78L05, 78L08, 78L09

[JenniferG]

I ordered 50 each of 78L05, 78L08 and 78L09 from Ali Express  With tax and shipping I think it came to less than 2 cents a piece.

I got them in the mail today and tested 3 of each of the above voltage regulators and here are the results.  Do you think they are decent enough?   I never bought semiconductors from Ali Express before.

I didn't put any smoothing caps on VIN/VOUT as suggested by manufacturer for this test.. I figured it wouldn't matter for this test?  I am not sure though.

[JenniferG]

This is the seller I bought them from. Order total was $4.40 including tax and shipping.  So 3 cents a piece.

https://www.aliexpress.us/item/3256801588953710.html

[wasedadoc]

All those look fine to me.  Suggest you have a look at datasheets to see the tolerances on the output voltages and the minimum input voltages required.

[JenniferG]

All those look fine to me.  Suggest you have a look at datasheets to see the tolerances on the output voltages and the minimum input voltages required.

Yeah I am still learning how to read them.  So new to this. Thanks I'll give a look at it

[MikeK]

The last test of each, though, is of no use since they all require almost 2V of dropout.

[wasedadoc]

The last test of each, though, is of no use since they all require almost 2V of dropout.

That is why I suggested to look at "the minimum input voltages required".

[jwet]

Just a note- the input and output caps aren't "smoothing" caps.  They are more correctly "bypass" caps.  Smoothing caps would be to reduce input ripple from the source, like 60 or 120 Hz.  Both of these caps are sized to tame high frequencies and transients.  They both improve transient response and stability in all but the simplest circuits.

For your DC only test assuming, you used short leads, etc., they aren't strictly required but their function is somewhat subtle and they should be used whenever possible.  This is just a note to better understand their function since you are posting as a beginner.

The purpose of the input cap, usually a .22 uF ceramic mounted very close to the input is designed to keep AC out of the regulator.  The data sheet usually says "required if the regulator is far from the source".  The idea is to bypass any high frequency content that may be on the input to the regulator.  Long input leads make the noise pickup worse.  Long leads also add inductance which raises the impedance of the source at higher frequencies.  Transient voltages on inductive sources create voltage fluctuations that are detrimental.   Keeping high frequency noise out and maintaining a low AC impedance at the input make the regulator's job much easier.  The regulator doesn't have enough bandwidth to get rid of high frequency noise present on its input.  The high frequencies can also get in and contaminate the voltage reference in the regulator which will then show up on the output.  The data sheet show the ac rejection of input noise- its not great past a Khz or so.

The output filter cap, usually .1 uF located close to the output enhances stability with transient loads.  Its basically a local bypass and keeps the output at AC ground even with widely varying loads.  Remember that internal to the regulator is a amplifier that is comparing the output to an internal reference and making corrections.    Bypassing the output enhances the stability of this control loop by providing a low AC impedance load to the output.  You won't see a 780x oscillate but an output cap will noticeably improve the dynamic response for step loads, etc.- overshoots and undershoots will be reduced.  This is a measure of relatively stability and this cap should also always be used.

The chip vendors make these capacitors "optional" to make the circuits more palatable to customers but in all but the very lowest cost or non critical circuits, they should be considered mandatory.

Most low dropout regulators (the 7805 does not fall into this category) absolutely require bypassing and the output bypass cap has to carefully selected to insure basic stability of the regulator.  The output cap is an important compensation component (dominant pole) in the circuit.

Have fun.

[Siwastaja]

Seem to regulate correctly, but they are still 99% likely fakes. Or I don't know if "fake" is a correct term, 78xx is manufactured by many manufacturers and only if you call it "LM7805" then it's a counterfeit; but I can manufacture SIWA7805 and sell it on Aliexpress under 78xx generic name.

But, having regulation during a simple test is just one thing. Other things to consider are true output current capability, true power dissipation capability, stability under more challenging loads, actual maximum input voltage under which the part does not blow up, short circuit protection, thermal protection, and lifetime generally. In fakes, these are often compromised.

