Roff

Jethro, I'm not going to try to decipher your netlist again. Post a schematic, as a .png or a .gif file. JPG schematics are fuzzy.

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Ron

jethro107

sorry
this is the jpg of the schematic

Attached Images

detector.JPG (34.9 KB, 17 views)

Roff

With +5V and 0V power rails, the input common mode range of the AD8009 is 1.2V to 3.8V. Your input signal is below that range.
The detector output is negative because the op amp bias current is 50uA.
You could use a negative supply to solve the common-mode range problem, but the bias current would still be an issue. Since your detector output has no DC information, perhaps you could AC couple it to the amplifier.
Since the AD8009 is a current feedback amplifier, you need to pay attention to the values of the feedback resistors, as well as the package, and not just their ratio. I have designed many video amplifiers using CF amplifiers, and it is not a trivial process. I doubt you can get a gain of 100 out of one stage while maintaining the bandwidth required to handle the detected signal in your example.
Why are you doing this? Is it homework?

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Ron

jethro107

thank you.
it'snt an homework it's a part of my master thesis but i'm not an Electrical engineer(i study informatic and telecomunication and I had only an electronic course for the basis of electronic) and for this, in this phase of my work, i have much problem to build the circuit and my tutor don't find the time for look my problem (he is a bastard if i must to use the right words). i'll try to use multiple stage of amplification

jethro107

hi,
i think that one stage amplifier it is the better solution because it coult be problem with the noise. so i think that an amplifier with output over 20-30 mV can be acceptable. i tested this solution but the problem is always the same: negative value of the amplifier input when i connect it to the detector circuit.
i tried to add a negative bias to the amplifier but the solution it's not good.

Roff

Do you really need to detect that fast damped sinusoid? It requires a wideand amplifier to amplify it. Most wideband amplifiers have high input bias current. If your modulation bandwidth were lower, you could use an op amp with very low bias current.
What information does the modulating signal carry?

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Ron

jethro107

no i must detect a gaussian pulse (like in uwb trasmission) but for the simulation i can use the damped sinusoid, it's similar.
i don't need to trasmit a particular information. i must insert this receiver in a system for positioning. i must calculate the time of flight of the pulse: the time between trasmission and the reception of the pulse.

Roff

So how wide is the Gaussian pulse? Did you understand the point of my previous post?

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Ron

jethro107

the uwb signal, that i must use, has 1 GHZ bandwidth. so the amplifier
ad8009 can be good i think

jethro107

the UWB bandwidth is equal to or greater than 500 MHz. so if the bandwidth is smaller than 1 GHz can be good

Roff

I don't think you will find an op amp that will give you significant gain at 500MHz. Here is a quote from the AD8009 datasheet:You probably understand the concept of gain-bandwidth product. It is not a linear function when using Current Feedback amplifiers, but it is still true that, the higher the gain, the lower the bandwidth.
I think you need an RF amplifier.

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Ron

jethro107

i think that:
The input signal to the amplifier has a bandwidth limited by filtering to the detector output(amplifier input). so, it should be possible to have a gain greater than 1 or 2 (respectively for 1 GHz and 700mhz bandwodht), but why does the input voltage become negative and i haven't an amplification for small(100mv) input signal?
i think that the use of the rf amplifier is good for my case but i don't understand why the amplifier doesn't work well.

Roff

Assuming the diode resistance >>500 ohms at 20uA bias, your bandwidth will be determined by the 500 ohm*100pF time constant, which I think is what you are saying. You said previously you needed 500MHz bandwidth. You don't even have 5MHz bandwidth. It does solve your amplifier bandwidth problem, though.
As I said, the reason for the negative output voltage is the amplifier's bias current. In the sim below, I reduced it somewhat by eliminating the RC filter after the detector filter. I then measured the bias current and added a current source to provide that amount of current (not a good idea in practice). In addition, I duplicated the circuit, but with a behavioral op amp which has nearly zero input current (Rin=500Meg). In this, I omitted the current source, since none is needed. From the sim, you can see that the two detector outputs (k and k1) are nearly identical, but the output offsets are different. This is due to the bias current in the inverting input, and the input offset voltage (2mV) built into the model for the AD8009.

Attached Images

	6GHz detector sch.PNG (26.8 KB, 9 views)
	6GHz detector waves.PNG (23.0 KB, 5 views)

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Ron

jethro107

hi thanks for the reply. i must calculate the power input to the diode. can you help me to find it?
tahnks

Roff

I don't know how to calculate it.

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Ron