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mosfet bridge driver

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ca178858

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Hi all-

I've been somewhat involved with electronics for a long time, but I've never been real good at analyzing analog circuits, and I was hoping someone could sanity check my work....

I'm trying to drive a network of red/green LEDs (they're the backlight for a keypad), the max current will be 240mA. The MOSFETs I'm planning on using are Fairchild FDC6420C, they're logic level with an n and p channel in the same package.

The two control lines will be connected to a PIC, I have lines to spare so I could drive Q1 and Q2 with different outputs, but with the extra MOSFET available (Q3-2) I used it as shown...

Other than the obvious question of 'will this work?' I have a few others:

-Does R2 need to exist? If it does whats a reasonable value?
-Should I tie the two control lines to ground with 10k resistors?

-On a completely unrelated note: would this circuit be suitable for driving a DC motor (if clamping diodes are added)?


Thanks for any help...
 

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R2 looks like a static load for PSU, you could get rid of it.

however, as it stands two problems

1) The LED's will not turn on. The diodes are in anti-series and thus no path exist for current to flow, no matter what polarity of voltage is applied. More than likely you will damage them since LED's do not have hight blocking voltages

2) there is nothing to limit the current IF the diodes were to conduct


best bet would be to put the LED's in anti-paralles with a burn-resistor for each one - that should do it
 
Looks like I chose a bad symbol for the schematic... The actual device has them anti-parallel...

The reason I through R2 in there was because if Q3-1 is off the source of Q1 and Q2 will be floating... I'm not sure if thats a problem.

Point taken about the current limiting resistors.
 
Why not do your logic in software? Unless I am missing something, you can save an FDC6420C and a resistor. See below.
 

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RonH,

Thats probably what I'll do. I had originally only wanted to use 1 of the 2 PWM modules available on the PIC for the LEDs and use the other one for something else.

In the case of this circuit it shouldn't really matter whether or not both MOSFETs are on at the same time (since they're current limited to with the specs), but if I was driving a motor with this circuit and didn't have R3 or R4, would the MOSFETs be in danger of causing a short in the time that they were turning off/on? Should I drive them individually to control the on/off times?
 
ahh someone actually concidering turn-on shoot-throughs - verygood

a simple hack would be at the gate of all the FET's put a resistor in series (should be there anyway) and then in parallel with that resistor a diode-resistor (in series arrangemen).

The orientation of hte diode will either slow down turn-on or slow down turn-off.

You really should look into a dedicated gate-drive circuit - a simple BJT push-pull will make hte switching performance better

There is a big thread about this somewhere here, have a look
 
Styx said:
ahh someone actually concidering turn-on shoot-throughs - verygood

a simple hack would be at the gate of all the FET's put a resistor in series (should be there anyway) and then in parallel with that resistor a diode-resistor (in series arrangemen).

The last schematic shows that R3 and R4 would limit the current should both transistors on one side turn on at the same time. It shouldn't be a whole lot more than what the circuit takes normally. Worst case possible would just be V+/R.
 
Oznog said:
Styx said:
ahh someone actually concidering turn-on shoot-throughs - verygood

a simple hack would be at the gate of all the FET's put a resistor in series (should be there anyway) and then in parallel with that resistor a diode-resistor (in series arrangemen).

The last schematic shows that R3 and R4 would limit the current should both transistors on one side turn on at the same time. It shouldn't be a whole lot more than what the circuit takes normally. Worst case possible would just be V+/R.
You missed his question about driving motors.
 
Oznog said:
Styx said:
ahh someone actually concidering turn-on shoot-throughs - verygood

a simple hack would be at the gate of all the FET's put a resistor in series (should be there anyway) and then in parallel with that resistor a diode-resistor (in series arrangemen).

The last schematic shows that R3 and R4 would limit the current should both transistors on one side turn on at the same time. It shouldn't be a whole lot more than what the circuit takes normally. Worst case possible would just be V+/R.

Well concidering I was the one that told him he will need those resistors for LED load.

With a machine in their place (ie just an inductor) shoot-though protection IS needed!!! 12V with a few hunder milliohms is alot of amps for little FET's

with the XOR the elimination of hard-shoot-through's is all but eliminated (leaving Single mode failure's) but the finite time to switch a leg will cause a short across the supply, be it with the FET's in their active region (and thus dissipate alot of power) or fully and thus take alot of current = alot of power, it is a eral problem that MUST always be addressed in H-bridge configurations with low-impedance loads!!!
 
the finite time to switch a leg will cause a short across the supply, be it with the FET's in their active region (and thus dissipate alot of power) or fully and thus take alot of current = alot of power, it is a eral problem that MUST always be addressed in H-bridge configurations with low-impedance loads!!!

Ok, so how is that normally dealt with? Driving each mosfet seperatly and turning one off waiting for the turn off time and then turning the other one on? Or can you tie their gates together and use a mosfet driver with its balanced turn off/turn on?

BTW- thanks for all the responses so far...
 
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