated to saturate with 5 volts on the gate, vice 10 for the conventional ones. I came across one from IRC that was logic level, rated 20 volts at 180 amps, & only $2.18!
 
Vgs(th) is the voltage at which the MOSFET begins to turn on. The three numbers (min, typical, max) give some idea of the distribution of values. A typical would be the mean of the distribution while the min and max might rpresent the 3-sigma points. The condition of 1 mA represents the current that defines the threshold Vgs(th) since there is no standard definition.

One thing to note is, that if it doesn't begin to tun on until 3V, how hard will it be turned on at 5V? For that you have to refer to the graphs. 400 mA in a TO-92 package seems like a bit of a stretch to me. I'd look further for a part in a TO-220 package.
 
Never mind the graphs because they show typical parts. You can't order a typical part, you might get a weak but passing one.

The tiny 2N7000 Mosfet has a max continuous current of only 200mA.
With 10V on the gate and 200mA through it then its max saturation voltage loss is 1V when it is cold and 1.8V when it is hot. It will be pretty warm with a current of 200mA.

With only 4.5V on its gate, it turns on poorly and has a saturation voltage loss of a little more than with a gate voltage of 10V.
 
audioguru said:
Never mind the graphs because they show typical parts. You can't order a typical part, you might get a weak but passing one.

The tiny 2N7000 Mosfet has a max continuous current of only 200mA.
With 10V on the gate and 200mA through it then its max saturation voltage loss is 1V when it is cold and 1.8V when it is hot. It will be pretty warm with a current of 200mA.

With only 4.5V on its gate, it turns on poorly and has a saturation voltage loss of a little more than with a gate voltage of 10V.
Of course that's what they show. It's more information then is contained in a line in a table. Why are you being so...ah prickly?
 
Papabravo said:
Of course that's what they show. It's more information then is contained in a line in a table. Why are you being so...ah prickly?
Parts are cheap.
I design circuits by using the minimum and maximum ratings so all my circuits work well. I don't have any circuits that don't work because the transistor passes the spec but doesn't have enough performance.

The yield from manufacturers varies as shown by the wide spread in the specs. Also, maybe somebody bought the entire "high specs" production before you got there.
 
audioguru said:
Parts are cheap.
I design circuits by using the minimum and maximum ratings so all my circuits work well. I don't have any circuits that don't work because the transistor passes the spec but doesn't have enough performance.

The yield from manufacturers varies as shown by the wide spread in the specs. Also, maybe somebody bought the entire "high specs" production before you got there.
And do you suppose that you are the only one who does that?
 
Absolutely use a MOSFET.

Look for Vgs(th) in the data sheet. That's the turn-on voltage. (2-4V typ)

Look for Gate-Source voltage in the ABSOLUTE MAX RATINGS section, or whatever it is called. This will be much higher than Vgs(th). (+/-20V or more typ).

A part we frequently use here at work is the IRF 7313. Dual N-channel enhancement mode mosfet in an SOIC8 package. Works fine with 5V out from a PIC. Just put a series R (100ohm) to isolate the PIC output from the gate capacitance of the fet, and a pull up or pull down resistor (10k) right at the gate so the FET is on/off at startup (before the pic takes control). We've got some other parts here too if you don't need the dual that I can recommend. But there should be lots of parts out there that will work.

ETA: Actually a much cheaper and better part to use would be this guy: NTD80N02-001

Do a search on that at digikey. They are 1.30ea. Very low on resistance, very high current. Work well with extremely low Vgs and Vds. Used them with 0.5V Vds and Id=4A with Vgs <3V here at work on a project. Worked fine.
 
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Papabravo said:
And do you suppose that you are the only one who does that?
I am with audioguru, I learned a long time ago when designing with bipolar transistors to use the minimum beta.

I started in electronics as a repair technician and worked my way up to engineering. Things I build for myself last like for ever. one comes to immediate mind, a digital clock with a MM5314N clock chip and flouresent display. Built about 24 years ago and still works fine.
 
