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Second thread, no: That wasn´t my thread to start with.Why post this second thread about it?
Emitter followers have too much voltage loss to be used as switches. They should be PNP transistors.
Darlington transistors have even more voltage loss. They should be P-channel and N-channel Mosfets.
...
Please, do you own tests before posting ¨theories¨;...At this point, please prove your ¨theory¨.
Aquamon, I suggest you do some theory proving of your own. Your circuit has some serious drawbacks.
First, your circuit is just a motor reversing circuit. I wouldn't call it an H-bridge, ...
Like I said, it works beautifully and I AM extremely happy with it.
That is all that really matters in the end.
SO, THERE!!!
You guys are way too young and probably....
Yeah, so much for asking for help....the time wasted on forum, could´ve done by now, myself.Hi Aquaman, you made laugh with that! Good point...and I am afraid, totally wrong!
You started a thread that seems to be going to nowhere....
Happy New Year!
Nothing on this site is 100% efficient...your voltage drops are inaccurate and you also need to remember CURRENT flow takes the EASIEST path...into consideration.Aquamon, I suggest you do some theory proving of your own. Your circuit has some serious drawbacks.
First, your circuit is just a motor reversing circuit. I wouldn't call it an H-bridge, because that usually means that it is being controlled by logic inputs, which yours is not. Yours requires that it be controlled by two external toggle switches or relays; one to turn it on/off (motor run/motor stop), and the other to select the motor direction (requires tying Rev to either the battery voltage, or to ground). Begs the question, if you need toggle switches (or relays) to control the circuit, why bother with the circuit; just control the motor directly from the switches?
Better still, just use a center-off, Double-Pole, Double-throw toggle switch. Only one switch, zero power wasted.
Second, a quick simulation shows how inefficient your circuit is.
View attachment 90096 View attachment 90097
With Rev tied low, with a 100Ω load, the voltage drop across the upper Darlington is 1.5V and the drop across the bottom one is 0.7V, leaving only 9.8V across the load (motor). With a battery voltage of 12V, that means you wasted 2.2V.
With Rev tied high, with a 100Ω load, the voltage drop across the upper Darlington is 1.9V and the drop across the bottom one is 0.7V, leaving only 9.4V across the load (motor). With a battery voltage of 12V, that means you wasted 2.6V. The difference is due to the additional drop across D1; why is it even there?
As suggested by Audioguru, contrast your motor reversing circuit to the simplicity and performance of what modern PFETS and NFETs can do:
View attachment 90098
Note that with such a light load (100mA), the voltage drop across the both top and bottom devices is a few mV. This circuit could switch several Amps and not break a sweat. Try that with yours...
Hi Aquaman, you made laugh with that! Good point...and I am afraid, totally wrong!
You started a thread that seems to be going to nowhere....
Happy New Year!
Yeah, so much for asking for help....the time wasted on forum, could´ve done by now, myself.