## How to pick and size a transistor for ON/OFF use...

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Hi,

I've been searching and reading what I can find on the Internet, and I've come up with:

The first page talks about using the hfe to calculate how much base current is required to supply a specific collector-emitter current.

My example is 260mA and 100hfe. According to the first page, you would take 260ma/100hfe and end up with 2.6ma to which you add 30% to come up with 3.38mA.

The second page had an interesting statement:

Quote:
A resistor RB is required to limit the current flowing into the base of the transistor and prevent it being damaged. However, RB must be sufficiently low to ensure that the transistor is thoroughly saturated to prevent it overheating, this is particularly important if the transistor is switching a large current (> 100mA). A safe rule is to make the base current IB about five times larger than the value which should just saturate the transistor.

They are saying that it needs to be oversized by a factor of 5 to prevent it from overheating. I don't really understand this because as long as the Ic is below the transistor max, why would it overheat from the base current not being large enough? Wouldn't the base current not being large enough just cause a lesser Ic ?

Then, I've read that a factor of 10 should be used for saturation...

Help me out here guys!!!

Thanks,

Alan

Hi,

Another question:

Take these two:

2N3904FS-ND
200mA Ic max

2N4401-ND
600mA Ic max

They are nearly the same price. Suppose you had a need for 3 of them. 30ma, 40ma, and 260ma. Why not use the 600mA Ic max model for all three? What downside is there?

Thanks,

Alan

alank2 wrote:
Wouldn't the base current not being large enough just cause a lesser Ic ?

Absolutely, the Ic would be less as you'd expect. But think about what's happening to the circuit at that point - you still have the same voltage across the transistor.. it's now causing a restriction to current flow (like a resistor! resistors create heat!). Your partially-biassed transistor is now operating in "active" mode and will dissipate heat. This will kill it if the heat byproduct isn't intentional.

without looking at the datasheets for your second question - there's no problem using a transistor with higher rated Ic everything else being equal. You may have a larger package (in which case, fitting the part may be an issue) and it may have lower gain meaning you need to pump more base current in - and your driver circuit needs to be able to do this.

Hi Geoff,

Thanks; I finally got this with some more reading tonight. Dropping small voltage across C and E = saturation and much less heat.

It seems like calculating the base resistor for satuation is an exact science. I will measure the C E voltage and see if I have enough base current.

Thanks,

Alan

alank2 wrote:
It seems like calculating the base resistor for satuation is an exact science. I will measure the C E voltage and see if I have enough base current.

I've never thought of it that way. Double the base current based on minimum hfe and you'll be fine. Note that hfe changes with Ic, so you'll need to get the value for your Ic, at least roughly.

The 2N4401 is a good general-purpose transistor in its size range. The 2N2222 (MPS2222 in plastic, I think) is about the same. Buy 'em in bags of 100.

The often accepted standard for "saturation" of a bipolar transistor is the point at which the beta drops to 10.

Thus, for hypothetical 260mA of collector current, you would need a full 26mA of base current to guarantee saturation (provided the collector load allows the Vce to drop to near-zero).

This standard for saturation holds from milliamperes of collector current to at least 10s of Amps of collector current.

Picking a transistor is not that simple. For max currents below 100mA or so, you can rely on the spec sheet absolute maximum collector current. This is usually defined by the physical current carrying capacity of the bond-wires and the die. In other words, what the point is where it behaves like a fuse.

At higher currents, you will find many transistors with a graph labeled something like "Safe Operating Area". That shows you the allowable combinations of collector current and voltage.

Jim

Until Black Lives Matter, we do not have "All Lives Matter"!

alank2 wrote:
It seems like calculating the base resistor for satuation is an exact science.
No, not in practice. Manufacturing tolerances for transistors are so high, that rules of thumb (like divide by five or ten to be on the safe side) are good enough.

Stealing Proteus doesn't make you an engineer.

Hi,

I did some tests last night.

BC517 darlington npn. 5V power source to parallel LED's to a very small 0.33 ohm resistor to the BC517 collector. Base current is 5V/1000ohms which is 5ma. I measured from C to E and expected around 1.4V but instead got 0.8V. Current is about 275ma.

I then looked at another thing I'm running off a BC517. A 20ma single LED. C to E voltage is 0.7V.

Since these are darlingtons shouldn't the C to E voltage be around 1.4V ???

Also, I tried a non darlington in place (labeled 842 MPS A06) on the 20mA single LED and it the C to E went from 0.7V to 0.04V.

Am I measuring or understanding wrong? Shouldn't it be 0.7V or 1.4V depending on type?

