How to make this heater work with transistor

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1

I'm trying to get a heater thingy (light bulb with resistor) turn on when the temperature drops below a certain point. The heater part has digital pin to base and ground to emitter. The power is 12V (6 D batteries) with low to emitter and high off of Collector.
Now I have a NTE 194 transistor. What happens is: when DO pin is HIGH the heater turns on a little bit but the transistor gets really hot really quick and I yank the wire.

Why is the transistor seem to be dropping all the power when what I want is the thing to saturate and drop as little volts as possible? I tried different resistance on the base like 10K, 4.7K and 680 ohm but it didn't seem to make a difference. The circuit worked great with an LED though.

So is there a better transistor to recommend or do you guys think I wired this thing up wrong?
Any advice would be great.
Thanks,
Kregg

2

Whats the resitance of the heater. Because the lower it is the more current will be needed to saturate the transistor and therefore the more base drive you will need depending upon the gain of the transistor of course.

In your case the transistor isn't turning on fully and is in the linear region and to get the transistor to fully saturate the current will have to be in your case Isat = (12-Vcesat)/R

where Vcesat is the saturation voltage of the transistor in volts and R is the resistance of your heating element in ohms.

Now for it to fully saturate the base current Ibasesat will have to be at the very least Ibasesat = Isat/[ch946]

where = [ch946] = Hfe = current gain of your transistor according to its datasheet.

So you will need to drive the transistors base a bit higher than that just to be sure. The simplest way to do this would be to try a darlington configuration which is this:

What is happening here is that the first transistor is driving the second. This has the effect of multiplying the gain, without going into a detailed explanation it means you can get away with less base current drive. You also get integrated darlingtons but I can't remember the part numbers.

But remember BJT's in saturation aren't completely "on" and still have a bit of "resistance" as evidenced by the saturation voltage and at high currents this can still get very hot, but since you are working with D Cells I assume the current won't be that great. Although I don't know how long batteries will last when powering a heating element.

If you can give the actual power rating or ideally the overall resistance of the heating element (light bulb + resistor) I can help you work it out. A schematic makes it easier.

Otherwise if in doubt try a darlington solution and see but also remember the darlington has a much higher saturation voltage because of two transistors in one package to expect less voltage "available" to your heating element.

3

Thanks Zageek,

The Resistance is 6 Ohm for the Heater element.

The spec had some different numbers for the different values so wasn't sure which ones to use, but came up with

((12-.15)/6)/30 = 0.0658

Thr hfe had different numbers on the spec sheet:
VCE = 5V, IC = 1mA 80
VCE = 5V, IC = 10mA 80 – 250
VCE = 5V, IC = 50mA 30

Which ones to use? the IC = 50mA looked closest to DIO Arduino 40mA output so I used that number.

So what you think geek?

4

I am working on an h-bridge motor driver. I am new to both the Arduino and electronics but I think an N channel mofset might work well for you.

Look at the IRFZ44. probably overkill but its less than $1.

5

Some progress to report:

Tried the darlington pair and the light is on creating some heat. Not enough, but some.

The transistor is still dropping 1.2 volts and gets very hot but seems to be holding together. Using the NTE 194 plus some other transistor that I have no idea what the spec is. I'll get another 194 and see if saturation can be achieved. I also have an N-FET that can be tried.

Thank you all for the help. The Venus Flytrap should be happy in its warm home (actually wouldn't dare let this thing run unless I'm sitting right next to it).

Kregg

6

HeHe

Well the lamp just burned out. time for plan B. At least it increased the temp by 2~3 deg F

Think I'll verify fire insurance first

7

I'm a little surprised that no one has pointed out yet that 6D batteries will only yield 9 volts not 12 volts.

8

and by 6 I of course meant 8
Eight being the correct number.

9

I tend to agree with rlwoodjr use the FET .. At 6 ohms and 12 volts you're trying to draw 2 amps. FETs tend to turn on at a lower resistance and therefore handle higher currents better.
Best of luck to you.

10

Darlingtons are not great at low voltages as they exhibit 2 diode drops, typically 1.5 to 2V, at saturation (here that means 4W dissipated) - most other configurations of 2 bipolar transistors do better (but won't fit in a 3-wire package). So a MOSFET is a good solution if its on-resistance is low enough (and its gate-threshold voltage is below 5V). For this application Ron should be < 0.1 ohms to avoid needing a heatsink.

Also note that bulbs when cold have about a tenth the resistance of their hot resistance, which might affect the design.

