Zachman
Well-known member
Why is impedance mismatching so dangerous?
The crux of the problem lies in the inductive nature of the output tranny. Inductive loads are pretty special things, since they STORE energy in a magnetic field. A property of this effect, as has been pointed out, is that the voltage can soar to levels above the supply voltage in the amplifier-- sometimes WAY above. You can't do that with any other kind of load other than inductive.
Now the transformer doesn't have an impedance of its own; it only reflects an impedance from one winding to another in proportion to the turns (or voltage-- they are the same) ratio squared.
So imagine that you've got an open secondary. This impedance is for all intents and purposes infinite. Thus, regardless of the turns ratio, the primary impedance is infinite as well. (leakage inductance and parasitic capacitance-- two unavoidable nasties of real-world trannies-- will limit this to some finite number less than infinity, but suffice it to say its really high.) This means that the primary will act like a constant current source, attempting to keep changes in currents through its windings to a minimum. This will be an important point later.
Operating into such a humungous load impedance will cause the plate to swing HUGE voltages, according to V=IR. Especially with tetrodes/pentodes, which are much better at cranking out current, the delta Ip will stay the same regardless of the load. Consider what happens when the R goes sky high.
Now, if the load were NOT inductive, the maximum possible voltage generated would be equal to the rail voltage. No problem. This is how it is in SS amps. But with tube amps, that's not the case.
The primary danger here is in the development of these extraordinarily high voltages, which can punch through winding insulation, arc over tube sockets, even arc inside the tubes themselves. Once an arc has struck you can be pretty sure it will happen again. And again.
This is not good. Probably the worst scenario is that the OPT primary arcs to the core, which is grounded, and that will cause mega current to flow. The OPT is toast, and the power supply will be too unless something stops that current in a big hurry.
So that's what can happen with too high a load. Admittedly, this is an extreme case scenario here, where you've got an OPEN secondary, and thus a very very high primary impedance to work into.
Notice above how I pointed out that tetrodes/pentodes will have a worse time of this than triodes. This is because of their much higher dynamic plate impedance, which can also be described as their being an approximation of current sources. The pentode will just keep cranking out plate current-- regardless of what potential the plate is. The electrostatic shielding effect of the screen grid continues to "pull" electrons from the cathode with the same force. Thus the plate current is largely independent of the plate voltage, a mark of high plate impedance. That current is what "builds up" (so to speak) when working into a high load impedance and generates the excessive voltages. It's almost as if you've got a constant current load (the unloaded OPT) on a constant current generator (the plate of the pentode)-- obviously if these two devices are "concerned" with currents, not giving a flip about the voltages involved, you can get some pretty crazy effects.
With a triode, the much lower plate impedance limits the extent to which the plate voltage will swing about uncontrolled. As the plate swings high, for example, the attraction of electrons from cathode to plate will increase due to the higher voltage. More electrons will be pulled to the plate, regardless of what the control grid is doing. More negatively charged electrons means less positive voltage, so the voltage is "automatically" decreased. This is a direct measure of plate impedance.
In fact, running a triode into a very high impedance is done all the time with interstage transformers, which generally are very lightly loaded. The inherent degeneration in the plate circuit keeps the peak voltages from becoming a problem. Actually, triodes "love" current loads of very high impedance-- the tube is operated in its most linear fashion and is free to do what it does best-- generate an output VOLTAGE.
You can think of the dynamic plate impedance of the tube as forming a voltage divider, with the inductive tranny between plate and B+, and the tube itself between plate and ground. Obviously, with the low plate impedance of a triode, the voltage cannot swing madly about. Now consider the very high plate impedance of a pentode, and how much higher those plate voltages can swing.
OK, that's the situation of too HIGH a load impedance. So what about too LOW of an impedance? Let's consider a dead shorted secondary on the OPT.
Now the primary presents a very low load to the tube, a low impedance, a vertical load-line. We will notice that the tables have exactly turned.
Since the triode's plate is like a voltage source, it will attempt to pass incredible amounts of current in a heroic attempt to make the plate voltage swing. Operating into a dead short, it cannot do this, so something eventually will give. The cathode will attempt to emit way more electrons than it can, and it will have a short, hot life.
The pentode, however, is more of a current source, so it will continue to pass the total plate current in accordance with the screen voltage and the control grid voltage. These have not changed with the alteration of the load, so the pentode will continue to merrily pump its current swings into a dead shorted load.
Take a look at some plate curves, if you need to. Find some for pentodes and for triodes. Better yet, find some for the same power pentode connected as a triode (g2 connected to anode).
