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I have got some LEDs, a breaboard, a 74HC08 IC, a 9 V battery, resistors of 10 and 100 ohms. I want to check the IC by connecting it to a breadboard but the IC input voltage is 5 V and current 8 mA. On calculation I found that if I connect around approx. 600 Ω of resistor then my battery would give a 5 V output to IC for its operation. My doubt is that if I have connected resistors to the input of IC i.e. Vcc then do I need to connect resistor to my LED as well or you could suggest some other way around? Currently I only have these items with me.

JRE
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Arjun Kumar
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    Does this answer your question? [how to reduce DC voltage using resistors?](https://electronics.stackexchange.com/questions/75448/how-to-reduce-dc-voltage-using-resistors) – jonathanjo Sep 29 '22 at 11:07
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    You have insufficient items with you to do the job properly. Get a voltage regulator. And you should be careful in specifying part numbers- a 7408 is quite different from a 74LS08 or a 74HC08, – Spehro Pefhany Sep 29 '22 at 12:20
  • I got your point but i watched some videos on youtube where i found people were connecting the IC with the 9V Battery directly without limiting the supply still the circuit was functioning properly and thanks for mentioning I have edited the IC's name . – Arjun Kumar Sep 29 '22 at 14:52
  • What you should do versus what someone on the internet got away with once are two different things. You need a voltage regulator. – winny Sep 29 '22 at 15:20
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    The 74HC series emulate older TTL family logic and MUST NOT be operated at a higher maximum supply voltage than shown in the specification sheet. Nominal operating voltage is usually 5V. Operation on 9V supply MAY work for random ignorant internet users - and may not, and will cause various seen and unseen issues. ALWAYS maintain pin voltages within datasheet specified voltage ranges. – Russell McMahon Sep 30 '22 at 00:41
  • A jb0 says - using a series resistor alone is a very poor solution. A resistor and zener diode will work OK if the resistor and zener values are correctly designed BUT a 5V regulator IC is not very costly and will be ongoingly useful as you experiment. – Russell McMahon Sep 30 '22 at 00:43

2 Answers2

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First, The 7408 is not a constant current device, therefore it will not always draw 8mA, so the 600-ohm resistor will not work. I'm not exactly sure how you found the 8mA in the first place because from the datasheet I found, it is from 11 mA with outputs high to 20mA with outputs low.

You do need the resistor for the led regardless. This is a good tool for calculating that resistor; however, I think you will run into some problems as the high level output current(I_OH) is only .8 mA.

jb0
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  • If using a 7408 or 74LS08, you can probably drive the LED directly from the output, as those versions have limited output drive current. A 74HC08, however, has much higher output current and would fry the LED, so you need a resistor there. – Hearth Sep 29 '22 at 15:03
  • @Hearth I'm curious - If just an LED (Let's just say 2V forward drop) is attached to the output, which is 5V, how is the other 3V dissipated, even if the IC has limited output current? Does the voltage on the IC output go down to lower the current? – jb0 Sep 29 '22 at 22:56
  • Yes, the voltage on the IC output goes down. The outputs can be considered as being in series with a resistor. – Hearth Sep 30 '22 at 00:06
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Using a series resistor isn't going to get the job done. The 74HC08 IC current will vary, depending on what it's doing and how much current its output pins are sourcing.

Rather than use 9V, you have a couple of options for making 5V to power your breadboard:

  • 3x 1.5V alkaline batteries (close enough to 5V to get the job done)
  • Hack a USB charger cable and use a power bank

If you choose the latter, the USB cable has 4 wires. The ones you want are Red (VBUS / +5V) and Black (GND).

If you still insist on using 9V, there are CMOS logic families that can do that. The CD4081B (quad 2-input AND) can run on 3 to 18V.

Or, you can use your 9V battery and take it down to 5V via a regulator like an LM7805. (Fun fact: the very first thing I ever built was a 5V supply, using a transformer robbed from an old clock radio, plus the 7805.)

But really, your life will be easier if you use 5V if you intend to play with small-scale logic.

Once you've done that, then use a resistor to limit the LED current. A good value range is 330 ohms to 1k ohms.

hacktastical
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