Sorry, I misread the circuit, I didn't notice the separate supply for the MOSFET. A 1Mhm: resistor will only make things worse so I don't think it's a very good idea.

It might help.


I don't know if they are any optocouplers fast enough. You might be able to buy one with a build in MOSFET driver.

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Remove the 1Mhm: resistors - they'll stop it working.

I've also noticed that you've used N-channel MOSFETs for the high-side drivers; this will mean that the MOSFETs will not fully turn on.

You either need to use P-channel MOSFETs for the high-side or drive the it with 75V and if you choose the latter you'll also probably need to protect the gate with a zener so it doesn't exceed 20V with respect to the source.

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You mean the top two N-MOSFET will not turn on..? Why is that..? I don't quite understand because I am new to MOSFET..

If I use P-MOSFET, that means that I need to trigger the gate with -Vth rite..?
How do I do that...?

Thanks..

 

Because N MOSFETs require the gate to be at a higher potential than the source. With your circuit the gate will be at 15V, therefore the source will flowat at about 5V to 10V giving a massive voltage drop across it and only a small voltage to the motor.

A P MOSFET will turn on when the gete potential is below the source, so if your motor requires 60V and P high side MOSFETs are used they will need 45V to 50V when on and 60V when off.

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Thanks for your explanation.. Initially, i do not really understand.. but after searching on the internet, i found MOSFET application note by DALLAS semiconductor.. Now I understand..
When you say "source flowing at about 5 to 10V", do u mean the V(DS) would be 60V - 10V which is 50V..? Did i understand correctly..? If that is the case then the power loss would be too much...

Can I ask, how to calculate the Voltage at source of MOSFET..? what formula should I use..?

If I use P-MOSFET for high-side driver, the source terminal (of MOSFET) would be connected to the 60V and the drain would be connected to the motor rite..? So, the current would flow in reverse (from source to drain) rite..?

I searched the internet and I found P-MOSFET for the circuit..
The datasheet can be obtained from here..
http://www.farnell.com/datasheets/44775.pdf
From the datasheet, there are 3 things i need clarification..
The first one is V[gs(th)]. The V[gs(th)] needed is -5V. How do I generate a negative voltage..? Is it negative voltage as in the current flowing from source to gate instead of gate to source (as in N-MOSFET)..? So, to solve that, I need to connect the source terminal (of MOSFET) to +ve of 5V supply and the gate terminal (of MOSFET) to -ve of the same 5V supply..?

Second thing is, the minimum V[gs(th)] is -3V, maximum V[gs(th)] is -5V and the V(GSS) is +/- 30V. How do I interpret these data..? Does it means that my gate voltage cannot exceed -5V..? How about the VGSS..?


Yeah.. i posted alot of questiong.... After searching the internet for days, i understand abit of MOSFET but there are still gaps, questions that i still cannot find answer to, things that are still gray and uncertain to me... Thanks for your time and your patience...

 

I made a mistake, meant the drain will be floating at 5 to 10V.

Yes, you need to pull the source 10 or so V below the drain before it will switch fully on.

Yes, the poor MOSFET will have 50V between the drain and source whilst carying the current.

Look at the graph of current vs gate source voltages on the datasheet.

Yes, a P-channel MOSFET is the reverse of an N-channel MOSFET, the current and voltages are all reversed.

I searched the internet and I found P-MOSFET for the circuit..
The datasheet can be obtained from here..
http://www.farnell.com/datasheets/44775.pdf
From the datasheet, there are 3 things i need clarification..
The first one is V[gs(th)].

You don't, all it means is the gate needs to be -5V with relative to the source in order for it to turn on.

No gate current flows in a MOSFET; the gate is an insulator.

The gate terminal needs to be 5V below the source (55V if the source is connected to +60V).

The gate threashold is the voltage on the gate at which the channel starts passing current, it isn't a very accurately controlled parameter so a range between 3V and 5V is given. Don't be fooled into thinking it will pass any current up to the maximum rating once the gate is above the threashold voltage. Look at the graph on the datasheet to find out what gate voltage is required for a given current.

