How are resistor power dissipation calculations performed?

Question

I'm designing a switching mode power supply to supply 150V at 10mA to a nixie tube that is rated for 150V and 3.5mA. The circuit diagram will be very similar to this design, with the main difference that I'm powering a single tube instead of 6.

If I understand correctly, the resistor is needed to limit the current flow, so as not to blow the tube. The resistor would have to be 42K Ohms (V/I or 150/0.0035). However, only the output current of the resistor is used in that equation. There are 6.5mA of current at 150V unaccounted for. Are these 6.5mA of current "addressed"/used in the power dissipation calculation (P=V*I) as the remaining current is dissipated as heat in the resistor? Or does the power dissipation calculation still use the output current of the resistor?

If the latter, what happens to the 6.5mA of current not used by the load?

Answer 1

You don't seem to understand the operation of a nixie tube at all.

You need the datasheet for your Nixie, since you don't specify it, here's a typical one of the old stock available on Ebay ...Burroughs B-5859 page 1 and page 2.

The critical section from page 2 is shown here:

I've added lines for the current (3.4 mA) and the tube segment voltage when alight. You can see from the voltage on the Y axis the supplied voltage is always well above the tube sustaining voltage.
The way each number in the tube operates is that the voltage rises above what is called the strike voltage, and once conducting this voltage falls to the sustaining voltage.

In the schematic you provided, notice that the power supply voltage is in fact 180 V DC.
With the tube shown above if you had a 170 V DC power supply and 130 V sustaining voltage you would have an anode resistor of:

(170 - 130) / 0.0034 --> 11.7k Ohm

If the sustaining voltage was 145 V, then the anode resister would be:

(170 -145) / 0.0034 --> 7.3k Ohms

So in each case only a small portion of the power supply voltage is across the anode resistor, the majority of the voltage is the sustaining voltage across the tube element that is on.

The sustaining voltage for the Nixie tube varies from tube to tube and also varies depending on temperature and exposure to light. (A tube is harder to start in the dark)

Additional info
Since you've now pointed out the tube you are using (IN-8-2), here's some more points to think about.
The current rating to get all the digits to appear constant brightness means that for each digit you need separate current control. If you don't do this then digits such as "1" appear very bright while an "8" appears much dimmer.
It's also a challenge when using digits and the decimal dot, since if this tube is not multiplexed, you have to design to ensure both the digit and dot can run together (you can do this by having a negative supply for the dot).

Since you typically only have a single anode resistor this means that you need to add separate resisters to the cathode pins to control the current individually.

Let's try an example, though I don't know the absolute currents required for each digit, so will propose guess values using the datasheet data.

From the datasheet:

1. We know the worst case firing voltage is 170 V, so lets set the PS at 180 V DC. The datasheet tells us not to go above 200 V DC.
2. We know the lowest sustaining voltage is 100 V DC.
3. The maximum current we'll set at 3.5 mA for the "8" digit"
4. Lets assume the "1" digit will half of the "8", so about 1.75 mA.

Let's assume also that the anode resistor is all that's needed for the "8" digit, we won't use a cathode resistor.

The anode resistor is then:

(180 - 100) / 0.0035 --> 23k Ohms with no cathode resistance required.
When the "8" cathode is pulled to ground, the voltage will be above the strike voltage. Once the digit is on the anode will be at 100 V DC (worst case), so there will be 80 V across the anode resistor (0.28 W, so I'd fit a 1/2 W resistor). If the sustaining voltage is higher for the tubes, say 130 V then you may have to recalculate the anode resistor, since this would drop the digit current.

For the "1" digit the resistance needed is:

(180 - 100) / 0.00175 --> 45k Ohms. Since we have 23k already in the anode the cathode resistor needs to be (45 - 23) --> 22k Ohms. When the "1" low side resistor is pulled to ground you'll have about 40 V across the cathode resistor and 40 V across the anode resistor.

Having built a bunch of clocks (though they were all multiplexed) based on Nixies in my misspent youth, my opinion is that it's best to use a constant current low side driver with one anode resistor setting the maximum safe current. It's then easy to use a 4 bit D/A to set the current values for each digit/dot as you multiplex through them, especially it the clock is MCU based.

Answer 2

The resistor would have to be 42K Ohms (V/I or 150/0.0035)

This isn't correct, yes that is Ohm's law but in this case we have to use it in a different way.

Let's take an example of just a simple LED and a resistor. We know that in a series circuit the current flowing through all components is the same but the voltage drop across each component can be different.

If we want to supply an LED with 3.5mA then that must mean that the resistor also has 3.5mA flowing through it. By simply doing \$R = V_{supply}/I\$ you aren't taking any of the component voltage drops into account.

If we take this example circuit, we can see that the voltage drop across the LED is 2.2V, this is known as the 'forward voltage' which can be found in the datasheet of any LED (this is the same as your 'sustaining voltage' for your nixie tubes).

We have our supply voltage of 5V and the voltage drop across the LED as 2.2V, therefore the voltage drop across the resistor must be 2.8V (\$V_{R1} = V_{supply} - V_{F}\$)

Going back to the earlier point that the current flowing through all components is the same we can now calculate the current flowing through the resistor at a given resistance now that we know the voltage drop across it by using Ohms law.
We said at the start that we want to supply the LED with 3.5mA so our resistance needs to be \$800\Omega\$ (\$R_{1} = V_{R1} / I\$ or \$R_{1} = 2.8 / 0.0035\$)

If we had simply used \$R = V_{supply}/I\$ we would have gotten a resistance of \$1428\Omega\$. Now, if we had used this resistor value, then the actual current flowing in the circuit would have only been \$1.9mA\$ (\$I = V_{R1} / R\$) as the voltage drop across our resistor would have still been 2.8V.

As a final note to remember, you can only use \$R = V_{supply}/I\$ accurately in a purely resistive circuit