WARNING! Anyone intending to make use of this technology should first be familiar with standard safety precautions for electricity, rocketry, and pyrotechnic materials. Neither the author, nor Tripoli Central California, nor the Webmaster, nor the Reaction Research Society will be responsible for any accidents or damages, as the use of this technology by the reader is beyond their control. If you do not agree, you are advised to stop reading now.

Note: Bob Dahlquist died a couple of years ago of unexpected health complications. He leaves this white paper behind as part of his legacy to the rocketry community. - Aerocon Systems, Editor, 2004

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CONSTANT VOLTAGE FIRING SYSTEMS

Member: Reaction Research Society and Tripoli Central California

To achieve the most reliable ignition and shortest delay when using carbon resistor igniters with a constant voltage source, the other resistances in the firing circuit should not add up to more than half of the igniter resistor's initial value. Then when the igniter's resistance decreases during firing, the sum of all the other resistances in the circuit will not exceed the resistance of the igniter.

When the igniter resistance is equal to the sum of all the other resistances in a constant-voltage circuit, the power dissipation in the igniter will be at a maximum.

(To clarify: By constant-voltage circuit, the author means a circuit in which the open-circuit voltage of the power source is constant, not the voltage across the load, or igniter resistor.)

In a well-designed system, the resistor undergoes a thermal runaway during firing. As the resistor initially conducts electricity and begins to heat up, its resistance decreases due to carbon's negative temperature coefficient. This causes more current to flow, which generates more heat, which causes the resistance to decrease further, which causes more current to flow, and so on.

When the resistor gets hot enough, pyrolysis converts some of the insulation to additional carbon in parallel with the original carbon film, further decreasing the resistance. This causes more current to flow, generating more heat, and so on.

If the other resistances in the circuit add up to more than the igniter resistance, this thermal runaway will be inhibited as the other resistances will control the current.

In addition, the energy transfer from the source to the resistor will no longer be at its maximum, as the other resistances in the circuit dissipate more power than the igniter. This will increase the ignition delay.

In choosing what value of resistor to use, the criteria are:
1. The resistor must draw enough current so that it dissipates at least 200 times its power rating initially.
2. The value of the igniter resistor, in ohms, must be at least twice the total of all other resistances in the firing circuit.
3. For shorter ignition delay, the resistor should dissipate 400 times its power rating when it is hot (assume its hot resistance is half the initial resistance).
4. You must use a standard value of resistance unless you are making your own resistors or ordering custom made resistors. Standard values are shown in Table 1.
DESIGN CALCULATIONS:
The following variables will be used in the calculations:
Ri = Initial resistance of the resistor (ohms)
Rh = Resistance of resistor when hot*
Rc = Sum of all circuit resistances except resistor
Rt = Total circuit resistance including resistor*
Pt = Total power dissipated in Rt (watts)*
Pr = Power rating of resistor (watts)
P = Power dissipated in resistor (watts)*
E = Open-circuit EMF (voltage) of power source
I = Current flowing in the circuit (amps)*
* Instantaneous values

Note: Rc includes the internal resistance of the battery or other power source; the contact resistance of all the connectors, terminals, relay or switch contacts, battery clips, and igniter clips in the firing circuit; and the resistance of the battery cables, firing leads, and igniter leads.

The maximum circuit resistance for a given circuit voltage is calculated as follows:

For a given value of Rc, maximum power transfer from the power source to the resistor will occur when Rh = Rc. Then Rt = Rh + Rc = 2 Rh. As these resistances are in series, the power dissipated in each will be proportional to the resistance. Thus, half of the total power will be dissipated in Rh and half will be dissipated in Rc.

Under these conditions, the total circuit power Pt required is twice the power required to pyrolyze the resistor:

Pt = 2 (400 Pr) = 800 Pr (Eq.4)

As Pr is 1/4 watt, Pt required is 800/4, or 200 watts.

Now we can calculate the total resistance:
Rt = E^2/Pt = E^2/200 (Eq.5)
Since Rc is half of Rt,
Rc = Rt/2 = E^2/400 (Eq.6)
And Rh = Rc = E^2/400 (Eq.7)
For less ignition delay, if possible, make your Rc half of the maximum allowable value:
Rc = E^2/800 (Eq.8)
The approximate initial resistance of the igniter resistor is twice our assumed hot resistance:
Ri = 2 Rh = E^2/200 (Eq.9)
This value must be rounded off to a standard value (see Table 1). It should always be rounded off to a higher value unless it is very close to a lower one.

Table 1: Standard 5% Resistor Values
Ý
 1 1.5 2.2 3.3 4.7 6.8 1.1 1.6 2.4 3.6 5.1 7.5 1.2 1.8 2.7 3.9 5.6 8.2 1.3 2 3 4.3 6.2 9.1
If you need higher values, multiply any of the above values by 10 or 100 or 1000.
Now calculate a value of current that will cause 400 Pr to be dissipated in the new value of Rh (which is 1/2 the new, standard value of Ri).
I = (P/Rh)^0.5 = (400 Pr/Rh)^0.5 = (100/Rh)^0.5 (Eq.10)
With Ohm's Law, we can calculate a value of total resistance such that the desired current will flow:
Rt = E/I (Eq.11)
The new value of Rc is then found by subtracting the new value of Rh from the new value of Rt:
Rc = Rt - Rh (Eq.12)
This is the new maximum value for Rc; if possible, make your actual circuit Rc no more than half of this maximum value, for less ignition delay.

