# LM318N comparator circuit

Discussion in 'Electronics Homework Help' started by HellasTechn, Jun 4, 2016.

1. ### HellasTechn

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Apr 14, 2013
Hello friends.

A friend of mine has this homework and needs help but have no internet acces so i post it on his behalf.

This circuit with the lm318N must trigger the buzzer when input voltage reaches 13 Volt.

Here is the circuit.

The transistor is a BD 649.

He used a variable DC supply and the buzzer keeps beeping and the lm318 heats up. also r1 and r2 get hot.

He probably burned the lm318 while testing the circuit but besides that do you think that this circuit can work ?

First i think that the BD649 is not a good choice.

I have almost no knowledge about theese circuits so... Any help welcomed.

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2. ### hevans1944Hop - AC8NS

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Jun 21, 2012
Increase the values of R4 and R5 from 4.7 ohms to about 10 kΩ.

3. ### chopnhack

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Apr 28, 2014
Hi Constantine, I am not sure if this will help your friend, but a quick search finds this:

4. ### hevans1944Hop - AC8NS

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Jun 21, 2012
This is a power Darlington good for eight amps or so. Why do you think it is a poor choice?

What does the loudspeaker symbol connected from emitter of BD649 to GND represent? Is this the "buzzer"? What voltage and current does this "buzzer" require? Why is no means provided to limit the base current to the BD649 when the LM318 output swings toward the positive rail?

Is your friend aware that there is an internal pair of inverse-parallel connected diodes across the inverting and non-inverting inputs of U? One of these diodes will conduct if the voltage difference exceeds about 0.6 V DC, and the current that ensues during conduction must be limited to less than 10 mA if this occurs.

Because of the voltage divider created by R2, R3, and Rv, the voltage at the non-inverting input of U will, initially, always be less than the voltage at the inverting input of U as POWER increases from zero toward some positive voltage. This will drive the output of U toward GND until the zener diode D begins to conduct at 7.2 V.

When D begins to conduct at a POWER voltage of about +7.2 V DC, the voltage applied to the inverting input of U becomes a constant +7.2 V DC while the voltage applied to the non-inverting input by the voltage divider continues to increase. When the voltage at the non-inverting input becomes greater than +7.2 V DC, as determined by the voltage applied to the POWER input and the resistance setting of Rv, the output of U will be driven positive toward whatever voltage POWER is when that occurs, hopefully at +13 V DC if Rv is set properly.

If during this procedure you have not created a voltage difference large enough to turn on the internal diodes connected across the inputs of U, nor exceeded the maximum current allowed through those diodes, the the output of U should switch from nearly zero to nearly the voltage applied by POWER, turning on the BD649, however with no provision to limit its base current.

5. ### HellasTechn

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Apr 14, 2013
it is a 12V buzzer that draws anly a few miliamps. That is why i think the bd649 was not the best choice.
S simple BC547 would be enough i think.

I think not. because he study engineering. This circuit is part of a lesson called "automation" he has no idea about electronics.
My original thought was that circuit uses it as reference voltage.

6. ### HellasTechn

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Apr 14, 2013
By the way thank you Hevans and chopnhack for your fast responses.

7. ### HellasTechn

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Apr 14, 2013
Here is an updated diagram.

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8. ### HellasTechn

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Apr 14, 2013
Correct. their value was in Kohm

9. ### hevans1944Hop - AC8NS

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Jun 21, 2012
It is a reference voltage, but only after POWER rises from zero to the 7.5 V DC zener voltage. When that happens, the voltage on the inverting input (applied through R5) should remain more or less constant as POWER increases from +7.5 V DC toward the "switching" voltage of +13 V DC. But at the moment zener diode D begins to conduct, to hold the inverting input constant, the voltage applied to the non-invertting input (through R4) is still less than +7.5 V DC because of the voltage divider R2/Rv. You can go to this website to use an on-line calculator to determine what value Rv should have to produce +7.5 V DC as the input (through R4) to the non-inverting input of U when POWER equals +13 V DC. Hint: about 9,3 kΩ will do it. When POWER slightly exceeds +13 V DC, the output of the op-amp should swing from near GND to near POWER, turning on the transistor Q.

Now all this assumes the differential inputs to U are never large enough to turn on the internal diodes connected across those two inputs on the LM318. Clearly that is not the case here. One of the diodes will turn on, and that will affect the voltage differential on the inverting and non-inverting inputs. However, the circuit probably should still work... probably. Your friend would be better off using a comparator instead of an op-amp for U.

10. ### HellasTechn

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Apr 14, 2013
Well as far as i know that is what their professor requested, so...

11. ### hevans1944Hop - AC8NS

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Jun 21, 2012
So, did your friend ever get the circuit to work?

12. ### HellasTechn

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Apr 14, 2013
Yes he actually did.
He blew a few zeners though, i dont know why.
I also told him not to use the transistor since the output could drive the buzzer without it.
Rv value was 9K2.

All well !
He sends his regards and Thanks.

hevans1944 likes this.
13. ### hevans1944Hop - AC8NS

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Jun 21, 2012
Pretty easy to blow up a zener if (1) you apply reverse bias far in excess of the zener voltage and (2) fail to provide a sufficiently large value of series current-limiting resistor.

Good call telling him to use the output of LM318 directly. Although I sometimes use op-amps as comparators, I cringe whenever I see anyone else doing the same thing. Does your friend now understand how the circuit works? Did he learn anything from this?

A further comment on driving buzzers: be careful the type you are using doesn't generate a huge back-emf that can easily damage semiconductor devices that fail to protect themselves. Helps to check this with an oscilloscope, just to be safe.

14. ### HellasTechn

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Apr 14, 2013
I am sure he did. I did !

I havent checked but since it survived

15. ### HellasTechn

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Apr 14, 2013
Back at school i remember we had a lesson about using opamps as comparators.

16. ### hevans1944Hop - AC8NS

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Jun 21, 2012
The problem is op-amps are designed to be used in closed-loop, negative feed-back circuits where the differential voltage between the two inputs stays small (because of the negative feedback). When used as a voltage comparator, the op-amp operates open loop and differential inputs can be several volts instead of microvolts or millivolts. Even if the op-amp has a high common mode voltage rating, it may not recover quickly to enter (briefly) a linear mode before changing output state as the differential input voltage approaches zero.

Nor are op-amps optimized for fast, logic-type, transition edges on their outputs, which is where a comparator excels and one of the reasons to use a comparator. The worst problem is making sure the op-amp output states are logic-compatible levels, and that oscillations do not occur during state transitions. In a typical comparator application of a high-speed (wide bandwidth) op-amp, this can be assured with positive feedback and the introduction of hysteresis in the switching points: