# How to calculate output collector power dissipation (optoisolator)?

#### Raymond Genovese

Joined Mar 5, 2016
1,653
This came from another thread and I am trying to make sense of it...can anyone please explain?

Given the circuit below:

Assume that the relay coil needs 40 mA @ 5V to energize.
The data sheet for the 6n139 is here https://www.mouser.com/ds/2/239/6N138S-TA1-1149909.pdf

The data sheet lists High output current – 60mA (p1) and a MAX average output current of 50mA.
Based on those characteristics, my original thinking was that the circuit will work - the NPN can energize the coil. [There is about 3.3 mA to the LED and the chip has a CTR above 200].

Then, I looked at this characteristic:
Output Collector Power Dissipation Po 100 mW MAX (p 8), and I thought, no, this will not work because 5V X 40 mA=200 mW.

Apparently I am quite wrong - I don't mind that so much, but I really don't like not being able to calculate the Po - how is it done, specifically, given these components? That is my question.

"Power dissipation: When a transistor conducts current between collector and emitter, it also drops voltage between those two points. At any given time, the power dissipated by a transistor is equal to the product (multiplication) of collector current and collector-emitter voltage. Just like resistors, transistors are rated for how many watts each can safely dissipate without sustaining damage. High temperature is the mortal enemy of all semiconductor devices, and bipolar transistors tend to be more susceptible to thermal damage than most. Power ratings are always referenced to the temperature of ambient (surrounding) air. When transistors are to be used in hotter environments (>25o, their power ratings must be derated to avoid a shortened service life."

I guess, in my mind, I was think that drop is going to be 5V. How do I calculate that drop? What measures are available in the data sheet to make the calculation?

#### Danko

Joined Nov 22, 2017
1,835
For IF=12mA; Vcc=4.5V; Io=24mA they guarantee C-E voltage drop (Logic low output voltage) from 0.2V to 0.4V.
In datasheet, on Fig.3, DC transfer characteristics, we can see for IF =3.3mA and Io=40mA, drop voltage about 0.47V.

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#### crutschow

Joined Mar 14, 2008
34,453
Saying the transistor dissipation is the transistor current times the supply voltage is only true if the load is zero ohms.

When a transistor (such as the opto output) switches a load, the transistor collector-emitter voltage is generally low (if the base current is sufficient to saturate the transistor). This is shown in the data date sheet as the on voltage or saturation voltage as Danko noted.
This means most of the voltage drop (and thus dissipation) occurs in the load.
The transistor dissipation is then just it's saturation voltage times the load current.

#### ebeowulf17

Joined Aug 12, 2014
3,307
For anyone not clicking through to the datasheet, here are the relevant snippets. I meant to answer this question after work, but others beat me to it!

#### ebp

Joined Feb 8, 2018
2,332
A caution in reading graphs in data sheets:

Graphs will usually show "typical" characteristics, which are sometimes quite different from the worst-case characteristics. The data sheet in question is quite complete in that it has numerous curves for each parameter, which is very helpful. but you may still need to study the details til your eyes bleed in critical applications. One example: Figure 4 shows the CTR peaks at about 2200, yet the tabular data says 2600 max. You have to take note of the test conditions. If you are lucky, you'll find all the answers in there somewhere, but not always. Data sheets are notorious for leaving out some detail that you just happen to be concerned with (e.g. semi-precision resistors might be specified with a temperature coefficient of resistance of ±25 ppm/°C - Okaaay, but if I have two 10 k resistors, how close can I expect their individual tempcos to be? An unhappy answer to that question may lead you to buying resistors that cost ten times or fifty times as much - see Vishay bulk metal foil resistors for some price sticker shock)

Different manufacturers may produce data sheets that are very different in detail for the same part number. I saw "Lite-on" and my first thought was "Their data sheets are awful." Not true at all here, it's probably about as good as you'll find. I think the original part came from Hewlett-Packard, but it might have been General Instrument Opto.

Optocouplers in general are pretty broad-tolerance beasts. A few types are reasonably tightly specified and you may have a choice of CTRs for the same basic part number with different suffixes. For lots of things, as long as the minimum CTR meets your requirements all is well. In something like a switch mode power supply where the optocoupler is part of the feedback path, variations in CTR mean variations in the gain of the feedback path, which has implications for the behavior of the whole closed loop.

One last note: Optocouplers made for high speed will have characteristics different from lower speed parts, other than just speed. They may be every bit as good for relay driving, just cost a good deal more. They may be inferior for higher currents because physically small transistors with low current ratings are typically faster than big ones.

One more last note: You won't get output saturation voltage of less than half a volt if you connect the output as a simple darlington. In that case the output "saturation" voltage (it in fact can't saturate) will likely be closer to a full volt. To get the low "true" saturation voltage you must take advantage of the split transistor design.

