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# OPA659: OPA659 about RF and CF value selection issues

Part Number: OPA659
Other Parts Discussed in Thread: OPA657, LMH6629, OPA857

Hi Team,

The customer is using OPA659. The customer uses  the Figure 39. Wideband, Low-Noise, Transimpedance Amplifier (TIA) of the datasheet reference circuit  for his

design. The CDIODE is 0.7pF in his deisgn. The customer would like to control the bandwidth at 100MHZ. The diode  that is in the Figure 39 can provide the

current from hundreds na to dozens  ua. And the input current frequency is from 10HZ to 30KHZ.

How to calculate the  RF and CF value in the Figure 39 when the bandwidth is controlled at 100MHZ?

Best Wishes,
Mickey Zhang
Asia Customer Support Center
Texas Instruments

• Mickey,

I am confused - you say input current frequency is 10 Hz to 30 kHz, but the bandwidth required is 100 MHz?

To get a theoretical understanding of how to design TIAs please refer to these 2 articles:

Within these articles there is a calculator which I use to come up with the following numbers. Note that when the calculator asks for input capacitance you should include the capacitance of the photodiode and the op-amps input capacitance.

Now lets get to the design:

OPA659: This has a Gain Bandwidth Product or GBW of 350 MHz. With 4.5pF of input capacitance you can only achieve a gain 1.5kOhm (Cf = 1.85pF) in order to get a closed loop bandwidth of 100 MHz.

OPA657:This has a Gain Bandwidth Product or GBW of 1600 MHz. With 5.9pF of input capacitance(4.5pF +0.7pF is from the opamp itself) you can only achieve a gain 4.2kOhm (Cf = 0.53pF) in order to get a closed loop bandwidth of 100 MHz.

LMH6629:This has a Gain Bandwidth Product or GBW of 4000 MHz. With 6.4pF of input capacitance(4.0pF +1.7pF is from the opamp itself) you can only achieve a gain 10 kOhm(Cf = 0.226pF) in order to get a closed loop bandwidth of 100 MHz. Stabilizing this amplifier could get a little tricky since it is not unity gain stable. You will need to simulate the circuit in TINA to get the exact gain. Also this amplifier is bipolar input so its current noise will be high which will degrade SNR.

OPA857: This is an integrated transimpedance amplifier with 2 switchable gains. This is probably your best bet. You can achieve 20kOhm of gain and a bandwidth of 105 MHz. The feedback cap. is integrated with the feedback resistor.

There is an EVM for this that you can purchase which includes the ability to add a photodiode directly at the amplifiers input for testing. Please see below

www.ti.com/.../tida-00978

-Samir

• Hi Samir,

Q1: For OPA659, why is the input capacitance 4.5pF? For the input capacitance of OPA659, is it CDIFF 1pF+ CCM 2.5pF+CD 0.7pD=4.2pF?

Q2: I have download the 0508.TIA_Calculator excel. From the excel, first, we need to get the GBP, RF and CIN value or the f-3dB, RF and CIN

value. But in this case, we don't know the RF value. How do you calculate the RF value?

Q3: The bandwidth required is 100 MHz for the customer. Does this mean the f-3dB is 100MHZ?
• Hello Mickey,

Q1: I added some parasitic capacitance from PCB, etc. I wasn't too worried about being accurate here because the OPA659 doesn't have sufficient bandwidth and is also too noisy. It really is not a good solution for a TIA application.

Q2:If you are using calculator 1, then you know the GBP and Input Capacitance based on the photodiode and the amplifier chosen. You then keep changing the value of Rf until you achieve the desired Closed loop TIA bandwidth.

 Calculator I Opamp Gain Bandwidth Product (GBP) 1600.00 MHz Feedback Resistance (RF) 100.00 kOhm Input Capacitance (CIN) 100.00 pF Closed-loop TIA Bandwidth (f-3dB) 5.05 MHz Feedback Capacitance (CF) 0.446 pF

Q3: Yes, the 100 MHz is the -3dB bandwidth. Of course for the OPA857 in a gain of 5 kOhm you would be fairly flat since its -3dB bandwidth is 125 MHz. Figure 3 with 1kOhm load in the OPA857 shows that you are within +/-1dB. If the customer is using this in a pulsed application then the peaking shouldn't matter.

-Samir