NPN Darlington circuit for 12V Diesel Glowplug

Question

I'm trying to build a very simple circuit to drive a glowplug. I am using a TIP122 NPN.

^{simulate this circuit – Schematic created using CircuitLab}

I wanted 0.017A going through the 12V plug. I estimated no more than 0.204W. I also assumed my NPN's HFE was 1000.

To find power dissipation \$12V \times 0.017A = 0.204 W\$.

Calculated base current with

$$Ib = \dfrac{0.017}{1000} = 0.000017A$$

$$R1 = \dfrac{4V-1.4V}{0.000017A} = 152941 \Omega$$

I hooked all this up, and the NPN got pretty hot. The glowplug started to get a little hot, probably around 100F by the time the NPN was around 200F and climbing fast.

I'm very new and have been more or less dealing with NPN saturation on simple devices like LED and Fans with no issues. I have done this exact circuit with a 12V relay I pulled from a car, but I was really just experimenting here trying to learn the limitations of an NPN.

Answer 1

You are calculating this wrong. You want 0.017 A going through a 1 Ohm device, so you must put 0.017 A * 1 Ohm = 0.017 V across it. Creating that kind of current control with a non-saturated grounded emitter NPN topology is pretty difficult since the Hfe is not a tightly controlled parameter for BJTs.

The correct way to do this is to put a current limiter resistor inline with the 1 Ohm device. Then when you saturate the device, according to the graph in the datasheet, the Vce(sat) at 0.1A will be about 0.6V, so you will have $$12 - 0.6V = 11.4V$$ across the pair.

Ohms law says you need a resistance of $$\frac{11.4V}{0.017A} = 670 Ohms$$

680 Ohms is a standard resistor value in that range and should give about 0.0168A in current.

Now on to the calculation of the base resistor. It is typical to ensure saturation by calculating as if the Hfe = 10. This ensures that you are fully saturating the transistor. So for a collector current of 0.017 A, you want a base current of 1.7 mA. The TIP122 Vbe(sat) with a collector current of 0.1A is about 1.25V so the resistor will have 4V - 1.25V = 2.75V across it. This results in:$$\frac {2.75V}{1.7mA} ~= 1.61 kOhms$$

^{simulate this circuit – Schematic created using CircuitLab}

Answer 2

As far as I can see you simply want to drive your glowplug.

By the circuit you are showing you would need about 12A through the 1 Ohm resistor for a fully-on driver.

If you assume a transistor gain of 250 the collector current vs. the base current dictates you need 12A/250=48mA flowing into the base of the transistor.

At 4V driving signal and assuming 0.85V Vbe drop, you need a base resistor of (4V-0.85V)/0.048mA ~ 65 Ohm

You stated the transistor is actually a darlington type. This means that when fully ON, the collector~emiter voltage drop is about 1V when fully ON. This means that out of 12V, the glowplug will only see 11V, the remaining voltage will be dropped on the transistor.

Considering all this, the powers will be:

P(battery)=12V*11A=132W

P(R1)=151mW

P(glow)=11V*11A=121W

P(transistor)=1V*11A=11W (HOT!)

You have to cool the transistor. the TO-220 package for TIP120 is NOT able to handle 11W of power with no heatsink. You will need a big one.

If you need to adjust the power, you should do it with PWM. This means quickly turning ON and OFF the transistor by cycling the input signal 0->4->0. But you have to do this quickly, probably in the kHz region.

If you plan to play with the glow current by varying the current through the glowplug in a linear fashion, the maximum power dissipation can be calculated by the max. power theorem, which states the maximum power dissipated in the transistor will occur when the impedance of the transistor matches the impedance of the glowplug.

In this case the max. power heating the transistor equals 36W (same power is also dissipated on the glowplug). This would require a massive heatsink. It's a waste, so don't do it.

Use a logic level mosfet and PWM it. this will do the trick, but you will have to include a diode (anode to collector, kathode to 12V) as a glowplug kicks back some voltage when you turn it off quickly.

All this is valid assuming you want to turn the glowplug fully ON. If you don't, the same principle applies, but be warned, that by setting the transistor base current to a fixed value and than expecting the collector current to stay stable is considered a poor design. This is probably why you found the transistor is dissipating more power as you expected. The gain was higher than you predicted, so more power was drawn from the battery.

Altough the collector current and base current are tied togather by the gain factor (called Beta for bipolar transistors), the transistor gain is by no means stable and should not be relied on. You can expect about 70% variation, affected by the collector current, temperature and frequency of operation. You must also take into account that the gain differs from batch to batch of manufactured transistors, even from the same factory.

Answer 3

If you are really drawing only 0.017 Amp from 12 volts, the transistor and glow plug together will only dissipate 0.2 watts, and will only be slightly (perhaps barely detectably) warm.

According to the TIP122 data sheet, the minimum DC current gain is 1000, with no mention of the maximum. You must have a very high gain transistor, and are drawing much more than your desired 0.017 Amp.

If the glow plug is intended to operate from 12 volts, you should supply enough base current to saturate the transistor.