Buck converter using logic level mosfet

[ian123]

Please help me with this design

I want to use a buck converter to convert a voltage input of 16-40v to 13.8 v. this will be controlled via a 16f628 via pwm.
The buck converter part is where my problem starts.

I am trying to make a buck converter using a logic level mosfet but seem to be having problems with the switching of the mosfet.

according to the spec sheet a logic level mosfet will switch on VGS(th) at 2V.
I have calculated the the buck converter to use the following components.
switching freq = 700kHz
induc ripple ratio = 0.3
capacitor voltage ripple ratio= 0.04
Max overshoot= 0.096
Max current = 500mA

there fore the duty cycle = 0.44
min inductance = 63.4 uH
Min Saturation current = 0.69 A
Min out put capacitance = 11 uF

when in practice the output current is way less than 500mA ie: it barely turns on a led also the voltage mesured across the output terminals is about 2V.

I have uploaded the test circuit here
at the moment i am using a function generator to produce a square wave.
If I use the TTL output on the generator then the circuit does not run, the TTL output is 5V .
If I use the normal output and increase the output to 10V then the circuit runs but with limited current ie: nowhere near 500mA
Please help

 

[ian123]

after some tests on the workbench
I have found that if I feed a negative 5v to the gate the circuit operates better
anyone who can shed some light on this please

 

[alec_t]

You have an N-FET in your circuit. If the desired source voltage is 13.8 then the gate voltage needs to be at least 15.8V (i.e. 2V more than the source voltage, since Vgsth is 2V) before the FET even begins to turn on (conducting only microamps).

 

[MrAl]

Hi,

As alec points out, the voltage at the gate must be 2v higher than the source to turn the MOSFET on. It looks like you are trying to do that through a diode and a floating power source which in itself might not be too bad, but the diode brings in problems.

First, the polarity must be such that it allows current to flow from the source to the gate with the gate being more positive than the source, and this means that the diode has to be connected with the ANODE to the source. Second, the polarity must be such that when the pulse goes to zero it must allow current to flow from the gate to the source, and this means the diode has to be connected with the CATHODE connected to the MOSFET source terminal.

We cant have it both ways, but if we remove the diode then we find that we can get that kind of conduction in either direction, so the diode must go.

Also, the MOSFET must be turned on and off quickly at a frequency as high as 700kHz. This means you'll need a decent MOSFET driver IC for the gate, one that can supply a good amount of peak current like 1 amp or more. It wont work very well trying to drive the MOSFET with a microcontroller chip pin alone.

 

[ian123]

Ok my understanding is that a logic level gate drive should be able to turn fully on with 5V??
that is why they make them no so.

If I understand it correctly then even though this is a logic level device I will still need 2 volts higher than the source to turn it on. if this is the case then on might as well just use a normal mosfet for this operation and drive it with a transistor
why then do they make logic level mosfets if you still need 2V above the source to drive it??

 

[ian123]

"It looks like you are trying to do that through a diode and a floating power source which in itself might not be too bad, but the diode brings in problems."


I am not sure where you see the diode driving the gate as the schematic shows the gate being connected to the a signal generator

 

[MrAl]

Hello,

That is correct to a point. They need 5v but for very fast switching (100kHz and up) they need a good driver circuit that can put out a high current for a very short amount of time.

They make logic level MOSFETs so you only need 5v to drive the gate, not 10v or more.

They also make P channel MOSFET's which require a different drive polarity. That might work better for you. Then the drain goes to your load and all you need is a zero volt signal to turn it on. To turn it off you either supply a positive voltage higher than the source or equal to the source.

I'm sorry i did not see the dot that connects the sig generator to ground. So you are driving the gate with a signal genrator in a normal fashion. In that case 5v is not enough as you now know, but maybe you could look into using a P mosfet.

 

[alec_t]

Ok my understanding is that a logic level gate drive should be able to turn fully on with 5V?

Correct. But that 5V must be between gate and source. In your circuit the 5V from the sig gen is between gate and ground.

 

[ian123]

ok so I have 2.8v between gate and source so how do i fix it?
if i want a micro processor to switch the mosfet on.

 

[ian123]

hey also make P channel MOSFET's which require a different drive polarity. That might work better for you. Then the drain goes to your load and all you need is a zero volt signal to turn it on. To turn it off you either supply a positive voltage higher than the source or equal to the source.


if i use a p mosfet then i would have to put a 10K resistor to pull up to the source voltage and then switch the mosfet on by supplying the low from the microprocessor. this means that while the mosfet is off there will be anything between 16 and 40 v on the output pin of the processor.

i am at a loss of how to make this work properly.

 

[crutschow]

To turn on the N-MOSFET you could use a bootstrap driver.
To turn on the P-MOSFET you could use an NPN transistor buffer from the processor with a resistor to V+. You need a low value resistor for fast turn-off of the FET or a two transistor push-pull driver for faster switching.

 

[ronv]

Since you are running at a high frequency you will probably need a high side driver like this:
http://www.irf.com/product-info/datasheets/data/ir2110.pdf
Even so it is kind of pushing it. How about a lower frequency and a larger inductor and capacitor?

 

[MrAl]

Hello again,

If you can drive the MOSFET from a microcontroller other than the one called "CPU" you can use this circuit.
If not we'll have to do it another way.

Note either way you may want to bring the switch frequency down to 70kHz rather than 700kHz as that will simplify the driver, unless of course you dont mind buying a driver IC chip made for driving MOSFETs.

Also note that if you intend to use the "CPU" to control the MOSFET then you need at least a temporary way to get power to it so it can start the buck circuit going. Before the MOSFET turns on for the first time there is no power to the "CPU".

 

[ian123]

Hi i can get supply for the cpu from the 13.8v line ( since there is a battery being charged on there) by putting one diode in the circuit. I have also adjusted the circuit to use transistors as mosfets now become more expensive if I have to use mosfet driver.

 

[MrAl]

Hello,

Not a bad idea really, to use a bipolar instead of a MOSFET. That makes it a little easier.

Since the battery can power the CPU then i guess you're ok as long as the battery always has some charge, with enough current to drive the base of the main switch transistor.

The resistor of 220 driving the base could dissipate a lot of power, like 7 or 8 watts, so a pretty big resistor would be needed. Maybe you dont need that much base current though. How much current did you intend on getting out of the regulator at the 13.8v point, is it still around 500ma ?

Check the main transistor to see that it can take the entire base current with the 40v upper input limit.

Most likely the circuit will not do well at 700kHz as mentioned before. A better design point would be 70kHz.

 

[crutschow]

.......................
Most likely the circuit will not do well at 700kHz as mentioned before. A better design point would be 70kHz.

Even 70kHz is high for a bipolar saturated power switch. I would go closer to 20kHz.

 

[MrAl]

Hi,

Well, 50kHz is very typical and efficiency not too bad, so 70kHz might be ok.