Grid Protection in G-G Service

Transmitting AØ for long periods on 10m with the load capacitor set for maximum C would result in very high grid-current and almost no RF output. Under such a condition it might be possible to overheat a grid. However, since most people tune an amplifier for maximum output--and maximum output virtually coincides with normal grid-current--very few people are likely to overheat a grid. Thus, complex electronic grid-protection circuits are seemingly unnecessary.

A disadvantage of electronic grid-protection circuits is that they are not effective against the most common source of grid damage--intermittent VHF parasitic oscillation.

Glitch Protection

During a major problem, the anode (plate) current meter and other amplifier components can be subjected to a large current surge as the HV filter capacitors discharge. The peak discharge current can exceed 1000a if a series resistor is not used to limit the short circuit current that can be delivered by the HV filter capacitors. The current limiting resistor is placed in series with the positive output lead from the filter capacitors. A wire wound resistor with a high length to diameter ratio works best. A 10 ohm, 10W wire wound resistor is adequate for up to about 3kV & 1A. For higher voltages, additional 10 ohm, 10W resistors can be added in series to share the voltage drop during a glitch. Wire wound resistors with a high length-to-diameter ratio are best for this type of service. Since about 1985, Eimac® has recommended the use of a glitch protection resistor in the anode supply circuit. Svetlana® typically recommends using a 10 to 25 ohm glitch resistor.

A HV current limiting/glitch resistor may disintegrate during a major glitch--so it should be given a wide berth with plenty of chassis clearance. If the chassis clearance is minimal, its a good idea to cover the chassis with electrical insulating tape. Glass-coated (a.k.a. vitreous) wire wound resistors are the most suitable type of resistor for this application. If a glass-coated resistor comes apart during a major glitch, it won't be throwing chunks of shrapnel around--like a less-expensive rectangular ceramic-cased resistor often does. Metal-case power resistors should not be used in this application. If a glass-coated glitch resistor is damaged during a glitch, it should be replaced with two such resistors in series to reduce the peak V-gradient per unit of length during a problem.

If the positive HV arcs to chassis ground--due to lint, a hapless insect, a VHF parasitic oscillation, or moisture--the negative HV circuit will try to spike to several kilovolts negative in the typical 1500W amplifier. In the real world, this type of glitch is not an uncommon occurrence. Anything that gets in the way of the negative spike may be damaged. Since the grid-current meter is normally connected between chassis ground and the negative HV circuit, the meter can be exposed to kilovolts at hundreds of amperes.

The easiest way to protect a current meter is to connect a silicon rectifier diode across it, or across its shunt resistor. Usually, only one diode {cathode band to meter negative} is needed in parallel with a DC meter. In some circuits, it is best to use two diodes in parallel [anode to cathode] with the meter movement to protect against positive and negative surges.

It may take more than one diode to protect a meter shunt resistor. A silicon diode begins to conduct at a forward voltage of about 0.5V. To avoid affecting meter accuracy, the operating voltage per glitch protection diode should not exceed 0.5V. For example, a 1 ohm shunt, at a reading of 1A full-scale, has 1V across it. Thus, two protection diodes in series would be needed to preserve meter accuracy. Similarly, if the shunt resistor for a 1A full-scale meter is 1.5 ohm, the maximum shunt voltage is 1.5V--so three diodes are needed.

Glitch protection diodes should not be petite. Big, ugly diodes with a peak current rating of 200a or more are best. Smaller diodes--and the meter they were supposed to be protecting--can be destroyed during a glitch. Suitable glitch protection diodes are 1N5400 (50PIV) to 1N5408 (1000PIV). In this application, PIV is not important. The 1N5400 family of diodes is rated at 200a for 8.3mS.

During an extremely high current surge, a glitch protection diode may short out--and by so doing protect the precious parts. Replacing a shorted protection diode instead of a kaput meter is almost fun.

To reduce the chance of the negative HV circuit spiking to several kilovolts, connect a string of glitch protection diodes from the negative terminal on the HV filter capacitor to chassis. At 200a, each diode will limit the surge voltage across it to about 1.5v. Typically, three diodes are needed--thusly limiting the negative spike to about 4.5 volts. Diode polarity is: cathode band toward the negative HV. With one simple wiring change, the same string of diodes can also protect the grid I meter and the anode I meter. This dual protection technique is incorporated into the Adjustable Electronic Cathode Bias Switch on Figure 7.

Design Considerations for Indirectly-Heated Cathode Tubes

A HV arc can destroy an indirectly-heated cathode tube. Here's how it happens: In some amplifiers, one side of the filament/heater is grounded. The cathode is connected to the negative HV circuit. If the negative HV spikes to several kilovolts, the cathode will often arc to the grounded heater. At a minimum, this breaks down the insulation between the heater and the cathode. Sometimes the heater wire burns out--and sometimes the cathode arcs to the grounded grid. Either way the tube is kaput.

