OT Protection at Output Question:

[jbrew73]

With a square-wave-like output the tubes are essentially switching between saturation and cutoff.
Without a load rapid switching into cutoff can generate high (kV) flyback spikes on the OT primary due to the energy stored in the (e.g. leakage) inductance.
A base load resistor across the output helps to damp the spikes by dissipating stored energy.
Another solution is to use catch-diodes on the primary.

What type diode would you recommend and how would you implement them?

[Helmholtz]

What type diode would you recommend and how would you implement them?

See the article by Tim Robbins I linked in my post above.

Here's an example with a newer Fender amp:
https://www.thetubestore.com/lib/thetub ... ematic.pdf

[R.G.]

I haven't looked at Tim's article, so this might be duplicative.
The root cause of OT failure is like that of any transformer, the insulation fails. In OTs, this is most likely to be either temperature rise or voltage spikes. As Helmholtz noted, you get spikes when you try to interrupt current in the leakage inductances suddenly. It's worse when this happens over and over again, which is what open output secondaries cause.

Output tube stages have a voltage gain that's expressable as the transconductance of the output tube times the reflected load resistance. An open secondary reflects as an open(ish) circuit and the tubes try to drive the primary inductance and leakage inductances. Huge voltage gain follows, This often causes hard square wave oscillation, and repeated high voltage transients.

Adding a secondary resistor amounts to trying to stop the oscillation by limiting how high the reflected plate load to the output tubes can go. This method has the advantage of being cheap, but the disadvantage of being hard to compute. As this discussion pointed out - how big or small a resistor do you need, and then how much power will it eat, as well as having to dissipate? The necessary resistance is very difficult to compute because amplifiers have different gain-phase margins before they oscillate. So a resistor that work on one amp may not work on them all. Still, good, cheap way to start.

Many amplifiers add stacks of diodes reverse biased to the plates of the output tubes. These are not used for their zener breakdown qualities. Instead, they clamp the downwards voltage on the primary when a positive transient is on the opposite output tube. Problem is, it works by transformer action, and leakage inductances are not clampable this way.

Another way to guard an OT is to put transient suppressors right across the primary. I've used MOVs (and mentioning this will get a reply about the capacitances of MOVs) with good effect. I have tried to kill an amp with MOVs on the OT primaries and could not do it. It's a third way to guard an OT that suppresses transients right where the spikes can do more damage.

[Raoul Duke]

Thanks for the great discussion all, very informative! I guess this solution has many more intricacies dependent upon the situation than I expected. I ended up putting a 10 watt WW 470R across the speaker jack (because I had one here) but plan to get a few higher values in to experiment with. The 470R had no noticeable effect on the sound to my ears FWIW.

RG, how would the computation for the “right” MOV value work out? Not being able to kill an OT with the MOV certainly speaks to reliability and is certainly worth experimenting with in my book.

Thanks again!

[R.G.]

Bear in mind that I only abused one amp this way. The fuzzy nature of how much leakage inductance an OT has, the stability margin and gain of the output stage and the type of output tubes all conspire to make calculation difficult.

Using MOVs or TVSS diodes at least lets you bound the problem. You want them to NOT conduct in normal operation first of all.

A P-P output stage will have a B+ and one side, then the other, is pulled toward ground by the "on" output tube. It can't get all the way to ground, maybe within 60-90V of ground. Transformer action pushes the plate of the "off" output tube higher than B+ by an equal amount.
So in normal operation, the output tube plates swing between perhaps 60-V above ground and (B+) + (B+ - 60V). Yikes that's hard to parse. Let's try an example. If you have a B+ voltage of 500V, and your output tubes could pull down to, say, 75V above ground. The "on" side of the OT then has 500V - 75V = 425V across it. The other side of the OT primary has an equal voltage across it, plus the B+ on the center tap. So the "off" side plate sees 500V + 425V = 925V above ground. The OT primaries see 850V across them. You want the MOV(s) to NOT conduct under these conditions. gh

That means that you could put one MOV across the entire OT primary, or one MOV on each half. Pick a MOV voltage that is comfortably higher than B+, and lower than a guessed transient. Most transformers have wire insulation that can take maybe 1kV or more across a winding, driven by the need to withstand line transients of 1500V to today's 4500V. So for a 500V B+, I'd probably choose something like a 750V rated MOV if I was using two, or a 1500V MOV if it went from plate to plate. I prefer the two-MOV version on some esoteric grounds.
I'd probably pick some 750Vac-rated MOVs and put one from CT to primary end on both sides.

[R.G.]

It struck me after I hit "submit" that MOVs across the OT would probably prevent output tube arcing too.

[Helmholtz]

Please read the Tim Robbins article.
It gives detailed advice on MOV selection starting from page10.