Thanks,

Alan

The base-emitter voltage, not collector-emitter, is about 1.4 for a Darlington. The collector-emitter path is still through one transistor.

Hi,

What significance does the base-emitter voltage have?

Thanks,

Alan

A nice feature of the 2N4401 is the way the gain holds up at high current levels.

Leon Heller G1HSM

Hi,

In another thread, someone was telling me that the BC517 Darlington I was using was taking some voltage away from the device I was switching. In my quick test I can see a difference of 0.7 vs 0.04. Is this because of the darlington pair?

Thanks,

Alan

Back on Darlington voltages -

Darlington transistors do not work well in saturation. When the primary current carrying transistor is saturated, the collector voltage of the input transistor is lower than its emitter, and it no longer provides the desirable gain. In that mode, the input transistor is not much more than a diode in series with the base of the second transistor. That is one of the reasons why it is so hard to get a Darlington into genuine saturation. The current gain of the input transistor helps, nicely, until the collector voltage gets down to about 1V, give or take a little. Then, it turns into mush.

Yes, in your post just preceding. That is because of the Darlington pair, for the reason I just described. I would not describe it, however, as "taking voltage away from the device I was switching". Instead, I would say that the (effective) on-resistance simply is not very low. It does, however, provide less voltage drop across the load.

Jim

Until Black Lives Matter, we do not have "All Lives Matter"!

jayjay1974 wrote:
The collector-emitter path is still through one transistor.

Is it? Base of output transistor is emitter of input transistor, and both collectors are connected together.

Therefore, darlington Vce is the Vce of the input transistor plus Vbe of the output transistor.

edit: Alan most basic electronics books or even Wikipedia should have an article about transistors and darlingtons. Then, there is Wikibooks about electronics.

http://en.wikipedia.org/wiki/Darlington_transistor

Hi Jim,

Thanks for the post!

Given that a uC pin can drive as much as 20mA, how do you decide what type of transistor to use based on a need to switch on or off loads requiring current such as:

30ma
300ma
3A
30A

Is there a point up to where a simple BJT isn't typically offered in enough gain and you have to move to a Darlington? Or something else? At some point is it better to move to relay?

Thanks,

Alan

Hi,

Jepael wrote:
edit: Alan most basic electronics books or even Wikipedia should have an article about transistors and darlingtons. Then, there is Wikibooks about electronics.

http://en.wikipedia.org/wiki/Darlington_transistor

Yes, I've been looking at them; just not understanding everything yet...

I saw this at that page:

Quote:

Another drawback of the Darlington pair is its increased saturation voltage. The output transistor is not allowed to saturate (i.e. its base-collector junction must remain reverse-biased) because its collector-emitter voltage is now equal to the sum of its own base-emitter voltage and the collector-emitter voltage of the first transistor, both positive quantities in normal operation. (In symbols, VCE2 = VBE2 + VCE1, so VC2 > VB2 always.) Thus the saturation voltage of a Darlington transistor is one VBE (about 0.65 V in silicon) higher than a single transistor saturation voltage, which is typically 0.1 - 0.2 V in silicon.

Are they talking about the VCE here in the last sentence?

Thanks,

Alan

Hi,

Take this for example:

http://www.fairchildsemi.com/ds/...

VCBO Collector-Base Voltage 45 V
VCEO Collector-Emitter Voltage 30 V
VEBO Emitter-Base Voltage 5 V

Does this mean the voltage between base and emitter must not exceed 5V, or the REVERSE voltage must not exceed 5V ? In my case I would have the emitter going to ground and the base switched by the uC.

Thanks,

Alan

Yes they are talking about VCE voltage.

There is no absolute best way of having 20mA of current at 5V (or 3.3V) available to drive currents you listed (30mA, 300mA, 3A, 30A).

There are also FETs, relays, contactors, optoisolators, solid-state relays, IGBTs and who knows what.

alank2 wrote:

VCBO Collector-Base Voltage 45 V
VCEO Collector-Emitter Voltage 30 V
VEBO Emitter-Base Voltage 5 V

Does this mean the voltage between base and emitter must not exceed 5V, or the REVERSE voltage must not exceed 5V ? In my case I would have the emitter going to ground and the base switched by the uC.

VEBO lists emitter first so it is VE-VB, meaning reverse voltage. And if you think the other way around, a forward VBE voltage of 5V is impossible, so is anything over about 0.7V for regular transistors.

Hi,

Jepael wrote:
There is no absolute best way of having 20mA of current at 5V (or 3.3V) available to drive currents you listed (30mA, 300mA, 3A, 30A).