At least getting the correct voltage is something, but my advice is, and this applies to beginners, hobbyists, and professional designers: working with circuits you have always enough work to do on your own design, and enough uncertainty fixing your own mistakes. Don't add extra uncertainty by using counterfeit Ebay crap, because that can be a huge time sink, because you never know what to trust.

[JenniferG]

The last test of each, though, is of no use since they all require almost 2V of dropout.

Yeah I know, I was just showing it

[TimFox]

Just a note- the input and output caps aren't "smoothing" caps.  They are more correctly "bypass" caps.  Smoothing caps would be to reduce input ripple from the source, like 60 or 120 Hz.  Both of these caps are sized to tame high frequencies and transients.  They both improve transient response and stability in all but the simplest circuits.

For your DC only test assuming, you used short leads, etc., they aren't strictly required but their function is somewhat subtle and they should be used whenever possible.  This is just a note to better understand their function since you are posting as a beginner.

The purpose of the input cap, usually a .22 uF ceramic mounted very close to the input is designed to keep AC out of the regulator.  The data sheet usually says "required if the regulator is far from the source".  The idea is to bypass any high frequency content that may be on the input to the regulator.  Long input leads make the noise pickup worse.  Long leads also add inductance which raises the impedance of the source at higher frequencies.  Transient voltages on inductive sources create voltage fluctuations that are detrimental.   Keeping high frequency noise out and maintaining a low AC impedance at the input make the regulator's job much easier.  The regulator doesn't have enough bandwidth to get rid of high frequency noise present on its input.  The high frequencies can also get in and contaminate the voltage reference in the regulator which will then show up on the output.  The data sheet show the ac rejection of input noise- its not great past a Khz or so.

The output filter cap, usually .1 uF located close to the output enhances stability with transient loads.  Its basically a local bypass and keeps the output at AC ground even with widely varying loads.  Remember that internal to the regulator is a amplifier that is comparing the output to an internal reference and making corrections.    Bypassing the output enhances the stability of this control loop by providing a low AC impedance load to the output.  You won't see a 780x oscillate but an output cap will noticeably improve the dynamic response for step loads, etc.- overshoots and undershoots will be reduced.  This is a measure of relatively stability and this cap should also always be used.

The chip vendors make these capacitors "optional" to make the circuits more palatable to customers but in all but the very lowest cost or non critical circuits, they should be considered mandatory.

Most low dropout regulators (the 7805 does not fall into this category) absolutely require bypassing and the output bypass cap has to carefully selected to insure basic stability of the regulator.  The output cap is an important compensation component (dominant pole) in the circuit.

Have fun.

For the 7800 series, the capacitor at the input is mandatory to prevent oscillation.
Years ago, a junior engineer forgot to install it (I usually used 0.47 uF/X7R) and the result was a strong oscillation at roughly 100 kHz.

[JenniferG]

These are rated at 100ma.  Would a constant 50ma dummy load be a good test for stability/heat? or is that too much? 

For a 50ma dummy load, using Ohm's law (I'm new to this) I guess the resistor value would be 100 ohms for the 5V regulator.  R = V/I.  R = 5/0.05; R = 100 ohms. P = 0.05*5 = 0.25 Watts.  1/4 watt.   So I could use a 100 ohm 1/4 watt resistor as dummy load for 50ma on a 5V regulator?  Or maybe I should go with 1/2 watt resistor?    I don't have any half watt resistors.. could I do two 1/4 watt 200 ohm resistors in parallel to divide the P used.. so they don't fry ?   I just learned ohms law a couple weeks back.

[Siwastaja]

Maybe shorting the output is a good test - 78xx series should have protection. Let it be for a few hours. It should run VERY hot, but after you remove the short, it should continue working like before.

[ledtester]

... could I do two 1/4 watt 200 ohm resistors in parallel to divide the P used.. so they don't fry ?   I just learned ohms law a couple weeks back.

Yes. Resistors of the same value in parallel will share the power equally.

If the regulators are rated at 100mA I would test them at that value. It's helpful to test them at lower values as well, but you should know what it's capabilities are before you put them into storage.