Attached Images

PMOSFET RFQP12PCha.GIF (50.2 KB, 15 views)

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Let me try to understand about N-MOSFET..
Lets say, the Drain is connected to the supply of 60V. According to the current vs gate-source voltage in the datasheet, when I make the V(DS) 40V as in the Source being at 20V, the MOSFET furns fully on rite (maximum current flow)..? This happen when V(GS) is greater that 8V..?

Am I understanding correctly so far..?

How would I make the Source voltage become 20V..?

Attached Images

N-MOSFET - transfer characteristics.jpg (38.4 KB, 5 views)

 

I came across this equation for N-MOSFET in wikipedia.

"VGS > Vth and VDS > VGS − Vth"

Is that what you are trying to say..?

 

The I_ds vs. V_gs curve is 2/3 the way down the page.

http://www.ece.gatech.edu/research/l...y/devchar.html

When operating a N channel mosfet as a switch, you want to operate in the triode region - Vgs as high as possible, Vds as low as possible - when the mosfet is fully on (i.e Vgs is as high as possible), the mosfet acts like a resistor with R = 1/(2K(Vgs-Vt)) [this is from the 3rd equation below the chart] The higher Vgs is, the better. Vds at that point is then governed by Ohm's law...

 

Thanks...
I think I finally understand the N-MOSFET I-V curve..
According to the I-V curve below attached below, lets say initially i supply it with V(GS) of 3.4V. When I increase the Drain voltage (from supply), V(DS) will increase until 0.5V. At the same time, I(D) will increase sharply until 3.1A. The resistance R(DS) at that time would be somewhere near 161mOhm (0.5V/3.1A). It is lower than R(DSon) which is 75mOhm because the V(GS) is still not high enough..

Lets say V(D) at that time is 30V. V(S) would be V(D) - V(DS) which is around 29.5V (30V-0.5V). At V(S) of 29.5V, it is way over the voltage of V(GS) and because of that N-MOSFET will not turn fully on.. Is that what you have been trying to say Hero999..?

When I continue to increase V(D), V(DS) and I(D) will continue to increase but not too much.. The higher V(D) is the flatter I(D) becomes until a point where it stop increasing..

In order for me to fully turn the N-MOSFET on, I need to keep on increasing V(GS). As V(GS) increase, I(D) and V(DS) will also increase because the MOSFET now acts like a resistor.. Is that what you mean hjames..? When that happen, V(S) will continue to decrease because V(D) is the same..
I need to keep on increasing V(GS) until V(S) decrease until it is below V(G).. When this happen, the MOSFET fully turns on..

In order to do this, I need alot of gate voltage. That is why N-MOSFET is not suitable for high side of the H-Bridge.. It is more suitable on the low-side of the bridge because the V(S) is always 0V..

Am I on the right track now..?

Attached Images

N-MOSFET (PHK5NQ15T).jpg (26.6 KB, 7 views)

 

You're overthinking the relationship between Id/Vds. Imagine the mosfet as a non-linear voltage controlled resistor. The resistance is some function of the gate-source - for a given Vgs, the mosfet will act like a very small value resistor, up to some current limit (when it enters saturation), and starts acting like a current source and dissipating a lot of power. For a higher Vgs, the resistance is smaller(lower switching losses), and the point where it enters saturation is higher(you can driver bigger loads).

Correct - In order to use a n-mosfets on the high side, you need to supply a gate voltage that is greater than the supply voltage that it is switching - since you want Vs of the mosfet to be very close to the supply voltage. There are lots of "flying capacitor boost" type circuits that do this (google it). N-mosfets are inherently cheaper and better, hence most high power drivers end up using all N-mosfets. If you can get P-mosfets in the ratings you need, then just use them.

 

Yes but you'll have 20V between the MOSFET's drain and source with the maximum current flowing, it will dissipate a huge amount of power.

The only way to make the MOSFET turn fully on and not dissipate a huge amount of power would be to take the gate to 75V.

If you want to use an N-channel MOSFET as a high side driver you want the source to be at the positive rail voltage; to do this you need to lift the gate above the positive rail voltage by the voltage required for it to pass the required current.

