RF PIN diode maximum current capability question for experts - Page 1
Forum section: electronics / rf-microwave

author: hcglitte
author_rank: Regular Contributor
timestamp: November 13, 2021, 02:44:52 pm
content: Hi,I have a question regarding maximum current through RF PIN diodes.For DC operation it makes sense that eg. 100mA (typical datasheet value) is the maximum current since there will be a high forward voltage across the diode, and the power dissipation is high.However, for RF currents I wonder if the same maximum current parameter still applies?This is because the voltage drop across the diode will be very small since the I-region is charged and acts as a low value resistor (typically below 1 Ohm), and hence the power dissipation of the diode is minimal - even though a high current may pass across the diode (i.e.. across the intrinsic layer).I wonder if the listed maximum current also applies to this scenario?Perhaps the current handling capability still is valid since this also depends on the area of the intrinsic layer?This is regarding a Schottky assisted PIN diode limiter design.Any help is appreciated.

author: TimFox
author_rank: Super Contributor
timestamp: November 13, 2021, 03:08:03 pm
content: The maximum RF current at a given DC bias current is frequency dependent: see a textbook for details.Basically, the DC current and storage time give you a stored charge in the diode, and a half-cycle of the integrated RF current (charge) must be less than the stored charge. At a given RF current value, that half-cycle charge is proportional to the period of the RF sine wave.

author: hcglitte
author_rank: Regular Contributor
timestamp: November 13, 2021, 04:59:13 pm
content: Hi TimFox,Thanks.As I understand it, you are talking about the negative half cycle of the RF current which must not be able to discharge the charge buildup in the intrinsic layer?According to literature:Q >> I_RF/(2*pi*f) : I_RF = assuming RMS current (not peak). f = operating frequency = 1000 MHz.where Q = I_f * tau : I_f = DC forward bias current. tau = carrier lifetime.Assume I_f = 10 mA and tau = 100 ns. Q = 1000 * 10-3 * 10-9 Coulomb = 1 nC.Assume the RF current being a sine wave with amplitude 200 mA. During the negative half cycle, the RMS current is 200 mA/sqrt(2) = 142 mA.Hence: 142 mA /(2*pi*1000 MHz) = 0.0226 * 10-9 C Coulomb = 0.0226 nC << Q = 1 nCStill, if the datasheet specifies a maximum current for the PIN diode of e.g 100 mA, will the example above hold?The DC current is 10 mA, but the peak conducting current across the PIN diode during the positive half cycle is 200 mA + DC bias current (and most likely 200 mA - DC current in the other half cycle)!

author: TimFox
author_rank: Super Contributor
timestamp: November 13, 2021, 05:03:51 pm
content: I haven’t read your equations carefully enough to check the math, but it seems correct. So long as the frequency is high compared with the reciprocal of the storage time, a small DC bias current allows for a much higher RF current—this makes PIN diodes very useful for RF power applications.

author: hcglitte
author_rank: Regular Contributor
timestamp: November 13, 2021, 05:16:16 pm
content: Ok, thanks. I was a bit worried since some datasheets specifies a maximum current - but then this apparently only applies to DC currents.

author: CaptDon
author_rank: Super Contributor
timestamp: November 13, 2021, 07:41:49 pm
content: I find it amazing how they use PIN diodes with no bias in a little section of WR90 waveguide to block the main bang of a 4kw magnetron from getting into the MMIC FET front end of ship radars. Granted they are placed after the circulators, but with close metal objects, arc faults, rain, snow and so forth it is possible for nearly all of the maggie power to be reflected into the PIN diode limiter. These replace the old biased (-700vdc typical) arc gap limiters. Some of the more powerful radars use pulse biased PIN diodes and some limiters still feature a biased arc gap followed by PIN limiters. Circulators did away with the need for Anti-TR unbiased arc tubes in the maggie transmit path. On the low side are the 4kw pulse 4w average power small craft radars, and then there are the 10-12kw ship and aircraft radars. The biggest I normally worked on (non-military) were 25kw 25w average. Then there were the military aircraft units that would cook you bringing about the W.O.W. (weight on wheels) safety cut-off switches!!! Cheers mates!!!

author: langwadt
author_rank: Super Contributor
timestamp: November 13, 2021, 07:49:31 pm
content: Quote from: CaptDon on November 13, 2021, 07:41:49 pmbut it is a short so it'll be reflected right back out

author: CaptDon
author_rank: Super Contributor
timestamp: November 14, 2021, 03:41:42 am
content: Anything above the PIN threshold probably acts like a bad mismatch and is returned to the circulator where it rotates to the next port which in this case would be back to the Mag. Mag goes to Antenna, Antenna goes to Receiver, Receiver goes to Mag. No need for Quarter Wave and Half Wave stubs anymore!! Wow, I am old enough to have worked on Raytheon and RCA stuff that used those specific length duplexer/diplexer specific waveguides and Magic Tees with balanced mixers. Early on during those un-initiated days I didn't know the metal shell of those 723A klystrons were +300vdc above the waveguide!!! Only needed to learn that lesson once!!! I bet that +300 only dropped to +299 when I saw the white arc of burning flesh, the smoke and the smell (And how deep an electrical burn actually is). Fortunately the current took a short path into my finger and out the soft flesh of my hand resting on the waveguide and didn't cross my chest!