[wasedadoc]

These are rated at 100ma.  Would a constant 50ma dummy load be a good test for stability/heat? or is that too much?.

Read the datasheet more carefully and note the first clause of  "If adequate heat-sink is provided, they can deliver up to 100 mA output current."

[JenniferG]

These are rated at 100ma.  Would a constant 50ma dummy load be a good test for stability/heat? or is that too much?.

Read the datasheet more carefully and note the first clause of  "If adequate heat-sink is provided, they can deliver up to 100 mA output current."

They are TO-92, figured they wouldn't use a heat sink.  I mean I see no way to attach one.  They look like a typical TO-92 small little transistor.

[wasedadoc]

These are rated at 100ma.  Would a constant 50ma dummy load be a good test for stability/heat? or is that too much?.

Read the datasheet more carefully and note the first clause of  "If adequate heat-sink is provided, they can deliver up to 100 mA output current."

They are TO-92, figured they wouldn't use a heat sink.  I mean I see no way to attach one.  They look like a typical TO-92 small little transistor.

Seems you are still reluctant to properly read the datasheet.  Look at section 3 of http://j5d2v7d7.stackpathcdn.com/wp-content/uploads/2016/07/78l05.pdf.  The TO92 version has by far the highest junction to ambient thermal resistance.

[Siwastaja]

Yeah, TO-92 can't be practically heatsinked.

You need to calculate the actual maximum current. They can't give you one number because it varies depending on your use case: namely input voltage and ambient temperature.

Example: Let Vin=12V, Vout=5V, and Ta = 40 degC

From the LM78L datasheet:
* TO-92 RthJ-A (beware, datasheet has typo here, it says Tja they mean RthJ-A) = 158.7 degC/W.
* Recommended operating temperature max 125degC
* Absolute maximum junction temperature 125degC

RthJ-A means that when the device dissipates one Watt of power, temperature difference between the semiconductor junction (J) and ambient (A) will be 158.7 degC. So if our ambient is at 40 degC, the junction will be at 198.7degC, well over the absolute maximum. So clearly you can't dissipate 1W.

How much we can dissipate? Given Tj_max = 125degC and Ta = 40 degC, maximum difference between junction and ambient is Tja = 85 degC.

Let's plug this into RthJ-A and solve for the watts: 85degC / (158.7degC/W) = 0.53W.

The linear regulator dissipates (Vin - Vout) * I. With Vin=12V and Vout=5V: P = 7V * I. Solve for I = 0.53W / 7V = 75.7mA.

And of course, if they say Tj=125degC is "absolute maximum", you don't really want to run it that hot. (125 degC is surprisingly low for an absolute maximum, though, so no need to add massive margin). But I generally calculate for Tj_max = 110 degC or something like that, assuming I'm sure I have every detail in calculation covered. If I'm unsure about something, then of course add more margin.

[JenniferG]

Seems you are still reluctant to properly read the datasheet.  Look at section 3 of http://j5d2v7d7.stackpathcdn.com/wp-content/uploads/2016/07/78l05.pdf.  The TO92 version has by far the highest junction to ambient thermal resistance.

"Still"?  I created this thread like 30 minutes ago.  No I haven't read it since.  I tried reading it earlier today.  I understand like 25% of what's on data sheets.. it intimidates me!  Thanks for the info though.

[BillyO]

There are heat sinks for TO-92 packages.  Just google to-92 heat sink.  There are flag style and there are clamp style.  Will you be using them in a PCB?  If so you can use large copper pours directly off the lead pads and mount the regulators so that they are tight against the board (shortest possible leads).  The clamp style heatsinks are the best and can be used in conjunction with a copper pour to bolt it to and thermal compound to get sufficient heat dissipation.  Also keep the input voltage as close as you can to the drop-out voltage at the current you want.  Vo+2.5V should be sufficient if the supply voltage has very little ripple.  This will reduce the dissipation to 250mW @ 100mA.

[Siwastaja]

There are heat sinks for TO-92 packages. 

