The Valve Wizard

How to design valve guitar amplifiers!

Spring Reverb Drivers

Information on driving spring reverb using valves (tubes) seems to be rather sparse, so the following is presented as a guide to valve amp designers. This article considers only the driver stage, and not the reverb recovery stage (which is usually a simple triode gain stage, unfortunately). More information on reverb tanks is also available at Rod Elliot's site.

How Spring Reverb works, in brief:
The input transducer of a reverb tank is a coil of wire wound around an iron core. The spring (which has a magnetic pole on the end) passes through an aperture in the transducer core. When a current is passed through the coil the changing magnetic field exerts a tiny twisting force on the pole, which twists the spring too, of course. The spring vibrations are then picked up in the opposite manner at the other end of the tank.
The most important thing to note here is that it is current which drives the tank.

How much current is required?
For the richest reverb sound, and for the lowest noise, the input transducer should be driven hard- almost to the point of saturation. Accutronics quote the saturation current of their cores as 3.5Amps per turn (rms), in other words; if there was only one turn of wire around the core we would need to drive it with 3.5Arms before saturation would occur (assuming sine-wave excitation). Other manufacturers' tanks are probably very similar. As more turns are added, the coupling of the EM field increases and the required current is reduced, so if there were 100 turns we would need only 3.5 / 100 = 0.035A. Accutronics quote this figure as the Nominal Drive Current, which is the amount of rms current required for full drive. Overdriving it will cause heavy distortion and is generally undesirable.
Interestingly, Rod Elliot has noted that the coils can be driven up to 10 times harder than the Nominal Drive Current, without saturating, but that the signal recovered at the other end of the tank is hardly increased by doing so, so there is not much point doing it.

The voltage required to achieve the Nominal Drive Current can be found from Ohm's law:
V = IZ
Where Z is the impedance of the coil, which is normally quoted at 1kHz.

For example, if a type A tank is rated as 8 ohms at 1kHz and has 124 turns;
I(nominal) = Amps per turn / number of turns.
I(nominal) = 3.5 / 124
= 28mA

So the required drive voltage is;
V = 0.028 * 8
= 0.2V
This implies that the power dissipated in the coil is 6mW. In fact, power dissipated in the coil is minimal because it is inductive, so voltage and current are 90 degrees out of phase, but this poor power factor also means that the 'apparent power' that must actually be supplied is more like a few hundred milliwatts at 1kHz. (The DC resistance of the coil is negligible, except at very low frequencies which we don't need- anything below about 200Hz is too muddy to be of use.)

Unfortunately, the input coil is an inductor so its impedance rises linearly with frequency. The coil's impedance will double with a doubling of frequency, so at 2kHz we would need twice the drive voltage to obtain the same drive current (and therefore twice the power). At 500Hz we would need only half the drive voltage (half the power).
Therefore, if we were to apply a constant audio voltage signal to the coil, we would find that current through the coil would be less at high frequencies than at low frequencies, so the reverb sound would end up being 'bass heavy' and would sound very dull indeed! Ideally then, the coil should be driven by a constant audio current signal, so that all frequencies are delivered into the springs at equal amplitude. Unfortunately valves are not well suited to this, but there are solutions.

The 'traditional' driver is simply a small, conventional power stage. A small input coupling capacitor filters out frequencies below about 200 - 300Hz, and the reverb coil is driven so hard that even higher frequencies are forced into it, one way or the other. This 'works', but is a considerable waste of power- and it is not unknown for the coils to burn out under such abuse (remember, we are unlikely to ever need more than 1W, but such drivers can usually deliver several watts!). No doubt this 'brute force' approach also compensates for the conventional, rather low-gain recovery stage.

This out-of-date method can be improved slightly by using a smaller coupling cap, so that the drive voltage to the power valve doubles with every octave, thereby compensating for the increasing impedance of the coil.
Obviously the headroom of the stage should be sufficient to accommodate the larger high frequency amplitudes, and the more headroom the better- a distorted reverb sound is rarely a good thing. Choosing the roll-off point for the input filter depends somewhat on taste. For good reverb sounds there is no particular need for strong frequency content above 1kHz as these can sound too 'tinny' or shrill. If we set the roll-off at 1kHz then, the frequency response of the reverb driver will end up being flat up to 1kHz, then attenuated at higher frequencies, which is fine. Normal rules apply for the design of the power stage, and normally an 8 or 10 ohm tank would be used, to suit a conventional output transformer, but any tank would do, provided the impedance ratio of the OT is suitable.
A simple circuit is shown [right], and is similar to many old designs. There is no screen bypass capacitor since excessive output power is not required. The input signal required for 1W output is about 4.5Vp-p at 1kHz, so the signal from the previous stage will probably need attenuating somewhat! At the roll-off frequency it is already attenuated by ^/2, so the potential divider in the schematic has been chosen to suit a 100Vp-p guitar signal from the previous stage. Other designs using different power valves will have different requirements of course. The lower resistor in the input divider could be made a potentiometer, but the reverb 'depth' control is normally placed on the recovery side of the tank, since this keeps the quality of the reverb good even at low levels (because the tank is always fully driven).

