Which Mosfet Spec is Min Output V?

danadak

Joined Mar 10, 2018
4,057
This part is characterized to run off 9 - 16V Vcc. Therefore the output
high level will be Vcc - (Iload x Ronhs), typically >> 3.5V in "normal"
applications.

The low output V will be = (Iload x Ronls), typically << 1V in "normal"
application.

because this is an hbridge the load path is

Vcc > High Side MOSFET > Load > Low Side MOSFET > ground

So the voltage across load becomes -

Vcc - (Iload x Ronhs) - (Iload x Ronls)


Regards, Dana.
 
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ebp

Joined Feb 8, 2018
2,332
You could place strings of diodes in series with the load. At 4 A, each diode would probably be around 0.7 V, but it would be wise to consult the datasheet for the diode contemplated. A diode designed for 6 A will have higher forward voltage at 4 A than one designed for say 15 A. You will need two strings of diodes connected in inverse parallel. This of course is inefficient due to the power dissipation in the diodes. The temperature coefficient of the forward voltage of the diodes will come into play but may be low enough to be ignored if the output voltage is not critical and/or the diodes are on heatsinks and kept at moderate temperature. The minimum voltage drop you would require is 1.5 V, which would mean 6 W of dissipation in the diodes. 1.5 V is a little high for two diodes in series, but if you start with 5.5 V and can accept a bit over 4 V it could work.

If and only if the load current at 4 V is always 4 A, you could use a single resistor in series with the load. Ohm's law is your friend.
 

Thread Starter

johnyradio

Joined Oct 26, 2012
434
Need a precise output V. But, if i can approx halve the output V with diodes or resistors, then i can adjust the supply V to precisely set the halved-output-V to whatever specific V i need. Right?

i think if i lose efficiency, that means losing amps, but i'm ok as long as this chip can put out the needed extra amps (load current + dissipation, correct)?

is there a more efficient way to step down the output V, without too much more complexity? An inductor?

Thx!
 
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ebp

Joined Feb 8, 2018
2,332
If you need precise output voltage you would need to run the whole thing closed-loop, comparing the output voltage to reference voltage and adjusting the input voltage accordingly - essentially wrap a voltage regulator around the bridge. This is not trivial.

No matter how the bridge is made, it will introduce some voltage difference between the supply voltage and the load. The question is how much difference can be tolerated, and if the answer is zero (never fully possible) then any circuit using an H-bridge will require closed loop control.

You really need to start getting your mind around some basic concepts. If you want 4 A into the load you will have to put 4 A into the bridge. The loss is as a voltage. see Kirchhoff's laws
 

crutschow

Joined Mar 14, 2008
34,406
Operate the bridge at 6V or more and use a PWM input to the bridge to get the desired load voltage (about 58% duty-cycle for 3.5V at the load from a 6V supply, for example).
That will give high efficiency.
 

Thread Starter

johnyradio

Joined Oct 26, 2012
434
If you need precise output voltage you would need to run the whole thing closed-loop
Fantastic reply! My plan is to use an off-the-shelf regulator of some kind at the input to the h-bridge. That's how i'll ensure stability. Whatever voltage drop i get across the bridge, i'll adjust the source regulator to compensate.

Operate the bridge at 6V or more and use a PWM input to the bridge to get the desired load voltage (about 58% duty-cycle for 3.5V at the load from a 6V supply, for example). That will give high efficiency.
Great idea! But isn't PWM a way to regulate current, rather than voltage?

Thx
 
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