How can I best reduce the current drawn by a PCB heatbed?

Question

I have a noob question about fairly basic circuit design (I think). I am building an incubator for making Tempeh at home, and want to use an old PCB heatbed I have lying around as the heating element. When I measured the resistance over the heatbed with a multimeter, it says the heatbed has a resistance of 2.3 ohms. If I understand correctly (please correct me if I'm wrong) that means the heatbed will draw 12V/2.3ohms = 5.217A. But my power supply is 12V, 2.5A, so I added an extra 21W, 12V load resistor to the circuit, assuming it would reduce the current to 12V/10.5ohms = 1.143A and not destroy my PSU.

The problem I have is that now the tiny load resistor gets very warm, but my heatbed doesn't heat up at all, which is not practical for what I'm making. The heatbed only needs to heat up to 35 Celsius (89.6 Farenheit), so I assume 2A should give me more than enough heat. I include a diagram here of how I've wired everything. Is there something simple I can do with my wiring to make the heatbed draw less than 2A (there needs to be enough for the arduino too) or will I just have to remove the load resistor and get a bigger PSU so the heatbed can draw the full 5.217A?

I saw someone mention PWM as a possible way to reduce current drawn by a DC fan in another question, but they didn't elaborate on how as there were other better alternatives to that specific question. My googling came to the conclusion that it would only reduce the average current, which means the heatbed would still draw 5.217A without the extra load resistor.

Answer 1

I'd say the only reasonable thing to do is to actually match your power supply and heater right from the start. Get another heater with suitable resistance, or a power supply with enough current capability.
Or get 2 more heaters and stack them on top of each other (connected in series).

After that, you can apply some form of PWM in order to regulate the temperature.

The reason your resistor heats up instead of the heater is because the resistor is of much higher value than the heater itself, so it will take most of the voltage drop and thus, most of the heat.
Any PWM type solution would still require the power supply to be able to deliver the same peak currents. That might be solvable with large enough capacitors, but it seems like a bulky and expensive way to go.

Or you could stick an inductor in series plus a flyback diode, but now you've essentially created a step down regulator.

Answer 2

First, your present setup is working correctly - just not as you intended. At 12 volts, your heat bed will pull 5.2 amps. That is about 60 watts. With the resistor in place, current drops to 1.1 amps, which provides about 3 watts. At 1/20 the original power it's not surprising that the bed doesn't get too warm. The dropping resistor is dissipating about 10 watts.

However, I think you have a much easier, cheaper solution available. Don't increase the current (although you definitely need a higher-power dropping resistor). Rather, buy some foam insulation board and make a box for your incubator. You can get foam board quite cheaply at any Lowe's or Home Depot. Start by making a 6-sided box with one layer, put your incubator in it and measure the temperature. If the temperature is too low, add another layer of insulation. If it's too high, punch some holes to allow cool air in.

Also, if you're going to build an enclosure, consider putting your (bigger) dropping resistor in with the heat bed. No sense in letting the extra heat go to waste. 12 volts at 1.1 amps is a total of about 15 watts. Better, get 10 82-ohm, 2-watt resistors, connect them in parallel, and use thermal epoxy to bond them to the back of the heat bed, spread out in a uniform spacing. If this provides too much heat, try cutting out one resistor at a time (which will increase the effective resistance and reduce the total power) until you get a result that you like.

Answer 3

First you need an isolated box. You can use a cooler (portable ice chest), the insulation that keeps the cold in will be just as effective at keeping it warm.

If the cooler is kept indoors, it won't need much power to stay at 35°C.

Suppose a surface of 0.5m2, corresponding to a cube with 30cm edges. With 2cm of 0.04 W/m.K polystyren insulation, this will require 1W per °C of temperature difference with the outside. So if you keep it indoors at 20°C and you want 35°C inside, count 15W.

So, basically, to play it safe, let's get double that, so 3x 15 Ohm 15W chassis mount resistors, run them at 10W each, bolted to an aluminium plate. That should draw a total of 2.4A. Put the plate on the bottom of the cooler, and put the tempeh dish on top. The temperature sensor should be on the plate, to make sure it doesn't overheat.

