Facebook
Google
GitHub
Linkedin
In order to measure current with an ammeter, you must open the circuit and put the meter in series with the circuit.
However, a DMM measures current by measuring the voltage drop across a resistor (applying Ohm's law). You can do the same since you have a resistor of known value in your circuit. If you measure the voltage across that resistor and apply Ohm's law, you can calculate the current.
So, for your 20 000 ohm resistor, if you measure (say) 40 volts across it
current = voltage divided by resistance (amperes, volts, ohms, respectively)
i = 40/20000 = 0.002 A = 2 mA
Because the voltage between the anode and cathode of a Nixie tube doesn't change a lot with current (within the normal operating range), you can use Ohms law and Kirchoff's voltage law to calculate the required resistance.
KVL says the sum of the voltages around a series loop is zero, so we have 180 V across the power supply, 140 V across the tube (anode to cathode), which would be taken as a negative value as we work out way around the loop, leaving 40 V across the resistor. You can find lots of info on KVL at AAC and elsewhere, so I won't dwell on it. Knowing the voltage across the resistor, you can now calculate the resistance using Ohm's law again. Let's say you want 6 mA, so
R = E/I = 40/0.006 = 6667 ohms
6800 or 6.8k (often written on schematics as 6k8) is the closest "E24" value. You can find info on the so-called "E series" of standard values on the web, too.
https://en.wikipedia.org/wiki/E_series_of_preferred_numbers
The way you were trying to measure current actually nearly short-circuited the tube because the current sensing resistor or "shunt" inside the meter is only a few ohms (or low currents or a small fraction of an ohm for higher currents.
We speak of current "through" something and voltage "across" something. I should have realized what you were doing right at the start.
However, a DMM measures current by measuring the voltage drop across a resistor (applying Ohm's law). You can do the same since you have a resistor of known value in your circuit. If you measure the voltage across that resistor and apply Ohm's law, you can calculate the current.
So, for your 20 000 ohm resistor, if you measure (say) 40 volts across it
current = voltage divided by resistance (amperes, volts, ohms, respectively)
i = 40/20000 = 0.002 A = 2 mA
Because the voltage between the anode and cathode of a Nixie tube doesn't change a lot with current (within the normal operating range), you can use Ohms law and Kirchoff's voltage law to calculate the required resistance.
KVL says the sum of the voltages around a series loop is zero, so we have 180 V across the power supply, 140 V across the tube (anode to cathode), which would be taken as a negative value as we work out way around the loop, leaving 40 V across the resistor. You can find lots of info on KVL at AAC and elsewhere, so I won't dwell on it. Knowing the voltage across the resistor, you can now calculate the resistance using Ohm's law again. Let's say you want 6 mA, so
R = E/I = 40/0.006 = 6667 ohms
6800 or 6.8k (often written on schematics as 6k8) is the closest "E24" value. You can find info on the so-called "E series" of standard values on the web, too.
https://en.wikipedia.org/wiki/E_series_of_preferred_numbers
The way you were trying to measure current actually nearly short-circuited the tube because the current sensing resistor or "shunt" inside the meter is only a few ohms (or low currents or a small fraction of an ohm for higher currents.
We speak of current "through" something and voltage "across" something. I should have realized what you were doing right at the start.
