Hoffman Amplifiers Tube Amplifier Forum
Amp Stuff => Tube Amp Building - Tweaks - Repairs => Topic started by: 22uf on October 25, 2023, 04:11:01 pm
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Sorry if this is a dumb one!
I want to put a sag resistor in my AB763 twin.
I've done some research and I know I need to place it between the output of the rectifier and the first filter.
I want to make it switchable, however I don't understand how I can do that with 400+v DC coming out of the rectifier?
Shopping around for switches, most are rated for 50v DC max, so I don't understand how I can do this safely?
Can anyone advise me how to do this? I am obviously not getting it as the internet is full of people with switchable sag resistors it seems! :icon_biggrin:
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my understanding; they drop the PA B+, causing the "clean swing range" to decrease, bring on clipping/compression "earlier"
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It’s by no means the standard that a switch should withstand 500v, but they do exist, case in point here. (https://www.mouser.co.uk/ProductDetail/CW-Industries/GTB4B13?qs=hkbal7gjZMzxAeq9AbBYpA%3D%3D)
Make sure you filter (most every parts supplier’s website allows you to) using the rating for DC. Either use an on/on (resistor on one side and straight through on the other, but an on/off would also work as a switchable bypass for the sag R (path of least resistance and all that).
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Aha! Ok so yes I have found a few rated for 500v DC.
That's super cool actually as it means I can also put the pre-amp dropping resistor in my 5F2A on a switch now also!
I didn't think you could get switches rated that highly ..
Sweet!
Thanks. Sorry for being a dunce.
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The DC across the switch will not be the full B+ voltage. It will only be the difference in B+ with the switch open or connected.
With a reasonable sag resistor value, that will probably be no more than 50V.
Or am I missing something?
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The DC across the switch will not be the full B+ voltage. It will only be the difference in B+ with the switch open or connected.
With a reasonable sag resistor value, that will probably be no more than 50V.
Or am I missing something?
I don't know (obviously). Do you have a link to something that better explains this concept? I'd really like to understand this stuff.
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The DC across the switch will not be the full B+ voltage. It will only be the difference in B+ with the switch open or connected.
With a reasonable sag resistor value, that will probably be no more than 50V.
Or am I missing something?
Sag resistor is not a factor. When the STBY switch is closed there will be zero volts across it. When the STBY switch is open there will be full B+ across it, ie, B+ on the rectifier side, zero on the filter cap side.
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The DC across the switch will not be the full B+ voltage. It will only be the difference in B+ with the switch open or connected.
With a reasonable sag resistor value, that will probably be no more than 50V.
Or am I missing something?
Sag resistor is not a factor. When the STBY switch is closed there will be zero volts across it. When the STBY switch is open there will be full B+ across it, ie, B+ on the rectifier side, zero on the filter cap side.
As I understand switch DC ratings, they define the maximum voltage and current across the switch when you flip it. Lower current typically allows higher voltage. Unfortunately, most switch data sheets only list DC ratings for high currents such as 10A. We don't really know what they can withstand when we're in the milliamp range.
I'm assuming the sag resistor and switch would be between the rectifier and the and main filter caps. The sag switch would short out the sag resistor when closed.
If the standby switch is open, there is only roughly 1mA across the sag switch (it's limited by the 220k+220k filter cap bleed resistors). If we assume a 200R sag resistor, that's 0.2V.
If standby is closed, the voltage across the switch is limited by I*R with regard to the sag resistor. Not sure what the typical current draw of a Twin is, but let's say 250mA, and we get a voltage across the switch of 50V.
The voltage across the switch would go up momentarily with the inrush current at turn-on, but for a very short amount of time. This would only be a problem if you flipped the sag switch at the exact moment of the inrush current.
The insulating rating of the sag switch would of course have to cope with full B+.
Please correct me if I'm wrong!
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Please correct me if I'm wrong!
You're correct. I misunderstood. I get it now.
Hoffman sells a 2 pole 3 position switch that will allow you to replace a standard STBY switch. This allows standby, sag, and full B+ options. I had it in my AC-15 for a while but found I was always using full B+ so I removed it.
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I've never messed with sag resistors. The immediacy of a diode rectifier and higher B+ voltage are my preference. Is there a functional reason why the resistors aren't placed on the AC /anode side of the rectifier diodes? It seems to me like the switches and resistors would both be less stressed. Is there something about the current draw on the DC side that behaves more like a sagging tube rectifier?
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I've never messed with sag resistors. The immediacy of a diode rectifier and higher B+ voltage are my preference. Is there a functional reason why the resistors aren't placed on the AC /anode side of the rectifier diodes? It seems to me like the switches and resistors would both be less stressed. Is there something about the current draw on the DC side that behaves more like a sagging tube rectifier?
Not something that I have explored myself.
But it is my understanding that the added resistors in some commercial circuits function as sag resistors (between the PT & diodes) ie: AC15C section below, 2- 82R/5W
Though I also recall HBP talking about adjusting PT secondary DCΩ with resistors to to replicate vintage transformer values, with the possibility of greater sag. Not sure what the primary intent was in the Vox
https://el34world.com/Forum/index.php?topic=30412.0 (https://el34world.com/Forum/index.php?topic=30412.0)
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... Is there a ... reason why the resistors aren't placed on the AC /anode side of the rectifier diodes? ... Is there something about the current draw on the DC side that behaves more like ...
You have to buy 2 resistors instead of one. That's likely the sole reason.