analog panel volt meter

To increase the voltage reading of a voltmeter, it is first necessary to calculate the value of the multiplier resistor, based upon the desired full scale voltage (E) and the full scale current rating of the meter (I). (A purist will also include the resistance of the meter in the calculation, but when calculating the multiplier in the multi-megohm range, and with a meter resistance of less than a few ohms, this resistance can be ignored without compromising meter accuracy.)

Then there are some other safety and reliability issues that must be considered. One example you mentioned is having a high voltage present at the meter panel.

To calculate the multiplier resistance of your first meter, such that it reads 5000 volts full scale, use Ohm's law, resistance (R) = voltage (E) divided by current (I).
For your 500 volt meter to read 5000 volts full scale, E = 5000 volts / .001 amps (1 mA) the result is 5,000,000, or 5 megohms. This is the total multiplier resistance. Since your meter has an internal resistor of 500 kohms, your additional resistance in series with the meter must be 5,000,000 minus 500,000, or 4,500,000, in other words, 4.5 megohms. Unfortunately, this will still put 500 volts at the meter terminals when you are measuring 5000 volts, which is undesirable, and has additional disadvantages, discussed shortly.

When calculating a meter multiplier, it is necessary to calculate the power dissipated in the multiplier resistor, as well as determine whether the resistors used can withstand the voltage across each one in the string of resistors.

In the case above, when measuring 5000 volts at 1 milliampere for full scale reading, Ohms law will calculate the power dissipated by either taking the voltage times the current, or the voltage squared divided by the resistance, or the current squared times the resistance. Each of the three will provide the same result. So let's take the simplest approach, 5000 volts times .001 amperes = 5 watts.

Since you want this meter circuit to be reliable, it is prudent to provide a multiplier resistor string that can dissipate at least double, and preferably four to five times the watts needed. Consider a series string of 5 megohms total resistance, at 20 watts. This could be accomplished by using 20 each 1 watt resistors at 1/20 of 5 megohms, in series, for an individual resistor value of 250,000 ohms, or 250K each.

Another concern when building a series string multiplier is the maximum voltage rating of each resistor. If you exceed this voltage, the resistor could fail open and cause the meter to read zero, or it could fail shorted or lower in resistance, causing the meter to read higher than the actual voltage. In the case of 20 resistors in series, 5000 volts divided by 20 equals 250 volts across each resistor, which probably is acceptable - but consult the resistor rating information to be sure. In summary, any combination of resistors in series, which provides a total of 5 megohms, and at least 15 to 20 watts dissipation, may be used. Just make sure enough resistors are used to avoid exceeding the voltage rating of each resistor.

Now to protect the meter, and to eliminate the 500 volts at the panel, the best option is to open the meter and remove, or short out, the internal multiplier resistor. Now the meter will have a fraction of a volt when passing 1 milliampere, reading 5000 volts from your supply.

It is best to place the multiplier resistors at the supply, and run a low voltage wire to the meter. To protect the meter, should any of the resistors become shorted, it is common practice to place two power supply diodes, back-to-back, across the meter. This diode pair is an open circuit with less than .5 to .7 volts or so across the meter terminals, and therefor will not affect the reading. But if the resistor string is compromised, such as shorting one or more resistors, the diodes will prevent excessive voltage across the meter, avoiding damage to the meter and providing a margin of personal safety. Better yet, place these diodes at the meter end of the multiplier string (at the power supply end of the meter wire), and then there will be no high voltage at the meter terminals, or at the meter end of the multiplier resistor string at the power supply, even if the meter is disconnected from the wire to the multiplier!

You cannot benefit from this diode protection scheme if the voltmeter has an internal multiplier resistor. The diodes must be in parallel with the low-resistance meter movement, and this is accomplished even with the diodes at the end of the resistor string at the power supply end of the meter wire.

Your second meter may not be a good candidate for reading 3000 volts full scale, due to the high current requirement to reach full scale. Calculate the multiplier resistor the same way you did for the first meter. R = E / I. E = 3000 volts. I = 20 milliamperes, or .02 amperes. R then = 3000 / .02 = 150,000, or 150 kOhms. At first glance, this appears to be just fine, until you calculate the power consumed (dissipated) by the meter multiplier resistors. W = E * I, or 3000 volts times .02 amperes, = 60 WATTS! This is excessive power wasted and generates a good deal of heat for a meter multiplier. However, if you were to design the power supply in a manner that this 150K was used as the bleeder resistor, then the meter could be placed at the bottom of the bleeder string and provide accurate readings without additional dissipation just for the meter itself. It is even more prudent to use the protection diodes in this case, but it is then necessary to be sure that the meter resistance is low enough that less than .5 volts is required across the meter terminals to read full scale at 20 milliamperes, such that the meter reading is not altered by the presence of the protection diodes.

One final comment. NEVER trust a meter reading to determine that the power supply capacitors are discharged. ALWAYS use a shorting device to verify and confirm without any doubt that the capacitors are discharged BEFORE touching anything in a high voltage power supply. If the multiplier or bleeder resistor feeding the meter fails, the capacitors will still have a lethal charge. Always work safely and cautiously around any device with high voltage exposed.