hi
this article says:
Does that mean, with complementary PN pair, you DON'T have to worry about dead time?
-thx
this article says:
Does that mean, with complementary PN pair, you DON'T have to worry about dead time?
-thx
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It depends on turn on and turn off times of the transistors.
Some have different turn on and off times.
I have never used complimentary pairs of mosfets just n-channel mosfets.
Some have different turn on and off times.
I have never used complimentary pairs of mosfets just n-channel mosfets.
P-ch and N-ch MOSFETs, compared at same Vds and Rds-on level, don't have the same Vgs-th, not same capacitances, not same transconductance, so gate drive has to be adjusted differently, in turn corresponding to "dead time adjustment". With 2x N-ch MOSFET dead time adjustment it is more commonly achieved by adjusting timing of a logic signal with logic gates and R-C-D stages. There is no way to escape from the dead time optimization step (without escaping from audio quality).
If you use a class d chip like the irs2092 that has programmable dead time built in.
The usual way of doing it is a resistor and diode in parallel to cause asymmetric turn on/off.
The usual way of doing it is a resistor and diode in parallel to cause asymmetric turn on/off.
Yes.
Or on the mosfet driver chip.
You could use something like ir2113 for driving mosfets.
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I understand your question such that If you used an NPN-PNP push-pull stage (base terminals connected to one another) and drove the two bipolar transistors into saturation (high <> low) dead-time and conduction overlap would not be an issue such as with FETs.
The starting up current of one bipolar would help removing the minority carriers from the other and switch it off.
The problem with bipolars is poor saturation, less fast switching an a more complex and inefficient drive circuit needing boot-strapping.
I'm not that good with semiconductor physics.
The starting up current of one bipolar would help removing the minority carriers from the other and switch it off.
The problem with bipolars is poor saturation, less fast switching an a more complex and inefficient drive circuit needing boot-strapping.
I'm not that good with semiconductor physics.
The problem with mosfets is the gate drive on the upper transistor needs to go above the b+ rail to ensure it is in saturation.
The irs2092 will provide this for you.
The irs2092 will provide this for you.
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