DOUT max. current supply / driving LED's / BLM

[This N°9]

After finishing the coding of the blm_x driver to drive a button-LED matrix, I was doing a lot of experiments to learn about the best way to drive it. My aim is to drive a soft-button matrix from sparkfun.com with RGB LED's. On the way some question marks passed my way, one of it was:

How much current is a 74HC595 able to supply? I read the datasheet(s), and had some problems to find the right interpretation of the numbers I found there:

Clamp Diode Current (IIK, IOK) ±20 mA

DC Output Current, per pin (IOUT) ±35 mA

DC VCC or GND Current, [glow=red,2,300]per pin[/glow] (ICC) ±70 mA

Phillips uses diffrent labels for the numbers:

input diode current ±20 mA

output diode current ±20 mA

output source or sink current

Q7’ standard output ±25 mA

Qn bus driver outputs ±35 mA

VCC or GND current ±70 mA

It looks like the 70mA is the max. sum current the 74HC595 can supply. There was a discussion about that in the programmers lounge (private access), I decided to continue it here. Some of the quotes from the discussion there:

TK:

A single 74HC595 output usually doesn't source more than 25 mA.

This:

The 70mA is not the max. current that the 595 can provide, but the shortcut-current (shorting a pin to gnd/Vcc). 35mA is the max. output-current, and 20mA (output clamp current) seems to be the current that can be fed in the 595 per pin (when pin is low). In one datasheet this value was clearly labeled as input current. The last I read from fairchild makes the 70mA-value more clear (per pin!):

Clamp Diode Current (IIK, IOK) ±20 mA

DC Output Current, per pin (IOUT) ±35 mA

DC VCC or GND Current, [glow=red,2,300]per pin[/glow] (ICC) ±70 mA

seppoman:

for me, this reads as "the 70mA IS the max. continuous current that either the GND pin or the VCC pin can carry - this can be partitioned to one or several pins, but a single pin can only carry 35mA max" i.e. as soon as all 8 are on, each can only take about 9mA. I agree the wording in different datasheets can be a bit fuzzy at times.

stryd_one:

Due to the mass confusion about this spec, SmashTV was kind enough to sacrifice a SR some time ago, and the results confirm that you are correct.

this:

Something's on my mind: if I connect 8 standard red LED's (Vf 2V, If 20mA -> 100R) through a 220R resistor array, the output current of a single pin will be ca. 5V/(100+220+142)R = 10.8mA

142R would be the output resistance of a single pin if the max. output current is 35mA. It could be even less(?). But 10.8 * 8 = 86.4, this is above the speced 70mA. I never noticed a change of brightnes when all LED's connected to a SR are turned on, but when these numbers are correct, the current would fall if the max. 70mA are reached, or the 595 would be in stressed state. Can somebody help me to understand this?

So, because of all this confuction, I did some experiments with the 74HC595 myself. Here are the results:

- Grounding a single output pin set to high (no resistor, other pins open): Ipin = 68mA

- Connecting a single output pin through 220R to ground (other pins open): Ipin = 20mA

- Connecting a LED through 220R (other pin's open): Ipin = 12mA

The next test result looks very interesting to me:

I connected all the pins set to high through 220R to ground, and measured the sum current from all pins to ground: Ipins-gnd : 142mA (!)

Then I let the 74HC595 in this state for some minutes, and measured the temperature: 36° C (free air).

The voltage Vcc came down to 4.8V

The results show: the 74HC595 is able to have a current of 142mA through Vcc without getting very hot or something, and all the pins are able to provide ca. 17mA at the same time (with 220R resistors)! But I just wonder for how long, and if the device will get damaged.

The other question mark that struck me: when I connect 8 LED's to a SR with 220R resistors (the standard MIDIBox solution to connect LED's), the total current through Vcc of the 74HC595 will be ca. 96mA.

This is more than the specified 70mA max. I really wonder about this number, especially because I read a shortcut-current of 68mA when grounding a single pin. This is suspiciously near to the 70mA. And: "DC VCC or GND Current, per pin (ICC) ±70 mA", why should they write "per pin" if this number would simply refers the max. current that can enter the Vcc terminal? Maybe it's rather the max. current that can flow from the Vcc terminal to a single driver! (?) And 35mA is the max. current that should leave a output terminal without stressing the driver to much? After all the experiments and datashits this would be the explanation that looks the most reasonable to me.

Why I'am so interested in this?

