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LTC5509 Standing D.C.

Thread starter #1
I'm using a couple of LTC5509 300 MHz to 3 GHz signal strength ics which output about 300 mV D.C. (with no signal input) to about 3v with maximum signal input.
What I am trying to do is to remove the 300mV when there is no signal to end up with 0V to approx 3V. I've tried using various diodes but the volt drop varies too much with temperature. I can't find a way to do this and would really appreciate some suggestions!

Nigel Goodwin

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Most Helpful Member
Use an opamp, and an offset of 300mV - basically an adder (although in this case you're 'adding' a negative value to subtract it). You can make the offset adjustable, and adjust the preset to zero the output.


Well-Known Member
As Nigel says - I think it is called a differential amplifier - your output is the difference of two input signals. If you use rail to rail opamps you shouldn't need to generate a negative voltage.


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Most Helpful Member
You need a R-R output amplifier. It must work at "VCC" voltage. The input common mode range must reach below 0.3 volts. (0V better).
You might want a R-R in & R-R out.
The resistors should have a ratio of 1:1.
You might want a RC in between the two ICs.

Thread starter #8
I've never been very good with op amps! I understand now - I need 0.3V on R3 to cancel out the 0.3V on the output. Any idea of the resistor values?


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Most Helpful Member
I did not look to see how much current the LTC5509 can output. I think a lot.
Do you have four 10k resistors? (1k to 100k)

You need 0.3V so if your VCC is good then make a voltqge divider. I assumed VCC=3.3V because the numbers came out good. The impedance at 0.3V is 1000 in parallel with 100 so about 90 ohms. If you have R3=10k then the circuit will see R3=10,090 ohms connected to 0.3V. The voltage divider could be 10k&1k. I am just trying to show how the divider adds a little resistance to the effective value of R3.

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