Two dumb questions about opamp lore and a solar pump controller circuit I found.

[Ed.Kloonk]

Hi.

Here is a schematic I found on Google images with no further detail. Most of it I can follow, but there are two things I don't quite understand...

1) What is the purpose of the diode(if it is not just for polarity protection) and the 10ohm resistor near the 12v input?

2) Why would you tie one of the op-amp inputs to the other unused op-amp? Wouldn't be more sensible to tie that input (pin 5) to ground since the pin next to it (pin 6) is grounded and pin four(gnd) is not so far away either?

 

[Psi]

Found some more info on that circuit here, might be useful

http://www.siliconchip.com.au/cms/A_30610/article.html

The diode maybe there to protect against back emf from the motor.

[Ed.Kloonk]

Ah. Thanks for that.

 

[Mechatrommer]

The diode maybe there to protect against back emf from the motor.

look like not.

[Psi]

The diode maybe there to protect against back emf from the motor.

look like not.

It would block reverse polarity produced from any back emf when the supply is cut. But yeah, its not located across the motor like one might expect.

[bilko]

I would say it is used to decouple the control circuit power supply from the motor power supply.

The 10 ohm resistor and reservoir capacitor provide a smoothed voltage to the LM393 comparator circuit. The diode prevents the control voltage draining back when the motor is switched on. Without the diode, the control circuit would switch off when the motor is switched on, causing the circuit to oscillate (depending on the size of the motor)

[Psi]

yeeh, that makes sense

[mikeselectricstuff]

LM393 is a comparator, not an opamp. tying comparator inputs together may cause oscillation, so they use inputs that will always be significantly different.

BTW the reason the diode is where it is is to avoid it needing to be big enough to carry the motor current. If connected backwards, the MOSFET's body diode would turn on, & the motor would run (backwards), but the diode protects the rest of the circuit.
The 10R provides some filtering, and may also be intended to act as a fuse, in case of a fault with the control circuit drawing excessive power. 

[scrat]

BTW the reason the diode is where it is is to avoid it needing to be big enough to carry the motor current. If connected backwards, the MOSFET's body diode would turn on, & the motor would run (backwards), but the diode protects the rest of the circuit.

I can't get this. How could the MOSFET's gate, or the high value resistors conduct the DC motor's current? With reversed DC input the current won't need to pass through the series diode, it will simply go on the MOSFET's body diode, as you mention.

I agree with yachtronics.

[mikeselectricstuff]

BTW the reason the diode is where it is is to avoid it needing to be big enough to carry the motor current. If connected backwards, the MOSFET's body diode would turn on, & the motor would run (backwards), but the diode protects the rest of the circuit.

I can't get this. How could the MOSFET's gate, or the high value resistors conduct the DC motor's current? With reversed DC input the current won't need to pass through the series diode, it will simply go on the MOSFET's body diode, as you mention.

They don't - maybe you misunderstood my comment - I was mentioning the reason why the reverse-polarity diode wasn't at the DC input to the whole circuit, which might be the more obvious place to someone reading a 'beginners' forum.

[Ed.Kloonk]

Wow. Thanks for all the replies. Fantastic information. I learned something several things today.

 

[scrat]

@mike: I see now... If the motor running reversed isn't a problem, you don't need protection there, of course.