How do the diodes protect the OT?

[trobbins]

Jaz, that's a pretty damn hi-fi spec OT with 200H PP and 0.002% leakage (ie. 50,000 ratio). Top line Williamson's from way back got to 13,000 ratio (Partridge CFB). A guitar OT coupling would be way down - probably about 0.995, but also likely lower capacitance as nowhere near the same interleaving, and lower inductance. I'd be thinking the model is fine as is for the purpose at hand, as core saturation is likely not an exacerbating issue.

[jazbo8]

I'm sure we can make the OPT model worse But for the purpose of our discussion, it's not that critical... With the secondary opened, there is very little power, but just how little, I don't know... Perhaps Alan will clue us in just how the transformer ceases to behave like one when there is no load when he gets up in the morning.

[Dave H]

Perhaps Alan will clue us in just how the transformer ceases to behave like one when there is no load when he gets up in the morning.

Mains transformers seem to transform just fine with or without a load.

[Alan0354]

I don't know what to think anymore. I am no transformer expert. The most confusing thing is the two winding of the primary is opposite, so even if you draw say 25mA, the field should mostly cancelled out except the mismatch as what Jazbo8 mentioned.

It is my believe ( don't know how to proof) that when you have open secondary, the effect of secondary should not be dominant factor at all because it is not drawing any power, energy is not transfer from primary to secondary. The transformer can be looked at like an inductor with center tap and connected as shown:



Forget the graphs, just the circuit this time

How does the circuit react when there's an input at A and B if you look at it as a center tapped inductor? If you can enter this into the model parameters, then I think you can get closer.


Also, the more practical question, Anyone used the MOV or TVS approach, and had tried strumming the guitar at reasonable level without speaker for say a minute and the clamping devices still alive? That will tell a lot about how much energy really stored in the OT.

[uneumann]

... when you have open secondary, the effect of secondary should not be dominant factor at all because it is not drawing any power, energy is not transfer from primary to secondary. The transformer can be looked at like an inductor with center tap ...

After following this thread and mulling it over - I decided to add my 2-cents and share what I came up with. At the risk of being too lengthy or mathy or missing something, I'll try to show a possible way through this. Sorry if I've screwed up something, but since no one is likely to actually test their amps with open secondaries, this may be the only option. I'll walk though what I consider a reasonable approximation of dissipated power.

On MOV vs diodes... maybe this discussion about power dissipation of MOVs is helpful as a start ... Metal Oxide Varistor (MOV).

Also - waveforms are cool - thank to LT for providing those clips - real data is high-value and they show how complex things really are. However, the question of power dissipation is simpler (I think) than fully understanding those traces. These approximations and simplifications seem reasonable to me, some of which were mentioned earlier in the thread.

1 - the secondary can be ignored since it's unconnected.
2 - a first-order approximation assumes only 1 tube conducts at a time and the other is open (cut off), which makes the whole primary an auto-transformer and easier to think about.
3 - conservation of power has to be observed - this is key to analyzing the problem and can't be overlooked. Diodes and MOVs lead to different voltages and currents (and their waveforms) that dissipate energy - but both have to dissipate it. The question I'm trying to answer is -- How much power do they have to dissipate?

Turn on one tube. The physics eq's for coil currents and voltages predict what happens. The voltage on the driving plate drops immediately, and then slowly rises while the current also slowly rises. If the current peaks at ~200ma (reasonable for a 6L6) the magnetic field in the core peaks there too. The DC coil resistance and tube saturation voltage determines the peak current.

The core field has an energy (like a charged cap). The equation for that energy (in Watt Seconds) = 1/2 (L I I) or 1/2 the product of L and coil current squared. Plug in your favorite OT primary inductance and peak primary current and you'll see what the energy is. A primary of 50H means if half (25H) is driven with 200ma, 1/2 WS is stored in the coil. That energy is released when the tube turns off and it has to go somewhere - you can't cheat energy conservation. If no current path is provided (diodes or MOVs), high voltage and leakage currents are created that will eventually create an easier current path (fried OT).

With MOV or diodes, the total energy over time is the problem. An occasional 0.5ws pulse can be absorbed easily. But if you have a continuous pulse series, say 100 Hz pulse rate, that's 50W continuous, unless I did something wrong in the math.

Any approximation is only useful if real conditions match assumptions. While the above is reasonable for a single pulse, fortunately it's not accurate for 100Hz repetitions since primary inductance (25H) limits primary current to increase slowly. So for a rapid train (say 100Hz) of short pulses (5ms), current will only reach a fraction of its long term peak. (Note this is only true for the open-secondary case where we ignore mutual inductance and the load - but that is our condition.)

Another formula tells how much current flows after an elapsed time. I(t) = (V/R) (1-e -(tR/L) ). R is the series resistance of the circuit. Its reasonable to approximate R = 1.5K to account for the DC saturated tube and primary series resistance. 1.5k is reasonable since it means that the steady state saturation current will be 200ma if we assume that ~100 volts of a 400v power supply are dropped across the saturated tube. Use the formula to compute the current after 5ms for a primary voltage of 300v.

