Few Clarifications on the IR2181

[electronicsLearner77]

Trying to decode the complete motor circuit from microchip and this is a continuation of the previous thread started, understanding the circuit related the IR2181 high side MOSFET driver. The screen shot of the circuit is below and i have attached the complete circuit.


If i see the data sheet and the pin diagram of the IR2181

Few of the things i understand are
If LIN is H/L then LO is H/L similarly if the HIN is H/L then HO is H/L.
Condition 1: If HIN is H -> HO is H -> Q5 is ON -> then DC+ voltage is applied at M1
If condition is met then
Condition 2: LIN is L -> HO is L -> Q6 is OFF.
Is my understanding correct ?
The Diode D9 is placed to avoid reverse connection of 15V? I am not sure why has not followed the same method to connect to Vcc, he directly connected the 15V.
I really don't understand what are Vs and Vb, what is the purpose of these two signals?

 

[Harald Kapp]

Condition 2: LIN is L -> HO is L -> Q6 is OFF.
Is my understanding correct ?

How do you arrive at this conclusion? Check it against this your statement:

If LIN is H/L then LO is H/L similarly if the HIN is H/L then HO is H/L.

Also have a look at the internal schematic of the IRF2181 on page 4 of the datasheet.

Note that the datasheet explicitly states that the high and low side channels are independent. Therefore it is mandatory that the signal generating circuit on the input side (pins 1 and 2) must ensure that only one channel (either high or low) is active at a time. If by error both channels were active, both Q5 and Q6 were on and would create a short circuit for the power supply.

 

[electronicsLearner77]

Sorry my mistake i wanted to say LIN is Low then LO is Low and hence Q6 is OFF

 

[electronicsLearner77]

The Vs and Vb are used for bootstrapping circuit it seems. I need to study on this, i have heard but never understood it.

 

[Harald Kapp]

Bootstrapping in this context means that the IR2181 generates a voltage higher than the supply voltage to ensure the high-side MOSFET (Q5 in the image below) is driven by a gate voltage substantially higher than the supply voltage. If you were to drive the gate of Q5 with DC+, then the source would be at DC+ - V_GS(sat). This would dissipate a lot of power in the MOSFET, possibly destroying it, and this power would not be available to the load.
By making Vgate > DX+, it is ensured that V_GS > V_GS(sat). The MOSFET will turn on completely, dissipating little power and full power will be available to the load.

The way this is achieved is by using a charge pump made from C37 and internal circuitry of the IR2181 and making use of Q5 and Q6.
When Q6 is on, C37 is charged to ~14 V via D9 and Q6_DS.
When Q6 is turned off and Q7 is turned on, the voltage at the intersection M1 will start to rise from ~ 0V towards DC+. Since the charge on C37 cannot be discharged instantaneously, the cvoltage across C37 will stay at ~14V. With M1 rising above 0V, the voltage on VB will thus also rise to VB = V(M1) + V(C37) which is greater than 15 V. Diode D9 will block discharge of C37 into + 15V. Therefore the supply voltage (VB - VS) of the high side driver will stay at ~14V and the driver will be able to supply V(HO) > DC+ to the gate of Q5.