Grounding one side of the heater is an invitation for cathode-to-filament breakdown. Instead, let both heater wires float. If the heater is fed through an c.40micro H bifilar RFC, one side of the heater can be wired to the cathode. Even though this arrangement can not protect against cathode-to-grid breakdown, it assures that the voltage between the filament and the cathode is unlikely to rise to dangerous levels.

Safety Devices


Manufactured amplifiers typically use a safety device to automatically short the +HV supply to chassis ground when the output section cover is removed. If the cover is removed before the HV filter capacitors have discharged, the resulting positive HV to ground short can damage the amplifier. In most g-g amplifiers, the only DC current path between the negative HV circuit and chassis ground is the grid-current meter and its shunt resistor. Even if the remaining charge in the HV filter capacitors is only 200V when the short from positive to ground occurs, without glitch protection diodes, the entire 200V appears across the grid-current meter shunt and the grid-current meter. Many potentially-fatal amperes can flow into the grid meter as the HV filter capacitors finish discharging. If the amplifier is accidentally switched on with the cover removed, rectifier diodes are a common casualty.

Automatic-shorting safety devices are not only dangerous to amplifier components, they can be dangerous to operators. It is dangerous to assume that an amplifier is safe to work on because it contains a safety device. Even though an amplifier's HV supply is shorted, if the amplifier is plugged in, its electric-mains circuitry is still alive and potentially fatal. Amplifiers are inherently dangerous. They should not be worked on casually--even if they have so-called safety devices.

The safest quick method of discharging HV filter capacitors is through a paralleled pair of wire wound resistors. The resistors limit the discharge current to a safe amount. In the unlikely event that one resistor opens, the remaining resistor will do the job. For the average 1500W amplifier, a paralleled pair of non-adjustable 1k ohm to 5k ohm, 50W resistors will do the job. Its always a good idea to check the anode supply voltmeter before putting your hands inside an amplifier.


Fuses have current and voltage ratings. For a fuse, the real test is opening safely--not operating without opening during normal operation. A fuse's maximum voltage rating is important. In some amplifiers, ordinary 250V 3AG fuses are casually used in circuits where they may be required to interrupt several kilovolts. Examples are the anode or cathode circuit. All's well until a problem occurs. When a 250V fuse attempts to interrupt a potentially-damaging flow of current in such circuits, the frangible link inside the fuse parts as it should. However, due to the available voltage, when the link melts, a metal vapour arc forms in its place. Metal vapour arcs typically have a voltage drop of around 20V--so the unsafe current will continue to flow in the circuit. At some point, the fuse will eventually explode--usually after serious damage has been done to other components.

During a glitch, circuits which normally carry low voltages can spike to several kilovolts. For example, cathode circuits normally see a maximum voltage of 30V to 100V. Thus, it might seem appropriate to use a 250V fuse to protect the cathode from excessive current. However, when a glitch occurs, several kilovolts can appear across anything that attempts to interrupt the flow of cathode current. The safest place to use ordinary 250V fuses is in the primaries of transformers.

Power Supplies

Ripple Filters

There are basically two types of DC filters: inductor-input / capacitor-output, and capacitor. Each type of filter has advantages and trade-offs.

Capacitor filters have good transient response. Since no inductor and resonating capacitor are used, the capacitor filter is simple to build, compact, cost-effective, requires no tuning and it is lightweight. The main disadvantage of a capacitor filter is that the capacitor is charged only during a small fraction of the waveform supplied by the transformer. No charging current flows until the instantaneous output voltage from the rectifiers exceeds the instantaneous voltage on the filter capacitor. This means that the transformer is either not loaded or severely loaded at different times during each cycle.

For example, with a electric-mains frequency of 60Hz, the duration of a half cycle is 8.333mS. Under load, using a capacitor filter, the capacitor charging time per half-cycle is typically only about 1mS out of the 8.333mS. This means that the ratio between output current and peak charging current can be 8 to 1. To combat I^2 R loss in a capacitor filter power supply, the transformer, all circuitry in the primary (including the electric-mains) and the filter capacitor should have low resistance. Capacitor filters are not appropriate for use with older-design transformers that were intended for use with inductor filters. Typically, such transformers have high winding resistance.

Inductor-input / capacitor filters can be of the resonated type or the non-resonated type. A non-resonated inductor tries to maintain a constant DC-current despite changes in the load current. This is the nature of any inductor. It always tries to maintain constant current by temporarily increasing the output voltage when the load current decreases suddenly, or by decreasing the output voltage when the load current increases suddenly.