You can use a MOSFET to PWM them with your microcontroller. Using a darlington like TIP120 is not a good idea, because the transistor will have about 1V across it, which means it'll need its own heat sink. A 5V logic level MOSFET is a much better choice.

Answer 4

I saw someone mention PWM as a possible way to reduce current drawn by a DC fan in another question, but they didn't elaborate on how as there were other better alternatives to that specific question. My googling came to the conclusion that it would only reduce the average current, which means the heatbed would still draw 5.217A

One approach you might take is to use LC filters with pulse width modulation. Although the current through your switching element would still be pulse width modulated, the current drawn from the power supply would be closer to the average current, with some ripple.

^{simulate this circuit – Schematic created using CircuitLab}

CircuitLab does not have a model for the TIP120 Darlington Transistor pair, so I emulated it with two discrete transistors. As others have pointed out, you may have better efficiency with a MOSFET transistor, rather than a Darlington. However, a) you may have the TIP120's on hand, and 2) a MOSFET with adequate rating may require a driver, which adds complexity to your circuit. So, I am using the emulated TIP120 in my example.

Similarly, there is no way in CircuitLab to use an Arduino, so that too is emulated in my schematic.

As you can see from the following diagram, the current drawn from the power supply is relatively constant, but the current in the load is pulse width modulated. The following diagram uses a 30% duty cycle.

Note: This circuit relies upon the pulse with modulation to limit the current. Use of this circuit with a duty cycle in excess of 30% may cause excessive draw upon the power supply. This may cause other components, such as the arduino to shut down, it may cause excessive current through the filter components, causing them to go into saturation or overheat, or, it may, if you are lucky, cause no problems whatsoever. However, it is NOT designed for such excessive current, so please don't push your luck.

Choosing Components

Capacitor C1 must deliver most of the current pulse to the load. It is therefore important that it be rated for high ripple, and have low ESR to prevent overheating. The following diagram shows the ripple current in C1

The ripple current for capacitors is usually rated as a RMS value. To find the RMS value, use this formula

$$I_{rms} = \sqrt{D*I_{on}^2 + (1-D)*I_{off}^2}$$

Roughly eyeballing this, at 30% duty cycle we get

$$I_{rms} \approx \sqrt{0.3*3.5^2 + 0.7*1.7^2} \approx 2.4A $$

The KTD101B476M99A0B00 has a rated rms ripple current of 4A, so that is probably an adequate capacitor.

Inductors L1 and L2 should have low resistance and be capable of handling the current. Say 1.8 A. 1140-222K-RC has about 0.5\$\Omega\$ resistance and a current capacity of 2.4 A, so it may be suitable for L1. For L2, 1140-470K-RC has a resistance of about 20 m\$\Omega\$ and a current rating of over 12 A, so it may be a little better than you need. On the other hand, L2 is on the side of the filter with significant high frequency components, and the datasheet for the 1140-470K-RC doesn't give any characterization at 31.25 kHz, the typical frequency of Arduino PWM outputs. So, perhaps the SHBC12-1R0A0051V is a better choice for L2.

The current through resistor R1 is initially high, but falls to a value that depends upon the resistance of L1. (The lower the resistance of L1, the less current will flow through R1 when the current settles.) Here is a chart of the current through R1 when the resistance of L1 is 500 mA.

In steady state, R1 might dissipate 800 mW. However, for a short time, during start-up, it will dissipate 1.4*8 = 11.2 watts. In your question, you mentioned that you used a 21 watt 8.2 ohm resistor. I would use that, although it is a bit more than necessary. However, since it is what you have on hand, you might as well use it. (If I were experimenting, I might personally try a 2 watt 8 ohm resistor, in the possibly mistaken hope that it would not be damaged by start up conditions. Or, if this was a production job, I might look through manufacturers' datasheets looking for resistors that are characterized for temporary overload conditions.)