How do I minimise transistor leakage current?

Other Hardware General Electronics
1

I have a circuit (see attachment - sorry about the focus) that is battery powered and needs to conserve the battery. It takes a momentary press of a push-button to short the collector and emitter of the transistor to power up the Tiny85 which then provides current to the base to keep the circuit powered until the Tiny85 turns off the base after a preset period. (another circuit runs off the terminals to the Tiny85).

The circuit runs fine but when it is off I still have ~3 micro-amps of power running through the transistor thereby slowly flattening the battery. What can I do to reduce this current draw to nothing or less than present. I do not want to use a relay.

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2

I've 2 questions
1- 3uA is not much current. What size battery are you using and how long do you expect this to last?

2- How long did it take to make the perfect circle for the transistor?

3

Hi Tinman,

  1. I'm using a 2000 ma cell that I want to last for a year on standby (obviously turning the circuit on from time to time will decrease this time). A small amount, I know, but the less power "wasted", the more there is left for the application. Any small effort fix such as using a FET or different transistor is a refinement worth having. I just don't know how to get there.

  2. 5 seconds; I have a very steady hand and a good eye........ (plus a 5 cent piece).

4

I think what you are seeing is the transistor being turned slightly.

Reasoning:

In the off state, ideally the ATTiny + and Ground are at the same potential (~ 5V) so the output pin on the ATTiny is also not too much below 5V. Hence the transistor is at least partly on.

Solution 1)
You could put a zener diode 2.7 to 3.3V zener diode in series with the 1k. This will reduce the voltage available on the base when the system if "off".

You should also add a resistor from the base to ground to bypass any leakage. Because we don't know the "off" characteristics of the ATTiny under these conditions you will have to play with the values.

Solution 2)

You could add a PNP pre-driver stage that would be pulled to ground to turn on. In the "off" condition you will not have to worry about the output characteristics of the output pin. This is a much better solution as I believe it will be more stable. Your current circuit and solution 1 rely on uncontrolled parameters of the µP.

BTW the PN100 should have in circuit leakage in the order of 10's of nano amps.

Good luck

John

5

When the transistor is off, MCU ground is 'high", and the input pin protection diode pulls the input pin to "ground - 0.65volt, giving the transistor (some) base current.

Can only be done with a two-transistor switch.
High side:
PNP transistor, emitter to battery, collector to attiny+, base via a (1k) resistor to collector of...
NPN transistor, with emitter to ground, and attiny output pin via 10k to base.
Start switch goes across collector/emitter of NPN transistor.
Flip everything upside-down, including NPN/PNP, if you want to switch low-side.
Leo..

6

Thanks guys. Do you mean like this?

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7

Almost.
1k between PNP base and NPN collector (as drawn), but not to attiny ground.
Attiny ground goes directly to ground, because only the positive supply is switched.
Maybe easier to understand with the attached diagram.
(forgot to add the momentary switch across the NPN transistor)

Might also need a (10k) resistor between emitter and base of the PNP transistor for minimum leakage.
Without that, any leakage of the NPN transistor is amplified by the PNP transistor.
Leo..
Switch.png

Switch.png

8

ATTiny85 should take less than 1uA of current while in Power down sleep with BOD and WDT disabled. It is probbbly less than self-discharge of the battery. Maybe you can even save power by sleeping the Tiny instead of cutting power - when button is pressed you resume the program with everything ready and save time (power) spent reconfiguring everything.

9

Or you could put a diode such as a 1N4148 in series with the base in your original circuit...

Allan

10

allanhurst:
Or you could put a diode such as a 1N4148 in series with the base in your original circuit...

Allan

Why does this should help? Until forward diode drop of the transistor base + ATTiny protection diodes + any diodes you add < Vcc the transistor should be (paritally) open. In fact I wonder OP got only 3uA of "leakage". Maybe the ATTiny is in power down sleep?

11

Ah - sorry. Didn't look at OP's circuit.

Be ok with a single fet ( 2N7000?) with a 100k pulldown on the gate.

Allan

12

Can't be done with a single transistor.
The input protection diode will try to phantom-power the MCU or drive the transistor.
Both will result in some power draw.
Leo..

13

lemming:
Hi Tinman,

  1. I'm using a 2000 ma cell that I want to last for a year on standby (obviously turning the circuit on from time to time will decrease this time). A small amount, I know, but the less power "wasted", the more there is left for the application. Any small effort fix such as using a FET or different transistor is a refinement worth having. I just don't know how to get there.

  2. 5 seconds; I have a very steady hand and a good eye........ (plus a 5 cent piece).

3uA for a year is 26mAh, completely insignificant here compared the cell's 2000mAh capacity,
you are worrying about the wrong thing. The self-discharge current of the battery itself is probably much
more, and that can only be changed by changing chemistries. What cell(s) are you currently using?

Note 'ma' should be mAh (milli-ampere-hours). The symbol for ampere is upper-case A. Capacity
is the product of current and time, so the units are different to current.

14

MarkT:
3uA for a year is 26mAh, completely insignificant here compared the cell's 2000mAh capacity,
you are worrying about the wrong thing. The self-discharge current of the battery itself is probably much
more, and that can only be changed by changing chemistries. What cell(s) are you currently using?

Note 'ma' should be mAh (milli-ampere-hours). The symbol for ampere is upper-case A. Capacity
is the product of current and time, so the units are different to current.

What Mark is doing here is calculating part of your project's power budget. It's actually very simple, you just multiply a bunch of numbers together to get the right conversion factor. The capacity used by a 3uA current over 1 year is calculated as follows:

Converting current and time to a capacity is as simple as multiplying the two numbers together.

3 uA * 1 year = 3 microamp-years. Now to convert microamp-years into milliamp-hours:

3 uAy * 0.001 microamps per milliamp = 0.003 mAy (milliamp-years)

0.003 mAy * 365 days per year = 1.095 mAd (milliamp-days)

1.095 mAd * 24 hours per day = 26.28 mAh.

Over a desired run time of 1 year, 3 uA of sleep current will use at most about 1.5% of your capacity. Is that significant enough to worry about? How much more extra run time can you get if you shave down this percentage?

This is why power budgeting is so important, because it lets you quantify which aspects of your systems are using how much power.

To calculate the capacity usage for states that you are in less than 100% of the time, you simply derate the above calculation by the percentage of operating time that you are in that state. Even if you sleep most of the time and are only awake briefly and unfrequently, the awake mode might still be the biggest power user if it uses much more current than sleep mode.

Suppose you're asleep 99.9% of the time, using just 3 uA of current. If awake mode uses more than 1,000x as much current as sleep (more than 3 mA), awake mode is the biggest power user in your system and is where you want to try and cut from first.

I have attached the Low-Power Design application note from Microchip for your study, which will explain many of the things you need to take into consideration when optimizing for battery life, and contains a great example of a power profile in Table 2 on page 7.

1416 - Low-Power Design Guide.pdf (274 KB)