When a conventional voltmeter is used to monitor the output voltage from a non-resonated inductor / capacitor filter power supply, the transient unregulation characteristic will usually not be detected because of the damped response in the meter movement. If a DC oscilloscope is used to monitor the output voltage while a string of around 5 WPM CW dashes are sent, the instantaneous output voltage swings can be easily observed. On make, the output voltage spikes downward. On break, the output voltage spikes upward. The amplitude and width of the spike depends on how much filter capacitance is used after the inductor and on the change in current. Upward and downward voltage spikes of more than ±50% are possible during a sudden load change on a non-resonated inductor / capacitor filter power supply.

The resonant DC filter maintains a fairly constant output voltage during rapid or slow changes in current demand--provided that a minimum current passes through the inductor. This minimum current can be the zero-signal anode-current [a.k.a. 'idling current'] of the tube itself. The inductor is resonated with a parallel capacitor. In actual practice, the value of capacitance used--as well as the bleeder resistance--is that which produces satisfactory voltage regulation. The resulting resonant frequency is usually slightly higher than double the frequency of the electric-mains. Resonant L/C pairs are available from Peter W. Dahl, Inc.

DC filter inductors come in two types, fixed-inductance and swinging-inductance. A swinging-inductor changes its inductance according to the current that is passing through it. Obviously, a swinging inductor can not stay tuned correctly with changes in current. Therefore, resonated-inductor filters can only use a fixed inductor.

The disadvantages of a resonant inductor filter are:

The advantages of a resonant-inductor DC filter are:

The resonant filter is used extensively by commercial and military amplifier manufacturers. Since a resonant filter demands much less peak power from the electric-mains than a capacitor filter demands, for 120V operation, where available power is typically much more limited than with 240V operation, a resonant filter is clearly the best choice. The resonant filter is also the best choice for high duty-cycle modes such as RTTY, FM or AM.

Rectifier Circuits


Transformers are available in two basic types: E-I (conventional) core and toroidal core. The E-I core is made from a stack of thin E-shaped and I-shaped iron plates. When placed together they form a rectangle with two windows for the windings. A stack of E-I rectangles make the completed core. The toroidal core is made from a continuous tape of grain-oriented material that contains iron and silicon plus other elements that increase the permeability of the core and decrease loss. This core material is known as Hipersil. Westinghouse Corp. was the original patent and copyright holder. Their patent expired decades ago. There are different grades of Hipersil tape. Grade 5 has the highest performance. Grade 22 has the lowest performance.

Higher permeability means that fewer turns are needed to achieve the required inductance in each winding. This means that larger diameter wire can be used. The end result is a transformer with low resistance and high efficiency. The Hipersil core is so efficient that the principal loss factor is the resistive loss in the copper wire. Hipersil core transformers are capable of producing extremely high peak currents. Thus, the Hipersil core transformer is ideally suited for capacitor filter power supplies.

It is difficult and time-consuming to thread a continuous tape core through the completed transformer windings--so someone came up with a faster way of uniting the core with the windings. Here's how it's done: The tape is wound on a form of the appropriate dimensions. The tape is spot welded together, removed from the form, and annealed at about 700 degrees C to relieve internal stresses. After cooling, the core is varnished and dried. Then the core is cut in half with a machine that makes a precise square cut. The faces of the cut are then polished flat. Thus, the halves of the core can slip into the completed windings, contact each other closely--restoring nearly perfect magnetic coupling between the halves of the core. The matched halves of the core are marked so that they can not be inadvertently mixed up with other core halves. The reunited halves of the core are held together tightly by steel bands like those used for binding heavy cartons and crates. If future access to the primary and secondary windings is needed, a Hipersil transformer can be disassembled by cutting the steel bands and removing the core halves.

Hipersil is no longer the most efficient type of core material. The new amorphous core transformer is starting to come into use by electric utilities. An amorphous core transformer is so efficient that if the secondary is unloaded and the primary is disconnected from the electric-mains, the collapsing magnetic field generates a voltage spike that can destroy the transformer. To avoid this problem, one winding is paralleled with a suitable voltage surge absorber.

Transformer Power Ratings

Transformers are commonly rated in maximum "volt-amperes" [VA]. Maximum VA are roughly equal to maximum RMS watts when the rated RMS current is flowing in each transformer winding and the transformer is operated from the rated input voltage at the design frequency. If the electric-mains voltage is reduced, the VA capability of the transformer decreases.

For SSB and CW operation, a lighter transformer may do the job just as well as a much heavier and more costly transformer. Manufactured 1500W amplifiers typically use a HV transformer with a continuous capability of roughly 600W--or VA. Such transformers are completely satisfactory for normal SSB operation. Such transformers are also capable of handling brief FM and RTTY transmissions--provided that the lower voltage tap is used.