- I worry about my (and all the other) midiboxes driving eight LED's simultaniously :D

- I want to know if I can drive my RGB LED's without having additional drivers at the source side.

I need ca 19mA for the red color to match with green / blue (they'r much brighter)

Maybe somebody can give me some comments and help me ease the pain in my brain?

EDIT:

some more test results:

- Ground 3 pins (no resistors), read current of fourth pin -> ground: I = 55mA

  Temperature of 74HC595: 45° C

- Ground 4 pins (no resistors), read current of fifth pin: 51mA

  Temperature of 74HC595: 53° C

But still alive.. 5 pins grounded sums up to a total current of 255mA. Here it starts to look critical to me. Now multiply 35mA with 8, you get 280mA.. the max. output current rating. 55° is still acceptable, now put the device in a closed box, and it will start getting critical. Does this make sense or not?

After all this I'm quite sure the 74HC595 can source 8 pins with 20mA each without any problems. What do you think?

[This N°9]

I contacted ON Semiconductor to get the definitive (?) answer:

Thank you for contacting ON Semiconductor.

The DC supply current listed in the maximum ratings is valid for whole device. Output current per pin is allowed to be max 35mA, but this is not for all pins at the same time, but just for one or two so the total current will not exceed 75 mA. Of course this is maximum rating and it is not recommended to use these limit values for standard operation of the device.

Indeed distrelec.com gives me a different / no answer (translated from german):

Thanks for your request.

We are sorry, we also just can try to interpret the datasheets / give an understandable translation:

..

"Supply Current" is the Current measured at the IC Supply terminal per Logic-Unit

Wanted to know what "Logic-Unit" refers to (translated from german and cooked down to the basic message):

Sorry we can't give you a definitive answer, so please refer the datasheet.

I love datasheets.

So what about 8xLED @ 12mA (with 220R at SR outputs) > 70mA ?

[seppoman]

I guess ON will know better than Distrelec what their chips can take ;) It's great that you did some tests with positive results, but if you wanna build something that's supposed to work without failure in all environmental situations and for several years, you should better accept what the datasheets of the manufacturers state and stay at least below 100mA.

So what about 8xLED @ 12mA (with 220R at SR outputs) > 70mA ?

the word is "duty cycle" - did you ever shake your head quickly while looking at a MB? the LEDs are only on part of the time so the average current is lower than calculated. If you're interested in numbers, you should ask TK.

S

[nILS]

the word is "duty cycle" - did you ever shake your head quickly while looking at a MB?

My head is always shaking really fast. Crack + coffee == bad combo ;)

[This N°9]

the word is "duty cycle" - did you ever shake your head quickly while looking at a MB? the LEDs are only on part of the time so the average current is lower than calculated. If you're interested in numbers, you should ask TK.

Duty cycle with standard DOUT's (no multiplexing)??? 12mA is exactly the current that I measures flowing through a LED connected to DOUT with 220R. If there would be a duty cycle, I would measure a lower value with a normal multimeter. And if you look at the tests I did, you see that the 74HC595 will not cut off at total current > 70mA. I scope will show the truth this evening..

Is it possible that the 70mA limit is only to guarantee a proper logic 1 level, and not to prevent the chip from damage? I wonder because:

I connected all the pins set to high through 220R to ground, and measured the sum current from all pins to ground: Ipins-gnd : 142mA (!)

Then I let the 74HC595 in this state for some minutes, and measured the temperature: 36° C (free air).

The voltage Vcc came down to 4.8V

[This N°9]

My head is always shaking really fast. Crack + coffee == bad combo

he word is "duty cycle" - did you ever shake your head quickly while looking at a MB? the LEDs are only on part of the time so the average current is lower than calculated. If you're interested in numbers, you should ask TK.

seppoman, it must have been the crack + coffe.. set a dout-pin to high, take a scope, connect it to the pin, watch your scope and turn the time/div button from 1uS up to 1mS, you will always see a straight horizontal line (don't shake your head while doing it).

Why should there be a duty cycle? The 74HC595 have latches, which hold the logic 1 until the data is shifted from the core to the registers. Latch enable will transfer the content of the registers to the latches, there's no time when the pin will be low.

Of course, if you have a multiplexed solution, there you have the duty cycle, but normally if you just want to connect some, say 16, LED's, you will do no multiplexing. So you will have a constant current from each output to the LED of 12mA (aproved by measuring it with 220R at outputs). 12x8 = 96, this is quite more than 70mA -> violation of the maximum ratings (if the technical assisten of ON tells the truth, assumed he has a clue and not just reads the datasheet too).