[Raoul Duke]

Thanks RG and Helmholtz! I’ve read the article but probably need to read it a few more times to fully get the message - it’s a little above me - but I’ll get it eventually, lol.

Great info and discussion - I really appreciate all the time taken to contribute!

[SamRE]

As I see it, the point is that to avoid the protecting resistor's resistance overly affecting the speaker's frequency-impedance characteristics, its resistance should be about 3x the speaker's peak impedance.
So about 30x the nominal speaker impedance.

Ideally, the resistor's power rating should be sufficient to accommodate the amp's maximum overdriven output power into that resistance.
It depends really on how long the user is expected to take to realise there's a problem with the speaker / its connection.
Many guitarists just keep turning it up when they can't hear anything.
I suppose worst case scenario might be 5 minutes cranked until they finally admit defeat

As a guideline, the cranked power output into a resistive load 30x the nominal load might be perhaps 1/3 of the rated power output (ie at the onset of clipping into the nominal load).

In my case I'd want the resistor to be connected whenever a speaker load is not and disconnected when a speaker isn't, so I can make measurements without the need for a loadbox. I'd be looking for a resistance about equal to the nominal speaker impedance with a power rating greater than the peak rated power wouldn't I?

[SamRE]

With a square-wave-like output the tubes are essentially switching between saturation and cutoff.
Without a load rapid switching into cutoff can generate high (kV) flyback spikes on the OT primary due to the energy stored in the (e.g. leakage) inductance.
A base load resistor across the output helps to damp the spikes by dissipating stored energy.
Another solution is to use catch-diodes on the primary.

Tim Robbins has put together a great article on OT protection:
https://dalmura.com.au/static/Output%20 ... ection.pdf

Thought I recognised the schematic style from somewhere. Is Tim Robbins on this forum?

[trobbins]

Yup

[SamRE]

Yup

Ah brilliant. I used your Eminar 70W bass doc as a guide to reverse engineer (And now Re-design) my own for years. I believe ours have the same OT.

[pdf64]

As I see it, the point is that to avoid the protecting resistor's resistance overly affecting the speaker's frequency-impedance characteristics, its resistance should be about 3x the speaker's peak impedance.
So about 30x the nominal speaker impedance.

Ideally, the resistor's power rating should be sufficient to accommodate the amp's maximum overdriven output power into that resistance.
It depends really on how long the user is expected to take to realise there's a problem with the speaker / its connection.
Many guitarists just keep turning it up when they can't hear anything.
I suppose worst case scenario might be 5 minutes cranked until they finally admit defeat

As a guideline, the cranked power output into a resistive load 30x the nominal load might be perhaps 1/3 of the rated power output (ie at the onset of clipping into the nominal load).

In my case I'd want the resistor to be connected whenever a speaker load is not and disconnected when a speaker isn't, so I can make measurements without the need for a loadbox. I'd be looking for a resistance about equal to the nominal speaker impedance with a power rating greater than the peak rated power wouldn't I?

We seem to be at cross purposes, waffling on about rather different things?

Yes your automatic dummy load's resistance should be the same as the nominal load impedance, and rated at least twice the nominal maximum power output.

I suggest not to rely on a jack socket's built in switching contacts for your application. They aren't reliable enough for my comfort.

[SamRE]

As I see it, the point is that to avoid the protecting resistor's resistance overly affecting the speaker's frequency-impedance characteristics, its resistance should be about 3x the speaker's peak impedance.
So about 30x the nominal speaker impedance.

Ideally, the resistor's power rating should be sufficient to accommodate the amp's maximum overdriven output power into that resistance.
It depends really on how long the user is expected to take to realise there's a problem with the speaker / its connection.
Many guitarists just keep turning it up when they can't hear anything.
I suppose worst case scenario might be 5 minutes cranked until they finally admit defeat

As a guideline, the cranked power output into a resistive load 30x the nominal load might be perhaps 1/3 of the rated power output (ie at the onset of clipping into the nominal load).

In my case I'd want the resistor to be connected whenever a speaker load is not and disconnected when a speaker isn't, so I can make measurements without the need for a loadbox. I'd be looking for a resistance about equal to the nominal speaker impedance with a power rating greater than the peak rated power wouldn't I?

We seem to be at cross purposes, waffling on about rather different things?

Yes your automatic dummy load's resistance should be the same as the nominal load impedance, and rated at least twice the nominal maximum power output.

I suggest not to rely on a jack socket's built in switching contacts for your application. They aren't reliable enough for my comfort.

Understood, you're talking about heaviest possible minimum load right? i.e. least possible power consumption to not instantly kill the OT?

[Helmholtz]

The lower the value of the safety resistor, the better it protects the OT.
But lower resistance means more dissipation ( as P = V²/R).
The more the resistor dissipates, the more power it steals from the output.
If the resistor takes 5W, the speaker gets 5W less.