I'm not looking for a single method to drive all these currents, it was more of a question of IF you had to drive these currents, what would you use in each case?

Thanks,

Alan

The "O" means these figures apply when the third pin (the one not named) is left open.

If these are listed in a "Maximum" rating section it indeed means the values aren't to be exceeded. In fact it means to stay well away from them.

Stealing Proteus doesn't make you an engineer.

alank2 wrote:

I'm not looking for a single method to drive all these currents, it was more of a question of IF you had to drive these currents, what would you use in each case?

Sorry I kind of meant that the driving method cannot be selected just based on certain current, as there are other important factors too.

It also depends if the load is inductive, resistive or capacitive, involves mains connections, is DC or AC, or if there is a need to switch the load on/off 100 times per year or 100 times per millisecond.

Hi,

ArnoldB wrote:
The "O" means these figures apply when the third pin (the one not named) is left open.

If these are listed in a "Maximum" rating section it indeed means the values aren't to be exceeded. In fact it means to stay well away from them.

Thanks for indicating what the O means. How do you know the direction that limit applies to? Or is it both directions because it isn't specified?

Thanks,

Alan

Hi,

http://www.softwareforeducation....

Says that when a mosfet: When fully turned on (saturated), the potential drop across the device is about 2 volts. (between the source and drain).

Is this a particular type of mosfet that would have such a large voltage between the drain and source, or is this typical of mosfets?

Thanks,

Alan

The hfe or beta of a transistor is equal to IC/IB.If a transistor needs to operate as an on/off switch must operate saturated.This means that the VCE is the minimum and IC=V/RLoad.IB=IC/hfe and the RB
is about RB=Von-0.7V/IB since beta is not exactly the same even with the same part numbers of transistors and theres also a small leakage current flowing from collector to base.

alank2 wrote:
How do you know the direction that limit applies to?
From the order of the letters.

VCBO = Collector -> Base with open Emitter.

alank2 wrote:
Or is it both directions because it isn't specified?
No, it is the indicated direction. A BJT is not symmetric. However, the reverse directions are typically not specified, because they aren't used in normal operation. Reversing the polarity of a BJT is rarely done, because its performance gets rather bad. In fact, I have never seen a circuit where a BJT was reversed by intention. I have just heard they exist.

Stealing Proteus doesn't make you an engineer.

alank2 wrote:

http://www.softwareforeducation....

Says that when a mosfet: When fully turned on (saturated), the potential drop across the device is about 2 volts. (between the source and drain).

Is this a particular type of mosfet that would have such a large voltage between the drain and source, or is this typical of mosfets?

That's a very misleading claim. The voltage drop across the device depends on the device, on Ids and on Vgs. Here's a cheap MOSFET that, driven from a 5V AVR output, would drop about 0.2V at 10A:

Why did you use the BC517 to drive a 20mA single LED? Simplily, using the 9013 NPN transistor to drive the 20mA LED is enough.

cheng_bei

Nobody can go back and start a new beginning, but anyone can start today and make a new ending.

Hi,
This is how I do my calculations.
If I was to drive a relay from a 5V micro where the relay was a 12V relay with a 50mA pulled in current.
I would use a BC547 (I have plenty), it has a max collector current of 100mA and a minimum hfe of 100.
The calculations I would use are -
IB (base current) is 50mA/hfe = is 0.5mA
RB (Base resistor) is (5V(Logic High)-0.6(VBE))/0.5mA
= a Base resistor of 9400ohm i.e. 10K
But to make sure the transistor acts as a switch (driven into saturation) and has minimal heat dissipation, I would double the base current i.e. use a 4K7 resistor.
Obviously other components may also be required depending on the type of load.
I have used this very simple calculation for twenty years doing all sorts of load interfacing from TTL, CMOS, Micro's to LED's, lamps, relays etc.
Some people may say that you do not get a full 5V swing from an output but the above works beautifully.
Higher currents, I use a BC337.

Neil.

Kind Regards,

Neil Wrightson.

Alan,

There are a couple of other things I "could" do when selecting a suitable transistor for a switched load. After doing it several times I get a hang for what works ... but it would be good practice for you until you understand what you are doing.

My starting point is the load current. For example if it is 1 amp, a Vcesat (the voltage between the collector and the emitter when the transistor is saturated) of say 0.6 volts will produce a 0.6 x 1 = 0.6 watt power dissipation in the transistor. Without heat sinking, that would rule out a lot of common transistors, especially if the circuit is likely to operate in a hot environment. So do a rough calculation of the likely heat load on your ideal transistor candidates.

