cyberfreak
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Will logic level MOSFETS turn off when 0V is applied to the Gate? Using this MOSFET as an example: http://www.fairchildsemi.com/ds/FD/FDS6961A.pdf
audioguru2
- Apr 6, 2004
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Hi Cyberfreak,
Your answer is there in the datasheet. 1uA is the maximum spec (for leaky ones that were made late on a Frday), most will conduct much less.
Most people would consider that a high-current Mosfet with a conducting current of only 1uA or less is turned off.
View attachment 36032
Your answer is there in the datasheet. 1uA is the maximum spec (for leaky ones that were made late on a Frday), most will conduct much less.
Most people would consider that a high-current Mosfet with a conducting current of only 1uA or less is turned off.
View attachment 36032
cyberfreak
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Hi AudioGuru
Thank you for showing me out that line on the datasheet. My Application is to switch a 12V Battery on or off by the logic NOR gate. Another interesing question arrises: There will be some voltdrop across the Source and Drain, due to Rds, when I apply 12V to the gate. Question is: Will the Source Voltage be, say 11.95V when 12V is applied to the gate and the 12V Battery connected to Drain? Reason I'm asking is that during the search for answers on the net, I've read that the gate voltage must be higher than the source, and in this case it's nearly the same.
Thank you for showing me out that line on the datasheet. My Application is to switch a 12V Battery on or off by the logic NOR gate. Another interesing question arrises: There will be some voltdrop across the Source and Drain, due to Rds, when I apply 12V to the gate. Question is: Will the Source Voltage be, say 11.95V when 12V is applied to the gate and the 12V Battery connected to Drain? Reason I'm asking is that during the search for answers on the net, I've read that the gate voltage must be higher than the source, and in this case it's nearly the same.
audioguru2
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Hi Cyberfreak,
I wish that you posted your schematic because I am just guessing that you are using the Mosfet as a source-follower. A source-follower can't switch current to a load unless its gate voltage is much higher than its source voltage when turned on. So for its source to be at 11.95V, its gate must be at maybe 23.95V!
The drain of a Mosfet is usually used as a switch so that its source is at a fixed voltage. You can use your N-channel Mosfet to pull the negative end of your load to ground* (or a P-channel Mosfet to pull the positive end of your load to a positive supply, but that would require you to invert your gate-drive logic).
* Ground the source. Connect the negative end of your load to the drain. Connect the positive end of your load to the positive supply. A positive voltage at the gate will turn-on your load.
If you use a P-channel Mosfet with its source connected to the positive supply, a positive voltage at the gate will turn-off your load.
I wish that you posted your schematic because I am just guessing that you are using the Mosfet as a source-follower. A source-follower can't switch current to a load unless its gate voltage is much higher than its source voltage when turned on. So for its source to be at 11.95V, its gate must be at maybe 23.95V!
The drain of a Mosfet is usually used as a switch so that its source is at a fixed voltage. You can use your N-channel Mosfet to pull the negative end of your load to ground* (or a P-channel Mosfet to pull the positive end of your load to a positive supply, but that would require you to invert your gate-drive logic).
* Ground the source. Connect the negative end of your load to the drain. Connect the positive end of your load to the positive supply. A positive voltage at the gate will turn-on your load.
If you use a P-channel Mosfet with its source connected to the positive supply, a positive voltage at the gate will turn-off your load.
cyberfreak
- Oct 7, 2004
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Hi AudioGuru
Please find attached my schematic.
As you can see, I must connect the Battery to Drain, so that the Battery's positive is isolated when the MOSFET is off. As you explained, that means I must use a P-Channel MOSFET instead and change my gate driver to an OR gate. But, how will I supply the MOSFET gate with 24V? Charge pump? How wil this charge pump look like?
View attachment 36033
Please find attached my schematic.
As you can see, I must connect the Battery to Drain, so that the Battery's positive is isolated when the MOSFET is off. As you explained, that means I must use a P-Channel MOSFET instead and change my gate driver to an OR gate. But, how will I supply the MOSFET gate with 24V? Charge pump? How wil this charge pump look like?
View attachment 36033
audioguru2
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Hi Cyberfreak,
Why are you using two P-channel Mosfets connected in series, with the top one with its drain and source backwards? ???
