Built circuit w/ timer IC connected to SHDN pin of a voltage regulator - not shutting down properly?

Thread Starter

Mahonroy

Joined Oct 21, 2014
406
Hey guys,
I built a circuit using a timer IC (TPL5111) connected up to the SHDN pin of a voltage regulator (MAX1724).
Here is a simplified schematic of the intended use of the timer IC:
timer_question1.jpg

The basic idea is that the device powers up, a microcontroller can do a few things, the microcontroller can indicate to the timer that its done doing whatever it wanted to do, and the device powers down for a few minutes with very little quiscent current drain... greatly increasing battery life.

I verified that the SHDN pin (EN pin) is held low for the duration of the timer, at which point its brought up to 3.3V to power everything up. The microcontroller does the task, then indicates to the timer that its done, at which point the SHDN pin is taken back to 0 volts again for the duration..... then repeat. This is good and is how its supposed to work.

My problem is that I can see the power rail gets 3.3V when powered up, but when powered down its held at a steady 2.8 volts. I can see that during this period of time current draw is still 4mA. It should be 0.0002mA when powered down (quiescent current of the voltage regulator in shutdown mode, and quiescent current of the timer IC).

Do you know what might be causing this behavior? Or how I should go about troubleshooting it? Any help/advice is greatly appreciated!

Here is my actual schematic for reference:
timer_question2.jpg
 

Deleted member 115935

Joined Dec 31, 1969
0
max1724 is not a ldo, its a boost switcher,

shut down does just that, it shuts down the switcher,
you still get current flowing via the coil , fets in the 1724.

works as expected.
 

Thread Starter

Mahonroy

Joined Oct 21, 2014
406
max1724 is not a ldo, its a boost switcher,

shut down does just that, it shuts down the switcher,
you still get current flowing via the coil , fets in the 1724.

works as expected.
I know it is not an LDO, its a boost switching voltage regulator. Shouldn't this work somewhat the same way?
I am not really following... the MAX1724 says its a logic controlled shutdown, and there is a built in N-channel power switch?
 

Alec_t

Joined Sep 17, 2013
14,280
This is what the 1724 datasheet says :-
"The MAX1723/MAX1724 enter shutdown when the SHDN pin is driven low. During shutdown, the body diode of the P-channel MOSFET allows current to flow from the battery to the output. V OUT falls to approximately V IN - 0.6V and LX remains high impedance."
That's why you see 2.8V on the output during shut-down.
 

Thread Starter

Mahonroy

Joined Oct 21, 2014
406
Thanks for the info guys. I didn't understand the difference between "shutdown" and "true shutdown", I was comparing so many different boost converters that I assumed the shutdown all worked the same, and missed that piece of the datasheet.

So with that in mind, I am thinking I can just use the MAX1795 instead. Its shutdown pin is the opposite though (high is off, low is on). Can I just use an inverter such as 74LVC1G14SE-7 to handle this? I am slightly worried that the inverter would end up using too much power when it is holding the pin high for shutdown?

It would be great to get an opinion on this so I don't make another stupid mistake, and thanks again.
 

Thread Starter

Mahonroy

Joined Oct 21, 2014
406
Yes. CMOS will dissipate only a few micro-watts.
Maybe this would be a better choise then?
NC7SV04P5X
Its for a battery powered device with long duration sleep modes. In sleep mode the only thing being powered is the timer IC (TPL5111), the boost regulator (MAX1795 in shutdown mode), and the inverter.
 

Thread Starter

Mahonroy

Joined Oct 21, 2014
406
Looks a better choice if you really need to shave off the odd uW.
Thanks again for all the help!
So I re-designed this thing, only using the MAX1795 instead, and an inverter for the SHDN pin since it works the opposite as before.

I am still getting a similar problem? When the SHDN pin is pulled low, this enables the MAX1795, and I see the 20mA @ 3.5 volts. When the SHDN pin is pulled high to 3.5 volts (same voltage as the battery supply, which is supposed to put the MAX1795 in shutdown mode), I see 5mA @ 2.55 volts.

How can this be?

Thanks again and any help is greatly appreciated! Below is the new schematic:
timer_question3.jpg
 

Thread Starter

Mahonroy

Joined Oct 21, 2014
406
Do you think this means that power is getting drawn from something external? Or is my shutdown circuit malfunctioning?
 

ian field

Joined Oct 27, 2012
6,536
I know it is not an LDO, its a boost switching voltage regulator. Shouldn't this work somewhat the same way?
I am not really following... the MAX1724 says its a logic controlled shutdown, and there is a built in N-channel power switch?
A buck regulator would break the current path from input to output by not energising the transistor.

A boost regulator would still pass some current via the flyback inductor.