There are, but the problem is the TO-92 package itself which is not designed to be heatsinked, the die is deep inside the plastic, and plastic has poor thermal conductivity, and surface area available for cooling is small. Manufacturers of TO-92 packaged parts therefore might not give value for RthJ-C. The heatsink probably helps a bit, but by how much, it's hard to say. In my opinion, if you need more dissipation than what you get from the TO-92 part given the RthJ-A as quoted, don't add a heatsink to it, change to a better package.

There are other packages that are fully plastic (for electrical insulation) but still can be heatsinked, but they have an internal metal tab and as thin layer of plastic as possible (while maintaining the electrical insulation) for at least semi-decent heatsinking capability.

[floobydust]

OP, there is a danger using name brand datasheets with chinese cost-optimized die-shrink IC's. It could be a copy or an original "improved" part. You could send some to Noopy (this thread) for die pictures to see what the part really is.
What I have found with sketchy 78xx regulators is they malfunction at high line, their output voltage starts to go up as you go above say 20-30V. Though, I can't remember if the minimum load spec was met when the drama happened. One way to make a cheaper IC is using a lower voltage process like in the LM1117 series, or sell rejects that don't quite work at the high end of input voltage.
But you have to be amazed at the low price.

[EPAIII]

Can't be practically heatsinked?

https://www.google.com/search?sxsrf=ALiCzsaeUVqwnOqCyEtOeS1KSg_Dlg4tZw:1669942294955&source=univ&tbm=isch&q=TO92+heatsink&client=firefox-b-1-d&fir=JpPInaLnYUMsqM%252CYAxXf7jhMUy_3M%252C_%253BC3trvQ1tJyVLjM%252CWw3saJtaRzBqDM%252C_%253BSXjAiYX8l7aauM%252C06grxW6JmGYPgM%252C_%253BPYeeGYp9Y2JLgM%252Cl14-IIyoCTMt5M%252C_%253BEuZoDXmClyyVRM%252Cwm57XPgqLjJuKM%252C_%253BKpe1BOJI3I2gVM%252CWMUKCzQTYB4HwM%252C_%253BLZBzFlwfmyqYsM%252CGsqtVv_-9dm8uM%252C_%253BmIjlFGgRyL_cMM%252CYe_LNPfPbwczEM%252C_%253BTi3qEgsPl505rM%252CdjC84Xld49ZRWM%252C_%253BhodvInlMlVMEuM%252CmybWFHMeaWJWTM%252C_&usg=AI4_-kSWIyPRxTvOKcGjvhTGfblIHyayyg&sa=X&ved=2ahUKEwjZyaTq29n7AhW0lGoFHdFVCYAQjJkEegQIGRAC&biw=1191&bih=616&dpr=1.4

I don't know about the "Ali Express", no name ICs, but TI seems to have no problems with 100 mA if the Input Voltage is within 4 Volts of the Output. The TI datasheet clearly shows that THEIR TO-92 (Z) package can handle at least 400 mW in "free air" with short lead lengths (0.125"). So a TI LM78L05, with no heatsink, and an input Voltage of 9 Volts and a 100 mA load would dissipate 4 V x 0.1 A = 0.4 W. And that would be withing the spec. for heat in the package. I have designed circuits with those numbers and never had any problems, in spite of no additional (beyond the normal copper foil leads) heat sinks.

http://users.ece.utexas.edu/~valvano/Datasheets/LM78L05.pdf

I have tested 78xx regulators with short circuits and no heat sinks for 12+ hours and they still worked like new. (note below) I would not hesitate to test a TI TO-92 LM78Lxx regulator with a 100 mA load, no heat sink, input V = output V + 4, and only short PC leads for as long as you wish. Or one by any other major manufacturer.

Now with 12 V as the Input, all bets are off. That's a 7 V drop and 7 V x 0.1 A = 700 mW. It WILL shut down. But all you need is a series dropping resistor if you must have 100 mA at 5 V from a 12 V supply.

If you are going to rip-off a component, you should meet ALL the original specs. That's my opinion and I am sticking to it. I do not buy from Ali Express and this only reinforces that practice.

