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Diode-less Power MOSFETS

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Ron H said:
Styx, the point is, that is not a rectifier (assuming the MOSFETs are on). Each leg conducts in both directions.

True

Question though?
Why are you after low-conduction drop? is it to try to get as much voltage from the supply

or is it to improve efficiency - ie less voltdrop => less power lossed in the switches.

If it is the efficiency you have to factor in the gate-drives of the switches.
 
Question though?
Why are you after low-conduction drop? is it to try to get as much voltage from the supply

or is it to improve efficiency - ie less voltdrop => less power lossed in the switches.

If it is the efficiency you have to factor in the gate-drives of the switches.
answer
yes , yes, true ...
maybe i'm not looking for a sync rectifier at all..
all i want is to be able to convert my 20V p-p generator output to DC, without loosing 12.8 W in the process..the frequency of the generator is (varible up to 100HZ)

**AC test:**
Voc =20Vp-p

(Loaded) is with a 1 ohm resistor
VLoaded =12Vp-p
average power:
(VmIm/2)= 6*6/2= 36/2 = 18W

**DC Test..:**
(a bridge rectifier setup) was used ,and a 3000uF cap across the output to smooth the wave..
VLoaded = 4V
I = 1300 mA (measured)
power 5.2W
 
Styx said:
The intrinsic diode present in MOSFET packages arew normally crap when used as free-wheel diodes in H-bridge arrangement, thus this method blocks the intrisic diode allowing you to use a better match (since intrisic diode characterists are nothing like their parent MOSFET - for starters they are more like Zeners)

You want the most efficient FWB but insist on using MOSFET's !!!
BJT have the lowest forward drop and conduction losses, hence if you look at the POWER-electronics table BJT's are still used when fully-controlled is need and very high current are needed to be controlled - their failing is their slow switching (still in the KHz range) and that to switch 1000A you need to at least source and sink 100A or there abouts

But if you drive the gate properly, you will never forward bias the body diode except perhaps for a mS of bootstrapping stage. This is the way to AVOID turning on the diode. So I don't know why you're harping on the poor recovery characteristics of a freewheeling diode that won't be used in that mode.

BJTs are not really great for a super-efficient circuit. The minimum voltage drop on a BJT is far higher than a decent MOSFET. The mess is the required base current. Depending on the BJT construction and temp, the gain to drive down to 0.3v vce(sat) may be pretty low thus an exceptionally high base current could be required. For both these reasons , and the lack of contraindicating factors such as high voltage, I think he is making the right choice with MOSFETs. Not that there aren't plenty of good uses for BJT synchronous rectifiers but probably not here.
 
williB said:
maybe i'm not looking for a sync rectifier at all..
all i want is to be able to convert my 20V p-p generator output to DC, without loosing 12.8 W in the process..the frequency of the generator is (varible up to 100HZ)

No, I think you're on the right track. It's not only a fun project but potentially useful experience, automotive regenerative braking for example needs to transfer a lot of rectified power this way. And it may improve your performance substantially.

Here's one other thought. There's no reason you can't make a FWB out of Schottky diodes. A Schottky diode has only 0.3v forward drop as compared to the 0.7 of a silicon diode, but that's only for fairly modest currents. Higher currents increase the Schottky voltage more than it affects silicon diodes. Sometimes as you approach the max current rating of the part the forward voltage may be greater than the forward voltage of a silicon diode of the same rating. The other issue is Schottkys have relatively poor reverse leakage blocking characteristics. Leakage increases with temp and overall current capacity of the part (larger junction area=more leakage).
 
williB said:
**AC test:**
Voc =20Vp-p

(Loaded) is with a 1 ohm resistor
VLoaded =12Vp-p
average power:
(VmIm/2)= 6*6/2= 36/2 = 18W

**DC Test..:**
(a bridge rectifier setup) was used ,and a 3000uF cap across the output to smooth the wave..
VLoaded = 4V
I = 1300 mA (measured)
power 5.2W

Oh yea, I would say in the DC test you have definitely overloaded the motor. Modelling the motor as having a linear output impedance (an oversimplified assumption!) the max power output would be when the voltage is 50% of the open circuit voltage, or 10v. At this point the efficiency is only 50% too.
 
williB said:
Question though?
Why are you after low-conduction drop? is it to try to get as much voltage from the supply

or is it to improve efficiency - ie less voltdrop => less power lossed in the switches.

