# Need to amplify 0,1-0,8mV range at least 100 times

• kurt
In summary: LM308) with the laser. In summary, the conversation revolved around the measurement and amplification of a small voltage output from a circuit using a digital voltmeter. The participants discussed using a differential amplifier circuit with LM308 and the possibility of using a laser to amplify the output, but it was deemed unreliable and potentially damaging to the LM308. Suggestions were given to use a split power supply and to be cautious of exceeding the maximum ratings of the LM308.
kurt
Hi,

I measure the output from my circuit between 0,1 mV to 0,8 mV at 200mV setting on my digital voltmeters. How can I measure it more sensitively at least 100 times, or amplify it 100 times so that I can know the values such as 0,427mV instead of 0,4mV?
I tried it using a differential amplifier circuit with LM308, R1=R2=1K and, R3=R4=100K, and 0-18VDC but it didn't work. Maybe something was wrong which I couldn't figure it out.

I appreciate any suggestions either on diff amp circuit or with any other methods.

kurt said:
Hi,

I measure the output from my circuit between 0,1 mV to 0,8 mV at 200mV setting on my digital voltmeters. How can I measure it more sensitively at least 100 times, or amplify it 100 times so that I can know the values such as 0,427mV instead of 0,4mV?
I tried it using a differential amplifier circuit with LM308, R1=R2=1K and, R3=R4=100K, and 0-18VDC but it didn't work. Maybe something was wrong which I couldn't figure it out.

I appreciate any suggestions either on diff amp circuit or with any other methods.

You will need to run the opamp off of split power supply rails (like +/-12V) if you want to amplify a signal that is near ground.

You can make a non-inverting amplifier from a op amp and to resistors. The equation which gives you the gain is Gain = 1 + R(top) / R(bot), where R top and R bot are resistor values. You can find a circuit diagram on google.

edmondng said:
i think you need more digits on the voltmeter if you want 0.427mV rather than 0.4mV.
Or like you said amplify it. Have you check the bandwidth of the chip and signal going in?

http://www.alldatasheet.com/view.jsp?Searchword=LM308

yes, i searched for voltmeters with more digits but they cost a lot. on the other hand i followed the link you gave to check the specs, but noticed an "amplifier for photodiode sensor" circuit, that was a direct solution so i built it with my fds100 photodiode and a red laser beam (ca. 620 nm), and it worked, w/o any power to my surprise, because i thought every opamp should work with some power. now i got the mV's in the range 26 mV to 42 mV. that is a much better resolution but i still need to try experimenting with it in different conditions.

thank you much edmondng for your fast response.

berkeman said:
You will need to run the opamp off of split power supply rails (like +/-12V) if you want to amplify a signal that is near ground.

thank you much berkeman for your reminding since i ran it first on +18V with no result. i applied the split power +/-18V (with four 9V batteries) and this time it amplifeid the 1,4 mV to 1,346V that is about 1000 times though it was supposed to amplify only 100 times with 100k/1k = 100. i think i need to check things if something is still wrong. now i am working on this circuit and another photodiode amplifer circuit, whichever gives any reliable results...

kurt said:
thank you much berkeman for your reminding since i ran it first on +18V with no result. i applied the split power +/-18V (with four 9V batteries) and this time it ...

Ouch. +/-18V is the Absolute Maximum Rating for the supplies on an LM308, and if you used fresh 9V batteries, you exceeded the +/-18V by about a volt. You should not be running parts at or near their Absolute Maximum Rating -- that's a good way to damage parts and run into other problems.