Secondly, realise that the Vcesat varies with load current and temperature. So read the datasheet and find the curves that show variation of Vcesat versus collector current and versus temperature.

Thirdly, find the curve that shows how the hfe gain varies with collector current and see what it is at your load current. Then you can calculate your base current.

Have fun.

Cheers,

Ross

Ross McKenzie, Melbourne Australia

As Ross points out, you want to figure out the thermal aspects as this can ruin your day if you ignore or get it wrong. Whilst datasheet can tell you the device can switch 50A, that doesn't give you a get out of jail card against the basic laws of nature. something to note is that mosfets have a RDSon spec - on resistance whereas a transistor will have a drop of around 0.6V. Teensy mosfets ( or high voltage ones) might have a RDSon of a few ohms, whereas more beefy ones have specs in terms of tenths or hundredths of ohms. It is an art to read datasheets ( it's an art to write them!).

Whenever possible, I tend to use MOSFETs instead of transistors, mainly due the heat dissipated at high current loads. At 10A, 0.6V VCE sat gives good 6W of heat dissipation. A beefy mosfet can have 0.01Ohms or less, that at 10A means 0.1V, or about 1W.

The good point about MOSFET's is that you activate them by voltage, not by current, so you need less current output of your control device (unless you are switching them really fast...). For LED's and such I use FDC303/304/6303/6304 or similar. But each case may differ from the previous, and there is not a general solution. Even, under certain circumstances, it could be much cheaper and finer solution to use something like ULN2803.

Take this information with a grain of salt. I tend to do my designs in a hurry and avoid calculations, so I usually do Ctrl+C, Ctrl+V from proven designs.

Guillem.
"Common sense is the least common of the senses" Anonymous.

Hi,

Thanks for all the excellent posts guys!!

Ross: So, the graph you are referring to is the one labeled:

Typical Pulsed Current Gain vs Collector Current

in this doc:

http://www.fairchildsemi.com/ds/...

It does say pulsed, but I don't see a non-pulsed chart to look at.

Thanks!!

Alan

Hi,

280mA
24mA
21mA

To keep from having a bunch of different parts, does it make sense to use the PN2222A for all 3?

Looking at the chart, it looks like hFe would be around 90 at 280mA. So if I use the above formulas:

280/90=3.11mA base current
(5-0.7)/0.00311=1382 ohms
half of that=691 ohms

So, 680 ohm.

Assuming I build a test circuit with this and check the Vce and it is low and all works well, my question is:

Would there be any downside to using the same base drive/transistor for the other loads even though they are less (21ma, 24ma)? This would allow me to just get a 6sip isolated resistor array with 3 680 ohm resistors...

Thanks,

Alan

Your thinking is good, but you've used the average hfe rather than the minimum. From the datasheet you posted, the minimum would be 40-50 (40@500mA, 50@150mA). So divide your resistor value by 2. Or, if you're just doing one or a few, you could just test the transistors beforehand and make sure you don't use any on the low end of the gain scale. Some DMMs will give you an hfe reading directly.

No harm in driving the other transistors with the same resistor value, other than wasting a few mA.

Alan,

As Kartman said, writing and reading datasheets is an art ... which the manufacturers use to their advantage to get you to buy their product. Caveat emptor!

The other thing to note from this datasheet is how the Vcesat changes from 0.3 volts to 1.0 volts. This is the maximum that the maker states. So for heat dissipation calculations work with maximum Vcesat figuress; for gain use the minimum hFE figures.

Sorry for the "longie"

Cheers,

Ross

## Attachment(s):

Ross McKenzie, Melbourne Australia

The reason they switch to pulsed testing at the higher currents is just because with Vce of 10V they would be burning out the transistor otherwise. But it is sneaky, since the hfe at lower Vce will be lower. The transistor cannot be operated at 500mA and 10Vce, so why test it there?

I'm confused. How can be Vce=10V if Collector-Emitter Saturation Voltage (to my limited understanding, Vce) is 1.0V?

Guillem.
"Common sense is the least common of the senses" Anonymous.

Guillem,

The topic became a little confused, but let me say this ...

If the transistor is being used as an amplifier and operating in its linear range, we could easily have a voltage from collector to emitter of 10 volts. Under those circumstances, the designer would still want to know what the hFE would be ... hence the specification. And if the designer wanted to have a collector current of 500 mA under that condition, clearly the life of the PN2222 transistor would be very brief because of the 5 watts dissipation.

However, the OP was interested in an "ON/OFF" operation and therefore really only interested in the saturated mode of operation ... and hence the Vcesat discussion.

But of course you already knew this :lol:

Cheers,

Ross

ps How is your little ear ring wearing daughter progressing? Precious times ...

