How to I half this output voltage!

[okami]

LM335 is a temp sensor, you feed it 0.4-5.0mA and it outputs a voltage representing the temp. ~3v at 25C then 10mv/C

I need the voltage range to be under 3.3v though! how can I half the output voltage? I tried a divider below with no success

The left circuit works correctly outputting ~3v. I tried added a voltage divider between the output and ground in the right circuit but I think its lowering the voltage to the LM335 to 2.225v! How can I do it without this problem?

Thx ^_^

 

[john monks]

Have you considered using a voltage follower like a transistor or an operational amplifier before your voltage divider?

 

[Harald Kapp]

No wonder, since the original R1=2kOhm your load (R1, R2 - btw: you should never give the same name to different components, better use R2, R3 in this case) draws too much current, bypassing the LM335.

Use higher resistance for the divider, e.g. 1*10kOhm. Note that you cannot then draw noticeable current from the divider's output. If you need more than a few hundred nA, add a voltage follower after the divider as John has suggested.

 

[Yaser43082]

Hi,

This is a really simple solution. You have the correct idea but your location of the voltage divider is in the wrong place. Lets look at your figure on your left:

R1 is connected to 5 Volts which is correct. To correct you problem, connect a 910 ohm resistor from R1. This should give you 1.56 volts out

 

[Harald Kapp]

connect a 910 ohm resistor from R1.

From R1 to?

 

[Yaser43082]

to ground

 

[KrisBlueNZ]

Yaser43082, your advice is bad. Please try to answer only questions that you are confident you fully understand.

Harald has explained the problem. The resistors in your voltage divider are too low. They are dragging the LM335 output voltage down. You need to use higher resistor values, such as 10k+10k as Harald suggested.

The LM335 needs at least 0.4 mA for its own operation. Assuming a maximum measurement temperature of 100 degrees Celsius, the LM335's output voltage will be up to about 3.75V. Allowing 0.2 mA for the voltage divider, you need at least 0.6 mA flowing through the resistor from the power supply, which has 1.25V across it; from Ohm's Law (R=V/I) its value should be less than 2.08k. I suggest 1.8k. For the voltage divider to draw no more than 0.2 mA at 3.75V its total resistance must be at least 18.75k so 2x 10k will be fine.

R1 (from +5V to LM335) = 1k8
R1 (top of voltage divider) = 10k (1% or better)
R2 (bottom of voltage divider) = 10k (1% or better)
Make sure the voltage divider resistors are identical and are located next to each other so their temperatures will track.

Edit:

You also need to make sure there is no DC load on the output of the voltage divider. I assume you're feeding an ADC? Can you give us the part number of it?

Also I would add a capacitor from the voltage divider output to ground (i.e. across the bottom resistor), located near the ADC. A good quality 0.1 uF or 1 uF part will do. Avoid multi-layer ceramic capacitors, especially ones with high dielectric constants (X7R, Y5V etc).