For a linear supply some good rules of thumb are:

1. Transformer RMS, AC secondary Voltage = final regulated DC Voltage. Example: Regulated 12 VDC supply requires a 12V transformer.

2. Use a full wave rectifier.

3. Filter cap = I / (2 x F x Vr) where I = max current, F = line frequency, and Vr = desired P-P ripple Voltage.

4. Voltage regulator IC with recommended external components.

That provides for things like the diode drops and the needed head-room in most linear regulators. And it works with a 6.5 Volt transformer for a 5 V regulator. If you need 12 VDC and 5 VDC, get a center tapped, 12 V transformer and use the center tap for the 5 V supply.

Note: Many years ago when I was designing my first power supply with a 78xx regulator I just could not believe what I read on the data sheet about short circuit protection. So, after several quick tests for an instant and for seconds and then minutes, I shorted it and went home. The next day when I arrived at work I truly expected smoke, but when I removed the short it sprang back to life like nothing happened. And it was only a bit warm, definitely not hot. You gotta love a good design!

Yeah, TO-92 can't be practically heatsinked.

You need to calculate the actual maximum current. They can't give you one number because it varies depending on your use case: namely input voltage and ambient temperature.

Example: Let Vin=12V, Vout=5V, and Ta = 40 degC

From the LM78L datasheet:
* TO-92 RthJ-A (beware, datasheet has typo here, it says Tja they mean RthJ-A) = 158.7 degC/W.
* Recommended operating temperature max 125degC
* Absolute maximum junction temperature 125degC

RthJ-A means that when the device dissipates one Watt of power, temperature difference between the semiconductor junction (J) and ambient (A) will be 158.7 degC. So if our ambient is at 40 degC, the junction will be at 198.7degC, well over the absolute maximum. So clearly you can't dissipate 1W.

How much we can dissipate? Given Tj_max = 125degC and Ta = 40 degC, maximum difference between junction and ambient is Tja = 85 degC.

Let's plug this into RthJ-A and solve for the watts: 85degC / (158.7degC/W) = 0.53W.

The linear regulator dissipates (Vin - Vout) * I. With Vin=12V and Vout=5V: P = 7V * I. Solve for I = 0.53W / 7V = 75.7mA.

And of course, if they say Tj=125degC is "absolute maximum", you don't really want to run it that hot. (125 degC is surprisingly low for an absolute maximum, though, so no need to add massive margin). But I generally calculate for Tj_max = 110 degC or something like that, assuming I'm sure I have every detail in calculation covered. If I'm unsure about something, then of course add more margin.

[mcovington]

The last test of each, though, is of no use since they all require almost 2V of dropout.

That is why I suggested to look at "the minimum input voltages required".

I wouldn't say it's useless.  It does provide some characterization, some confirmation that the tested chips are alike (not totally different chips labeled with the same label) and might be interesting to compare with known-genuine ICs.

[james_s]

That all looks pretty reasonable to me. Can't say I've ever used a 7808 though and I've used a 7809 maybe twice in my life. Do you have a specific application in mind for those?

[jwet]

The important concept with linear regulators is that they dissipate power by the equation "voltage drop x output current".  Dropping 2.5v at 100 mA is .25w.  Keep the drop comfortably above the minimum but not a lot more, it just turns into heat.  With a large drop, say 20v, you would want to keep the current down below 25 mA or so. (.5W).  High temps are the enemy of all components- reduces their life considerably.  780x regulator are very robust but you can't change the physics of what's going on.

Also the power rating for resistors is generally where their temperature will rise 100 deg C over ambient.  At room, this would be 125C at rated output, quite hot.  I would derate R's at least 50% from the standard wattage- so use a 1/2 W where you will dissipate 1/4w- this keeps the rise down to 50 deg C or 75C at ambient.  You can parallel and series R to spread out the heat.  Two 1/4w 50 ohm R's in series or two 1/4w 200 ohm R's in parallel would give you 1/2w 100 ohm load.  Either configuration will share the load equally.