Check the initial power dissipation in the resistor; it should be at least 200 Pr. Begin with the new Rt:

Rt = Ri + Rc (Eq.13)
The initial current, based on the new Rt is:
I = E/Rt (Eq.14)
And the initial power dissipation in the resistor is:
P = I2R (Eq.15)
Since Pr is 1/4 watt, we want the initial value of P to be at least 50 watts. If it is not, re-check your calculations or have someone else check them for you.

EXAMPLE 1:

The power source for this example is a 12.6 volt lead-acid battery.

Rh = E^2/400 (Eq.7)

Rh = (12.6)^2/400 = 158.76/400 = 0.3969 ohms

Ri = 2 Rh = 0.7938 ohms (Eq.9)

The nearest standard value is 1.0 ohms. Therefore the new Ri will be 1.0 ohms and the new Rh will be 0.5 ohms. The required firing current is now:
I = (100/Rh)^0.5 (Eq.10)

I = (100/0.5)^0.5 = (200)0.5

The square root of 200 is 14.14; therefore I = 14.14 amps.

The new total resistance to allow this current to flow is:

Rt = E/I (Eq.11)

= 12.6/14.14 = 0.891 ohms

The new maximum Rc is found by subtracting the new Rh from the new Rt:
Rc = Rt - Rh (Eq.12)

= 0.891 -0.5 = 0.391 ohms

For less ignition delay, use about half that value for Rc, or about 0.2 ohms if possible.

Checking the initial power dissipation in the resistor:

Rt = Ri + Rc (Eq.13)

= 1.0 + 0.391 = 1.391 ohms

I = E/Rt (Eq.14)

= 12.6/1.391 = 9.06 amps

P = I2R (Eq.15)

= (9.06)^2 * 1.0 = 82.06 watts

Re-checking with Rc = 0.2 ohms:
Rt = 1.0 + 0.2 = 1.2 ohms

I = 12.6/1.2 = 10.5 amps

P = (10.5)^2 * 1.0 = 110.25 watts

EXAMPLE 2:

The power source for this example is a 28-volt lithium-manganese dioxide battery.

Rh = E^2/400 (Eq.7)

= 28^2/400 = 784/400 = 1.96 ohms

Ri = 2 Rh = 3.92 ohms

The nearest standard value is 3.9 ohms. The new Ri will be 3.9 ohms and Rh will be 1.95 ohms.

The firing current required is:

I = (100/Rh)^0.5 (Eq.10)

= (100/1.95)^0.5 = 51.2820.5 = 7.16 amps

The total resistance for this current to flow is:
R = E/I (Eq.11) = 28/7.16 = 3.91 ohms
The maximum circuit resistance outside the igniter is:
Rc = Rt - Rh = 3.91 - 1.95 = 1.96 ohms (Eq.12)
For less ignition delay, use 1 ohm Rc if possible. Remember that Rc includes the battery's internal resistance; therefore its internal resistance must be substantially less than 1.96 ohms, and less than 1 ohm if possible.

This time, let us assume that the battery internal resistance is 2/3 of an ohm and that the remainder of Rc is also 2/3 ohm. Rc is then 1.33 ohm. Now the total initial resistance is:

Rt = Ri + Rc = 3.9 + 1.33 = 5.23 ohms
The initial firing current is:
I = E/Rt = 28/5.23 = 5.354 amps
The initial power dissipation is:
P = I2Ri = (5.354)^2 * 3.9 = 112 watts
The total resistance when the resistor is hot is:
Rt = Rh + Rc = 1.95 + 1.33 = 3.28 ohms
The predicted firing current becomes:

I = E/Rt = 28/3.28 = 8.54 amps

The predicted power dissipation in the hot resistor is:

P = I2Rh = (8.54)^2 * 1.95 = 142 watts.
These power values indicate that the igniter resistance, or the circuit resistance, or both, could safely be increased slightly. It would be preferable to increase the igniter resistance slightly rather than the circuit resistance, to leave some margin for any unplanned increase in battery internal resistance.

You are invited to try this for yourself; change Ri to 4.3 ohms and calculate the power dissipations, then increase the battery internal resistance 50% and re-calculate. To learn more, increase Ri to 4.7 ohms and repeat the same calculations with the original battery internal resistance and with it increased 50%.

The final resistor value to use should be determined by test-firing the brand of resistors you intend to use, in a range of values bracketing the calculated optimum value.

Don't make the mistake of ordering 1,000 resistors without first ordering 10 and testing them to make sure they will work in this application.

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