(Read 5128 times)
May 11, 2018, 01:10 pm
When you put the gate to 0V/5V manually, only 8721 is working ok ?
show schematic used for testing.
May 11, 2018, 04:22 pm
: May 11, 2018, 04:22 pm by Marciokoko
Sorry, here it is...
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May 11, 2018, 04:53 pm
You have no load.
Parallel to diode - LED with 1k resistor., can be only resistor if you have a oscilloscope.
9V battery will not give perfect results if you want to put the big load; I =10A or more.
It is anough for start.
what is the frequency on D5, or just slow LOW/HIGH - 0V/5V ?
May 12, 2018, 03:33 am
So I looked at the Drain Ia vs Vgs graphs for each and they all give between 50-80A. The one that yields the most (120A) is the IRLB4132 and the one that yields the least is the IRF510 at 1A, which is expected since its the only one without the L designation:
IRLB4132 - 120A
IRLB8721 - 80A
IRL540 - 50A
IRF510 - 1A
I want to figure out why 8721 works but not the others. I thought the Id vs. Vgs would at least be consistent, but IRLB4132 seems to be able to conduct more and IRL540 and FQP seem to be able to conduct just about the same.
Just saw your post mod #55. You are looking at a graph of Vds vs Id which is a bit confusing to me. From what I understood of the article I mentioned, the graph they look at is Vgs vs Id. This makes sense to me because it seems to tell me that at a certain Gate voltage, I will get a certain current from drain to source. I dont quite get the idea behind the Vds vs Id. Ill go back to looking at mosfet videos, but for the time being, as MarkT said, RDSon is what I need to look at to use a mosfet as a switch. So how can you explain why only 8721 works reliable and the others dont if their RDSon values are quite similar?
Im thinking there must be something about my setup that is making these not work properly because they all seem to be logic level mosfets (except for IRF510):
May 12, 2018, 03:46 am
your Voltmeter has resistance more than 100k, a good voltmeter in megaohms, you can not have proper working any transistor with 100k resistor in Drain, you are talking 30A and you may have 3mA
May 12, 2018, 04:27 am
I don't understand
May 12, 2018, 05:09 am
Google - mosfet control
May 12, 2018, 05:22 am
I think I figured it out. It might be the unsteady 5v supply to the Arduino.
May 12, 2018, 05:32 am
: May 12, 2018, 05:37 am by ted
Motor resistance 1 ohm , this is the load, on your diagram you don't have it. You changed it by multimeter 1 megohms.
look at - Simple Power MOSFET Motor Controller
May 12, 2018, 04:38 pm
A MOSFET, like all semiconductor devices, has leakage current. If you don't want that
to confuse a voltage measurement, use a load that will take more than the leakage current.
A voltmeter is not such a load, its very high resistance. Try a 1k resistor as a load, then you'll
see a more realistic "off" voltage.
For instance if the MOSFET has 1uA of leakage, that will show as 0.001 volts across a 1k resistor,
but as 10V across a 10Mohm voltmeter. The 1k resistor makes it clear that the device is off.
Mechanical switches (at low voltage) have leakage currents when off millions or billions of times less
than this so you never have to worry about adding such a load to measure mechanical switches.
Note that a MOSFET where the drain isn't connected can be checked by a multimeter in resistance
mode - just measure resistance from drain to source. Off should be megohms or upwards, on is
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Over 50 years of Electronics and Programming experience!
May 12, 2018, 06:01 pm
: May 12, 2018, 06:14 pm by ReverseEMF
Quote from: Marciokoko on May 12, 2018, 04:27 am
I don't understand
And allow me to clarify.
Things to know:
All volt meters [including multimeters set to the DC Volts or AC Volts range, and Oscilloscopes] have internal resistance. The better ones [usually that means,
] have higher resistance, because, ideally, we want this resistance to be as high as possible, so the meter is less likely to
the voltage it is measuring. Imagine you are trying to read the voltage across a resistor [in a circuit]. Let's say it's a 100k resistor in series with a another 100k resistor, with 5V across the two. In this case, there will be 2.5V across each resistor. If I put a 100k resistor across one of the existing 100k resistors, the voltage across that parallel pair of resistors will change to 1.67V! A
change! If that other 100k resistor [the one we put across the existing 100k resistor] was a Voltmeter, then perhaps you can see the problem. The meter should have read 2.50V. But, instead, it read 1.67V!!
Now, if instead of 100k, we place a 1M resistor across that 100k resistor, the voltage across that resistor pair will be 2.37V. Still not the same as 2.50V, but closer. So, a Voltmeter with an internal resistance of 1M will do a better job of reading that voltage, but it's still wrong. What if the meter has an internal resistance of 10M? The voltage will read 2.49V. Much better.
