Driving PIN Diode at RF in Reverse Bias - Page 1
Forum section: electronics / rf-microwave

author: Lyuben
author_rank: Newbie
timestamp: May 12, 2025, 03:06:16 pm
content: I have an IC chip with a PIN diode on it that I am trying to drive in reverse bias (at -5V DC voltage) at either 300MHz or 600MHz with a 10Vpp voltage sinewave (so voltage goes from -10V to 0V). The p and the n contacts of the diode I have wirebonded to a PCB that I designed myself to have 50 Ohm characteristic impedance coplanar waveguides and have put an SMA connector for the signal input from my signal generator. I measured the complex impedance with my NanoVNA but I couldn't de-embed the gold traces on the IC itself, the wirebonds, and the PCB traces, so their impedance is also included in the measurement. I measured:Z(300MHz) = 2.32 - j*24.8 Ohm andZ(600MHz) = 9.15 + j*33.1 OhmThe problem that I am facing is that I don't want much current to pass through the diode or traces/wirebonds since heat will be dissipated which can burn and potentially destroy them. Yet, I need to have the voltage swings across the p and the n-contact of the diode.Is it possible to design a matching network that allows for that? My initial thought was to put a matching network in between my device and the signal generator that is actually high-impedance and doesn't serve the purpose to match things to 50Ohm. I would then have unwanted reflections but perhaps there is a way to isolate my generator/oscillator from them (put a buffer?). However, wouldn't doing such a thing cause my diode's bandwidth to go down? (would the diode still be able to follow the 600MHz change in voltage?)P.S. I might be able to use GS RF probes to isolate and measure the impedance of the diode alone, but this still doesn't solve my problem since the wirebonds and gold traces on the IC I need to deliver voltage signal in a non-experimental test environment.Any advice will be greatly appreciated!========================Just to clarify some things. The PIN diode is always reverse-biased at negative 5V DC voltage. It is not meant to go in forward bias. The diode itself is an electro-optic phase modulator in the optical sense. Applying reverse-bias voltage across the p-i-n junction causes the refractive index of the semiconductor material locally to change. This change is necessary for an optical signal that passes through it.I am interested in applying the correct voltage amplitude across it, while maintaining low current as I believe it can heat my IC circuit, burn, or damage it. For modelling purposes I found that modeling the diode as a capacitor is easier (I believe it makes sense since the diode is never used in forward bias anyways). The impedance at DC voltage is of the order of MegaOhms (so no current at DC except for some very small leakage current).What I'm worried about is the displacement current induced by the RF voltage source due to the high swing of the signal (10Vpp). I'm afraid that it can cause the gold traces leading to the diode to heat up, and perhaps even the diode itself to heat up significantly. Below you can find a schematic of what I believe is a reasonable equivalent circuit. The PIN diode is C3. I have also attached the sim results from LTSpice and you can see the current through the PIN diode, as well as the current through the series resistance (gold pads), etc.Btw, the impedance measurements are of the red box (PCB transmission line, wirebonds, IC traces with parasitics). I couldn't de-embed the PIN diode from the rest to measure only its own impedance. Ofc, I tried my best to come up with some reasonable rounding-up for the effects of the additional thingies that I listed in the brackets. So for the wirebonds I took 3nH, for example.

author: RoV
author_rank: Regular Contributor
timestamp: May 13, 2025, 05:19:07 pm
content: The problem is not clear to me.Can you post a schematic diagram and a few photos of your prototype and measurement setup?

author: EggertEnjoyer123
author_rank: Frequent Contributor
timestamp: May 13, 2025, 06:31:57 pm
content: You don't need to do this.I think your misunderstanding is that the PIN diode acts like a normal diode, and that if the DC + RF voltage is too high, then the diode will conduct for part of the RF cycle. However, this actually doesn't happen. A certain amount of charge is needed to turn on the PIN diode. With the RF signal, a large current flows, but for a very small amount of time, and the amount of charge removed is very small (and it gets added back on the negative cycle). Therefore, the diode doesn't turn on, even if the DC+RF is positive for an instant. With DC, even a small current over a long enough time can remove all the charge and cause the diode to conduct.Basically a PIN diode is like a normal diode but slow to turn on/off, and therefore it isn't affected by RF.

