Author Topic: dropping voltage activation circuit (Read 12864 times)

Richard Crowley · « **Reply #25 on:** December 30, 2015, 04:02:49 pm »

Quote from: Simon on December 30, 2015, 03:58:05 pm

Yes I was thinking that too but I'm trying not to overly confuse a newcomer to electronics.

Good point. But one of the things that I like about this forum is the balanced coverage of theoretical AND practical aspects of circuit design in the Real World.

Simon · « **Reply #26 on:** December 30, 2015, 04:06:16 pm »

Well maybe the OP can tell us what solution he is leanig towards. I also expect that diode voltage drops are not guaranteed to be the same from diode to diode but that should not matter if this is a one off.

Simon · « **Reply #27 on:** December 30, 2015, 04:08:14 pm »

I diodes are used to raise the ground a "load" resistor needs connecting to it as the voltage on it will also change with the current going through it which at the low current the sensor uses can also be unpredictable. So say 5mA is passed through it the voltage drop will be mosre stable than with a few hundred uA and a varying temperature.

J4e8a16n · « **Reply #28 on:** December 30, 2015, 04:28:09 pm »

Quote from: Simon on December 30, 2015, 04:06:16 pm

Well maybe the OP can tell us what solution he is leanig towards. I also expect that diode voltage drops are not guaranteed to be the same from diode to diode but that should not matter if this is a one off.

The OP (if it is me) is leaning, if he can't succeed in a virtual ground, towards a diode.

JPD

alsetalokin4017 · « **Reply #29 on:** December 30, 2015, 04:50:01 pm »

From the Texas Instruments data sheet for the LM35:

Discuss.

Richard Crowley · « **Reply #30 on:** December 30, 2015, 05:03:29 pm »

Quote from: alsetalokin4017 on December 30, 2015, 04:50:01 pm

Discuss.

Yes, either of those schemes would seem appropriate for @J4e8a16n's application.

(OP = Original Poster = J4e8a16n)

Richard Crowley · « **Reply #31 on:** December 30, 2015, 05:06:23 pm »

Here is a case of the Celsius scale being less practical than Farenheit. At least for measurement around the freezing point of water. Unless you have a more complex split/dual power supply (like +/- 5V, etc) working with signals very close to zero/ground becomes rather problematic.

poorchava · « **Reply #32 on:** December 30, 2015, 05:14:21 pm »

I would use a cheap RRIO opamp like MCP6002 to amplify the voltage x20 or so and then use the other half of the opamp as a comparator (with some hysteresis added) with TL431 as reference.

Sent from my HTC One M8s using Tapatalk.

Simon · « **Reply #33 on:** December 30, 2015, 05:27:01 pm »

Use 1 or two diodes and connect a resistor from the diodes to Vcc such that 5-10mA flow, don't use a schotky diode, you will of course need a thermometer to calibrate the setup and it may not be repeatable on other identical units you build.

As i said before waiting until 1 degrees before heating may be too late as you have to consider that your heater may take time to overcome the temperature drop and if it's not powerful enough the water will freeze anyway.

In air conditioning we don't like the cold heat exchanger to freeze as it stops the air flowing, but we don't wait for the cooler to get to 1C we cut it off at 3C

J4e8a16n · « **Reply #34 on:** December 30, 2015, 08:22:46 pm »

I will take youre advices.

JPD

J4e8a16n · « **Reply #35 on:** January 05, 2016, 04:05:03 pm »

Hi,

In LTSpice the comparator lm311 does not switch from 0 to 6 volts.

Thanks for following.

JPD

Simon · « **Reply #36 on:** January 05, 2016, 05:42:40 pm »

The negative of U2 need tonnecting to the negative supply (real ground) and the negative of the sensor to the virtual ground. Of course the potential devider will also need taking to the negative supply (real ground) as you are so close to the virtual ground with your desired trigger point youmight as well connect two resistors of equal value, one from pthe potitive rail and one to the negative rail and then put a trimmer in between them of cuitable value so that the wiper can select the trigger point and have a good amount of resolution in the area that interests you.

Richard Crowley · « **Reply #37 on:** January 05, 2016, 07:09:58 pm »

Quote from: J4e8a16n on January 05, 2016, 04:05:03 pm

In LTSpice the comparator lm311 does not switch from 0 to 6 volts.

Perhaps you missed this....

Quote

It seems pretty unlikely that a 311 could possibly pull down against a 1 ohm resistor.

If R6 is really 1 ohm. Then the output can never go low. R6 is simply wrong.

Simon · « **Reply #38 on:** January 05, 2016, 07:14:13 pm »

ah yes you probably need a transistor, that way you don't try to use a tiny pull up resistor that the comparator can't cope with.

J4e8a16n · « **Reply #39 on:** January 05, 2016, 08:23:46 pm »

Like that?