Ross McKenzie, Melbourne Australia

@Ross:
:twisted: Yes, I was aware about many of the issues regarding Vce and such, although it has been a long time since the last time I had to calculate this values. I was refering mostly to the two lowest lines of the image you had posted just before, and only to point out the relative 'obscure' values that one can find in datasheets. As had been written before, reading and writing datasheets is an art. They can be soo confusing.

After some time, I tend to read only some values and try them under the intended conditions. No simulations, no complex calculations. There are many real variables that can affect the whole performance (PCB design and proper layout, capabilities of the whole design for heat removal, air flow, heatsinks, real curent and voltage conditions, wiring, etc). My everyday experience showed me by the hard way that the 'worst conditions' at lab environment is far much better than the best real life environment.

BTW, really precious times. The best experience I ever had/have. :D Not too much time for other things, but it doesn't matter, I enjoy each minute I share with her.

Guillem.
"Common sense is the least common of the senses" Anonymous.

Hi,

Thanks everyone for the great posts and information. I was using a 1K base drive with a BC517 darlington for my three switched loads - 270ma, 25ma, and 22ma. I didn't realize the darlington was taking so much of the voltage from the component I was trying to drive. The LED and Buzzer were probably not affected too much, but my LCD backlight already has a very high forward voltage around 4.2V which required the use of a pricier and very small 0.33 ohm resistor to control the current at 270ma. This did work, but my biggest gripe was how sensitive it was to voltage drops from 5V. There was no room for error and even a drop to 4.95V would cause a major change in backlight brightness and current.

After evaluating the PN2222A which also happens to come in a part that is already formed at 0.1" spacing, I have changed to it. I am using a 330 ohm base drive which allows 11-12ma of current from the uC pin. My lcd backlight is now controlled by a 2.4 ohm resistor which was much cheaper and it isn't nearly as sensitive to voltage drops as it was before. Vce is 0.2V instead of 0.8V. This has upped the uC pin current from 3.5ma or so to 11ma or so.

I am quite happy with the change and that I now have a better understanding of how to size and pick transistors!

Thanks,

Alan

You would be even more happier with a FET driving the backlight instead of BJT. Much less voltage wasted over the transistor, and virtually no current taken from microcontroller IO port except during switching FET on or off.

So now that you have learnt BJT transistor and BJT darlington operation, take a look at N-channel enhancement mode field-effect transistors.

Hi,

Do you need a base drive resistor with a MOSFET? Or can the uC pin drive the gate directly?

Thanks,

Alan

Is it better to have a resistor. As the gate is basically a capacitor, large currents can flow, so you can for a very short while overload the uC pin driver.

There are also little capacitors between the gate and the drain and source. Quick and large transients of the load can propagate to the gate to the uC pin.

Keep in mind a BJT will have a positive temperature coefficient and may go into thermal runaway.
A fet will have a negative thermal characteristic ( source drain resistance increases with temperature thus limiting drain current) and will tend to be self regulating in terms of thermal runaway.

Hi,

Jepael wrote:
You would be even more happier with a FET driving the backlight instead of BJT. Much less voltage wasted over the transistor, and virtually no current taken from microcontroller IO port except during switching FET on or off.

I know, old thread, but I've got a question:

Let's say you supply 5ma to the base of a BJT and it is in saturation.

Let's say you change the base resistor and now 9ma is flowing through the base and it is still in saturation.

Does the extra 4ma go to the load? Is it wasted in heat?

Thanks,

Alan

One of your questions 2 pages ago asked why stock 3 transistors with 10,100,and 1000ma collector current and I bet the answer is: back in the heyday of transistor design (70s?)they kept making them higher gain, higher current, higher power, etc etc. Notice the 3904 is heftier than the 2222, and the 4401 is heftier than the 3904 (2N numbers keep going up!), so I think if they are all the same price, just get a bunch o 4401s.

Imagecraft compiler user

alank2 wrote:

Let's say you supply 5ma to the base of a BJT and it is in saturation.

Let's say you change the base resistor and now 9ma is flowing through the base and it is still in saturation.

Does the extra 4ma go to the load? Is it wasted in heat?

Thanks,

Alan

That would depend how you have connected the load, doesn't it?

For a simple case of NPN transistor, E to gnd, C to load and B to microcontroller, it is wasted in heat, if 5mA was enough already to get low enough Vce voltage. Extra base current can make Vce smaller though, so you have to calculate if you actually consume less power in total if more power is fed to base. Should be easy calculate, P=U*I, I=U/R etc.

Why don't you use MOSFETS!!!