energy detection

[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?

 

[jethro107]

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?

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?

 

[jethro107]

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

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:

Small Signal Bandwidth (–3 dB)
1 GHz, G = +1
700 MHz, G = +2

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.

 

[jethro107]

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.

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]

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.

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.

 

[jethro107]

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.

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

 

[Roff]

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

I don't know how to calculate it.

 

[jethro107]

I don't know how to calculate it.

ok thank you!