Hi guys, I designed a simple circuit here to use an Arduino Pro mini controlling a heater rod for my personal project. My heater rod is rated at 24V, 25W. I'm controlling the heater rod using a BJT transistor, FMMT617TA which is some leftover parts from my previous project.

After I got my circuit board designed and made, I only realized that the FMMT617A has a Collector-Emitter Breakdown Voltage (BVceo) of 15VDC when I'm doing testing. To my surprise, everything seems to be working fine. My arduino pin 11 is giving 3.7V, I'm measuring Vce = 2.7V, voltage across heater = 20.9V, current draw of circuit is at 1.1A. I was expecting that the FMMT617TA transistor would be burnt, as I'm applying 24VDC to drive the circuit.

I'm confused here. My understanding is BVceo is actually the maximum allowable voltage that can be used for the transistor to control the load, and normally we should always chose a transistor that has BVceo > system voltage (for this case, I should be using a transistor with BVceo > 24VDC). So did I misunderstood the interpretation of BVceo?

 

You might just be lucky with Vceo, but I don't believe Vce=2.7V and I=1.1A on a SOT23 transistor. That's almost 3 Watts, and a SOT 23 would melt at 3W.

 

There is no guarantee that a particular transistor will break down at any particular volatge above its maximum rated VCEO , Just a guarantee that it won't break down below that voltage.

 

You might just be lucky with Vceo, but I don't believe Vce=2.7V and I=1.1A on a SOT23 transistor. That's almost 3 Watts, and a SOT 23 would melt at 3W.

Yes, I'm sure that's what my multimeter is showing me. The transistor doesn't get very hot right away, but you can see the temperature is rising as time moves on.

There is no guarantee that a particular transistor will break down at any particular volatge above its maximum rated VCEO , Just a guarantee that it won't break down below that voltage.

So do you mean if I were to power up my circuit using 40VDC, the same transistor will still work? That doesn't sound right, does it?

 

At 40V it might work, but the odds are lower. You can always measure BVceo, the "o" means that the base is open circuit. Just connect it to a high voltage with a high resistance and measure the collector-emitter voltage.
My guess is that your Vce measurement was 0.27V, which would be more appropriate for a saturated transistor.

 

Agree with Ian0

It is really good form to make sure you don't exceed the component's ratings, and this is particularly true for components used in things for other people or customers.

I once worked for a company, kind of a large one with huge market. Unfortunately the design engineers were not very well educated and their culture was to trim every little bit of cost out of a product that they could get away with. I once pointed out one engineer that he needed to use a 1/2 watt resistor in a certain part of a circuit. He looked at me and said "Do you know that will cost xx (pennies) more?" Eventually field failures put the company on the brink of bankruptcy. The layoffs were painful all around.

Know the component's specifications and stick to them, leaving some margin for safety.

 

So do you mean if I were to power up my circuit using 40VDC, the same transistor will still work? That doesn't sound right, does it?

That's not what Dick said.
It may or may not work at 40Vdc with your particular transistor.
The rating only tells you the lowest voltage it's guaranteed to work, not the highest.

 

It’s about probability and statistics. If you bought 100, then 25 might work at 30V and 5 at 40V. The actual breakdown voltages will follow a normal distribution, but we are not told what the mean and standard deviations. The quoted figure for Vceo is probably (mean-6*σ).

 

At 40V it might work, but the odds are lower. You can always measure BVceo, the "o" means that the base is open circuit. Just connect it to a high voltage with a high resistance and measure the collector-emitter voltage.
My guess is that your Vce measurement was 0.27V, which would be more appropriate for a saturated transistor.

I see. However, I wasn't trying to operate my transistor in saturation state. I'm using my arduino to operate the transistor with PWM (I think I forgot to highlight this in my original post, sorry about this!), in return to control the temperature output of my heater rod. So the transistor was never fully in a saturated ON state. It's working and I'm getting 20.8V across my heater rod, with Vce = 2.7V.

That's not what Dick said.
It may or may not work at 40Vdc with your particular transistor.
The rating only tells you the lowest voltage it's guaranteed to work, not the highest.

Understood. So I guess I probably got lucky here with this transistor? In that case, we should always use a transistor with the BVceo > system voltage to prevent potential failure in the system. Am I right?

 

Hi guys, I designed a simple circuit here to use an Arduino Pro mini controlling a heater rod for my personal project. My heater rod is rated at 24V, 25W. I'm controlling the heater rod using a BJT transistor, FMMT617TA which is some leftover parts from my previous project.

After I got my circuit board designed and made, I only realized that the FMMT617A has a Collector-Emitter Breakdown Voltage (BVceo) of 15VDC when I'm doing testing. To my surprise, everything seems to be working fine. My arduino pin 11 is giving 3.7V, I'm measuring Vce = 2.7V, voltage across heater = 20.9V, current draw of circuit is at 1.1A. I was expecting that the FMMT617TA transistor would be burnt, as I'm applying 24VDC to drive the circuit.

I'm confused here. My understanding is BVceo is actually the maximum allowable voltage that can be used for the transistor to control the load, and normally we should always chose a transistor that has BVceo > system voltage (for this case, I should be using a transistor with BVceo > 24VDC). So did I misunderstood the interpretation of BVceo?

Datasheet has thermal specfications- junction temperature being the key. Everything you do has to be manipulated around that figure.




At the end of the day, one way or another, including any derating, you're staying within these parameters:

625mW Total Power Dissipation. Figure out how you're staying within params, rather than guessing or having other guess.

 

I see. However, I wasn't trying to operate my transistor in saturation state. I'm using my arduino to operate the transistor with PWM (I think I forgot to highlight this in my original post, sorry about this!), in return to control the temperature output of my heater rod. So the transistor was never fully in a saturated ON state. It's working and I'm getting 20.8V across my heater rod, with Vce = 2.7V.

If you're running it from PWM, then it MUST be either off or saturated, otherwise there's no point in using PWM.
Do you mean that the average Vce is 2.7V?

 

Personally I would use a logic level N channel MOSFET like an IRLZ44N. It's on or it's off.

 

Breakdown voltages are many times non destructive, if the current is limited. A zener diode uses that as a controlled parameter. If you don;t limit the current, bang. Analog IC's have this thing with the substrate diode Vin can not exceed Vss or Vdd by a diode drop. Funny thing is when the IC is unpowered Vcc and Vdd are zero.

 

Personally I would use a logic level N channel MOSFET like an IRLZ44N. It's on or it's off.

Unless the gate voltage is ~2V then it would be partially on, just like a bipolar transistor with not enough base voltage to saturate it.

 

If you're running it from PWM, then it MUST be either off or saturated, otherwise there's no point in using PWM.
Do you mean that the average Vce is 2.7V?

I'm measuring with a digital multimeter, so I guess the voltage shown should be average Vce?

Personally I would use a logic level N channel MOSFET like an IRLZ44N. It's on or it's off.

Yeah, I'm changing to MOSFET. I think that's the better choice for handling load that has high current draw. I probably should change the footprint from SOT23 to something else too