Hello, I need to create a non inverting op amp to be read from a odm d69-30 lcd meter. The voltage is produced by a photodiode which produces 0.38V when hit by the laser. The meter claims it works with voltages between 0-19V but it only starts working around 9V. I am using a ltc1050 op amp its pins are 2.-, 3.+,4.V-,6.Vout, 7.V+. I am powering the amp with a 9V battery so it's rail power is 0-9V I have used the op amp calculators only to find myself not going near my intended target voltage. Plz help me design it

 

Welcome to AAC. A schematic would help us to help you .

 

use the + input for non inverting.

 

Search 'photodiode amplifier schematic' for dozens of example schematics. You will quickly notice that they all are variations on the same theme. The photo diode is connected reverse biased. The way a photodiode works is that its reverse leakage current increases when it is exposed to light. Typically this signal is processed with a transimpedance amplifier - current input, voltage output. That is why the diode is connected to the inverting input without the normal series input resistor.

ak

 

Check page 13 of datasheet for circuit.

http://cds.linear.com/docs/en/datasheet/1050fb.pdf

 

use the + input for non inverting.

Just keep in mind that if you use the + input, the math for calculating the gain will be different.

 

Sorry I haven't been on I had finals to deal with... So I looked at page 13 and built the photodiode amp. However, the main problem is the voltmeter that I am forced to put on this circuit, it draws its power from the probes and claims to be able to go from 0.5V-19.99V but has no datasheet that I can find or reference number. I can't seem to power it from the circuit even though I am now getting 4.5V from my normal voltmeter, I believe this is because I need to amplify the current from photo amp circuit. As I said before this is D69-39 panel voltmeter and it works with a 9V battery, so I am assuming it claims that it can read 0.5V it needs to be at a more appropriate current and since a 9V battery can be read it most likely needs to be amplified to 500ma(current of a 9V). So I was working out a current amplifier today which the diagram is below.

 

Okay so after hours of websurfing I finally found some specifications on this little bugger voltmeter: http://chinameter.en.alibaba.com/pr...69_30_LCD_DC_Digital_Voltage_Panel_Meter.html its the 3rd DC choice 19.99V and it's required amperage is 19.99mA which I am not getting close too .

 

19.99 mA sounds more like the high end of a readout range than an actual power requirement.

ak

 

Where are you getting specs that claim it's powered by the same source it's measuring. Those models all describe needing a power supply, unless you've got a "Special power source as your request" model that does that.

Also, I see no mention of current requirements on that page. Where are your getting that? Did the meter come with specs that made claims about where it gets power from or how much it needs? If so, please upload those. So far, I can't make sense of this.