ULN2003A: ULN2003A - The input voltage to turn off the output

Dice,

Lets see. Typical input resistance is 2.7k + 7.2k + 3k = 12.9k. We are looking for off so let assume there is no base emitter current.
The input drive is 0.7V + 5.1k, Expected input pin voltage is 0.7V * (12.9k / (12.9k + 5.1k) = 0.5V
The expected input current is 0.5V / 12.9k = 40uA

The ULN2003AI has a 30uA minimum for off current (at 105C) , where 'off' means 500uA of output current flow.
Therefore there may be some output current at higher temperature.

Does the 0.7V input voltage also change with temperature?

Do all the output drive 5V resistive loads?

Hello Ronald,
Thank you for your reply.

I have two answer and a question.

Firstly, I answer your two questions.

(1)
0.7V input voltage does not change with temperature.

(2)
All the outputs drive 1kohm resistances connected by 5V.

Secondly, I have a question about your answer.
"The input electric current expected at input voltage 0.7V is 40uA",
In this case, please teach me the maximum output electric current (Ic).

Best regards,
Dice-K

Hello Michael,
Thank you for your reply.

(1)
I understand that it is necessary to make an input electric current less than 30uA.
If the input voltage is less than 0.54V, is the series resistance OK at 5.1kohm?
0.54V/(12.9kohm + 5.1kohm) = 30uA

(2)reminder question
"The input electric current expected at input voltage 0.7V is 40uA",
In this case, please teach me the maximum output electric current (Ic).

If the output current is small, the system works normally.
Therefore I want to know the value of the output electric current.

Best regards,
Dice-K

Dice-K,

(1) Yes, you are correct.

(2) if the input current is 40uA, based on characterization I have done previously, it is very unlikely that the device will allow even a small amount of current. At the highest temperature, I found that the input current to turn the device ON was approximately 50uA, suggesting that very few if any devices will be on at 40uA. In the event that they are ON, the current will be small.
Measuring exactly how small would be very difficult because as I mentioned, when I test devices on the bench, they will all be off at 40uA.

While we can't guarantee that they will all be off, I can say will very high confidence that they should be, and those that do turn on will only allow a limited current at the output.

I hope that this helps to clarify.
Best,
Michael

Hello Ronald and Michael,
Thank you for your replies.

I have a question about your answers.

"The ULN2003AI has a 30uA minimum for off current (at 105C) , where 'off' means 500uA of output current flow."

"The key here is to look at the specification for Ioff.
Ioff minimum specification is 30uA, so it should be designed to meet this minimum specification."

Do these answers adapt to ULN2003APWR?

ULN2003APWR     (Ta:-20C~70C)
ULN2003AIPWR    (Ta:-40C~105C)

Best regards,
Dice-K

Hello Michael,
Thank you for your reply.

Though the temperature range of ULN2003AIPWR is different from ULN2003APWR, I understand that the specification of Ii(off) is same.

ULN2003APWR     (Ta:-20C~70C)       Ii(off)=30uA
ULN2003AIPWR    (Ta:-40C~105C)      Ii(off)=30uA

If my understanding is incorrect, please give me a reply.

Best regards,
Dice-K

Hello Michael,
Thank you for your reply.

It is the matter of Ii(off)=30uA.
Is this an electric current that includes the electric current of the resistance and the base electric current to a transistor of the internal circuit?

In the case of the circuit of the customer, when VI=0.7V (series resistance =5.1kohm), an electric current of 39uA flows only by internal resistance.
How much is the base electric current?

In the data-sheet, Ii(max)=1.35mA at Vi=3.85V.
Because the internal resistance consumes about 0.3mA, is this base electric current 1.05mA?

Best regards,
Dice-K

Hello Michael,
Thank you for your reply.

Please teach me a base current with Fig8 and fig6.
For example, when output current is 200mA, Ii is 350uA.
When Ii is 350uA, Vi is 2.5V.
Because internal resistance is 12.9kohm, the base current is 350uA-(2.5V/12.9kohm)=156.2uA.
Is this calculation correct?

Best regards,
Dice-K

Hi Dice-K,

Ultimately, the important information is the input current vs. the output current. This will determine the amount of input current needed to drive a specific load, and the amount of current going into the base of the transistor will be a part of that.

If you did need some additional information, we have some data (below) which plots the DC current transfer ratio (beta) vs. IOUT. As evidenced by the graph, the base current is going to depend on the output current.

Thanks,

Alek Kaknevicius

Hello Alek,
Thank you for your reply.

My customer has a question about the following sentence.

"This will determine the amount of input current needed to drive a specific load,
and the amount of current going into the base of the transistor will be a part of that."

For example, by the given graph, when the output current is 10mA, hfe is 80 at Ta=-40C.
Therefore the input current 'Ii' is 10mA/80=0.125mA.
I think that this 'Ii' is 'input current needed to drive a specific load'.

Does this 'Ii' mean 'internal resistance current + base current + α'?

Doesn't this 'Ii' mean 'internal resistance current + base current'?

Best regards,
Dice-K

Hi Dice-K,

I agree with your definition, Ii is the 'input current needed to drive a specific load.' This is the total current flowing into one of the input pins for a specific output current. 

Why does the customer need to know how much of that leakage is internal resistance current and how much of it is base current? Isn't the only concern the amount of input current needed for a specific output current?

Thanks,

Alek Kaknevicius

Hello Alek,
Thank you for your reply.

I talked with my customer,
So, there is one question.

When Vi is 3.85V, Ii(max) is 1.35mA.

Does this mean that ULN2003A can input up to 1.35mA?
or
Does this mean that ULN2003A needs 1.35mA input to make output?

Sorry for repetitive questions.

Best regards,
Dice-K

Hello Alek,
Thank you for your reply.

My understanding is as follows.

 When 'Vi is 3.85V' and 'the output load is zero' and 'COM pin is floating', the maximum input current is 1.35mA.

 ULN2003A does not need 1.35mA input to make output.
 (Figure-8 shows this.
  For example, when Input current is 250uA, Output current is 100mA.)

Is my understanding correct?

Best regards,
Dice-K

Hi Dice-K,

Yes, your understanding is correct :).

Thanks,

Alek Kaknevicius

how much current on each leg if one is turned on, if 4 are turned on, or all are turned on, and, is it possible to have 16v on the input legs, and 5 volt on the signal/activation legs?
thanks

Hi Andrew,

Yes, it is possible to have 16V on the output pins (xC) and 5V on the control pins (xB). Could you please elaborate on your question about current on each leg? Figures 4 and 5 in the datasheet do a very good job of explaining the amount of current which can be applied on each channel if a certain number of channels are active.

Thanks,

Alek Kaknevicius

Hi Alek, I would like to get 2 amps total out of such a device, sustained, for long periods possibly, so, can the ULN2003A do it?, or is there a better transistor array SMD chip that can do it?, so, with regards to the legs, can they all provide 500mA when all switched on?. from the fig 4-5 they can only provide about 100mA when all on, is this right?
thanks

and I forgot to ask again, the chip has pull-down resistors built in?, so I don't have to add pulldown resistors in my circuit?
thanks again!