Why does linear mode produce more heat? - Page 1
Forum section: electronics / beginners

author: mvno_subscriber
author_rank: Regular Contributor
timestamp: January 11, 2021, 02:44:21 pm
content: I'm trying to understand why, in a transistor, operating it in linear mode produces more heat than in saturated mode.If I envisage it as a voltage controlled resistor, I would assume the heat dissipation would be a function of how much is current passed through it. But this doesn't seem to be the case. Additionally, the fixed voltage drop of course makes it behave very differently from a normal resistor.My googling skills are seemingly horrible; would anyone care to, in hobbyist terms, explain to me how this works? And are there any basic formulas I can use to calculate heat dissipation in a BJT transistor?

author: SVFeingold
author_rank: Frequent Contributor
timestamp: January 11, 2021, 02:59:30 pm
content: P=VI works just the same in a BJT. In saturation the voltage drop across the transistor is about as low as it's going to get. The heat dissipationisa function of how much current is passed through it. What makes you believe that it isn't? The transistor is more or less operating as a variable (current controlled) resistor in order to maintain a given output voltage or current, whichever is being regulated.As for basic formulas, P=VI pretty much tells you what you need to know. It's not so straightforward to calculate as something like a plain resistor, but the idea is the same. If you're using it in something like a linear regulator then you have some idea of your input/output voltages. That difference multiplied by the current at any given time is going to be the power that the transistor needs to dissipate.

author: drussell
author_rank: Super Contributor
timestamp: January 11, 2021, 03:00:30 pm
content: When used like a switch, a transistor will ideally be either "hard on" or "hard off", spending very little time partially on, in the "linear region."When it is fullyon, the effective resistance of the transistor is very low, so thetransistor itselfwill be dissipating little power. When it is fullyoffit will be dissipating no power since no current is flowing, the effective resistance is infinite.When you turn a transistorpart way on, it can be thought of as having some intermediate effective resistance between the value it would have if it were full on (very low) and full off (very high). Because it is now has some significant effective resistance, there will now be significant power dissipated by the transistor itself, depending on what the net voltage drop is across that effective resistance by whatever current is flowing through it.

author: dferyance
author_rank: Regular Contributor
timestamp: January 11, 2021, 03:02:11 pm
content: It is a function of how much current is passed through along with resistance. When saturated the resistance is very low. P = I2R. So assuming current is the same, it is all about resistance. But what if current isn't the same? Well yeah, a short circuit can get quite hot through the transistor. So it absolutely matters what the rest of the circuit is like, what kind of load.To try a car analogy. Breaks turned off, no heat. Breaks fully locked, well heat in the tires but not in the breaks. Breaks partially on, lots of heat.I'm sure someone else here has a much better physics-based answer, but this is how I understand it in my head.

author: langwadt
author_rank: Super Contributor
timestamp: January 11, 2021, 03:03:51 pm
content: Quote from: mvno_subscriber on January 11, 2021, 02:44:21 pmP = U*I

author: Peabody
author_rank: Super Contributor
timestamp: January 11, 2021, 03:26:36 pm
content: Dissipation is equal to the current times the voltage drop across the transistor, not just the current. When the transistor is off, the voltage drop is at a maximum but there's no current. When it's fully on, there's lots of current, but almost no voltage drop. When it's partially on, that's when things heat up.

author: rstofer
author_rank: Super Contributor
timestamp: January 11, 2021, 08:10:07 pm
content: Look at VCESat in the datasheet. It will usually be about 0.2V. That is the minimum voltage drop when the device is in saturation. Now you know the voltage drop across the transistor, all you need to do is multiply by the current and you have the power being dissipated. Let's say it's 1A so we have 0.2 Watts.Now let's assume a Vcc of 12V and the transistor in the linear region so there is, say, 6V across the transistor (and 6V across the load). If we were still dealing with 1A, the power dissipation would be 6W or about 30 times higher.The first example is a switch while the second example might be a Class A Amplifier.Where things go badly wrong is with linear power supplies and high transformer voltages. Assume 24VAC from a transformer and about 33V DC before the regulator (we're talking about the iconic 30V 3A PS). Then assume the PS output is 30V so that only 3V is dropped across the pass transistor. For a 3A load, this is just 9W. But change the output voltage to 1V and 32V is dropped across the pass transistor. For a 3A load we now have to dissipate 96W. That's insane! What happens in real linear power supplies is that the transformer has multiple secondary taps so the pre-regulator voltage can be dropped as the output voltage is adjusted downward. The funny part is, the lower the output voltage, the higher the internal dissipation!Just some fun with Ohm's Law...

author: mvno_subscriber
author_rank: Regular Contributor
timestamp: January 12, 2021, 09:05:37 pm
content: Ok, so the resistive element of interest would be the voltage drop between collector and emitter, times current flowing through. So I have a look at a random NPN, 2N3904:https://datasheet.octopart.com/2N3904%28FSC%29-Fairchild-datasheet-7285131.pdfThe VceSat seems to indicate that the voltage drop goes down until about 10mA, after which it increases almost exponentially. What does this mean, then? I assume it's not going nuclear at 100mA. It's supposed to handle 200mA collector current. How should I read this?

author: Zero999
author_rank: Super Contributor
timestamp: January 12, 2021, 09:59:11 pm
content: Quote from: mvno_subscriber on January 12, 2021, 09:05:37 pmThe saturation voltage is only 0.15V, at 100mA, giving a collector power dissipation of only 15mW, even if it's 10 times that, at 200mA, which is unlikely, the collector power dissipation will only be 300mW, hardly nuclear.Note that 200mA is absolute maximum rating. It's not a good idea to run it continuously at that current. The fact that none of the specifications in the data sheet are given for collector currents above 100mA, implies it's a bad idea to use that part for more than 100mA.

author: radiolistener
author_rank: Super Contributor
timestamp: January 13, 2021, 06:55:42 am
content: Quote from: mvno_subscriber on January 11, 2021, 02:44:21 pmYou can imagine linear PSU as voltage divider, where R1 is a transistor in the linear PSU and R2 is a load resistance:Linear regulator set R1 value in such way to keep required voltage on R2.As you can see, heat dissipation will depends on the current and R1:I = Uinput/ (1/(1/R1+1/R2))P = UR1* IUR1= I * R1P = I^2 * R1As result, high current through a load leads to a high heat dissipation power on the transistor regulator.SMPS works like impedance transformer by using PWM with L/C or transformer components. Transistor on SMPS has R1 for very short period of time (during each transition ON/OFF), this is why it is more efficient. But as a side effect it produce a lot of noise, because it works in pulse mode at variable frequency. Each pulse consists a lot of harmonics and it may lead to an issue for sensitive analog circuits and equipment.