#### hobbyist

Joined Aug 10, 2008
892
Apparently I am quite wrong - I don't mind that so much, but I really don't like not being able to calculate the Po - how is it done, specifically, given these components? That is my question.
In general:
Find on the data sheet the Pmax. then using your current value you calculated to drive the load, divide that current value into the Po.max, value on the data sheet, to determine the maximum voltage allowed across the transistor, then design your circuit to have a voltage (VCE) to be much lower than that, to ensure not burning out the transistor.

In specific:
as in the case of your application, you already know you want to drive the transistor into its saturation region if possible with the load.
To use the opto as a switch, so your looking to drive the transistor into the millivolt region.

#### Raymond Genovese

Joined Mar 5, 2016
1,653
Thanks all for the valuable and very informative information. I suppose that I should have been able to figure this out on my own, but I appreciate the assistance - and most importantly, I think I understand.

Except for a few instances when going straight from a schematic, I have only used these as switches. All of what has been said reiterates what I have read a many times - when using an NPN as a switch, make sure it is in saturation mode. Now, it makes clearer sense.

Thanks all, again.

#### Danko

Joined Nov 22, 2017
1,835
One more last note: You won't get output saturation voltage of less than half a volt if you connect the output as a simple darlington. In that case the output "saturation" voltage (it in fact can't saturate) will likely be closer to a full volt. To get the low "true" saturation voltage you must take advantage of the split transistor design.
Recommended in datasheet circuit provides low saturation voltage:

#### ebp

Joined Feb 8, 2018
2,332
Thanks for pointing that out Danko. I didn't even notice that in the data sheet as I whizzed by looking for the tables and graphs.

Raymond, you don't necessarily want to saturate a bipolar transistor used as a switch - it depends on what you're doing (electronics would be soooo much simpler if the phrase "it depends on ..." didn't come into it so often.) Usually in a common-emitter switch it will saturate unless you take quite elaborate steps to prevent saturation (look up "Baker clamp"). Saturation actually significantly slows turn-off speed (totally irrelevant for switching a relay; big issue when bipolars were common in switch mode power supplies). A saturated transistor will have a lot of excess carriers in the base region and to turn it off you have to get the carriers out of there. Darlington transistors are slow to turn off not because they saturate but because there is no way to actively remove carriers from the base of the "output" transistor - you just have to wait until they recombine "naturally". Emitter-coupled logic, which at one time was used extensively in mainframe computers, was built with bipolar transistors that didn't saturate because of the configuration, which made it much faster than logic families like TTL where the transistors saturate. Schottky TTL is faster than regular TTL because the transistors aren't saturated as much (think of them as just damp, not saturated - actually don't, I made that up ). If you overdrive an op-amp it often results in some of the transistors saturating and it takes a long time, relative to the normal speed of the amp, to recover. Data sheets for high-speed amps will sometimes have numerical and graphical info on overdrive recovery.

The free simulation packages that are around are a great way to play with some of this stuff, as long as the models have sufficient detail. Doing it on the bench with real parts can be quite tricky.

#### crutschow

Joined Mar 14, 2008
34,453
To illustrate epb's comment about the effect of saturation on switching speed, here's an LTspice simulation showing the large difference in saturation turn-off delay between the common 2N3904 general purpose transistor, and the 2N2369 high-speed switching transistor, for a base current of 1/10th of the collector current.
The 2N2369 has specific doping (I believe it's gold) to cause the excess carriers to rapidly recombine and essentially eliminate the recovery delay.

The 2N2369 has a delay of only about 30ns (not really visible on the plot), whereas the 2N3904 has a delay of about 3.3μs (over a 100 times longer).
As expected there is not significant difference in the turn-on delay.

#### ebp

Joined Feb 8, 2018
2,332
Oooh, spiffy, crutschow! Thanks. Are the transistor models you're using something others can get free or cheaply?
Without putting a lot of effort into it, can you show us the 3904 with a bit of negative swing on the base, or with a Baker clamp or anything else you can think of that either limits saturation or sucks carriers out.

People who are learning to use simulation who happen to be wandering by: Note how crutschow names nets (signals) in the schematic and how those names appear on the plot. This makes the plots a lot easier to follow than the default of Nxxxx. Also note that he has selected the X and Y factors really well to show us everything with good resolution, instead of showing us 10 identical cycles or using vertical scales twice the signal amplitudes. People post sim plots with nets unnamed and bad choices for axes, and it makes it hard for us to see what's happening.
Those who aren't wandering by are free to leave their nets unnamed and run for any time that suits their fancy. (It always make me chuckle when I'm in a store and there is an announcement "If so-and-so is still in the store would you please return to ...")