[seppoman]

well ok, I didn't try this shake head thing lately, either I mixed something up or MIOS behaviour has changed at some time. btw duty cycle doesn't necessarily mean multiplexing, you could turn on the outputs for 2 ms and off for 2 ms and you've got a duty cycle ;)

Anyway, I guess this proves that the 595s can take some more current than the datasheet says. Still I'd suggest not to drive them much further over the limit or else a failure would become more probable.

S

[This N°9]

Sorry that I may fall on your nerves by beeing so sticky to this.. but if a current > 70mA would really be a condition that could damage the 74HC595, then the standard MIDIbox solution for driving LED's would be in serious troubles.

I really don't beleive that this is the case, as somebody until now would have noticed this, and my 74HC595 experiments go far beyond the 12, even far beyond 20mA per pin, and the device is still alive.

Ok, I did not test for hours or days, but even at 200mA supply current, which is a case that will never be used to drive LED's or additional logic devices, the 74HC595 did not get warmer than 53° C after ca. 10min.

I assume that the maximum rating is to ensure a proper logic-1 level of 5V at the outputs, that's not relevant when driving LED's. At 20mA at 7 outputs, the voltage of the 8th output drops to 4.86V (Vcc=4.93V), maybe this could cause instable contitions for some logic families (bad slew rates etcetc.). The device does not overheat or something, and I don't thing that the bond wire will melt under those conditions, and the current specs of the single outputs will not be violated, so what?

Because I'am very stubborn, I also contacted fairchild semicondutor's support. They have a very nice knowledge base & asking-questions site (not this we-want-to-sell-but-we-don't exist-thing), where you can get a member and ask questions. I'am waiting for the answer from there now.

Besides, some people in the arduino community also use exactly the same approach (74HC595 + 220R) to drive LED's : http://www.arduino.cc/en/Tutorial/ShiftOut (But you have alway to be careful with those DIY punks..  :) )

[This N°9]

Funny, fairchild semi gives a different answer (the one I wanted :D ):

Replied By: MH Eng

Response: Hi Matthias,

The max rate of DC output current and the DC Vcc and GND current is only for one pin.

Thus, if the DC Vcc current per pin is 70mA max, so 8 X 70mA = 0.56A ( max ). In your application, there are only 12mA per pin, 96mA would not damage the part.

I beleive each register unit will have it's own bond wire to the Vcc/GND terminal, so the only part that has to take the total current is the pin / terminal itself.

So, happy news, it's possible to source/sink the full current rated for one pin (25mA sink, 35mA source).

[ilmenator]

Basically that means that there are different types of 595s out there, with specs depending on manufacturer.

Funny, fairchild semi gives a different answer (the one I wanted   :D ):

Unfortunately, this merely means that it is unsafe to rely on the Fairchild specs, as they exceed what is (commonly?) specified elsewhere. So, when you published schematics, you'd have to further specify the brand of 595s that this schematic is valid for, which is bad, because it is common for mail order shops to provide customers with seemingly compatible devices from other brands.

[This N°9]

Basically that means that there are different types of 595s out there

Are you sure? I beleive there are common specifications for 74HC logic family types. My assumtions is that ON and other manufacturer just missed the "per pin / logic unit" information in their specs. Besides my tests don't include fairchild devices, and the results look more like the answer from fairchild would make more sense. When 70mA is really the limit for the whole device, there must be a bottleneck somewhere, this can only be the bond wire or a common Vcc / GND line on the silicon. This would show either in a current cutof @ 5V when you reach some current value, or the device would get very hot. Both is not the case, even if I drive *a lot* more current than 20x8 mA (see test results in my first posting).

These are all speculations so far, I fed this info back to ON, and hope their answer will clear the issue. I will also ask fairchild if their per-pin spec just covers their own devices. I think this is safer than rely on speculations.

[ilmenator]

... I think this is safer than rely on speculations.

That is very true!

[This N°9]

I started a new discussion on sparkfun's forum. I think "max. power dissipation" could be the magic word(s):

http://forum.sparkfun.com/viewtopic.php?p=72352#72352

basicly, if you multiply the voltage drop from Vcc to the drivers, at a given output-current each driver, with the total current Vcc-> drivers, you get ca. the power dissipation, as the input's and logic do not need a lot of current. At 20mA each output, I measured a V-drop of ca. 0.75V -> 120mW Pd, upper limit is 500/600/750 in plastic DIP, depending on manufacturer. 1° C / 10mW is the temperature derating, the device got 36° C @ 8x20mA. makes sense at a room temperature of 20-25°C.

