# Concept of “resistance” and “power dissipation” in a transistor - Explanation?

Ok, so I've got way more experience with mosfets than BJTs. I'd become very accustomed to the idea of mosfets having a specific 'on' resistance, which would allow me to calculate actual heat generation and operating conditions for the mosfet quite easily.

I understand a BJT doesn't have an actual 'on' state. Instead, it has a diminishing gain as you raise base current further and further. What I'm wondering about is how to get an idea of how much resistance there is in a BJT -or- how much it will heat up when I operate it in various conditions. I'm guessing that if I have a resistive load that allows 5 amps (for instance) and I am switching it with a BJT, then biasing the BJT on as high as possible will minimize heat generated in the BJT because you open the channel as wide as possible. But what I find awkward is the fact that the gain approaches zero once you get above a certain 'grey area'. So how do I figure out what is practical? Do I just have to look at the gain curve graphs on the datasheet and pick some reasonable range?

To calculate power dissipation in a bipolar transistor you just need to know the collector-to-emitter voltage when it's passing the current that you want to control. This may be called the saturation $V_{CE}$ in the data sheet, and it will typically be less than 1 V, perhaps as low as 0.3 V (assuming that your providing enough base current for the transistor to be conducting really well). Multiply that voltage by the current being switched and you have your power dissipation.

Just FYI, there is no "channel" in a BJT. Bipolar transistors work by a completely different mechanism than field-effect transistors. Also, the notion of a fixed source-to-drain resistance for an FET is a highly simplified model and is really just an approximation of the true behavior.

• So the lower Vce constitutes lower power dissipation for a given current? – JamesHoux May 26 '13 at 2:08
• Yes, that's how it works. – Joe Hass May 26 '13 at 2:11

The power in any electrical circuit is the product of current and voltage:

$$P = IE$$

This holds for any individual component (resistors, MOSFETs, BJTs) or entire circuits (cell phones, radios, computers). In the case of resistors, or things that look close enough to resistors (like MOSFET channels), then voltage and current are related by a constant (the resistance), so we can calculate one from the other, so we only need to know any two of the voltage, current, and resistance:

$$P = IE = I^2 R = E^2/R$$

For everything else we either need to know both the current and the voltage, or have some other function that allows us to calculate one from the other. However, power is still just the product of current and voltage.

For BJTs, we can usually neglect the power from the base current because it's much smaller than the collector current. If so, then all we really care about is the collector-emitter voltage $V_{ce}$ and the collector current $I_c$.

The datasheet will give the details about how $V_{ce}$ relates to $I_c$. Here's BC547:

You can see that if you hold the base current constant, then $V_{ce}$ is not a linear function of $I_c$, as it would be if the collector-emitter was a resistance. You can also see that if you hold $I_c$ constant, then increasing the base current can make $V_{ce}$ lower, up to a point.

For example, let's say our base current is $50 \mu A$, so we use the lowest line. Say collector current is $10mA$. It looks like those lines intersect at about $0.25V$, so the power is:

$$0.25 V * 10mA = 2.5mW$$