11

"Also note that bulbs when cold have about a tenth the resistance of their hot resistance, which might affect the design."

Actually I didn't calculate in the bulb resistance at all. The bulb dropped 4.5 volts and 1.5A so I calculated:

(12-4.5)/1.5 = 5 ohm

is that calculation even close?

I'm going to try the MOSFET next. Sounds cool

12

The bulb dropped 4.5 volts and 1.5A so I calculated:

You can only measure the voltage drop of the bulb if you have a series resistor to measure things against. How did you know it took 1.5A? You are probably measuring the drop in power from the power supply which will tell you something about the output impedance of the power supply but not much else.

13

Hi Gigatropolis forgot about this thread.

Ok to answer your question as to which Hfe to use, you must use the one that is closest to the value of current that will be flowing through the collector in your circuit.

Since you are going to be switching the heating element the current will flow from the battery(positive terminal), through the element then into the collector and out the emitter and back into the battery(negative terminal). This will be Ic.

So Ic = (12-0.9)/6 = 1.85A (I chose 0.9V for the saturation voltage of the transistor)

So when the transistor is on the current flowing through the collector -emitter will be approximately 1.85A ( approximately because the saturation voltage will probably be slightly different)

Now since a transister is sort of a current controlled switch you will need a certain current in the base to get 1.85A to flow ie to "close the switch" in this particular case. Ok

Looking at the values of Hfe vs IC we see that for Ic 50ma we have a current gain (hfe) of 30. Since we are way above Ic=50ma we are actually at Ic = 1850ma or 1.85A we will use Hfe = 30. The gain usually gets lower for higher currents.

A current gain of 30 means that whatever current flows in the base, the collector current will be 30 times more. To get 1.85A flowing through the collector, in otherwords to turn this girl on you need at least 1.85/30 = 61.6ma (rounded off) flowing into the base.

So with 61ma flowing through the base you will have at least 30x61ma= 1848ma ( because of rounding off earlier) flowing in the collector. To play it safe rather push a bit more than 61ma into the base, push say 100ma into the base. That will defnitely turn it on.

But remember that once the transistor is on fully the collector current is no longer Hfe x base current. because the device has turned fully on and can't lower its "resistance" anymore. In that region between off and full on(saturation) the relation of Ic = Hfe x Ib holds true and this is a linear relationship, meaning that when you increase the base current by an amount, the collector current also increases by an amount directly proportional to the change in base current but by an order of magnitude greater as determined by the gain (Hfe) This linear relationship is why this region of operation is called the linear region as opposed to when the transistor is fully on (saturation) and fully off(cut off).

In the linear region it acts like a variable resistor and this means it will run hotter because of I2R losses so you want the device to be fully on. Thats why you try to supply more current to the base than the Ib required to take the device into saturation.

So far I can tell you this isn't going to work with the Arduino because you need at least 100ma which is greater than the spec of the ATMega which can only source 40ma.

Remeber the digital out of the arduino is going to drive the base. Another thing to remeber is that you also don't want to overdrive the base and damage the transistor. It will sort of work if you give 40ma to the base but the maximum current in the collector will be 1.2A and the transistor won't be fully on so it will overheat and waste power.

So you must use a darlington configuration. Try using two of the same transistor wired as darlington to give you a bigger effective gain and treat the whole thing as one transistor just with a bigger gain and higher Vce sat ( saturation on state voltage) and all the other calculations apply.

Note however that because the darlington configuration has a higher satuturation voltage you will likely get slightly less current flowing.

With regards to the MOSFET as mentioned earlier, that would be the ideal solution because mosfets have a very low on state resistance, but to drive a mosfet is not as simple as it looks, because while in theory mosfets have a high input impedance they have a nasty parasitic capacitance and this makes driving them harder, you also need a higher voltage to drive them and 12V is pretty tight it would be better to have 15V.

If you wish to use a mosfet try a logic level mosfet, they usually have an "L" in their part number, such as IRL540. These have a kind of driver built in and can run directly off TTL logic, your mileage may vary with these though. Or you could you a mosfet driver IC but these are sometimes harder to source.

14

I think I learned more from this thread than anywhere else.

Thanks Zageek for your detailed explanation on transistor theory and passing on your experience.

Thanks all for your replies. Actually starting to get somewhere with this project, and a lot more fun since I'm not totally lost in space.

It got up to 80 deg f today so forgot why I was making a heating element in the first place. Thinking more of making a fan with a door that opens and closes to control the temperature of the terrarium.
Too much fun. Wish I had more time for this stuff.