As the out-driver's resistance is no constant and will raise at raising current, 35mA source current per output makes sense in the context of max Pd.

EDIT: in other words, if the IC does not overheat, I see no reason why it should get damaged

[This N°9]

I am currently torturing our little friend 74HC595 a bit more, to let it talk about it's current drive capabilities, as I have still no answers from tech supports at fairchild, ST micro and ON providing contradicting information:

condition: total output current = 270mA (33.75mA each output)

after ca. 10min:

temperature: 41° C (surrounding air ca. 21° C)

voltage drop Vcc->output terminals: 1.46V

this results is a power dissipation of 394mW, well below 500mW (consolidated max. rating different manufactueres)

[This N°9]

I got the answers from ST,ON and fairchild:

ST:

Dear customer,

the mentioned current is for the complete IC. If you like to drive LEDs, plese refer to the product selector on our website under:

http://www.st.com/stonline/stappl/productcatalog/app?path=/comp/stcom/PcStComOnLineQuery.showresult&querytype=type=product$$view=table&querycriteria=RNP139=648.0

Here you will find a list of our LED driver Portfolio.

Best regards

fairchild:

I am not sure what is the other manufacturer part that you use for your application, but what my opinion is if the part is the cross reference of Fairchild part, then it should be the same of current limit per pin.

ON:

This part is not capable of sourcing or sinking much more than approximately 10 ma.  The Voh and Vol specifications list 7.8 ma at a Vcc of 6 V.  We do not have a logic IC that is capable of a 35 ma output.  You  need to consider other parts such as Mosfet drivers.

[TK.]

You have to differ between how much current can a 74HC595 source/sink w/o the risk for damage, and how much current can a single 74HC595 output source/sink w/o influencing the current through other outputs (resp. w/o violating the logic-level specs, but this isn't for interest here anyhow).

On my original 4x16 Duo-LED Matrix design, there is a noticable influence once more than 8 LEDs are sinked through a single 74HC595 at the same time. Therefore I recommented sink drivers.

Comment to answer from ST: they are right - if you are searching for a perfect solution, use a chip which has been created for driving LEDs.

Comment to answer from Fairchild: they are right - they are speaking about a current limit through a single pin, they don't mention the influence between outputs, but you haven't asked for this...

Comment to answer from ON: they are right, but Voh/Vol isn't for interest while driving LEDs (in other words: miss-using a logic chip)

Best Regards, Thorsten.

[This N°9]

You have to differ between how much current can a 74HC595 source/sink w/o the risk for damage, and how much current can a single 74HC595 output source/sink w/o influencing the current through other outputs (resp. w/o violating the logic-level specs, but this isn't for interest here anyhow).

That's exactly my point of view too. In my tests, where I sourced 33mA from each output (which is not necessary in LED applications, as a LED should not be driven beyond 20mA, regardless of colour), I measured a voltage drop of 1.46V from Vcc to output pins. Of course this violates logic level specs, but LED's don't care about logic levels.

1.46V * 264mA -> power dissipation 385.44 mW (uppper limit is specified beyond 600/750mW, depends on manufacturer), so I guess the chip will not be damaged under those conditions.

On my original 4x16 Duo-LED Matrix design, there is a noticable influence once more than 8 LEDs are sinked through a single 74HC595 at the same time. Therefore I recommented sink drivers.

Note that I never planed to sink that much current directly through 74HC595. With a current of 12mA each LED (this is a fact when driving LED's with 74HC595 & 220R resistors), you would have a sink current of 96mA. This is simply not possible, the datasheets specify a sink limit for a sinlge pin of 20mA. Even when grounding a high pin, you will have only a current of 68mA. I guess this will be aboout the same for sink current, I never tested this, because it's not interesting for my application. For sinking, I use the ULN2803, they are cheap and I just need one even for a huge matrix. The driver chips I found are more expensive, and complicate the whole design, and I would need several chips for a matrix. So I would like to use the 74HC595's directly just for sourcing.

Comment to answer from Fairchild: they are right - they are speaking about a current limit through a single pin, they don't mention the influence between outputs, but you haven't asked for this...

This is not totally correct. My exact question was, if it is possible to source 20mA each output simultaniously without damaging the device. My original question to fairchild:

Is it possible to draw 20mA each output pin (at the same time) from the 74HC595? My aim is to drive LED's.

