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Hi, I recently used this H bridge in my lab and it worked perfectly but I would like some insight as far as how the bjts work. I was told they are my drivers but I am not even sure what that means. I used one potentiometer to change the voltage in one bjt and another potentiometer for the other bjt. Thank you
how does this H bridge work?
- Thread starter Thevenin99
- Start date
Hello, and welcome to AAC.
A p-mosfet works by applying a low voltage to its gate in relation to the voltage that it's switching. As shown in the following diagram.
P-fets are useful when one intends to drive a load from its high-side. N-fets, on the other hand, are more adequate for driving a load from the low side of the circuit.
What the bjt transistors in your diagram are doing when they're activated by allowing current to flow through their base, is that they bring down the voltage between the p-fets source and its gate. Thus activating the p-fet.
Imagine removing the BC547 transistor and 100 ohm resistor at the left of your diagram. The p-fet at the upper left would always be switched off, since there would be no voltage difference between its gate and its source, due to the presence of the 10k resistor connecting them together.
Now, if you were to put back the transistor and 100 ohm resistor, and then activate the transistor, current would start flowing through the 10k, 100 ohm resistor and the transistor's collector and emitter. This will result in a voltage drop between the p-fet's source and its gate, allowing it to conduct current between its source and drain.
A p-mosfet works by applying a low voltage to its gate in relation to the voltage that it's switching. As shown in the following diagram.
P-fets are useful when one intends to drive a load from its high-side. N-fets, on the other hand, are more adequate for driving a load from the low side of the circuit.
What the bjt transistors in your diagram are doing when they're activated by allowing current to flow through their base, is that they bring down the voltage between the p-fets source and its gate. Thus activating the p-fet.
Imagine removing the BC547 transistor and 100 ohm resistor at the left of your diagram. The p-fet at the upper left would always be switched off, since there would be no voltage difference between its gate and its source, due to the presence of the 10k resistor connecting them together.
Now, if you were to put back the transistor and 100 ohm resistor, and then activate the transistor, current would start flowing through the 10k, 100 ohm resistor and the transistor's collector and emitter. This will result in a voltage drop between the p-fet's source and its gate, allowing it to conduct current between its source and drain.
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Also, I don't know how your application is supposed to work, but it makes no sense to me using a pot at the bjt's collector. In an H-bridge configuration such as the one you've mentioned, you want all the transistors to conduct at their full capacity. Especially the mosfets.
Thank you very much! you are rigjht, in our circuit we connected 2 potentiometers, each of them to the gates of the NMOS and the voltage sources were also connected each to the bases of the bjts.
Yes, in this case it's the low side mosfets that would be best to apply a PWM signal to. This is because the 10k resistors on the high side would hinder the p-fets from quickly discharging their gates.