If it is the efficiency you have to factor in the gate-drives of the switches.
answer
yes , yes, true ...
maybe i'm not looking for a sync rectifier at all..
all i want is to be able to convert my 20V p-p generator output to DC, without loosing 12.8 W in the process..the frequency of the generator is (varible up to 100HZ)

**AC test:**
Voc =20Vp-p

(Loaded) is with a 1 ohm resistor
VLoaded =12Vp-p
average power:
(VmIm/2)= 6*6/2= 36/2 = 18W

**DC Test..:**
(a bridge rectifier setup) was used ,and a 3000uF cap across the output to smooth the wave..
VLoaded = 4V
I = 1300 mA (measured)
power 5.2W
Willi, 3000uF is not an adequate filter cap for a one ohm load at 100Hz. The ripple from an ideal full wave rectifier will be 6V p-p! The average DC voltage (what you would read on a DC meter) will be 7.28V.
This also assumes that the generator has zero output resistance.
The ripple will be worse at lower frequencies.

Below is a plot of the waveform. Remember, this is with ideal diodes.
 

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Oznog said:
williB said:
**AC test:**
Voc =20Vp-p

(Loaded) is with a 1 <a href="#">ohm</a> resistor
VLoaded =12Vp-p
average power:
(VmIm/2)= 6*6/2= 36/2 = 18W

**DC Test..:**
(a bridge rectifier setup) was used ,and a 3000uF cap across the output to smooth the wave..
VLoaded = 4V
I = 1300 mA (measured)
power 5.2W

Oh yea, I would say in the DC test you have definitely overloaded the motor. Modelling the motor as having a linear output impedance (an oversimplified assumption!) the max power output would be when the voltage is 50% of the open circuit voltage, or 10v. At this point the efficiency is only 50% too.
I dont know the output impedence , but the resistance is 1.3 ohms..
 
Ron i got the same result on the scope , as i got in simulation..
It wasnt enough capacitance to adaquatly filter the output..your right..
 
Willi, if your generator has a center tap that you can ground, I can post a design (and results) that uses two MOSFETs and a dual comparator (but not just any dual comparator). I have simulated it, but I have not built it. It also doesn't work well under light loading (the comparators oscillate). It might work for you, if you can figure out what a realistic load is (not one ohm :) ).
 
williB said:
Yes , i can get a center tap from it no problem..!!
i'd really like to see the design..
It's on my computer at home. I'll try to post it tonight.
 
could someone simulate this .. i'm pretty excited ...
its so simple ...
it works as well with a battery as a resistor...
 

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williB said:
could someone simulate this .. i'm pretty excited ...
its so simple ...
it works as well with a battery as a resistor...
Did you sim it? I did, and, as I expected, the transistors have 2.6V peak across them when they are conducting. This is not because of ON resistance, but because of the gate threshold voltage.
 
williB said:
you checked the output ? right?
yes i did sim it..
As I said, I did sim it, and yes, I checked the output. You would be better off with diodes.
 
williB said:
i think i want a second opinion.
Willi, in the waveform window you can right-click on the name of the waveform and a dialog box opens which will allow you to enter an algebraic expression. Plot the voltage across each of the upper series pairs of MOSFETs. I got 2.6V (peak) when forward biased.
Also, keep in mind that the maximum reverse gate voltage is 12V. The only thing protecting you from breakdown is the 1.3 ohm generator resistance and the 1 ohm load resistance. If the load resistance goes up, you will break down the transistor gates.
I hope you can get a second opinion.

Edit: I think I was wrong about the gate breakdown.
 
Try replacing the upper transistors with p-channels, as below.
 

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Ron H said:
williB said:
you checked the output ? right?
yes i did sim it..
As I said, I did sim it, and yes, I checked the output. You would be better off with diodes.
I just redid the sim with MBR20100CT diodes ..
The diodes beat the Mosfets by 0.5 A with a load of 1 ohm..
the diodes scored 8.0 A while the Mosfets got 7.5 A..
 
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