Current = 300/1500 (1 - e-(.3)) = ~50ma. That means a single 5ms pulse creates ~62 mw sec of energy. Taken at a continuous rate of 100Hz that's ~6.25w continuous dissipation. Now factor in that both primary halves are doing this, to get you ~13w. This is not unmanageable, but it's also not insignificant for either diodes or MOVs.

The eq's are non-linear, so changes in the circuit parameters will impact the result.
Primary current has a squared relationship to power. Reducing the (tR/L) term exponentially decreases current. So, if we raise the frequency to 1KHz, we get t = 500us pulses and each pulse drops to 5.9ma. But, since we have 1000 of them per sec, overall power is at 5.9w - almost the same as the 100Hz case.

This is worst case in that it uses square wave signals. Sine waves should produce lower currents and power. Although it's rough, it's a method or estimate you can tweak to your own transformer and circuit, without risk of smoke and fizz.

[trobbins]

Good addition to discussion. Simulation or some controlled test may improve your estimate of achieved plate current after a period of time, given that many things are changing during that time, such as plate resistance, grid voltage and even incremental inductance.

A portion of the plate current may also come down via a reducing grid voltage (ie. energy dissipated in that plates circuit) depending on the rate of change of grid voltage (ie. sine is very soft dv/dt after peak). This relates to the loading that each plate puts on its half-winding during the transition to when a protective device would start consuming the remaining energy. As grid voltage passes through its idle level (ie. signal voltage at zero crossing), the effective plate resistance is probably about 10k (400V at 40mA) for each plate.

There will also be a smallish level of energy consumed by the effective capacitance of each half-primary as it is changed by at least 400V.

The devil is in the detail, but it is all good to mull over, and gives some insight as to what really is worse to do if secondary is unloaded.

[Alan0354]

1 - the secondary can be ignored since it's unconnected.
2 - a first-order approximation assumes only 1 tube conducts at a time and the other is open (cut off), which makes the whole primary an auto-transformer and easier to think about.
3 - conservation of power has to be observed - this is key to analyzing the problem and can't be overlooked. Diodes and MOVs lead to different voltages and currents (and their waveforms) that dissipate energy - but both have to dissipate it. The question I'm trying to answer is -- How much power do they have to dissipate?

Turn on one tube. The physics eq's for coil currents and voltages predict what happens. The voltage on the driving plate drops immediately, and then slowly rises while the current also slowly rises. If the current peaks at ~200ma (reasonable for a 6L6) the magnetic field in the core peaks there too. The DC coil resistance and tube saturation voltage determines the peak current.

The core field has an energy (like a charged cap). The equation for that energy (in Watt Seconds) = 1/2 (L I I) or 1/2 the product of L and coil current squared. Plug in your favorite OT primary inductance and peak primary current and you'll see what the energy is. A primary of 50H means if half (25H) is driven with 200ma, 1/2 WS is stored in the coil. That energy is released when the tube turns off and it has to go somewhere - you can't cheat energy conservation. If no current path is provided (diodes or MOVs), high voltage and leakage currents are created that will eventually create an easier current path (fried OT).

With MOV or diodes, the total energy over time is the problem. An occasional 0.5ws pulse can be absorbed easily. But if you have a continuous pulse series, say 100 Hz pulse rate, that's 50W continuous, unless I did something wrong in the math.

Any approximation is only useful if real conditions match assumptions. While the above is reasonable for a single pulse, fortunately it's not accurate for 100Hz repetitions since primary inductance (25H) limits primary current to increase slowly. So for a rapid train (say 100Hz) of short pulses (5ms), current will only reach a fraction of its long term peak. (Note this is only true for the open-secondary case where we ignore mutual inductance and the load - but that is our condition.)

Another formula tells how much current flows after an elapsed time. I(t) = (V/R) (1-e -(tR/L) ). R is the series resistance of the circuit. Its reasonable to approximate R = 1.5K to account for the DC saturated tube and primary series resistance. 1.5k is reasonable since it means that the steady state saturation current will be 200ma if we assume that ~100 volts of a 400v power supply are dropped across the saturated tube. Use the formula to compute the current after 5ms for a primary voltage of 300v.

Current = 300/1500 (1 - e-(.3)) = ~50ma. That means a single 5ms pulse creates ~62 mw sec of energy. Taken at a continuous rate of 100Hz that's ~6.25w continuous dissipation. Now factor in that both primary halves are doing this, to get you ~13w. This is not unmanageable, but it's also not insignificant for either diodes or MOVs.

The eq's are non-linear, so changes in the circuit parameters will impact the result.
Primary current has a squared relationship to power. Reducing the (tR/L) term exponentially decreases current. So, if we raise the frequency to 1KHz, we get t = 500us pulses and each pulse drops to 5.9ma. But, since we have 1000 of them per sec, overall power is at 5.9w - almost the same as the 100Hz case.

This is worst case in that it uses square wave signals. Sine waves should produce lower currents and power. Although it's rough, it's a method or estimate you can tweak to your own transformer and circuit, without risk of smoke and fizz.

This is pretty much what I had in mind and tried to describe until last night, that I look at one tube at a time.