author: Lyuben
author_rank: Newbie
timestamp: May 14, 2025, 10:04:26 am
content: I modified the original question to include more details. So if I understand correctly, I shouldn't be worried about the PIN diode heating up? Still, I'm worried about the displacement current. There has to be some internal resistance in the diode, however small, that I am afraid can heat up. Also, what about the gold traces leading to the diode? I picture some current swishing around due to the positive and negative cycles but wouldn't that cause the pads to heat up as well. The traces are 20um wide on one device and 10um wide on another. Both are 2um thick. That's why I am looking at some way to reduce the current while maintaining the voltage swing.

author: whollender
author_rank: Regular Contributor
timestamp: May 14, 2025, 02:49:18 pm
content: It seems reasonable to be worried about the PIN diode heating up, since your impedance measurements would seem to point to a pretty high current level.It looks to me that you'll end up with about 140mA RMS into your device, but I don't see a way of reducing that without also reducing your voltage swing at the device.You may just need to adjust your voltage swing if you see device failures.

author: Lyuben
author_rank: Newbie
timestamp: May 14, 2025, 03:11:56 pm
content: This is also what I calculated and what I fear. So the goal of my question here was to see if there is any smart solution (call it a trick) that I could use, that I didn't think of. Of course, if nothing can be done - reducing the voltage swing is the only option.

author: Odysseus
author_rank: Regular Contributor
timestamp: May 15, 2025, 12:20:32 am
content: Not possible. Your requirements are in direct conflict with each other (reduce current, but maintain voltage). We cannot circumvent ohms law.You can measure and model your circuit such that you can estimate the power dissipation of each component and see at what voltage level it becomes an issue.Maybe more practically you can get your hands on a good IR camera with appropriate macro lenses and just keep and eye on temps.Another simple method to avoid smoke is to raise the applied voltage in steps, monitoring the behavior of the EOM for temperature induced drift.You can operate at a lower frequency, and work your way up, keeping an eye out for thermal issues on the way.Can you keep the duty cycle low? Just pulse RF for 1us every 1ms? Semiconductor devices are characterized way beyond their steady state limits this way.

author: EggertEnjoyer123
author_rank: Frequent Contributor
timestamp: May 15, 2025, 05:18:45 am
content: Are you missing a 50 ohm (I assume the source has impedance)?The PIN diode should reflect most of the power back at the source.

author: Lyuben
author_rank: Newbie
timestamp: May 15, 2025, 09:39:43 am
content: Thank you for the advise! I will try to characterize the device with low duty cycle as you proposed. I can put a trigger from the signal generator to my optical measurement equipment such that I run the RF signal for 1-2 periods and do my measurements synchronously. It shouldn't heat up significantly, I think. I might be able to get an IR camera as well.Regarding the termination - yes my signal generator has 50Ohm output impedance. The absolute impedance |Z| at 300MHz as you can see is about 25Ohm. How should I proceed to match my circuit? I think reflectionless matching with low VSWR is what I should be looking at. I calculated the LC values necessary to bring it to 50Ohm at 300MHz, the problem is that I don't know the exact frequency I should operate the diode at. By rough calculations it should be 300MHz with some possible deviation in the range +/- 20MHz. The change in impedance in this 40MHz region is not much but enough to make the circuit mismatched. That's why few days ago I was looking at some autotuner designs. Perhaps something with a UHF varactor inside for the tunable capacitance that I characterize beforehand and an RF power detector with a microcontroller to create some feedback. Just some thoughts of mine, it's a bit out of the scope of the question.

author: EggertEnjoyer123
author_rank: Frequent Contributor
timestamp: May 17, 2025, 12:19:15 am
content: I assumed you were using the PIN diode as an RF switch, which doesn't seem to be the case.With a 50 ohm impedance source, you can't get a lot of voltage across the PIN diode. If the PIN diode were replaced with a 2 ohm resistor, it's obvious that the voltage across the PIN diode is going to be nowhere near 10V (because of your 50 ohm source). With a normal 50 ohm load you get half the voltage across the diode. With a 2 ohm load you get 2/52 of the voltage, which is 4% roughly of the source voltage.The way to do it (assuming you don't care about bandwidth) is to just match to 50 ohms at your frequency of interest (this will be narrowband), and just reduce the output power accordingly.