If a power supply's DC output voltage drops more than about 10% under modulation, it's a fairly safe assumption that a more capable transformer is needed. Of course, not using enough filter capacitance or excessive electric-mains resistance can also cause poor regulation.

Transformer Current Ratings

Increasing the current in any conductor causes a square-law increase in the amount of power dissipated in the conductor. Since P=I^2 x R, doubling the current causes a 4 times increase in dissipation. This is an especially important consideration with transformers because they have considerable difficulty dissipating the heat that is generated deep inside their windings. This problem is compounded because copper has a positive, resistance versus temperature, coefficient. Thus, as the copper heats up, its resistance increases--which increases the dissipation--which increases the resistance, et cetera. This can lead to thermal runaway and transformer failure.

If a transformer has a secondary rating of 1A RMS, it means 1A with a resistive load. If connected to a rectifier and DC filter, the 1A rating does not necessarily apply. For example, if a fullwave-bridge rectifier, resonant filter circuit is used, the RMS current rating can be multiplied by at least 1.2. The DC output voltage will be about 0.85 times the RMS voltage. If a fullwave-bridge rectifier, capacitor-filter circuit is used, the loaded DC output voltage will be about 1.3 times the RMS voltage. A 30% increase in voltage sounds good, but obviously you don't get something for nothing. The trade-off for the increase in voltage is a decrease in current capability. The high peak current demanded by the capacitor filter translates into a substantive current capability decrease. .

Any formula for converting a transformer's RMS current rating to a DC output current rating is bound to be problematic due to the large number of variables. Here's a rule-of-thumb that is fairly accurate. If, after about an hour of typical operation, the outside of the transformer is uncomfortably hot for one's thumb, the internal parts of the transformer are probably deteriorating. Reducing the average load current slightly will greatly reduce transformer heating because of the square-law relationship between current and power dissipation. For example, reducing the current by 30% will reduce winding dissipation by about 50%.

There is a simple, reasonably accurate, 2-step approximation for determining the safe SSB, maximum current rating for a specific transformer for use with a capacitor filter and a full wave bridge rectifier. A slightly different approximation is used for a full wave voltage-doubler. These approximations are based on the DC resistance and the AC-voltage of the transformer's secondary winding. These approximations are useful when shopping around surplus stores or swap meets. All that's needed is an ohm-meter and a clip-lead. The clip-lead is used to short the primary of the transformer. This dampens the inductive voltage spike that occurs when the ohm-meter is disconnected.

The fullwave-bridge, capacitor filter approximations are: Multiply the secondary winding resistance by 70 to find the minimum intermittent load resistance that can be placed on the power supply. To find the DC output voltage under load, multiply the secondary RMS-voltage by 1.3. To find the safe intermittent current rating for SSB service, use Ohm's law and divide the output voltage by the minimum load resistance. For a more accurate evaluation, use the appropriate graphs in this book.

For example, a 2000V RMS secondary winding has a DC-resistance of 60 ohms. A full wave bridge rectifier, capacitor filter, circuit will be used. The safe, minimum, intermittent load resistance is approximately 70 x 60 ohm = 4200 ohms. The approximate voltage delivered under load would be 1.3 x 2000V = 2600V DC. Thus, the maximum intermittent load current is 2600V ÷ 4200 ohms = 0.62a.

Another approximation can be used to find the amount of filter capacitance needed. The approximation is 50,000 divided by the minimum load resistance. In the above example this is 50,000 ÷ 4200 = 12micro F.

For a full wave voltage-doubler, capacitor filter, power supply, the SSB-service approximations are: Minimum intermittent DC-load resistance equals 300 times the winding resistance; DC-output voltage, under load, equals 2.5 times the secondary RMS voltage.

For example: A 1000VRMS transformer has a winding resistance of 10 ohms, the minimum load resistance for full wave doubler operation would be 300 x 10 ohms=3000 ohms and the output voltage would be 2.5 x 1000V=2500V. The maximum intermittent load current is 2500V ÷ 3000 ohms = 0.83A.

The amount of filter capacitance needed for each half of the full wave voltage-doubler circuit is approximately 200,000 divided by the minimum load resistance. In the above example each of the two capacitors should have a minimum of 200,000 ÷ 3000 = 67micro F.

There is more to transformer performance than secondary resistance. If a Hipersil® core is used, core loss is minimal and the maximum intermittent power capability increases. Primary resistance is another factor to consider since it is effectively in series with the electric-mains resistance. Electric-mains resistance can cause a voltage drop problem if the amplifier is a fair distance from the service entrance box and you are using a capacitor filter power supply. One solution is to use larger diameter wires than the electric code requires. Another solution is to install the power supply near the service box and bring the HV DC to the amplifier.