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Andy1845c said:
I need to switch a 400mA load with a PIC. Would it be better to use a logic level MOSFET or a Darlington? Are there alot of advantages one way or the other or is it more a matter of preferance in this case?

More a matter of personal preference than anything else - and certainly commercially its far more common to find bipolars than FET's used for such a purpose.
 
Thanks everyone.

Speakerguy - thanks for the recommendation!
 
Papabravo said:
And do you suppose that you are the only one who does that?
Many guys design circuits using the typical spec's. I guess they throw away the ones that don't work. It must be frustrating to make only one and it doesn't work because of the minimum spec's.
 
Back to the original question........
Originally Posted by Andy1845c
I need to switch a 400mA load with a PIC. Would it be better to use a logic level MOSFET or a Darlington? Are there alot of advantages one way or the other or is it more a matter of preferance in this case?
There are a lot of advantages and disadvantages to both.

A logic level MOSFET can cost twice as much.
The MOSFET may require less math.

A transistor will generally be cheaper.
You need to do more math.

For a logic level MOSFET, I presume that the data sheet already carefully describes the ON resistance and max current at the proposed gate drive level. (Don't look at threshold voltage, it tells when the device begins to turn on, this is not when it becomes useful.) Although a commercial circuit will need much more analysis, for one-off this might be all you need to know.

For a transistor, you need to drive current into the base. If you want the transistor to have a minimum on-voltage, you need to consider the saturated beta (gain). Darlingtons have an advantage in this area, because their overall gain (beta) is higher, but their best-case on-voltage isn't as good.

Now, you have a PIC and a 400 mA load. If you choose a transistor or darlington it needs to have an absolute max rating of a lot more than 400 mA, because their gain tends to drop off badly after about 1/4 to 1/2 their rating. Then the choice is between a darlington or transistor. In your case, you have 20 mA of drive and a 400 mA load. This calls for a saturated gain of 20. Rule of thumb is to use 10 for a saturated gain on a single transistor. So, unless you choose a super-gain transistor, you don't have enough current out of the PIC to drive it.

Again rule of thumb (you can change this if you truly understand the data sheet) is about 100 for the saturated gain of a darlington (10 * 10). In this case the PIC has plenty of juice. Give the darlington base 4 mA or more and it will well handle 400 mA.

Driving the base of the darlington: Your PIC is probably 4.5V, give or take, when it's high. The darlington base is about two diode drops (1.3V?) above ground. That means you want a resistor that conducts more than 4 mA with 3.2V across it. A common value like 680R seems to be good; it provides enough base drive without overloading the PIC.

Summary:

If you choose a MOSFET with a current rating of 400 mA or more at a gate voltage of 5V (or less), you're done - at maybe 50 cents. Your ON voltage will be defined by the MOSFET ON resistance.

If you choose a darlington (or a pair of 2N2222's) and a 680R resistor, you're done (after doing the calculations above) for about 25 cents. Your ON voltage will be about 1.2 to 1.8 volts with a good design.
 
The ULN2003 has seven channels that can drive 500mA/channel, and the voltage drop of the ULN is around 0.7V from 100mA too 250mA

I havent tested it further as I havn't had the need too...

It simply requires a logic voltage level to switch an earth too the output, eg;

**broken link removed**
- GND would be applied to Pin 8 on the ULN2003
bullet
- Pin 9 on the ULN2003 is connected to the positive supply for inductive loads such as motors and relays, eg;

**broken link removed**
 
mneary said:
If you choose a darlington (or a pair of 2N2222's) and a 680R resistor, you're done (after doing the calculations above) for about 25 cents. Your ON voltage will be about 1.2 to 1.8 volts with a good design.

No need for a darlington for a 400mA load, and no need for much in the way of calculations either - assuming it's a PIC that's doing the driving?, just roughly calculate the resistor to give 10-20mA to the base - 220 to 470 ohm would be fine.

It's a good idea to fit a resistor for an FET as well, as otherwise if the FET goes S/C it may well kill the PIC as well - and it's likely to go S/C if you only specify a 400mA device for a 400mA load!.
 
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