Using a resistor as a fuse in the HP 3048A option K23 DC blocking filter - Page 1
Forum section: electronics / projects

author: matthuszagh
author_rank: Frequent Contributor
timestamp: March 01, 2022, 02:20:21 am
content: Full disclosure: I've cross-posted this on the HP-Agilent-Keysight Groups.io (https://groups.io/g/HP-Agilent-Keysight-equipment/topic/how_does_r1_act_as_a_fuse_in/89439547). But, the question (using a resistor as a fuse) is more broadly applicable, so I'm also posting it here.I'm looking at the schematic for the HP 3048A option K23 DC blocking filter (service manualhttp://www.hparchive.com/Manuals/HP-3048A_option_K23_DC_blocking_filter.pdf). I've attached a simplified version of this schematic for convenience. Note that I've omitted the circuitry related to reverse polarity (switch and diode protection). But, I don't need that to address my question.The manual states that the 51.1ohm R1 acts as a fuse protecting the DC blocking filter (C3-16) from a large input DC voltage (K23 is rated for +/- 30 V):QuoteC3-16 are rated for 50 VDC. How does this fuse action work and how can we ensure it's reliable?The description of R1 in the manual is:RESISTOR 51.1 1% .05W TF TC=0+-100However, if you look up the part on the keysight part finder (https://www.keysight.com/my/partDetail/0757-0394), the description says it's a 0.5 W part, not 0.05 W. I don't know which is correct.Anyway, if we're using the resistor as a fuse, presumably we're relying on excess power to destroy the resistor and count on it failing open. There's no DC current path through R1. But, the startup ramp of a DC voltage is AC. Because there's no DC current path, we need to rely on a large, short-duration excess of power rather than a steady excess of power to blow R1.Current can either flow through C1 or L1 out of R1. C1 will present a low impedance, but only at high frequencies (|Z_R1|=|Z_C1| at 663MHz). L1 will slow the voltage rise across C3-16. Presumably, the goal is for R1 to blow open before the voltage across C3-16 exceeds the voltage rating. If I make the initial ramp rate very fast I can greatly exceed the 1.6V R1 rating (in the 0.05 W case) or 5 V (in the 0.5 W case), but only for a very short time. How can I be sure to destroy the resistor? And, if I destroy it, how can I be sure it will blow open? The NASA reliability design handbook (MIL-HDBK-338B) gives the failure mode distribution for film resistors as (p. 7-198):| failure mode | probability ||------------------+-------------|| open | 0.59 || parameter change | 0.36 || short | 0.05 |While 5% is low, it's not low enough (IMO) to feel like we can count on this outcome. Also, it seems like it would be easy to circumvent this mechanism and still expose the blocking cap to an overvoltage condition. For instance, we could supply a slow ramp that ultimately exceeds the 50 V rating.There's a limiter across the output, but no proper fuse to blow if one of these diodes starts to conduct. It seems like the HP 11848 or 3561 (instruments used downstream of this DC block) input is pretty exposed in this case.Thoughts? I initially thought that R1 might be a fuse resistor, but the keysight part finder pretty clearly shows it as a regular thin film resistor. Given that this was designed by HP, I'm reluctant to think this is a bad design and feel like I must be missing something.

author: bob91343
author_rank: Super Contributor
timestamp: March 01, 2022, 04:56:16 am
content: It may be a fusible resistor. You can recognize them by the light blue sandy finish. I suspect 0.5 W rating.