First the pentode case: look at the way the curves lie on the page. Imagine a horizontal load line (infinite load, open secondary) drawn on the graph. Notice how the pentode doesn't look like it would work this way-- an infinitesimal control grid voltage change would produce a gargantuan change in plate voltage. The tube is NOT happy. Now imagine a vertical load line (zero load, shorted secondary). The pentode's peak current for a given control grid voltage doesn't change much at all-- the vg1=0 plate curve is nearly horizontal for most power pentodes, cutting right across all of the various plate voltage points. It doesn't matter what Vp is at all-- no matter where you draw that vertical line, the peak plate current is pretty much the same. The tube is happy.
Now the triode case: imagine the horizontal load line now. Notice how the plate voltage is almost PERFECTLY proportional with respect to control grid voltage. No matter which tube you try, or what current you draw that horizontal line at, it will be VERY linear. The tube is happy. Then consider the shorted output tranny case, with a vertical load line. Notice how the vg1=0 curve will produce a humungous plate current since the plate curves are so much "steeper" than the pentode's case. The tube is NOT happy.
What the heck does all this mean? Well, hopefully you aren't running ANY tube amp into a shorted or open load... Since no pentode is a perfect current source, and no triode is a perfect voltage source, the actual characteristics are somewhere between the two idealized cases. As LV pointed out, you're much better off running a pentode amp into a lower load impedance than it expects. For those of you with triode output tubes, or a triode switch, you're better off running into a HIGHER load impedance. If you don't see why by now, reread this essay. It's also a good idea to take a high value power resistor, say 470R, and hard wire it right from the OPT secondary's 16R tap to ground. This will dissipate a very small amount of power under normal conditions, but will limit the extent to which the primary impedance will tend towards infinity in the case of a disconnected load.
For what it's worth, I've been deliberately "mismatching" load impedances by one tap for years. In other words, either a 4R or a 16R load on an 8R tap, and so on. This small mismatch will limit output power and will change the clipping points of the output tubes, but will not damage anything in a properly designed amplifier. Keep in mind that a higher load impedance in a pentode amp will put additional stress on the screens, so you may want to have at least 1k stoppers installed. A lower load impedance will cause more plate current to flow, and if you're running the tubes at the edge of acceptable quiescent plate dissipation that may push them over the edge into the red zone. If you've got an old vintage amp you'd hate to see get damaged, by all means, play it safe and don't mismatch at all. But if you're wondering about how it sounds, and the amp's got good trannies in it, then mismatch away. Just keep it within ONE TAP please, for safety's sake.
-Ken Gilbert, by way of PMG
The crux of the problem lies in the inductive nature of the output tranny. Inductive loads are pretty special things, since they STORE energy in a magnetic field. A property of this effect, as has been pointed out, is that the voltage can soar to levels above the supply voltage in the amplifier-- sometimes WAY above. You can't do that with any other kind of load other than inductive.
Now the transformer doesn't have an impedance of its own; it only reflects an impedance from one winding to another in proportion to the turns (or voltage-- they are the same) ratio squared.
So imagine that you've got an open secondary. This impedance is for all intents and purposes infinite. Thus, regardless of the turns ratio, the primary impedance is infinite as well. (leakage inductance and parasitic capacitance-- two unavoidable nasties of real-world trannies-- will limit this to some finite number less than infinity, but suffice it to say its really high.) This means that the primary will act like a constant current source, attempting to keep changes in currents through its windings to a minimum. This will be an important point later.
Operating into such a humungous load impedance will cause the plate to swing HUGE voltages, according to V=IR. Especially with tetrodes/pentodes, which are much better at cranking out current, the delta Ip will stay the same regardless of the load. Consider what happens when the R goes sky high.
Now, if the load were NOT inductive, the maximum possible voltage generated would be equal to the rail voltage. No problem. This is how it is in SS amps. But with tube amps, that's not the case.
The primary danger here is in the development of these extraordinarily high voltages, which can punch through winding insulation, arc over tube sockets, even arc inside the tubes themselves. Once an arc has struck you can be pretty sure it will happen again. And again.
This is not good. Probably the worst scenario is that the OPT primary arcs to the core, which is grounded, and that will cause mega current to flow. The OPT is toast, and the power supply will be too unless something stops that current in a big hurry.
So that's what can happen with too high a load. Admittedly, this is an extreme case scenario here, where you've got an OPEN secondary, and thus a very very high primary impedance to work into.
Notice above how I pointed out that tetrodes/pentodes will have a worse time of this than triodes. This is because of their much higher dynamic plate impedance, which can also be described as their being an approximation of current sources. The pentode will just keep cranking out plate current-- regardless of what potential the plate is. The electrostatic shielding effect of the screen grid continues to "pull" electrons from the cathode with the same force. Thus the plate current is largely independent of the plate voltage, a mark of high plate impedance. That current is what "builds up" (so to speak) when working into a high load impedance and generates the excessive voltages. It's almost as if you've got a constant current load (the unloaded OPT) on a constant current generator (the plate of the pentode)-- obviously if these two devices are "concerned" with currents, not giving a flip about the voltages involved, you can get some pretty crazy effects.
