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### Topic: Newbie puzzled by H-Bridge (Read 19335 times)previous topic - next topic

#### Njay

#30
##### Jun 15, 2012, 04:57 pmLast Edit: Jun 15, 2012, 05:00 pm by Njay Reason: 1
So how am I supposed to apply 6V to both bases while still keeping 9V through the collector circuit?

You're not . To turn the PNP transistor OFF, you need to stop the current from flowing from the emitter to base and out. This means that the voltage at the base cannot be lower than at the emitter (current always flows from higher voltage to lower voltage, just like water at different heights, you see?). You can't do that directly with 6V, because the emitter sees 9V and 9V is always bigger than 6V and therefore current will flow and this will turn the PNP transistor ON. A "converter" is needed, and that's one of the reasons why there are other transistors in an Hbridge besides the "4 switches". If your bridge ran at 6V, then you could use the control 6V signal directly, since with a PNP base at 6V and its emitter also at 6V there's no current flowing.

The simplest "translator" (voltage level translator) is this, a bipolar transistor "inverter":

If you apply 0V at Vin, the transistor is OFF and it's like it's not even there; in this case there's 10V at its colector, due to the resistor (actually depends on how much current flows in the resistor, but let's forget about that now and focus on concept). The voltage at the transistor's colector is the supply voltage, be it 10V as in the example or be it whatever you supply it with.

Now, if you apply let's say 5V to Vin, this will turn on the transistor (current will flow from base to emitter, since base voltage will be higher than emitter voltage) and it will connect its colector to GND.

As you can see, you turned 0V into 10V and 5V into 0V. That's why it's called an inverter and it performs "voltage level translation".

This is not the best way of doing it for an hbridge and there are a few more little details, but I'm just trying to show the concept.

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