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>>Using input drivers as emitter followers brilliantly removes 2 
>>resistors but is probably aimed at situation when Vmotor = Vdrive. 
>>Comes unstuck when Motor supply is substantially higher than drive 
>>voltage, as in this case with Vbat = say 6.5v (or even 6v) and Vcc = 
>>5v.

> I think you're wrong, Russell.  The input transistors are operate as 
> constant current sinks so long as they don't saturate.

It depends how you run your H Bridge.
My concept of H Bridge driving is that you run everything saturated by 
default unless the deign specifically prevents that happening. (My 
design (circuit soon, fully described in prior posts) so far in fact 
doesn't always run the bottom switch pair in saturation but this is a 
fully defined state).

With Bob's beautifully economical circuit (attached), IF you can 
control your drive levels to limit current (and there's no easy way to 
do that here) then you can run the input driver so it doesn't 
saturate.

BUT if you take the driver base to 5v PIC (actually Epson 4 bit) 
driven then the input emitter rises to 5v-Vbe1.

This pulls the top driver base down to V+-Vbe2. The input driver tries 
to saturate and its collector is essentially at 5-Vbe1. As it's hard 
connected to the base of the upper driver which is also essentially 
saturated then 5-Vbe1 ~~= V+-Vbe2. If Vbes are equal (only approx as 
the upper driver carries far more base current than the input 
transistor) then we must satisfy  5=V+  OR artificially force the Vbes 
to fit. When V+=6.5v and driver v = 5v then we have about 
6.5-5-0.7-.0.8 = OKish. ie the Vbe's may be a bit stretched from where 
they'd like to be and excess base drive is provided. Much more than 
that and you have "problems" (Robocop style). eg a 3 volt input and a 
6v motor drive would not be nice. Add one resistor and Mr Ohm looks 
after things for you.


        RM 

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