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# I am using more than 400 pieces of ad8009 in the circuit. However, it is impossible to determine if the chip is broken. The signal I get experimentally has a 4 V bias voltage. (Output pin) However, the output without a signal has a different bias. Another problem is that there is amps that works normally if i proceed with the next experiment. Circuits incorporate all voltage pins and use + 5 V and - 5 V.

• Hi,

Sorry to hear you're having problems. Could you please attach a schematic?

Thanks,

Kris

• Hi,

Before we proceed, you are beyond the operating Output Voltage Range of AD8009 which is only +3.8 V. Some parts might have blown up becuse of overstress.

Could you give the following values:

+HV

R1

R2
C1

What is the amplitude and the width  of the current pulse from the photodiode? Could you give us the part number or the terminal capacitance of the photodiode, please?

Regards.

Jino

• And this board is made of PCB.

For example, an input to 1 C-1 is a small current signal.(photo detector signal)

We build this signal to be stable and amplified.

However, some outputs have failed to produce correctly since the test.

The original signal has a 4 volt bias and has a negative signal. (It's the same signal as the picture.)

But some amps have a - 4 V bias and the signal is not measured.

Is this a burnt amp?
However, the operating voltage is measured normally. (+ 5 V, - 5 V)

Does it mean that the amps is dead when other amps - 4 volts bias is measured?

• Thank you for the information. Could you reattach your schematic with the photodiode? A picture would do.

• Thank you for your interest and support.

+HV  =  +55V (Reverse voltage for pothodiode)

R1   =   1K

R2    =  100M
C1    = 0.1nF

- amplitude and the width  of the current pulse from the photodiode

And the photodiode we use is MPPC array of Hamamatsu. (Here is datasheet /  http://www.hamamatsu.com/resources/pdf/ssd/s13361-3050_series_kapd1054e.pdf ) we connected like this.

• No circuit adds photodiode. Only the input of the amplifier is the cathodes of the photodiode. The above circuit is a circuit used to amplify a photo-diode array per pixel. Anodes just connect to the ground.

• Hi,

With all the component values you have in your application, AD8009 couldn't make it. See below:

1. The AD8009's crossover frequency (fc) at RF = 1 kohms is ~ 180 MHz.

2. The APD's terminal capacitance in conjunction with RF will make a fp (pole) = 1/ (2pi*RF*Cs). Cs = 320pF, so fp = 497.36 kHz.

3. We set fx = sqrt (fp * fc), fx = sqrt (497.36k * 180 M) = 9.46 MHz.

4. Signal Bandwidth can be determined the fz( zero) which is dependent on RF and CF (1/(2pi*RF*CF)). To allow a maximum TIA bandwidth with a 45deg phase margin, let fz = fx. Therefore, CF = 1/ (2pi*9.46 MHz * 1k) = 16.83 pF. Set CF < 16 pF.

5. In your current design, you are limiting the signal bandwidth to 1.59 MHz. And the input pulse you have, it won't make it.  You can refer on page 30 of ADA4622 datasheet for the details about the computations. With that being said, you have to modify your circuit.

2. Use this circuit. With a AD8099, 100 ohms to meet the 20ns requirement. Vout = 50 mA * 100 = 5 Vpp. Set the output Reference (2.5V) such that we are sure that the amplifier is within operating range. Output will swing from +2.5 V to -2.5 V. 3. These are the responses. Hope this helps.

Regards,

Jino

It is so difficult to modify the chip that it is about to change the resistance.