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MOSFET Issues

mrShrimp

Jan 20, 2014
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Hello,

This is my first post on Electronics Point, and I am by no means a circuit guru, so please bear with me if I seem to make rookie mistakes in my posts :).

I am having trouble using an n-channel enhancement-type MOSFET (part no. is NTE2996) to use one voltage source (PS1) to switch current coming from an external power supply (PS2) on and off. I have attached a diagram of the circuit that I have been testing so far, which I will make references to from now on.

Before I explain the problem, a bit of context might be helpful. The circuit that I am trying to switch is for a 3D printer's heated bed. The printer's old heated bed was inadequate, and I am trying to install a new one, but the current that the new one draws is too much to run through the 3D printer's circuit board directly, hence the MOSFET. Also, the power supply that the 3D printer runs off of is inadequate to supply the current needed for the new heated bed, so that is why there are two power supplies in my diagram.

PS1 in the diagram really refers to the two leads on the 3D printer's circuit board that were being used to heat the old heated bed, so the circuit is a bit more complicated than my diagram shows it to be, since "PS1" is not directly connected to the actual power supply that the printer is using. I can usually control the voltage coming out of PS1 via software, and that is how I am switching the MOSFET.


Okay, now for the problem. When I disconnect PS2 and replace it with the leads of an ohmmeter, the circuit works as it should. When I use the printer software to turn off PS1, the gate-source voltage is 0V, and the drain-source resistance (measured on the ohmmeter) is extremely high. When I use the printer software to turn on PS1, the gate-source voltage is 12V, and the drain-source resistance is around 1 ohm, as it should be.

When I connect PS2 in lieu of the ohmmeter, however, the circuit does not work as it should. Even though I do not have PS1 turned on via the printer software, the gate-source voltage reads 12V, and the printer's heated bed (R2) heats up on its own. I am not sure if the current from PS2 is leaking through from the drain to the gate, or if it is somehow causing PS1 to turn on when it is not supposed to.

When I removed the MOSFET from the circuit, the resistance between the drain and the gate read as infinite with the positive lead on the drain and the negative lead on the gate, which is as it should be (the two are physically isolated within the MOSFET, aren't they?). Oddly, when I connect the ohmmeter the other way with the MOSFET disconnected (plus on the gate and minus on the drain), the drain-source resistance becomes very low. I have some theories for why this happens, but I am not entirely sure of them. I also don't know exactly how the 3D printer circuit board switches PS1, except that it is also done using a MOSFET, so it is hard to tell what is causing the gate-source voltage to measure 12V when I am not telling PS1 to turn on.

Could I please have some help solving this problem?


Thanks for reading my rambling post!
- mrShrimp
 

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Harald Kapp

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Welcome on the forum.

Your circuit looks correct. The behaviour you see is inexplicable, unless the real circuit differs from the schematic. A possible source of confusion is the body diode in the MOSFET. This diode is connected cathode to drain, anode to source (it can't be removed, it is due to the manufacturing of a MOSFET). This diode can cause current to flow through the load (R2) in two cases:
1) PS2 is reverse polarity, i.e. + to source, - to R2
2) PS2 is AC, not DC

By no means, however, does this explain why Vgs is 12V. To exclude any leakage currents, reduce R1 to 10k. Even a value as low as 1k shouldn't harm the circuit as PS1 originally was supposed to drive the old heater.
 

mrShrimp

Jan 20, 2014
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Hi Harald,

I double checked the circuit, and it is set up just like the schematic shows it to be. I switched out R1 for a 10k resistor, but nothing seemed to change. I think I just discovered something that could help me better understand the problem, however.

Just a minute ago, I was about to test the circuit with a diode between the R1 lead attached to the positive lead on PS1 and the gate on the MOSFET to see if anything was leaking from the drain to the gate, or if the 12V reading was coming from PS1 itself. I had the voltmeter ready to measure the voltage directly on the PS1 leads when I was plugging in PS2's power cord, and I noticed that as soon as the safety ground on the PS2 power cord touched the receptor on the outlet, the voltmeter read an output of 12V from PS1's leads.

Does this mean that the problem arises from the fact that the two power sources are grounded to the same outlet? I feel confident that nothing is leaking in the MOSFET circuit now, but I'm afraid that the problem might be related to the fact that "PS1" in the diagram is not directly connected to the power source it draws from.
 