Simon · « **Reply #40 on:** January 05, 2016, 08:34:22 pm »

The transistor is wrong, connect a 4.7K resistor from Vcc to the op amp output, connect an NPN transistor, not a PNP, base to comparator output, emitter to negative collector to the relay, the other end of the relay to Vcc

J4e8a16n · « **Reply #41 on:** January 05, 2016, 09:23:21 pm »

Does not go to zero.

Richard Crowley · « **Reply #42 on:** January 05, 2016, 09:33:43 pm »

What is "U3" for?
What does "U4" do? It appears to be connected to nothing.
Where are the other two pins from U1 (ground and power)?

The reference voltage for your comparator (U2) is ~ 5.8V.
It seems highly unlikely that you will ever get 5.8V out of the LM35 (U1)
So we would not expect the comparator (U2) to ever change state.
Not clear what R1 is for?

I would be looking at the output from U2 directly before attempting to add any downstream circuit elements.

Simon · « **Reply #43 on:** January 05, 2016, 09:42:10 pm »

The best thing is to make this on a breadboard for real

J4e8a16n · « **Reply #44 on:** January 05, 2016, 10:09:20 pm »

They did not have a model with + and - but the lm35 will produce the voltage from .temp -55 to +85 degrees or so.

U4 works. It will provide 3.7 volts (in the real circuit) for the lm35.

I thought U2 would switch from 0 to Vcc when the + input would be lower than the - input. It is to say 30 mV. The plus input does go lower than the minus input.

Richard Crowley · « **Reply #45 on:** January 05, 2016, 10:56:53 pm »

U4 will output 6.00V with the circuit you show. How do you get "3.7V"?
If V5 is assumed to be providing exactly 12.00V, then R2 and R8 exactly divide that in half to make 6.00V
We don't know exactly how you are using the 6V output from U4?

We have no clue how you have either ground or power for the LM35 connected because nothing is shown in the diagram.

I still don't know what U3 is for, or what U4 is for either?

If V5 is assumed to be providing exactly 12.00V, then the voltage divider R4/R5 is creating 5.81V
Unless the output from the LM35 can go above and below 5.81V, U2 will never change state.

I don't see how the LM35 can possibly go anywhere near 5.81V.
But then I don't know where ground and power are connected.

J4e8a16n · « **Reply #46 on:** January 06, 2016, 12:00:33 am »

The dotted lines shows (in the previous post.) where the lm35 will be connected in a real circuit. The resistor is not shown.
The LM35 requires 4mA. If ther is a resistor between it and Vcc it will make a voltage divider. Be it 5.8 or 6volts does not matter.
In the image nothing say that the comparator inputs has to be higher than the output. It can be two sources of voltage, no? One to 7, 8 volts and the ouput can be connected to 12 volts going to the same ground as the 7, 8 volts? No?

I will do a real circuit....

JPD

Richard Crowley · « **Reply #47 on:** January 06, 2016, 02:55:35 am »

Quote from: J4e8a16n on January 06, 2016, 12:00:33 am

The dotted lines shows (in the previous post.) where the lm35 will be connected in a real circuit.

Sorry, a terrible idea. A couple of diodes in series is a much more sensible and reliable solution.

Quote

The LM35 requires 4mA. If ther is a resistor between it and Vcc it will make a voltage divider.

No That is not the way to use the LM35 (or any other IC, for that matter)!

Did you miss Figure. 18 in Reply # 30? If you want to offset the output of the LM35 up into a usable range you put something to offset the output UP from zero/ground. Putting anything in series with the supply pin does the OPPOSITE of what you need!

Quote

Be it 5.8 or 6volts does not matter.

Sorry, I don't know what that means?

Quote

In the image nothing say that the comparator inputs has to be higher than the output. It can be two sources of voltage, no? One to 7, 8 volts and the ouput can be connected to 12 volts going to the same ground as the 7, 8 volts? No?

The comparator simply compares the two inputs. It switches one way or the other depending on whether one pin is higher than the other or vice-versa.
If the + input sees 6V, then the comparator will turn OFF when the - input goes higher than 6V, and it will turn ON when the - input goes lower than 6V.

Connect the power input of the LM35 directly to the power bus (12V or whatever).
Then connect the ground pin of the LM35 through a couple of diodes to ground (as shown in Fig. 18 in Reply #30)
The diodes will OFFSET the LM35 voltage UP to a more manageable signal level of around 1.4V at zero degrees Celsius.
Then use a pot (like your "U3") to adjust the setpoint/threshold voltage into the + pin of the comparator.
You can set the output of the pot at around 1.5V so that your compartor switches the relay ON just above freezing.


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Author Topic: dropping voltage activation circuit (Read 12864 times)

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