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Obsolete bridge rectifier replacement

throbscottle

Well-Known Member
Thread starter #1
One of the bridges in my oscilloscope melted and I need to find a replacement. Have an in-line one in there at the moment but it needs a little square or round one to fit the board. The original bridge was made by Silec, long since absorbed by Thompson which is now part of ST, part number is "110 B6". Doesn't exist it appears.

Anyway, it's a small square case 0.35" per side and 0.2" tall with 0.2" pin spacing, I'm guessing 150V 2A minimum rating. (based on 65v rms into 50vdc out, and the fuse is 1A) PCB holes are 1mm.

There is such a vast range of bridges out there, can anyone suggest a suitable replacement?
 

unclejed613

Well-Known Member
#3
you could make one out of diodes. 1N4004 diodes would do it just fine.
 

throbscottle

Well-Known Member
Thread starter #5
I think it's a little close to fit 4 diodes in. That Vishay bridge looks about right though. I hadn't noticed the 2A versions whilst looking at it's siblings.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
#6
I think it's a little close to fit 4 diodes in. That Vishay bridge looks about right though. I hadn't noticed the 2A versions whilst looking at it's siblings.
4 x IN400x diodes take very little space, and a handy technique for where space is tight is to fit two diodes on top of the PCB, and two underneath.
 

dr pepper

Well-Known Member
Most Helpful Member
#8
I was going to say I'll send you one of these for nowt, but none are that physical size, plus these are salvaged so could be blown anyway.

20181011_125228[1].jpg
 

gophert

Active Member
#11
1N4007 is 1 amp put 2 in parallel you get 2 amps. These are so cheap I bought a 100 for $2 free shipping.
It feels like that should be true but... it is just like putting two LEDs in parallel on one current limiting resistor. Unless both diodes have the same forward voltage drop (exactly), more current will go through the device with the lower voltage drop. As more current goes through, the forward voltage rises so not all current travels though the device with the lower Vf. One could always hope the two devices have close Vf but HOPE Is not my preferred engineering strategy. All Diodes are cheap (even 2A diodes) That the OP may as well buy the right one instead of settling for double-stacked 1N4007 diodes.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
#12
It feels like that should be true but... it is just like putting two LEDs in parallel on one current limiting resistor. Unless both diodes have the same forward voltage drop (exactly), more current will go through the device with the lower voltage drop. As more current goes through, the forward voltage rises so not all current travels though the device with the lower Vf. One could always hope the two devices have close Vf but HOPE Is not my preferred engineering strategy.
Very true, many commercial items do parallel diodes though, and they DO fail - it's one of the potential faults you can spot just looking at the circuit, another is putting resistors in series or parallel, likewise that's another common failure point.

If you do parallel diodes you should place current balancing resistors in series with each one.

However, as far as this bridge goes, using 1A diodes gives you a 2A bridge anyway - and the ready made bridges will use 1A diodes as well for their 2A rating (if they used 2A ones they would rate it at 4A).
 

gophert

Active Member
#13
However, as far as this bridge goes, using 1A diodes gives you a 2A bridge anyway - and the ready made bridges will use 1A diodes as well for their 2A rating (if they used 2A ones they would rate it at 4A).

Is this really the case? Only two diodes conduct current at any point in time and those conducting diodes are in series with each other so there is no current sharing in a bridge.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
#15
Is this really the case? Only two diodes conduct current at any point in time and those conducting diodes are in series with each other so there is no current sharing in a bridge.
Yes, the positive half cycles go through one pair of diodes, and the negative half cycles through the others, so they share the current 50/50.
 

unclejed613

Well-Known Member
#17
the OP stated the circuit had a 1A fuse, so it's really not requiring 2A diodes. 1A diodes will suffice. the only real problem i can see is if the charging current of the filter cap is well over 1A for more than a second or two. i've seen that happen in voltage multipliers, and the particular failure i saw was because the charging current was 3 or 4 times the rated current of the diodes being used
 

Nigel Goodwin

Super Moderator
Most Helpful Member
#18
the OP stated the circuit had a 1A fuse, so it's really not requiring 2A diodes. 1A diodes will suffice. the only real problem i can see is if the charging current of the filter cap is well over 1A for more than a second or two. i've seen that happen in voltage multipliers, and the particular failure i saw was because the charging current was 3 or 4 times the rated current of the diodes being used
Look up the spec of the diodes, the 1A rating is continuous, but they have huge surge current ratings (for charging the electrolytics) - if it was an issue the circuit would already have a surge limiting resistor (or thermistor) in the circuit.

I can't beleive there's so much discussion over a little bridge rectifier, swat four diodes in, job done :D
 

unclejed613

Well-Known Member
#19
I can't beleive there's so much discussion over a little bridge rectifier, swat four diodes in, job done :D
i agree, in the time it takes to read this thread, one could have the diodes in place and the unit operating again. 9 technicians, 10 opinions...
 

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