The bottom Mosfet by itself will work fine (its source connected to the positive supply) but the supply for the Cmos gate driver IC must be at the same supply voltage. If the logic supply voltage is less than the load voltage then the P-channel Mosfet will never turn off. In that case an N-channel Mosfet with its drain pulling the negative end of the load to ground will turn it on with a "logic-level" gate voltage.*
A P-channel Mosfet turns-on when its gate is negative from its source voltage. So when the Cmos output goes to ground, the Mosfet turns-on, and when the Cmos output goes to the positive supply (the same voltage as the source), the Mosfet turns-off.
What max frequency is your circuit switching at? If it is too high for the low-current output of the Cmos driver to charge or discharge the gate capacitance, the Mosfet will smoke then melt!
*Speaking of melting, we should go back to your other post about "logic-level Mosfets". With a 5V logic supply (why use only 5V for ordinary Cmos that works fine and even better with 12V?) which might actually be only 4.5V, your tiny N-channel Mosfet has a max Rds of 0.14 ohms. With 1.8A through it, it will dissipate 454mW. It will be much cooler with a 12V gate voltage.
BTW, what is your load? On your schematic it looks like a thermistor, maybe a hot-wire anemometer? Maybe not, that is somebody else's post. ;D
Why are you using two P-channel Mosfets connected in series, with the top one with its drain and source backwards? ???
The bottom Mosfet by itself will work fine (its source connected to the positive supply) but the supply for the Cmos gate driver IC must be at the same supply voltage. If the logic supply voltage is less than the load voltage then the P-channel Mosfet will never turn off. In that case an N-channel Mosfet with its drain pulling the negative end of the load to ground will turn it on with a "logic-level" gate voltage.*
A P-channel Mosfet turns-on when its gate is negative from its source voltage. So when the Cmos output goes to ground, the Mosfet turns-on, and when the Cmos output goes to the positive supply (the same voltage as the source), the Mosfet turns-off.
What max frequency is your circuit switching at? If it is too high for the low-current output of the Cmos driver to charge or discharge the gate capacitance, the Mosfet will smoke then melt!
*Speaking of melting, we should go back to your other post about "logic-level Mosfets". With a 5V logic supply (why use only 5V for ordinary Cmos that works fine and even better with 12V?) which might actually be only 4.5V, your tiny N-channel Mosfet has a max Rds of 0.14 ohms. With 1.8A through it, it will dissipate 454mW. It will be much cooler with a 12V gate voltage.
BTW, what is your load? On your schematic it looks like a thermistor, maybe a hot-wire anemometer? Maybe not, that is somebody else's post. ;D
cyberfreak
- Oct 7, 2004
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Hi AudioGuru
Answer to your first question: Don't know really why I chose to use 2 Mosfets in series. Maybe because I've read somewhere that the diode inside the Mosfet can help stopping any leakage current. Something like that. I realized before my last post that it is not really necessary.
Please find attached a new drawing, illustrating a final dawing what we were discussing. You will see I've changed the Logic driver to an OR gate, CD4071, and P-Channel Logic level MOSFET or an Enhanced Mode Mosfet, NDT454P. The Battery supply voltage is +-12V, where it also powers the Logic circuit, to decide when to switch the battery on or off. Applying a 0 at the Gate will turn the MOSFET on, hopefully hard enough to supply my load with 1.8A of power. And a High (12V) to switch the Mosfet off. Do I still need to apply a higher voltage to the Gate than Source?
There is not really a frequency speed.. It is switch on and off at certain condition from the OR gate logic. It will stay on or off continuously at certain conditions.
My load is GPRS/GPS circuitry with my own uP for Asset Tracking, with a charging circuit for solar panels. That's why I need to switch my battery backup on or off at certain conditions.
Will the current driven by the OR gate be enough to drive the Mosfet. Current according to the datasheet is 2.25mA. Mosfets are voltage devices, but current must charge/discharge the gate capacitance, as you mentioned.
http://www.fairchildsemi.com/ds/CD/CD4071BC.pdf
View attachment 36034
Answer to your first question: Don't know really why I chose to use 2 Mosfets in series. Maybe because I've read somewhere that the diode inside the Mosfet can help stopping any leakage current. Something like that. I realized before my last post that it is not really necessary.
Please find attached a new drawing, illustrating a final dawing what we were discussing. You will see I've changed the Logic driver to an OR gate, CD4071, and P-Channel Logic level MOSFET or an Enhanced Mode Mosfet, NDT454P. The Battery supply voltage is +-12V, where it also powers the Logic circuit, to decide when to switch the battery on or off. Applying a 0 at the Gate will turn the MOSFET on, hopefully hard enough to supply my load with 1.8A of power. And a High (12V) to switch the Mosfet off. Do I still need to apply a higher voltage to the Gate than Source?