Maybe a capacitor coupled buck/boost topology would solve the problem.
 

Thread Starter

Mahonroy

Joined Oct 21, 2014
406
Can't tell, since the external thing isn't shown/specified.
Here is the rest of the circuit. Everything runs off that 3.3v rail that the MAX1795 generates. If a DC power connector is plugged in, the batteries are disconnected and the 3.3v rail is generated from the R-78E3.3 instead. The battery power goes to the TPL5111, MAX1795, Inverter, and Channel 0 of the ADC, so I couldn't imagine power being drained from elsewhere when the MAX1795 is in shutdown mode.

circuit_question_lowpower.jpg
 

Alec_t

Joined Sep 17, 2013
14,280
One possibility is that the unwanted current drain is via IC1. Its datasheet doesn't indicate what happens if the output is driven by an external voltage when the input is 0V. But of course there shouldn't be any external voltage there if the MAX chip is in true shut-down.

Edit: Another possibility is that the 74AHCT inverter isn't working as you expect since, according to TI, "AHCT devices should operate over a range of supply voltage from 4.5 V to 5.5 V". You could check the voltage on the inverter output pin to verify this.
 
Last edited:

Thread Starter

Mahonroy

Joined Oct 21, 2014
406
One possibility is that the unwanted current drain is via IC1. Its datasheet doesn't indicate what happens if the output is driven by an external voltage when the input is 0V. But of course there shouldn't be any external voltage there if the MAX chip is in true shut-down.

Edit: Another possibility is that the 74AHCT inverter isn't working as you expect since, according to TI, "AHCT devices should operate over a range of supply voltage from 4.5 V to 5.5 V". You could check the voltage on the inverter output pin to verify this.
I didn't use the "74AHCT" inverter, I used the low power one "NC7SV04P5X". (sorry I just noticed the schematic text is not updated)
I verified that the inverter is pulling the SHDN pin high (3.5 volts) for enable, and pulling to 0 volts for shutdown mode.

I will de-solder IC1 (R-78E3) for now to remove that variable and report back what I find.

Thanks again for checking this out!
 

Thread Starter

Mahonroy

Joined Oct 21, 2014
406
Ok I de-soldered the IC1 (R-78E3) and nothing changed. Still getting about 5mA when shutdown mode is enabled.
FYI, I am taking the current reading at the batteries.

However... I tried with only a single battery and the thing now works. These batteries are AA (Energizer Ultimate Lithium) and the damn things are reading 1.75 volts each. So I guess the 3.5 volts is too much for the MAX1795 and the shutdown does not work.... ugh. I was hoping the 2x batteries together would be under 3.3 volts, but that is not the case.

So I'm guessing that after the batteries discharge a bit then everything will work properly, but until that happens current draw will be higher.

Do you think I should try and incorporate something to drop the voltage slightly? Or would you just leave this as is?

EDIT:
I found 2x batteries that were reading 1.475 (2.95 together), tried them and I get the same problem with the 5mA current draw in shutdown. So it seems like there is heavy current draw until the batteries are really drained? Man I am stumped
 
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Alec_t

Joined Sep 17, 2013
14,280
Can you isolate the 1795 from all loads except a known dummy load, connect the shut-down pin directly to the battery only, and see if the problem persists? That should establish if the 1795 itself is misbehaving.
 

Thread Starter

Mahonroy

Joined Oct 21, 2014
406
Can you isolate the 1795 from all loads except a known dummy load, connect the shut-down pin directly to the battery only, and see if the problem persists? That should establish if the 1795 itself is misbehaving.
Sure, I can dremel out the trace and put a resistor to create a load. I will also de-solder the timer and connect the SHDN pin to battery. I was looking through the datasheet and could not find a minimum load specified, what would you recommend? Something like a 100 ohm resistor between the 3.3v output and ground should create a 33mA load would that do?
 

Alec_t

Joined Sep 17, 2013
14,280
I've never used a 1795, but 33mA sounds reasonable for testing, given that the datasheet specifies a >100mA capability.
 

Thread Starter

Mahonroy

Joined Oct 21, 2014
406
I've never used a 1795, but 33mA sounds reasonable for testing, given that the datasheet specifies a >100mA capability.
Hi Alec,
I built a PCB with only the MAX1795, the input capacitor & inductor, the 2 output capacitors, and a 100 ohm resistor for a load. Like this the thing works perfectly. I get 40mA current draw when SHDN is grounded, and when I connect SHDN to battery I get only 0.01mA current draw. This was with 3.5 volts input too. I tested with a single battery of 1.75 volts the shutdown continues to work properly.

So I guess the next step is that I should solder on the timer, inverter, and those components corresponding circuitry? Is there something I should try first you think?

Thanks again for the help!
 
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