Harald Kapp

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Do you say that PS1 and PS2 are not isolated but connected to ground?

Try 2* 9V batteries instaed of the power supplies to ensure the circuit is really working correctly. Then replace the batteries one afte another by the power supplies and check for any stray current.
 

mrShrimp

Jan 20, 2014
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I'll try that as soon as I can tomorrow.

Something else I noticed while testing the circuit was that the gate-source voltage would read read 11.5V or 10V when the safety ground of PS2 was connected not only when PS2 was fully connected, but also when I disconnected either one of its leads.

When I disconnected PS2's positive lead, the gate-source voltage was around 11.5V (which is the output voltage of PS1) upon the connection of PS2's safety ground, and when I disconnected PS2's negative lead (with only the positive lead connected), the gate-source voltage was around 10V upon the connection of PS2's safety ground.

I am not sure how this would work, but since PS1 and PS2 both connect to outlets in my house, could that be linking them together and causing the problem when I plug them in?
 

Harald Kapp

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Normally I'd expect the 12V side to be completely insulated from your house's mains. However, to make absolutely sure use batteries.

One other idea:
What does the voltage on PS1 look like if nothing is attached? PS1 may actually be a PWM modulated voltage which would allow the control circuit of the 3D printer to finely control the power. The pwm signal may never be completely off. It may show small pulses in the "low" setting of the heater. These pulses may be enough to charge the Gate-Source capacity of the MOSFET so you see a voltage there. An oscilloscope would ve valuable here.

One more idea:
It is easier to turn power on and off by switching the negative side of the power (gnd), not the positive side, see schematic.
attachment.php

In that case the gate of your external MOSFET is permanently connected to +12V and by connecting the grounds of the two power supplies you effectively short circuit the output transistor of the heater control. If that happens, you could use a P-MOSFET instead of the NMOS. The P-MOSFET would have to be inserted with source into the positive leg of the heater power supply, Gate is connected to the "-" output of the heater control, drain goes to the heater. The other leg of the heater goes to GND.


A leackage from drain to gate is most unlikely.
 

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mrShrimp

Jan 20, 2014
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In reply to your previous post, yes, PS1 and PS2 have their grounds connected, which is, I think, unavoidable with an n-channel MOSFET. I tried the circuit (not the same wiring, but the same MOSFET and circuit layout) on a breadboard using an Arduino Uno to switch the gate voltage, and the 5V line on the microcontroller to supply the drain source-current across a resistor, and the circuit worked as it should. As far as I can tell, this was the same as the circuit using the two PSUs, since the ground on the microcontroller was shared between the gate and source pins, as is the ground between the PSUs.

Unfortunately I was unable to go to the electronics store today and get a p-channel MOSFET with an adequate current capacity, but I should be able to run a test with one before the weekend.

As it turns out, the bed heater pin is a PWM pin. The controller is a RepRap Melzi, which is based on the Arduino Leonardo. The wiki page on the Melzi says that it uses pin 10 for the bed screw connector, which is what I am using for PS1. Pin 10 on the Leonardo is a PWM pin, so there you go!

When I removed the resistor from gate to ground, I was in fact experiencing a gradually rising gate-source voltage without sending any software signals to the printer telling it to do so. It stopped at 4V (the threshold voltage of the MOSFET), reset back to 0V, and repeated the process. If the PWM is never truly off, then this is solid evidence of its effects. I think whatever harm it could cause is mitigated by the resistor from gate to ground, though. The gate-source voltage is a constant 0V with that resistor attached and with PS1 turned off.

I'm sorry, but I didn't understand this suggestion of yours: "It is easier to turn power on and off by switching the negative side of the power (gnd), not the positive side, see schematic. ... In that case the gate of your external MOSFET is permanently connected to +12V and by connecting the grounds of the two power supplies you effectively short circuit the output transistor of the heater control."

What distinguishes switching the negative side of the power versus the positive side, and is the schematic showing the p-channel MOSFET switching the negative or positive side of the power?

Thanks for all your help so far!
 

Harald Kapp

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I'm sorry, but I didn't understand this suggestion of yours
That was not a suggestion, just an explanation for my assumption that the negative side of the heater supply is switched. Q1 in my example may be a bipolar transistor or a MOSFET. My schematic shows a possible configuration of the controller's output circuit to explain why connecting grounds will short circuit the output transistor and feed 12V to the gate.