When a MOSFET is turned "OFF", it is never, really
all the way off
! There is a small amount of what is called "leakage current" [i.e. this is not a perfect world
]. That leakage current behaves like any other current passing through the Drain to the Source [or vice versa, depending on if it's an N-Channel or P-Channel device, and on what current convention you adhere to--suffice it to say, that a current flows]. If the resistance at the Drain [i.e. from the Drain to the supply rail (+5V in this case)] is small, this leakage current will be so miniscule in comparison, it can, in most cases, be ignored. But, if the resistance is very large, then the leakage current will develop a significant voltage drop across that resistor.
For instance, if the Drain resistor is 10Ω, and the Drain leakage current [when V
is 0V] is 20µA, then the voltage across the Drain resistor will be: 20µA * 10Ω = 200µV In most cases that can be ignored. BUT, if the Drain resistance is, say 100k, then the voltage across the Drain resistor will be: 20µA * 100k = 2V! And that's when the MOSFET is supposed to be OFF! What if that Voltmeter is
used as the Drain resistor
[i.e. Negative Probe at the Drain, and the Positive probe at +5V], and say, that Voltmeter has an internal resistance of 1M. The voltage across the meter, when the MOSFET is OFF, will be: 20µA * 1M = 20V -- well, actually, what will happen here is this: Because there is only 5V in the circuit, 15 more volts won't
appear. Instead, the 1M resistor will limit the current such that most of the 5V will be dropped across it. In other words,
20µA will flow -- more like 5V / 1M = 5µA. And notice the implication of that. When the MOSFET is OFF, there is 5V across the Voltmeter. AND, when the MOSFET is ON, there is also 5V across the Voltmeter! In other words, NO CHANGE! In other
words, there is no change in the
If you try to measure the Drain, with a Voltmeter connected
the MOSFET [Positive probe on the Drain, and Negative probe on the Source], with
going from the Drain to the Positive Supply [i.e. +5V in this case], then when the MOSFET is ON, you will read 0V. And, when the MOSFET is OFF, you will also read 0V. This is because there is nothing to pull that Drain up to the Positive Supply Voltage. It's like if you connected a resistor to ground, and left the other side of the resistor open. All you are ever going to read is 0V, no matter what the value that resistor is [OK, yeah, if it's a very large resistance, you might read some stray voltage due to antenna effects, but none of it will be coming from the Positive Supply]! The MOSFET is the same -- the channel is a resistor, that varies in value based on what voltage is applied to the Gate. If one end of that channel is
, there will, always, only be
zero volts across it
So, that's why you need a "load" resistor on the Drain of your MOSFET,
you apply a meter to the Drain. What you are measuring, with your meter, is the Voltage Divider effect of the series resistances: The resistance of the
in series with the resistance of the
. As the resistance of the MOSFET channel, changes, it changes the ratio between
. For instance, if the
is 10Ω, and when the MOSFET is ON, the
is, say, 25mΩ, the voltage across the MOSFET will be (using the voltage divider formula):
[25mΩ*5V]/(25mΩ+10Ω) = 12.5mV
The voltage across the MOSFET, when it's OFF -- with a 20µA leakage current -- will be:
5 - 20µA*10Ω = 4.9998V
Now, why didn't I use the
voltage divider formula
to come up with that value? Because, when the MOSFET is OFF, it's not clear what the channel resistance [i.e. the resistance from the Drain to the Source] actually is. BUT, we do have an idea of what the
will be, because that is listed in the datasheet [and that value, BTW, is the
case -- or
, so it might, actually, be less].
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Please DON'T Private Message to me, what should be part of the Public Conversation -- especially if it's to correct a mistake, or contradict a statement! Let it ALL hang out!!
Design and Repair of industrial control systems.
Electronics Engineer/Industrial Control
May 13, 2018, 02:03 am
: May 13, 2018, 02:03 am by TomGeorge
Put a 1k resistor where you will put your load.
Then measure the voltage;
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Everything runs on smoke, let the smoke out, it stops running....
May 13, 2018, 02:18 am
ok it was indeed the unsteady 5v supply to the mcu that i guess wasnt delivering the 5v to the gate. I plugged it into my computer instead and it worked consistently, every time, with all ll mosfets and with the irf510 i got 6.5V out of the 9.7 possible.
So I guess I need to add a 5v booster to the incoming from the lipo battery to the mcu.
May 17, 2018, 12:10 am
ok im back because it didnt seem to be quite that. It turns out when I touch the back of the mosfet is when it works.
Jun 12, 2018, 01:06 am
Why could that be? The back plate of the mosfet is connected to Drain isnt it? So why would that happen?