I read in the datasheet "DC VCC or GND Current, per pin (ICC)", this indicates that 20mA each output will exceed this limit, or is this value per driver / logic unit?

I did some tests regarding input/output current with the 74HC595, which aproved that the 74HC595 is able to provide this source, but I don't know if it will get damaged under those conditions.

The whole discussion thread can be found here: https://fairchildsemi.custhelp.com/cgi-bin/fairchildsemi.cfg/php/enduser/myq_idp.php?p_sid=l7MUBgxj&p_accessibility=0&p_redirect=&p_lva=&p_sp=&p_li=&p_iid=77963&p_created=1241624385&p_allorg=1

Comment to answer from ON: they are right, but Voh/Vol isn't for interest while driving LEDs (in other words: miss-using a logic chip)

Yes, true. My only concern is, if 20mA each output simultanously will damage the device. Of course it's an abuse of the original usage aim of the chip, but if it works and does not damage the IC, I don't care :)

To make more clear what my aim is: if it's possible to source 20mA per output (simultanously), and ..

- ..this way of driving the LED's will not affect brightnes if one/several LED's are on (which seems to be not a problem, as I alredy drive the LED matrix like this)..

- ..will not damage the chip..

- ..will not make the input logic (registers / latches) unreliable..

..this would be the most easy/cheap way to drive the matrix.

In either case (special source drivers or not), I'll use the ULN2803 for sinking.

[This N°9]

....the influence between outputs....

I did some more test to have a clear answer to this:

74HCH9H, 220R at each output:

- one pin through 220R to ground: I_single = 19.9mA

- all eight pin's to ground, single pin current I = 19mA --> I_total = 152mA

Max. current drift @ ca. 20mA (max) each output: 4.5%

This is acceptable, you will hardly see a visible change when switching on 1 or 8 LED's.

I also measured the voltage drops:

- one pin through 220R to ground : U_dropsingle = 0.6V

- all pin's through 220R to ground : U_droptotal = 0.61V

With those numbers I could calculate:

- The total power dissipation when all pin's are sourcing ca. 19mA:

P_tot = U_droptotal * I_total = 92.7mW

- Assuming a simplified model of the current flow through a common resistor, and the single drivers (simplified to constant resistors in parallel), I could calculate the common resistance, through which all the current of all drivers has to flow, and the resistance of the single driver (only @ ca. 19-20mA each output!):

V_cc---R_common----R_driver---- Out 1

                       Â¦

                       Â¦

                        -- R_driver---- Out 2

                       Â¦

..

..

I_single = 19.9mA (current for a single pin sourcing)

I_total = 152mA (current for all pins sourcing)

U_dropsingle = 0.6V (V_drop from V_cc to a single output sourcing)

U_droptotal = 0.61V (V_drop from V_cc to outputs when all pin's are sourcing)

R_single = U_dropsingle / I_single = 30.15R

R_total = U_droptotal / I_total = 4.013R

R_single = R_common + R_driver   -----> R_common  = R_single - R_driver

R_total = R_common + R_driver / 8  ---> R_common =  R_total  - R_driver / 8

R_single - R_driver = R_total  - R_driver / 8

7/8 * R_driver = R_single - R_total

R_driver = ( R_single - R_total ) * 8/7 = (30.15R - 4.013R) * 8/7 = 29.8..R

R_common = R_single - R_driver = 30.15R - 29.8..R = 0.35R

Power dissipation at the common resistor (V_cc-pin, bond-wire, and the common way on the silicon to the supplies of the single drivers) if all pins sourcing:

U_dropcommon = U_droptotal * (R_common /  R_total) = 0.053..V

P_common = U_{dropcommonres} * I_total = 0.053V * 152mA = 8.09..mW

Power dissipation of a single driver (all outputs sourcing 19mA each):

P_driver = (P_tot - P_common) / 8 = ( 92.7mW - 8.09..mW ) / 8 = 10.57mW

[This N°9]

I updated fairchild about the condradicting answers, and provided a link to this forum thread ca. a week ago.

Now they marked the request as "solved" without further comments, so do I. You can read all the answers, numbers, test results in this thead. Make yourself a picture and decide yourself.

My personal opinion:

it is possible to drive 8 LED's at max. 20mA each with one 74HC595, without damaging the IC, but I won't recommend to go beyond 20mA * 8. The advantages of using additional drivers are: less power consumption (less heat dissipation), more stable current supply (without: current drift max. 4.5% @ 8*20mA).