What distinguishes switching the negative side of the power versus the positive side
In principle: nothing.
In practice: switching the high side requires a PMOS, which is more expensive than a comparable NMOS, or you need a so called high side driver which increases the circuit effort.

is the schematic showing the p-channel MOSFET switching the negative or positive side of the power
Which schematic do you refer to?
 
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Sergey

Jan 26, 2014
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written so much that I do not catch on ... :) But a mistake in connecting mosfet I noticed. that would be necessary to fully open the gate of the mosfet energized for about 15 volts more than the supply voltage (voltage passing through the source and drain). to reliably and quickly close mosfet must submit to the gate of -15 volts (ie 15 volts less than the supply voltage). Yours! Not much kick if offtopic ...
 

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Harald Kapp

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15 volts more than the supply voltage (voltage passing through the source and drain).
This is not necessary. It is Vgs that counts source is at 0V, so a gate voltage of 12V makes for Vgs=12V, enough to open the NTE2996 (Vgs(th) = 4 V max.).

to reliably and quickly close mosfet must submit to the gate of -15 volts
If you need to remove the gate charge very fast then a negative Vgs helps. In this simple switching circuit Vgs = 0 V is enough to close the the NTE2996 (Vgs(th) = 2 V min.).
 

mrShrimp

Jan 20, 2014
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Sorry for not replying earlier. School work has been piling up and I've had little time to work on this. In response to your last post Harald, I was referring to the schematic that you posted.

I ran a test with a p-channel enhancement-type MOSFET instead of the n-channel MOSFET, but I noticed something was wrong before I turned PS2 on. The way I connected the p-channel MOSFET was to hook up + on PS1 to source and - to gate to achieve a negative gate-source voltage.

With the printer power supply turned on (PS1 represents the output of the MOSFET that the printer uses to control the heated bed power, not the leads of the printer PSU directly, so at this point PS1 is reading 0V even though the printer's circuit board is being supplied 12V) and with PS2 unplugged from the wall socket, everything was looking normal. As soon as I plugged PS2 halfway in, however, the printer's power supply (not PS1) dropped to 7.5V from 12V. I ran the same test twice more with only one of PS2's leads connected each time so as not to damage anything (which I eventually ended up doing after some different tests; only the MOSFET was damaged though) and I found that the printer power supply drops to 7.5V when PS2's minus lead is connected to source in the configuration I described.
 

Harald Kapp

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I have to confess your description is a bit hard to understand.

It sounds like your two power supplies are somehow connected. You could measure that using an Ohmmeter.

You can isolate PS1 from PS2 using a photocoupler like this circuit shows:
attachment.php
.
Now PS1 and PS2 no longer share a common connection. For this purpose you can use any kind of photocoupler, the CNY17 is only an example.
 

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mrShrimp

Jan 20, 2014
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Sorry for the late reply, but we finally got the heated bed working! We printed our first ABS print yesterday, and it experienced almost no curling.

We optically isolated the power supplies as you suggested, and it worked after an initial mishap. The working circuit powers a phototransistor with PS1, which works with a resistor of a constant value to act as a controlled voltage divider to turn a MOSFET gate high or low. PS2, in addition to powering the heated bed through the MOSFET channel, provides the voltage that is divided by the phototransistor to switch the MOSFET on or off.

The heated bed does not heat up as quickly as it is advertised to, which could have something to do with the switching circuit, but it is still MUCH better than our previous one.

Thank you for your help!
 

rdtsc

Mar 19, 2014
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Hello, note that from the datasheet for the NMOS device used, http://www.nteinc.com/specs/2900to2999/pdf/nte2996.pdf on the second page of the specs, it's input capacitance or Ciss is 3210pF or 3.2nF. This is common for many high-power mosfets and is the result of essentially putting hundreds of smaller mosfets into one case, as to multiply current capability. (Everything is a trade-off in electronics...)

What this means for you however, is that it will take significant energy to switch it on and off, since each time, the signal is combating the charge stored in the gate. (3.2nF.) The faster it is switched on and off, the more energy the gate will consume overcoming this stored electric charge. If you are trying to send it a PWM signal, you will almost certainly need some kind of driver, such as one of Microchip's Power MOSFET drivers. Read one of those datasheets for more info.
 
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