With a triode, the much lower plate impedance limits the extent to which the plate voltage will swing about uncontrolled. As the plate swings high, for example, the attraction of electrons from cathode to plate will increase due to the higher voltage. More electrons will be pulled to the plate, regardless of what the control grid is doing. More negatively charged electrons means less positive voltage, so the voltage is "automatically" decreased. This is a direct measure of plate impedance.
In fact, running a triode into a very high impedance is done all the time with interstage transformers, which generally are very lightly loaded. The inherent degeneration in the plate circuit keeps the peak voltages from becoming a problem. Actually, triodes "love" current loads of very high impedance-- the tube is operated in its most linear fashion and is free to do what it does best-- generate an output VOLTAGE.
You can think of the dynamic plate impedance of the tube as forming a voltage divider, with the inductive tranny between plate and B+, and the tube itself between plate and ground. Obviously, with the low plate impedance of a triode, the voltage cannot swing madly about. Now consider the very high plate impedance of a pentode, and how much higher those plate voltages can swing.
OK, that's the situation of too HIGH a load impedance. So what about too LOW of an impedance? Let's consider a dead shorted secondary on the OPT.
Now the primary presents a very low load to the tube, a low impedance, a vertical load-line. We will notice that the tables have exactly turned.
Since the triode's plate is like a voltage source, it will attempt to pass incredible amounts of current in a heroic attempt to make the plate voltage swing. Operating into a dead short, it cannot do this, so something eventually will give. The cathode will attempt to emit way more electrons than it can, and it will have a short, hot life.
The pentode, however, is more of a current source, so it will continue to pass the total plate current in accordance with the screen voltage and the control grid voltage. These have not changed with the alteration of the load, so the pentode will continue to merrily pump its current swings into a dead shorted load.
Take a look at some plate curves, if you need to. Find some for pentodes and for triodes. Better yet, find some for the same power pentode connected as a triode (g2 connected to anode).
First the pentode case: look at the way the curves lie on the page. Imagine a horizontal load line (infinite load, open secondary) drawn on the graph. Notice how the pentode doesn't look like it would work this way-- an infinitesimal control grid voltage change would produce a gargantuan change in plate voltage. The tube is NOT happy. Now imagine a vertical load line (zero load, shorted secondary). The pentode's peak current for a given control grid voltage doesn't change much at all-- the vg1=0 plate curve is nearly horizontal for most power pentodes, cutting right across all of the various plate voltage points. It doesn't matter what Vp is at all-- no matter where you draw that vertical line, the peak plate current is pretty much the same. The tube is happy.
Now the triode case: imagine the horizontal load line now. Notice how the plate voltage is almost PERFECTLY proportional with respect to control grid voltage. No matter which tube you try, or what current you draw that horizontal line at, it will be VERY linear. The tube is happy. Then consider the shorted output tranny case, with a vertical load line. Notice how the vg1=0 curve will produce a humungous plate current since the plate curves are so much "steeper" than the pentode's case. The tube is NOT happy.
What the heck does all this mean? Well, hopefully you aren't running ANY tube amp into a shorted or open load... Since no pentode is a perfect current source, and no triode is a perfect voltage source, the actual characteristics are somewhere between the two idealized cases. As LV pointed out, you're much better off running a pentode amp into a lower load impedance than it expects. For those of you with triode output tubes, or a triode switch, you're better off running into a HIGHER load impedance. If you don't see why by now, reread this essay. It's also a good idea to take a high value power resistor, say 470R, and hard wire it right from the OPT secondary's 16R tap to ground. This will dissipate a very small amount of power under normal conditions, but will limit the extent to which the primary impedance will tend towards infinity in the case of a disconnected load.
For what it's worth, I've been deliberately "mismatching" load impedances by one tap for years. In other words, either a 4R or a 16R load on an 8R tap, and so on. This small mismatch will limit output power and will change the clipping points of the output tubes, but will not damage anything in a properly designed amplifier. Keep in mind that a higher load impedance in a pentode amp will put additional stress on the screens, so you may want to have at least 1k stoppers installed. A lower load impedance will cause more plate current to flow, and if you're running the tubes at the edge of acceptable quiescent plate dissipation that may push them over the edge into the red zone. If you've got an old vintage amp you'd hate to see get damaged, by all means, play it safe and don't mismatch at all. But if you're wondering about how it sounds, and the amp's got good trannies in it, then mismatch away. Just keep it within ONE TAP please, for safety's sake.
-Ken Gilbert, by way of PMG
