H-Bridge confusion

Discussion in 'Homework Help' started by local, Sep 8, 2008.

1. local Thread Starter New Member

Sep 8, 2008
1
0
Hey guys,

I am a long time reader and first time poster. But seeing all the help people give hopefully someone is able to help me out with the troubles I am having understanding this H-Bridge circuit. I am wondering what the point is of using resistors R2 and R3 (voltage divider). When I simulated this circuit on LTspice and removed R3 it did not make any difference in the output current.

Also if someone is able to provide some insight on how the low side MOSFETS are being controlled by the PNP that would greatly help me out.

Here is the link to the website.

I hope I am not violating any forum rules by posting this link. I did not see anything against posting links to external websites in the code of conduct.

Thanks guys.

2. mik3 Senior Member

Feb 4, 2008
4,846
63
The PNP transistors are off so no voltage is applied on the gate of the MOSFET's. When you apply a voltage less than 11.3 on the base of the PNP then it starts to contact and thus turning on the MOS.

3. SgtWookie Expert

Jul 17, 2007
22,183
1,728
R2 is the pullup resistor that turns off MOSFET Q2 when Q1 is not conducting. It's really too high of a resistance; power MOSFETS should have very brief rise/fall times on their gates.

R3 limits the current through Q1's collector when it's first turned on. Since the initial gate voltage will be 12v and R3=100 Ohms, the maximum current flow through Q1's collector will be 12/100=120mA. This is a realistic limit for a 2N3904. However, the base resistor is so large (47k Ohms) that Q1 will never reach saturation; base current would be around 0.1mA, and maximum current flow through Q1's collector would be perhaps 14mA if the hFE was around 150. As you can see, the base resistors should be reduced in value considerably.

Taking a look at ONsemi's datasheet for the 2N3904, you'll note that for Ic=100mA, hFE has a guaranteed minimum of 30. So, to figure out the base current we'll need, divide 100mA/30 = 3.333... mA.

Let's say you're driving this from a microcontroller (uC). A uC might supply 10mA to 25mA of current @5v from it's output pins. Since R=E/I, and E=5 (Vdd/Vcc of the uC) then R=5/.00333=1500 Ohms (rounded off). That's fortunate, because 1.5k is a standard resistor value.

Now if J1 (on the right side) is jumpered on pins 2-3, when Q1 gets turned on it will pull the gate of Q2 low, and also pull the base of Q3 low, causing Q3 to conduct, and charge the gate of MOSFET Q4 via R5.

This provides a complete path for current from +12v via MOSFET Q2, through M1, and then through MOSFET Q4 to ground. The motor runs in the forward direction.

Does this make sense?