But in this case, we have to look at both tube at the same time. As you pull the plate of the first tube low by increasing the current, the current of the plate of the second tube is decreasing.....which result in the voltage of the plate of the second tube increasing. This in turn cause the plate of the first tube to go below the 100V saturation voltage due to the opposite direction of the CT winding. In fact the voltage on the second plate cause the first plate to go below 0V. This is proven by LT and other people already.

But I think you gave the best theory so far. I cannot say I agree 100%, but I cannot point to any particular part that I cannot agree. You described how the energy is being input to the OT by increasing the current on one side.


I am going to elaborate from where you left off:

1) So you increase current on the first tube and charge the OT as you described.
2) Simultaneously, the other side of the winding is trying to raise the voltage until the MOV turns on. The energy from the first tube is dumped into the MOV as the current increase in the first tube as you described.
3) This means the OT never store energy. It just transfer from the first tube ( as you increase the current) to the MOV on the side of the second tube.

I think people that know how to set up the model should try simulate the OT with no load as a center tap inductor like I've shown before.

From the above, if you have a continuous signal for a few seconds as I kept saying before, you are going to burn the MOV or TVS as they are designed to absorb high instantaneous power.........one non repeating pulse, they are not designed to take on continuous power. Each MOV or TVS can handle less than 5W. So if you pump high power for a few seconds, you can risk burning the device.

So the question is to people that use MOV or TVS to do high side clamping, have they even try driving the amp with reasonable signal without speaker for say 1 minute and the clamping device survive? If the MOV or TVS survives, this model is still not good enough.

[jazbo8]

Another formula tells how much current flows after an elapsed time. I(t) = (V/R) (1-e -(tR/L) ). R is the series resistance of the circuit. Its reasonable to approximate R = 1.5K to account for the DC saturated tube and primary series resistance. 1.5k is reasonable since it means that the steady state saturation current will be 200ma if we assume that ~100 volts of a 400v power supply are dropped across the saturated tube. Use the formula to compute the current after 5ms for a primary voltage of 300v...

We have a high voltage NOT high current situation when the output is unloaded... I still do not see how the "high current" is generated, Alan used 25mA, you are using a whopping 200mA in your analysis! With no load, the pentode has very high gain AND very high ra, i.e., nowhere near the 1.5k you used in the assumption.

[trobbins]

Alan, the sim model is fine for the transformer - perhaps if you firstly read up on transformer theory and apply known valve electronics understanding to the issue then that may be a help.

[jazbo8]

I think people that know how to set up the model should try simulate the OT with no load as a center tap inductor like I've shown before.

Why would you want to do that? Please look up how the transformer models are made in the first place. You are trying to re-invent a poorer wheel, if that's what you want to do, no one can stop you - more power to you brother...

[Dave H]

If you can believe it this is what LTSpice says my 2 x EL84 amp does with no load (top plot) and no load + plate clamp diodes (second plot).

[Alan0354]

We have a high voltage NOT high current situation when the output is unloaded... I still do not see how the "high current" is generated, Alan used 25mA, you are using a whopping 200mA in your analysis! With no load, the pentode has very high gain AND very high ra, i.e., nowhere near the 1.5k you used in the assumption.

He actually use 50mA for 100Hz and 5.9mA for 1KHz using current equation of inductor. As for me, I use switching idle current which the lowest is 25mA for 6V6. I just calculate using switch between 25mA to 0mA. This is all initial approximation.

I don't necessary agree with his number. But at least he did try to go through the logic step by step. It is very obvious that nobody here so far know the answer. I treat it as a process to derive the answer.

[Alan0354]

Alan, the sim model is fine for the transformer - perhaps if you firstly read up on transformer theory and apply known valve electronics understanding to the issue then that may be a help.

As I said, transformer is designed to work within certain parameters, when you have open secondary, you cannot exactly use the transformer model. When we were developing switching HV supplies, my engineer burned quite a few transformers, literally crisped. That was because of the loading is out of range.

If you have a better explanation, I am all ears. Join in and try to explain the process. At least me and Uranium really try to look into the MOV.

[Alan0354]

Why would you want to do that? Please look up how the transformer models are made in the first place. You are trying to re-invent a poorer wheel, if that's what you want to do, no one can stop you - more power to you brother...

As I said, transformer is designed to work within a certain parameters. When you have open secondary, that is out of the normal operation.

Dave H just did the simulation using LTS. I take that it's using a transformer model. It only glitch to 1KV. I can assure you if that is the real situation, this whole subject will never come up because this will NEVER burn the OT or tube if people forget to plug in the speaker.

Seems like we have two school of thoughts, both me and Uranium tends to believe the OT behaves out of the transformer model because of conservation of energy, that if you have open secondary, the secondary has very little effect on the primary. You and Trobbins want to keep the transformer model.

If I have time today, I am going to put the diode clamping to ground and measure the current pulse with the scope. That should give some insight on the mechanism.

I have my design inside the KMD that has power scaling. I wonder if I bring the voltage down, would it be safe to run without speaker and measure the plate directly. But I really hate to take the chance of burning the OT....worst, burning the scope.

[jazbo8]