Gold doping sounds reasonable to me, but I really don't know much about that stuff. Gold doping is certainly used in common switching diodes. It makes them fast with the penalty of higher reverse leakage current. I've used 2N3904s as low-leakage diodes (B-C junction). They are slower than1N4148s but much cheaper than most low leakage diodes. plus it eliminated a line from the BOM, since they were used elsewhere on the board.

There is an AAC member who is something of an electronics celebrity and could probably teach us a good deal about the stuff.
Ron M, are you out there?

#### Raymond Genovese

Joined Mar 5, 2016
1,653
Thanks for pointing that out Danko. I didn't even notice that in the data sheet as I whizzed by looking for the tables and graphs.

Raymond, you don't necessarily want to saturate a bipolar transistor used as a switch - it depends on what you're doing (electronics would be soooo much simpler if the phrase "it depends on ..." didn't come into it so often.) Usually in a common-emitter switch it will saturate unless you take quite elaborate steps to prevent saturation (look up "Baker clamp"). Saturation actually significantly slows turn-off speed (totally irrelevant for switching a relay; big issue when bipolars were common in switch mode power supplies). A saturated transistor will have a lot of excess carriers in the base region and to turn it off you have to get the carriers out of there. Darlington transistors are slow to turn off not because they saturate but because there is no way to actively remove carriers from the base of the "output" transistor - you just have to wait until they recombine "naturally". Emitter-coupled logic, which at one time was used extensively in mainframe computers, was built with bipolar transistors that didn't saturate because of the configuration, which made it much faster than logic families like TTL where the transistors saturate. Schottky TTL is faster than regular TTL because the transistors aren't saturated as much (think of them as just damp, not saturated - actually don't, I made that up ). If you overdrive an op-amp it often results in some of the transistors saturating and it takes a long time, relative to the normal speed of the amp, to recover. Data sheets for high-speed amps will sometimes have numerical and graphical info on overdrive recovery.

The free simulation packages that are around are a great way to play with some of this stuff, as long as the models have sufficient detail. Doing it on the bench with real parts can be quite tricky.
Oh Great, now I have to learn LTspice Thanks much for all the info - have bookmarked the thread.

In the case of the circuit I originally posted - my understanding is that I do want the transistors saturated...and, as I have drawn it, that is what is expected to happen - right?

#### hobbyist

Joined Aug 10, 2008
892
In the case of the circuit I originally posted - my understanding is that I do want the transistors saturated...and, as I have drawn it, that is what is expected to happen - right?
In your circuit yes, you are looking to use your transistor as a switch, however can saturation be expected, not exactly.

From my experiance, of experimenting with transistor switches, I found, there are two things needed to put the transistor into saturation.
emitter more neg. than the base, and the collector approaching, neg. with respect to the base, as both diodes are approaching forward biased,.

The base current needs to be substantial, around 1/10th of collector current, and the collector load needs to have a high enough resistance to cause a large voltage drop across it, so as to reduce the VCE, to around 200mV range.

Is the relay coil high enough resistance in the collector load, I don't know.

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#### ebp

Joined Feb 8, 2018
2,332
You don't strictly need saturation here, but it is a reasonable objective to saturate the transistor. Let's just ignore the issue with a Darlington for the moment.

The speed issue is totally irrelevant when driving a relay, so there is no advantage to avoiding saturation on that account. Saturation does reduce the power dissipation in the transistor somewhat. Though that isn't really an issue here either, you might just as well gain the benefit. I think the most compelling reason to drive the transistor well into saturation is the whole issue of variation in CTR of the coupler. If you drove the input with just enough current to get the transistor turn on "gently," degradation over time or change with temperature could lead to having significant voltage across the transistor when it is intended to be ON. If you select a coupler that requires that you drive the input to close to its maximum allowable current, it would usually be wise to select a different coupler an/or add an external device at the output for more current gain. One of the great things about optocouplers with transistor outputs is that the are "floating" so you can used them as either "high side" or "low side" drivers. Optocoupled triac drivers a great boon for a class of circuit we're not supposed to talk about here.

You can't saturate the output transistor in a Darlington because its base drive comes from its own collector. It is sort of a negative feedback effect - as the collector voltage falls the base begins to be starved of current, preventing saturation. With that particular coupler, you can take the collector of the phototransistor to V+ via a resistor and in that way drive the output truly to saturation. Most couplers with darlington outputs don't give you that option. If anything other than the collector and emitter are pinned-out it is normally the base.

Getting simple circuits up and running with packages like LTSpice is pretty easy. I have never used it but I have used the TI equivalent. I could put together and run simple circuits without even resorting to the documentation. There are hooks. One notable one is that m means milli whereas M or Meg is used for ... But what's 9 orders of magnitude between friends? I'm sure several of the people at AAC would send you some files that would help get you started.