Although the previous circuit produces reasonable results, it is not without its shortcomings. It was assumed that the coil presented an 8 ohm load (or whatever) more or less all the time, like a loudspeaker. Unfortunately it doesn't, as discussed, so at frequencies above and below 1kHz the reflected impedance to the valve will be higher and lower respectively, resulting in reduced power output. So the actual frequency response at the coil will be a gentle rise up to 1kHz, then a slightly steeper drop at higher frequencies. This is perfectly good for many users, but for a flatter response there is an obvious modification; use a triode. Triodes are much more forgiving of variations in load impedance. Although they deliver maximum power into an impedance equal to one to two times their internal anode resistance, a doubling or halving of this does not result in as much power loss as with a pentode / tetrode. Of course, triodes are much less power efficient (less than 25% in class A versus almost 50% in pentodes), but we don't actually need much power anyway.
The schematic [right] shows the EL84 in triode mode. The input sensitivity is now about 5Vp-p, distortion is mainly glorious 2nd harmonic and the final frequency response is closer to the desired one. Using an EL84 to deliver a few hundred milliwatts is quite a significant waste of power. An ECC81 (12AT7) or ECC82 (12AU7) would be much more efficient (and small), and many amps have used them.
Some budget amps even use a single ECC83 (12AX7) (which is capable of delivering about 200mW into a 100k load impedance), but it has a reputation for producing a weak sound. This is due both to its limited output power, a weak recovery stage and the fact that a 50k to 100k output transformer needs a very high primary inductance in order to produce low frequencies well (like 50 Henrys!), so manufacturers usually end up using a lower primary impedance, which really results in even less power. In theory, a high impedance, low inductance OT could work in our favour since it will automatically provide the necessary first-order, high-pass filter, but finding the right operating conditions to achieve it is probably more trouble than it's worth.

Cheap current source drivers:
Low impedance reverb tanks seem to be persisting in valve amp design, which makes little sense since valves are not well suited to driving low impedance loads. By taking advantage of high-impedance tanks (usually intended for op-amp drivers) we can build much simpler valve drivers. A constant-current source can be approximated by using a voltage source and placing a large resistance in series with the coil. Provided the resistance is large compared to the coil's impedance (ten times, say), the coil's varying impedance will make almost no difference to the current flowing through the resistor and coil over a wide bandwidth. The circuit [right] illustrates how this might work.

A paralleled pair of ECC82s has an internal anode resistance of about 5k, so for maximum power the load should be about 10k. In this case the load is formed by the parallel combination of Ra and Rl, which are both 22k, making 11k in total (ignoring the coil which is negligible and the output coupling capacitor which is large). Taking a bias point of roughly -6V, the AC loadline [below] indicates that at full 12Vp-p input we obtain about 12mAp-p swing (shared between the two load resistors) making 6mA p-p or 2.1mArms through R2 which is about perfect for an Accutronics type F tank (1.4 to 1.9k)!

Obviously the drawback of this method is that we must use both triodes to obtain sufficient power since no OT is used. A larger triode like an ECC99 or even a triode-strapped pentode might be used on its own. Since the stage now functions like a current source, all the frequency shaping can be done at the input. In this case, C1 and C2 attenuate frequencies below about 300Hz and above 2.6kHz respectively. R1 simply allows the coupling capacitor to discharge if the tank is disconnected.

For driving medium impedance tanks (100 to 300 ohms, say) a more 'up-to-date' solution would be to use an SRPP. An SRPP can deliver significant current into relatively heavy fixed loads. A total load of about 2.2k large enough to constitute a constant current source, but low enough to allow the valves to dump plenty of current into it. The optimum value for both cathode resistors is:
Rk = (ra + 2Rl) / mu
Using an ECC82 yields:
Rk = (10000 + 4400) / 19
= 758 ohms,
820 ohms being a close standard from the E12 range.

The quiescent current is given by:
Iq = HT / 2(ra + muRk)
Iq = 300 / 2(10000+ 19*820)
5.9mA
And since the SRPP can only operate in Class A, the peak current delivered into the load is 2Iq per triode, making 23.5mAp-p in total, or about 8.4mArms- more than enough for a medium impedance tank. The maximum input signal before clipping is simply 2Iq * Rk, which is a bit less than 10Vp-p. The necessary filtering can be done at the input, and the input level can be limited to give the right nominal current for the actual tank being used.
In this circuit the heater voltage will also need elevating by more than 30V.