# how to determine comparator resisters

Discussion in 'General Electronics Chat' started by arthur92710, Jun 27, 2008.

1. ### arthur92710 Thread Starter Active Member

Jun 25, 2007
307
1
Hi guys!
I need some help with a comparator circuit.
How do I find out what resisters to use?
Here is an example

I used 2 quad's and 8 leds to make an led thermometer. The lowest voltage I will have is 2.73v and the highest is 3.13v .5v for each led. each led is for 5c.

Thanks!

Apr 5, 2008
19,914
4,141
Hello,

In your example the voltage divider is dependend on V+.
This can be changed by replacing R1 with a current source.
To get 2.73 Volt , you can calculate the resistor this way.
R = U / I . If I = 10 mA (for example) the resistor will be 2.73 / 0.01 = 273 Ohm.
Each step has a voltage 0.05 Volts. so the intermittend resistors are all 0.05 / 0.01 = 5 Ohm.

Greetings,
Bertus

Dec 19, 2007
14
0
4. ### arthur92710 Thread Starter Active Member

Jun 25, 2007
307
1
But what if the current is not 10mA?

Dec 19, 2007
14
0
--------
I= R/v
--------

6. ### beenthere Retired Moderator

Apr 20, 2004
15,808
295
The relationships in bertus' examples still apply. Do the math using the actual (or desired) current.

7. ### arthur92710 Thread Starter Active Member

Jun 25, 2007
307
1
but then I need to know resistance...
so it keeps going around not knowing either the resistance or voltage.

Beenthere:
how would I find out the actual current without knowing the resistance?

Im so confused now.. XP

Apr 5, 2008
19,914
4,141
Hello,

The 2.73 Volts resistor will be 2.73 / I .
So if I = 20 mA R = 2.73 / 0.02 = 136.5 Ohm.
If I = 5 mA R = 2.73 / 0.005 = 546 Ohm.
The same goes for the intermittend resistors.
So if I = 20 mA R = 0.05 / 0.02 = 2.5 Ohm.
If I = 5 mA R = 0.05 / 0.005 = 10 Ohm.

Greetings,
Bertus

Dec 19, 2007
14
0

Apr 5, 2008
19,914
4,141
Hello,

Here I made a little schematic.
The values may not exact, but will bw close.

Greetings,
Bertus

File size:
70.5 KB
Views:
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11. ### arthur92710 Thread Starter Active Member

Jun 25, 2007
307
1
I had the same circuit except the 2 diodes and transistor. What are they for?
I did some tests in multisim and if I change the 1k resister, the one after the 2 diodes, to 150 it works right!
Can I use a different diode and transistor?
Thanks!

EDIT**
Actual I replaced the whole diode transistor thing with a 190ohm resister.

The circuit works when I replaced the LM335 with a pot and used a pot on the voltage divider for tuning, so I will have to get some more temperature sensors. What would be better? A LM335 or a thermocouple?

Last edited: Jun 27, 2008

Apr 5, 2008
19,914
4,141
Hello,

As I told you when the powersuplly changes (this can happen when you use a battery as powersupply), the voltages for your comparators will change too.
Thats why I recommended a constant current source.
http://www.jefspalace.be/analog/input stage/constant current source.php
The only differecnce with mine is that this source goes to powersupply.
I have changed this by mirroring it and use a PNP transistor.

Greetings,
Bertus

13. ### arthur92710 Thread Starter Active Member

Jun 25, 2007
307
1
I dont understand how that works. If you have a voltage of 5v and a constant current of 1A, then if you use ohms law your resistance should be 5ohms. But what if you use a bigger resistor? like 10ohms, would the voltage become 10v?

And in the third picture,why connect ground to the battery positive?

Last edited: Jun 28, 2008

Apr 5, 2008
19,914
4,141
Hello,

I was concentrating on the firtst 2 pictures.
I also do not understand te tekst of picture 3 , they are using other numbers for the resistors that do not match the picture.

I have redrawn my thermometer schematic with the new current source.

Greetings,
Bertus

File size:
73.3 KB
Views:
18