NE555 vs ICM7555 timer power usage? - Page 1
Forum section: electronics / beginners

author: ElectricPower
author_rank: Frequent Contributor
timestamp: November 07, 2021, 07:23:20 am
content: I have made a circuit that will hold my 5V powerbank awake with giving it 100mA pulses every 25. second. I have copied the schematic from Mr.Carlsons Lab on Youtube.The circuit works perfectly and i have tested it with my power bank. All good so far.Since i did not had IC7555 timer when i started the project, i sed an NE555 with the same pinout. Booth timers work fine. I just swap them out.I use the IC7555 timer because Mr.Carlssons lab says that this timer uses mutch less power (current) than the old trusty NE555 timer. And there is here i'm a little confused...When i measure current draw between 5V from power bank and the VCC input pin on booth timers i get more or less the same current draw. I meassure 2.5mA in idle with ICM7555 and 1.8mA idle with NE555. So actually the old timer uses less current.Do i meassure it wrong?

author: jpanhalt
author_rank: Super Contributor
timestamp: November 07, 2021, 08:51:10 am
content: Quote from: ElectricPower on November 07, 2021, 07:23:20 amAre you sure what you are measuring? Is there ANY load? Idle current for the ICM7555(Philips datasheet) at 5V is about 75 uA. For the LM555 (Texas Instr), it is 3 mA. Both are presumably without any load.

author: Kleinstein
author_rank: Super Contributor
timestamp: November 07, 2021, 08:55:05 am
content: Depending on where you got the chip from, I would consider that the CMOS chip could be broken or possibly a fake (mislabled NE555).

author: ElectricPower
author_rank: Frequent Contributor
timestamp: November 07, 2021, 09:06:41 am
content: Quote from: jpanhalt on November 07, 2021, 08:51:10 amYes, i'm measuring with one probe on the 5V INPUT and the other probe on the VCC input on the Timer. I power thw circuit with my RD6006P power supply set to 5V. There is no load on the circuit. Just the LED that indicates the pulses every 25. second.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: November 07, 2021, 09:07:15 am
content: Quote from: Kleinstein on November 07, 2021, 08:55:05 amYes, that is an possible explanation....

author: ElectricPower
author_rank: Frequent Contributor
timestamp: November 07, 2021, 01:59:56 pm
content: Is there a possibilty that i measure it wrong?

author: ledtester
author_rank: Super Contributor
timestamp: November 07, 2021, 02:19:08 pm
content: Quote from: ElectricPower on November 07, 2021, 01:59:56 pmWhat are you using to measure the current?

author: ElectricPower
author_rank: Frequent Contributor
timestamp: November 07, 2021, 02:23:09 pm
content: Quote from: ledtester on November 07, 2021, 02:19:08 pmI'm using a multimeter in mA mode and i measure with one probe to the 5V line and the other on the VCC pin to the timer.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: November 07, 2021, 02:47:03 pm
content: I thinking that this is straight forward, and i think i have been screwed by the seller. Not the end of the world, but i Will order som new ones and hope for the best.

author: Ian.M
author_rank: Super Contributor
timestamp: November 07, 2021, 02:58:19 pm
content: A CMOS 555 will have a rail-to-rail output, but when even slightly loaded, the best a bipolar 555 can do is reach two diode drops below its supply rail. Take the suspect 555 and tie its Trigger and Threshold pins low and its Reset pin high to set its output permanently high, load Output with a 10K resistor to Gnd, apply an appropriate supply voltage, and measure the voltage drop from Vcc to Output. If its a small fraction of a volt, you've got a genuine CMOS 555, and if its over 1V its a 'classic' bipolar 555.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: November 07, 2021, 03:02:40 pm
content: Quote from: Ian.M on November 07, 2021, 02:58:19 pmI think that the "high" current draw in idle with no load speaks for it self. I think this is fake timers. I think i just order som new onesThe cost is wery low and it's just worth the time i think.