There is not really a frequency speed.. It is switch on and off at certain condition from the OR gate logic. It will stay on or off continuously at certain conditions.
My load is GPRS/GPS circuitry with my own uP for Asset Tracking, with a charging circuit for solar panels. That's why I need to switch my battery backup on or off at certain conditions.
Will the current driven by the OR gate be enough to drive the Mosfet. Current according to the datasheet is 2.25mA. Mosfets are voltage devices, but current must charge/discharge the gate capacitance, as you mentioned.
http://www.fairchildsemi.com/ds/CD/CD4071BC.pdf
View attachment 36034
audioguru2
- Apr 6, 2004
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Hi Cyberfreak,
Now I am really confused. Your schematic doesn't show a load to ground.
It shows instead an external power supply, +12VI, which is used as a backup charger on a rainy day when the solar cells can't charge the battery. Here's how I think it works:
1) The Mosfet is turned off and the Cmos gate is powered by the battery through the PTC resistor. Your very low current load is powered from Vbatt too.
*But if the battery falls low enough (10 or 11V) the Mosfet will begin to turn on because Vbatt will be negative to +VI.
2) Something detects that the battery needs charging and also detects that the solar cells are impaired. Therefore something applies logic 0 to both OR gate inputs causing its output to go to logic 0. Therefore the Mosfet turns on.
3) With the Mosfet turned on, +VI charges the battery through the PTC resistor. The charging current is limited to 1.8A by the PTC because when the current gets too high, the PTC heats and increases its resistance.
*But the moment the Mosfet turns on, the PTC is cold and the charging current will be unlimited, blowing the Mosfet.
*Another problem is that the charging voltage, +VI, must be much higher than the nominal battery voltage in order to charge. Having that votage difference, the Mosfet never turns off.
*Most problems above are cured by having the Cmos gate powered from +VI.
The only problem that remains unfixed is the current surge through the Mosfet and cold PTC.
Now I am really confused. Your schematic doesn't show a load to ground.
It shows instead an external power supply, +12VI, which is used as a backup charger on a rainy day when the solar cells can't charge the battery. Here's how I think it works:
1) The Mosfet is turned off and the Cmos gate is powered by the battery through the PTC resistor. Your very low current load is powered from Vbatt too.
*But if the battery falls low enough (10 or 11V) the Mosfet will begin to turn on because Vbatt will be negative to +VI.
2) Something detects that the battery needs charging and also detects that the solar cells are impaired. Therefore something applies logic 0 to both OR gate inputs causing its output to go to logic 0. Therefore the Mosfet turns on.
3) With the Mosfet turned on, +VI charges the battery through the PTC resistor. The charging current is limited to 1.8A by the PTC because when the current gets too high, the PTC heats and increases its resistance.
*But the moment the Mosfet turns on, the PTC is cold and the charging current will be unlimited, blowing the Mosfet.
*Another problem is that the charging voltage, +VI, must be much higher than the nominal battery voltage in order to charge. Having that votage difference, the Mosfet never turns off.
*Most problems above are cured by having the Cmos gate powered from +VI.
The only problem that remains unfixed is the current surge through the Mosfet and cold PTC.
cyberfreak
- Oct 7, 2004
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Hi AudioGuru
How my design fits together is the following: +12VI is connected to my load and my load return(gnd) is connected to the same ground as my battery. Another external 12V PSU (in-case no solar cells avialable) is connected to +12VI and same ground. My logic sense if there is an external PSU present or not, to turn the MOSFET on or off. The solar cells is connected to VBAT (which have a higher voltage ,round 14V,with trickle, bulk and float charge capability) and is also sensed by my logic. All the parts of the design work. It is just the MOSFET to keep the battery off or on at that conditions that I can't get to work. MOSFETs is not my strong point.
How my design fits together is the following: +12VI is connected to my load and my load return(gnd) is connected to the same ground as my battery. Another external 12V PSU (in-case no solar cells avialable) is connected to +12VI and same ground. My logic sense if there is an external PSU present or not, to turn the MOSFET on or off. The solar cells is connected to VBAT (which have a higher voltage ,round 14V,with trickle, bulk and float charge capability) and is also sensed by my logic. All the parts of the design work. It is just the MOSFET to keep the battery off or on at that conditions that I can't get to work. MOSFETs is not my strong point.