Don't get bogged down with equasions, look at the graphs on the datasheet.

Look at the graph, there are many conditions which will give you a 0.5V voltage drop.

For a voltage drop of 0.5V, you'll require drain current of about 3.1A with a gate-source voltage of 3.6V, the gate voltage (relative to 0V) will be 29.5 + 3.6 = 33.1V or about 1.3A with a gate-source voltage of 3.4V, the gate voltage (relative to 0V) will be 29.5 + 3.4 = 32.4V.

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Thanks for all the time spent.. I understand it finally..
P-MOSFET function based on the same principle but in reverse form..
I think I am getting there..
Thanks again..

 

You've got it.

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Traditional H bridges indicated by diagrams uses 2 N and 2 P channel POWER MOSFETS, The reason for using the 2 different types is because these 2 types responds differently,

It is important to know that:

N type MOSFETs when Given a +5V to +20 Volts (High) at the gate, they will turn On, and connect your circuit.

P type MOSFETs when grounded or LOW at the gates they will turn On.(Yes, it would blow up if there is no resistance at the Gate)

So if you had an incomming signal that was either 1 or a 0, High or Low, only 1 N and 1 P type mosfet will turn on, thus if you refer to the diagram, the MOSFETS that turn on will be diagonal from each other. If your input signal changed, the other 2 diagonal MOSFETS will conduct while the 2 MOSFETS you were just refering to will turn off.

This is traditionally used so that you will never have a SHOOT thru, or shorting out the H BRIDGE, Even a Micro second will hurt your circuity at high frequencise.

I would Prefere to use N-TYPE MOSFETS ONLY Because they have a significantly higher Amp rating than P-Type mosfets, You can do this with just the same, but You will need to use A LOGIC INVERTER, Hex invert. You would have 1 incomming signal either high or low, and the signals will divert to the MOSFET DRIVER that operates the top of the MOSFETS according to the generic diagrams, and the other signal is diverted into the LOGIC Inverters. The output on the Logic Inverters will then reach a second MOSFET DRIVER that drives the bottom set of MOSFETS.

This works well because only the Bottom ROW of MOSFETS Give off Heat, as the bottom row will turn on slower due to the Lagg from the Logic gates. This is also very efficient because it also has no chance of shooting thru. Due to the lag in the Logic Gates.

I would also stay away from PWM signaling. Instead I would use a linear Voltage regulator. This is because PWM constantly turns on and off the MOSFETS, these switching times are turned into wasted energy in forms of Heat.

I would recommend setting up several N type MOSFETS, 12 is a good number in parallel so you reduce the load on each MOSFET and dissapate Heat better. Also for better reliability.
I use IRF3704ZPBF N-TYPE MOSFETS from International Rectifyer? (47Amps at 25C temp / 56Amps at 100C also PBF means Lead Free)
This is the one I use because it has a fast Switching time, and low Gate Resistance. But it is limited to 18 volts maximum, You should search up on their website for a higher voltage rating up to 1000V MOSFETS.

As for MOSFET DRIVERS I used with my IRF3704Z a pair of TC4426PBF(Leadfree) Mosfet Drivers that provides 1.5Amps of Current for the MOSFET Gate if needed in case the Gate resistance is high. The MOS Drivers are also PDIP =D

On my H-Bridge, I had 2 resistors hooked up to the H Bridge so that there is resistance between Gate to Source and Gate to Drain. This way, when power is off, any noise or watever will not damage your motor, Or if the signal is somehow borken or not getting thru, the H-Bridge simply wont turn on. The resistance cant be too big or else it will reduce the effects of the MOSFET Driver(in other words, increase the load on the MOSFET drivers which makes respond slower.

Should you have any questions Please don't hesitate to contact me Via Email

Jun Zheng,
Gr 12 Senior Engineering Student.
David Thompson Secondary
Vancouver BC. Canada
FIRST Robotics member
Team # 1346


I created an account when I read thru this havoc, I felt frustrated hahaa...the need to answer ppls questions!