Quiescent current much, much higher than specified (in JK flip-flops). Are manufacturers lying?

Thread Starter

aromring

Joined Jul 22, 2015
28
Hello EEs,

I have ordered the following JK flip-flops:
http://www.ti.com/lit/ds/symlink/sn74hc109.pdf
(the SN74HC109 version) and
https://media.digikey.com/pdf/Data Sheets/ST Microelectronics PDFS/M74HC112.pdf
encouraged by the advertised low quiescent supply current (Icc) at room temperature: 4 uA in the first (page 4) and 2 uA in the second (also page 4). In any case, both manufacturers promise that in absolutely worst case the Icc should not exceed 40 uA.
To my dismay, the actually measured Icc was 3-4 orders of magnitude higher than that! First, I've made a vanilla measurement with just power pins connected, as illustrated in the first circuit in the attached figure:
jk-flip-flop-testing.png
Then I noticed the "VI = VCC or GND" note in "Test condition" column. Well, fair enough: with inputs flapping in the breeze the flip-flop may be oscillating internally which would explain extra energy draw. Therefore, I connected both as intended (second circuit). Essentially, all the inputs are now tied to Vcc and both Q and Q' are now stable (LOW and HIGH, respectively, upon powering up). It didn't matter though: the ammeter still showed Icc = 1000-12000 uA for the first JK and 1000-25000 uA for the second. In both cases the current was swinging wildly between these ranges.

Are both manufacturers lying through their teeth in the data sheets? ;) Or is it me who is doing something wrong?

(I want to use the JK as a Q toggle driven in actual circuit by an external clock rather than the switch SW1 used here for testing.)
Thank you in advance for your valuable experience in this matter.
 
Last edited:

WBahn

Joined Mar 31, 2012
24,688
Have you connected up BOTH of the flip flops as shown in your right hand diagram? I can't tell for sure what the "both" is that you are referring to -- whether it is both inputs and both outputs on just one of the flip flops, or if it is referring to both flip flops.
 

Thread Starter

aromring

Joined Jul 22, 2015
28
Have you connected up BOTH of the flip flops as shown in your right hand diagram? I can't tell for sure what the "both" is that you are referring to -- whether it is both inputs and both outputs on just one of the flip flops, or if it is referring to both flip flops.
"Both" means "both flip-flops".
 

Papabravo

Joined Feb 24, 2006
12,389
On the package. Any unconnected input which is not in a defined state will cause an increase in quiescent current. This increase could be several orders of magnitude. The value of Vcc will also have an effect. Read the datasheet carefully to understand what Vcc is used in the manufacturer's test. For the record, I am not aware of any manufacturers, that have successfully lied through their teeth on datasheets. I would be damn careful of my facts before I threw wild accusations around.

Still it is possible that the provenance of the parts you have may be questionable. They may even be factory rejects recovered from the bin and resold by unscrupulous distributors.
 

AnalogKid

Joined Aug 1, 2013
8,102
Are both manufacturers lying through their teeth in the datasheets? Or is it me who is doing something wrong?
It's you.

The part is decades old, with probably a billion sold in that time, used by tens of thousands of designers. It's you.

There is an undefined/unconnected pin in both drawings between the Q and -Q outputs. What is that?

Parametric characterization of a CMOS device is a very tricky thing. What are you using to measure the current? What is it's series impedance in the Vcc (actually, Vdd) connection. Your schematics show no decoupling of the power rail after this series impedance. What is the DC voltage on the Vcc pin after the ammeter?

I suggest you replace R1 and R2 with 0 ohm connections, and move C1 to the Vdd pin, keep its connections between Vdd and Vss or GND as short as possible, and try again.

ak
 

Thread Starter

aromring

Joined Jul 22, 2015
28
It's you.

The part is decades old, with probably a billion sold in that time, used by tens of thousands of designers. It's you.

There is an undefined/unconnected pin in both drawings between the Q and -Q outputs. What is that?

ak
Ah, there is no 10th pin. The "pin" between Q and Q' is an artifact of how Circuitlab.com defines custom part. For measuring current I'm using a digital multimeter.

Thank you for all the valuable suggestions; I'll try them tonight.
 

#12

Joined Nov 30, 2010
18,076
Your digital multimeter cannot resolve 1 uA accurately.

ak
I have a trick to fix that. Place the voltmeter in DC volts and connect it in series with the chip supply pin. If the meter has an impedance of 1 meg, a microamp will show up as 1 volt on the meter. Losing a whole volt can be problematic in a 5 volt circuit but a suitable resistor can be placed in parallel with the meter to change the range of measurement. For instance, 10K in parallel with the meter will require 100 ua to display a volt on the meter.

With my excellent meter, I can measure to 1 na with little difficulty.
 

Thread Starter

aromring

Joined Jul 22, 2015
28
Yup, that was me. I have transferred everything to another breadboard, carefully reseated all components, and now I'm measuring Icc = 0-5 uA. One of connections must've been loose.

I am sorry for raising false alarm. And for accusing ;) poor manufacturers. This is the first time I'm dealing with JKs and through this embarrassment I've gained valuable experience. Wildly varying Icc = unconnected input pin, no matter how well it all looks.

Thank you all for replying!
 

AnalogKid

Joined Aug 1, 2013
8,102
Wildly varying Icc = unconnected input pin, no matter how well it all looks.
Just so you know, the probable cause of the high static current is not internal oscillation. CMOS parts have totem-pole input, internal, and output stages. A totem pole circuit has one pull up transistor and one pull down transistor stacked up in series. If only one is on at a time, no problem. but if the inputs are floating, both can be partially on at the same time, causing high (for CMOS) static current. In early CMOS parts this could be fatal, but in newer ones it just causes high static current and ambiguous output states. External oscillation is possible, because it takes very little capacitance to form a feedback network when the input impedance is 1 billion ohms.

ak
 
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