# Double Peak Phenomenon in Transimpedance Amplifier Simulation

I'm simulating a transimpedance amplifier circuit in order to replicate a phenomenon that my team is experiencing. The phenomenon can be seen below in the image circled in red:

My photomultiplier here in the simulation is being triggered at 10 ns by the voltage source V1, and you can see in that big peak before the red circle, our signal is getting amplified by the amplifier. However, we're also getting a small peak, as shown in the circled red part. I don't understand why this is happening, but I know that it becomes more prominent as you reduce the feedback resistance, R1. Here's a view of it at 50 ohms:

If I increase the feedback resistance too high, the signal gets clipped, and I'm trying to avoid that. I don't think that this is necessarily an op-amp exclusive problem, as I see this issue using different op-amp models as well, but why is this occurring, and what can I do about it?

EDIT: In response to one of the replies, here is the output signal when the feedback resistance is 100 ohms:

Here, you can see clipping as it reaches 1.13 volts. This is the situation that my team is trying to avoid; rather than have the signal be clipped, we want a pulse, like how the previous images show (though without that second peak shown in red).

EDIT: As suggested, I performed a small signal analysis of my circuit by injecting a small 15 mA AC current source (max current limit of the SiPM) into the circuit:

Thanks to a comment, I performed the initial analysis wrong by using an AC voltage source, so I ran it again with the current source. I chose the direction of the source based on how the current would flow through the SiPM (it seems that changing direction of the source just inverts the phase), and the DC trigger didn't seem to have a noticeable effect on the circuit (based on running it with/without the trigger). I see about 5 peaks in the graph, but what does it mean in regards to the circuit?

EDIT: Sorry that I didn't mention this earlier, but I'm using a model of the SensL MicroFJ-60035-TSV SiPM for my circuit.

• A frequency-domain simulation (AC sweep) of your circut should give you a hint why this happens. Inject an AC current into the IN1 node. Mar 27 '18 at 14:18
• Sorry, but if it's not clear, the phase of LOAD1 is the same as at OUT1. R2 doesn't affect the phase. Mar 27 '18 at 20:32
• @JonathanS. I added a plot of my AC sweep in the circuit, though I'm having trouble understanding the results. The way it is now with the AD8014, it starts out at -6 dB so that means there's no gain? It's the same if I start the frequency at 1 Hz instead of 10 kHz, never goes above 0 dB. What does that mean? Mar 28 '18 at 17:46
• You did the AC sweep wrong - you have to use a current source, not a voltage source. Disconnect your photodiode and replace V4 with an AC current source. The value in dB is irrelevant for this simulation, you have to look for peaks in the response. Mar 29 '18 at 11:27
• That peak at about 150MHz in the frequency response is the ringing you're seeing, it's probably because the phase shift of the OpAmp begins to rise at that frequency, turning the feedback on the inverting input (slightly) positive. You should be able to compensate this, maybe by splitting R1 into two resistors of 12.5 Ohms and adding a small capacitor to ground in the middle. It's hard to figure out why exactly the peaking occurs, so you have to play around a little and watch the frequency response. (The capacitor should be really tiny, 0.47pF or so probably) Mar 30 '18 at 0:50

## 2 Answers

The AD8014 data sheet clearly shows that you will get significant peaking of the output signal when you have a feedback resistor that is this low. I would be considering a resistance that is at least 100 ohms.

If I increase the feedback resistance too high, the signal gets clipped, and I'm trying to avoid that.

At the moment with a 50 ohm feedback your peak signal is 900 mV. With 100 ohm it's going to be more like 1.8 volts so, given that the AD8014 is capable of producing $\pm$ 3.4 volts at its output, you should be OK. If the amplitude is too high for subsequent stages and causes clipping of those other stages then use a resistor attenuator after the front-end.

• Thanks for the reply. I posted an image of my simulation when the feedback is 100 ohms, and it's getting clipped at 1.13 volts. My team is trying to avoid this; we're trying to get peaks like previous images show instead of a clipped signal. I'll take a look at your suggestion of a resistor attenuator, what what do you mean by after the front end? Also, wouldn't adding more resistance by means of an attenuator slow the signal down? Mar 27 '18 at 14:50
• @user101402 just to see what happens, try also putting a resistor in parallel with the photodiode to shunt some of the PD current away. Maybe 100 ohm to 1 kohm. It seems to me that your photodiode current is quite large. Mar 27 '18 at 15:01
• Thanks for the reply. The photodiode current is rated for a maximum of 15 mA as stated by the manufacturer's datasheet. We're using a SensL MicroFJ-TSV-60035 SiPM. Mar 27 '18 at 15:04
• In response to another comment, I performed a small signal analysis on my circuit using an AC voltage source at 180 mV, the approximate voltage of the SiPM when active. That little dip at the end seems strange, and I'm not sure how this explains the phenomenon. Mar 28 '18 at 13:22
• Sorry for the constant replies. Jonathan S. mentioned that I performed my analysis wrong, that I should use a current source instead of a voltage source. Thus, I reperformed my AC sweep with a current source with the current amplitude based on the max current expected from the SiPM. Mar 29 '18 at 21:25

High speed transimpedance amplifiers are very sensitive to tiny parasitic inductance of the input line (few nanoH from just few mm wire). That causes 'ringing' or even oscillations.