Instead, if you are constrained to using 9V batteries for your power supplies, just run the LM308 off of +/-9V. That will give you about +/-7V of useful operating voltage for the LM308's input and output voltages (see the datasheet).

kurt said:
so i built it with my fds100 photodiode and a red laser beam (ca. 620 nm), and it worked, w/o any power to my surprise, because i thought every opamp should work with some power.

i don't think you build it right. it may 'seem working' but not correctly. the output must be coming from another path. i believe all op amps need power to work

typical supply for op amp are +-12V, +-15V, +-5V depending on supply voltage rating op amp of course.

edmondng said:
i don't think you build it right. it may 'seem working' but not correctly. the output must be coming from another path. i believe all op amps need power to work

Shining the red laser on the photodiode is generating a photocurrent, and the OP is probably just seeing the results of the photocurrent going wherever it can with the circuit unpowered (like through the I->V conversion resistor).

berkeman said:
Ouch. +/-18V is the Absolute Maximum Rating for the supplies on an LM308, and if you used fresh 9V batteries, you exceeded the +/-18V by about a volt. You should not be running parts at or near their Absolute Maximum Rating -- that's a good way to damage parts and run into other problems.

Instead, if you are constrained to using 9V batteries for your power supplies, just run the LM308 off of +/-9V. That will give you about +/-7V of useful operating voltage for the LM308's input and output voltages (see the datasheet).

On one pair I exceeded by half the volt and it was just below 18V on the other pair. Right... then i will run the LM308 off of +/-9V batteries on monday and see the readings again. Thank you much...

edmondng said:
i don't think you build it right. it may 'seem working' but not correctly. the output must be coming from another path. i believe all op amps need power to work

typical supply for op amp are +-12V, +-15V, +-5V depending on supply voltage rating op amp of course.

i built the circuit exactly the same way as given on the datasheet. first i supplied power with +/-18V thinking that it should be fed somehow. i got the constant output +18,11V with the laser blocked or unblocked. then i removed all the batteries, and the reading was about 26mV on daylight in the room with laser beam blocked, and rose up to 46,2mV with the red laser beam. then it responded consistently each time repeated even for the in between values. on the diagram http://www.alldatasheet.com/datasheet-pdf/pdf/70230/LINER/LM308.html it was given that Vout = 10V/uA. so i think that converts micro ampers generated from the photodiode into voltages.
thank you much...

## 1. How can I amplify a 0,1-0,8mV range by 100 times?

There are a few ways to achieve this, depending on your specific needs. One option is to use an operational amplifier (op-amp) circuit, which can amplify small signals by a factor of 100 or more. Alternatively, you can use a programmable gain amplifier (PGA) or an instrumentation amplifier, which are specifically designed for amplifying small signals with high accuracy.

## 2. Can I use any amplifier to achieve this level of amplification?

No, not all amplifiers are capable of amplifying such a small range by 100 times. It is important to select an amplifier with a high gain and low noise characteristics to ensure accurate amplification of the signal.

## 3. Is there a limit to how much I can amplify a signal?

Yes, there are limitations to how much a signal can be amplified. Beyond a certain point, the amplified signal may become distorted due to the limitations of the amplifier or the input signal itself. It is important to carefully choose the appropriate amplification factor for your specific signal.

## 4. What are some potential applications for amplifying a 0,1-0,8mV range by 100 times?

This level of amplification is commonly needed in biomedical and scientific research, where small electrical signals need to be accurately measured and analyzed. It can also be useful in electronic instrumentation and sensor applications.

## 5. Are there any potential drawbacks to amplifying a signal by 100 times?

Yes, there are a few potential drawbacks to consider. Amplifying a small signal by such a large factor can also amplify any noise or interference present in the signal, resulting in a less accurate measurement. Additionally, it may require more complex and expensive equipment to achieve such high levels of amplification.

• Engineering and Comp Sci Homework Help
Replies
1
Views
1K
• Electrical Engineering
Replies
22
Views
3K
• Introductory Physics Homework Help
Replies
10
Views
1K
• Electrical Engineering
Replies
4
Views
2K
• Engineering and Comp Sci Homework Help
Replies
13
Views
2K
• Electrical Engineering
Replies
15
Views
7K
• Electrical Engineering
Replies
138
Views
23K
• Engineering and Comp Sci Homework Help
Replies
24
Views
2K
• DIY Projects
Replies
27
Views
5K
• Engineering and Comp Sci Homework Help
Replies
10
Views
3K