I am trying to analyze the operation of a BJT H-bridge configuration. I have searched through these forums and Google and can't seem to find an answer to my question.
It is my understanding that it would be desirable to have the "ON" transistors in the saturated state, to provide maximum supply voltage to the motor. This is also a general practice whenever using BJTs as switches. They should operate at cutoff and saturation.
I tried to simulate a simpler circuit, ignoring the two "OFF" transistors and only looked at the two "ON" transistors. Suppose I wanted to have my bridge supply 1 amp of current with a 5 volt supply (arbitrarily chosen). I used a 5 ohm resistor as the load to simulate the motor. If both transistors saturate (Vce <= 0.2 V), then the load should receive about 4.6 V. In simulation the load receives only about 1 or 2 volts.
If I lower the base resistances to increase base currents, it improves to about a maximum on 2.8 V across the load. If I increase the load resistance to about 50 ohms, then I can get about 4.7 V across the load, but obviously the current is not what I want.
I've read for saturation that Vbc should be forward biased, or Vb > Vc. This explains why increasing the load resistance causes the bridge to go into saturation, since the voltage drop across the resistor gets large enough for Vc to become less than Vb. Though, with such a small load resistor it takes a lot of current to make an appreciable voltage drop. If beta is a worst case of 50, then an Ib of 20 mA should be enough to provide 1 A at the collector. However, even when I make the base resistance ridiculously low (10 ohms), I still do not get anywhere near supply voltage across the load.
Does this mean that it is not possible to saturate the transistors with such a small load resistor?