HP 6023A PSU again :) This time R1, which is a 20ohm 7W fusible. The relay pulls 1s after power is applied and I assume that the general idea is to take the egde of the rather substantial inrush into C2, C3, C4.


Since it is fusible it will blow in case something goes wrong at the input filter or later. Now the closest I can find is Yageo FKN7WSJR-91-22R from Digikey which is 22 ohms, but not stocked. OK, what I have there now are three 68ohm in parallell, but I want to stick to the original as far as possble. So what would be better? 22 ohm 7W or two 10ohm 3.5W of I can find such a beast?

Edit: When I compare to another HP unit (HP6038A), it says "0811-3667 res 20 5% 7W", no mention of fusible?

Edit: I tried inrush limiting NTCs as per suggestions below in answers and comments. This is what the input current looks like with a fixed resistor of 22.6 ohms:

23 ohm regular

The first cycle is 20A or more. Unsure as we are outside the linear region of the transformer. With a 25 ohm NTC:

25 ohm NTC

Two 80 ohm 1,6A (https://www.mouser.fi/ProductDetail/871-B57236S800M51) in series:

160ohm NTC

This was with the PSU set to output 100W into a resistive load. You can see the kick when the relay kicks in and then the load turns on. So, a 80-160ohm NTC?

It is rated for an Imax of 1.6A, but that is steady state, right. This NTC will only have any current through it for 1s (or if the relay fails). The label on the instrument says 850VA max and the mains fuse is 8A. 850VA at 230V is 3.7Amps. Should I select an NTC that is rated for 3.7amps? I guess not as there are not any large enough ones with that rating.

I feel more inclined to follow the table on page 3 of the datasheet where they discharge through the NTC. Or would the Ametherm MS22 12104 which is rated 120ohms 4A be a better choice?

  • \$\begingroup\$ Does it need to be through hole or SMT? \$\endgroup\$ – Voltage Spike Sep 25 at 19:28
  • \$\begingroup\$ The 0811-3667 is the HP part number for the "res 20 5% 7W" \$\endgroup\$ – Huisman Sep 25 at 20:03
  • \$\begingroup\$ @VoltageSpike I think through hole as the Yageo FKN7WSJR-91-22R is also through hole \$\endgroup\$ – Huisman Sep 25 at 20:07
  • \$\begingroup\$ If space allows, you could put two FW70A10R0JA in series (10\$ \Omega \$, 7W) (Mouser stock) \$\endgroup\$ – Huisman Sep 25 at 20:19
  • \$\begingroup\$ Yes, it is through hole. Forgot to say and yes, there is ample space on the PCB. \$\endgroup\$ – AndersG Sep 26 at 6:36

Using 10W resistors or more is preferred over looking for non-std values.

I would consider an ICL to be a better solution with a 20 to 30 Ohm range. With 600uF and 20 Ohms you have a 12ms time constant yet the relay time is 1 second so a slower RC time constant is workable with 30 Ohms reduced stress on the Caps and still full charged after 20x Tau =RC . Observe the max rated current for the holding current as less important as the Relay bypasses the R after 1 sec but may offer less protection if the system power cycles on/off rapidly in a few seconds.


  • \$\begingroup\$ OK. So if I go this route, B57236S0250M0 which is 25 ohm and rated 2.5A and as you say. It will only have to endure for 1 sec normally as the relay closes after 1s. \$\endgroup\$ – AndersG Sep 26 at 8:47
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    \$\begingroup\$ The actual current will depend on the initial resistance V(t)/R. Since the Cap and Diode ESR are much lower. As it heats up in xxx ms the resistance will rise as the current drops and the voltage drop with no load applied yet after 1 sec should reach the peak line input. It is hard to calculate and depends on the thermal conduction by its location. But the result is much less surge and somewhat self regulating. Your precise results may vary the optimum value , but the parts are cheap and test verification is recommend with a variety of larger values. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 26 at 16:28
  • \$\begingroup\$ I guess you are right. Yes, the parts are cheap and I might try that, even if I am leaning towards a standard 20ohm 7W wirewound \$\endgroup\$ – AndersG Sep 26 at 17:15
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    \$\begingroup\$ Both will work , but cap. MTBF depends on square of RMS current * ESR * time. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 26 at 19:59
  • \$\begingroup\$ And time is short, but current might be large. Thus it would make sense to select a slihtly larger NTC to lower the inrush current. OTOH did probably HP overspec the caps as they did with everything in those days when quality still was paramount :) \$\endgroup\$ – AndersG Sep 27 at 7:27

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