author: Terry Bites
author_rank: Super Contributor
timestamp: November 07, 2021, 05:19:17 pm
content: The genuine article should draw less than 100uA when the led is inactive. Off brand timer or ciruit error maybe to blame but if you are meauring with a DMM (or worse a USB monitor) you won't get the average reading you need. What does the supply current measure with the LED disconnected?

author: ElectricPower
author_rank: Frequent Contributor
timestamp: November 07, 2021, 05:28:20 pm
content: Quote from: Terry Bites on November 07, 2021, 05:19:17 pmSupply current that shows on the display is 0.0025A. In idle when the led is off.

author: jpanhalt
author_rank: Super Contributor
timestamp: November 07, 2021, 06:16:25 pm
content: Perhaps it's obvious to others, but not to me. Just what do you mean by "idle." In that circuit, the output is high (LED = on) while the capacitor is charging. When it is low, the discharge transistor must be turned on and will require current to its base. What do you see when the capacitor, discharge pin, and LED are disconnected?

author: ElectricPower
author_rank: Frequent Contributor
timestamp: November 07, 2021, 06:22:10 pm
content: Quote from: jpanhalt on November 07, 2021, 06:16:25 pmWith idle i mean no load. Power is on (5V) and i measure 2.5ish mA if i place one probe on the 5V input an the lther on the VCC pin on the timer. The circuit is an copy of the circuit made by Carlsons Lab in this video:https://youtu.be/lOL3ic8msas

author: StillTrying
author_rank: Super Contributor
timestamp: November 07, 2021, 06:26:47 pm
content: "Supply current that shows on the display is 0.0025A"Are you using the 10 Amps range.

author: jpanhalt
author_rank: Super Contributor
timestamp: November 07, 2021, 06:37:42 pm
content: Quote from: ElectricPower on November 07, 2021, 06:22:10 pmThat doesn't answer the question. The output of a 555 is high (LED on) while it is "idling." If that is the case, then you may be measuring the LED current.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: November 08, 2021, 01:20:11 am
content: Quote from: StillTrying on November 07, 2021, 06:26:47 pmNo, this is the current that shows in the lab power supply display. My multimeter shows 2.5mA.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: November 08, 2021, 01:23:49 am
content: Quote from: jpanhalt on November 07, 2021, 06:37:42 pmThe circuit is not on when the led is off. Look at the schematic postet over. When the pulse is given every 25 second the led lights up to indicate the circuit work and the circuits draw 92mA.

author: ledtester
author_rank: Super Contributor
timestamp: November 08, 2021, 02:45:31 am
content: You might be interested in the "CS555" -- draws only 5uA:-https://www.jameco.com/Jameco/Products/ProdDS/2146433.pdf-http://www.customsiliconsolutions.com/downloads/Revised%20Standard%20products/CSS555_App_Note1_Serial_Interface.pdf- An explanatory video:https://youtu.be/5fJvhQCE1ow

author: ElectricPower
author_rank: Frequent Contributor
timestamp: November 08, 2021, 02:51:57 am
content: Quote from: ledtester on November 08, 2021, 02:45:31 amMaybe. But i think that a 2mA ish draw is not bad from an power bank in real life. If it charges the phone with 1000mA or 1002mA i dont care. But yes, i'm going to order a genuin timer

author: ledtester
author_rank: Super Contributor
timestamp: November 08, 2021, 04:15:44 am
content: Ok, here's another idea...The TPL5010/5110/5111 chips are timers which consume only 35nA - cost is about $1.With a single external resistor they will produce a 50ms pulse at a programmed period (determined by the resistor). With a simple RC-mosfet combo you can stretch that pulse to the 2 seconds that Mr. Carlson's design implements.

author: StillTrying
author_rank: Super Contributor
timestamp: November 08, 2021, 09:19:59 am
content: I suppose yet another way to check if it's a CMOS or bipolar 555 is to disconnect the RST from +5V and connect RST to 0V through the mA meter. A bipolar RST passes about 0.5 mA to 0V, a CMOS about 0 mA.I'll be surprised if it turns out to be the 555.