Sudden drop in supply voltage when activating a 50mA load

Thread Starter

Zurn

Joined Mar 4, 2019
73
Hi everyone,

A few months ago I received some excellent advice in designing the following circuit:

Burn-Wire-Circuit.png

It's purpose is an ultra-redundant/fail-safe way to activate an anchor release system for an oceanographic instrument. It has been tested and, so far, works very reliably. (If you question the rationale for this circuit, the full thread can be found here). HOWEVER...

The 15V supply you see on the above diagram also powers a number of other circuits in the device. During the activation of the MOSFET there are about 1-2 ms where the supply voltage drops down to about 5V, which I believe is due to the inrush current of the sudden activation of the anchor release system. It's enough to cause some of the instruments to reboot, which isn't an option.

Before the MOSFET is activated, the system runs at about 356mA. The anchor release load (or "burn wire" as I've been calling it) only adds about 50mA to the load. However, during those first few ms, this must be MUCH larger. Unfortunately, I lack the tools to properly measure this inrush current (clamping meter, etc).

So my question is: what would YOU do to stop this from happening? I have imagined/researched a number of options, including adding capacitance to the MOSFET drain to make a soft start (my anchor release system certainly doesn't have to turn on/off instantly), changing my MOSFET switch to a fancy load switch IC, limit the current available to burn wire load (thermistor?), and adding voltage regulation to the instruments being effected by the drop (going to do this regardless).

I should warn you: expect all suggestions to be met with further questions.
 

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danadak

Joined Mar 10, 2018
3,716
You may be seeing the gate charge current transient of the MOSFETS sucking a lot
of current out of uC output pin. Use a low value R between uC output pin and gate
of MOSFET to see if that helps. Use a scope to help in evaluation monitoring supply.

The gate R will slow turn on of MOSFET so there is a tradeoff. Also MOSFET will spend
more time not hard on hence dissipating more power. So try 50 ohms, maybe then 100
ohms to see if that helps.


Regards, Dana.
 

Danko

Joined Nov 22, 2017
843
Electrodes in sea water are supercapacitor (20uF on every sq. cm of electrode surface).
You may lower charging current by increasing time of MOSFET activation.
1574987773203.png___________________EDIT: 1575051982898.png
"The burn wire is a simple and effective sacrificial anode that upon disintegration by electrolysis will release the tension in the retaining hose clamp, allowing the MicroXM to slip out freely. Upon the trigger of a release signal, a positive electric potential (13–16 V via a 10V resistor) is applied to the burn wire, with respect to a cathode in contact with seawater, and the electrolysis disintegrates the burn wire, which frees the MicroXM. The burn wire is insulated from seawater except for a 3-mm section. The electrolysis is concentrated into this short and exposed section, and disintegrates the wire in 70–100 s. A release signal is triggered upon (i) hitting the ocean bottom, (ii) losing communication with the MicroXM continuously for more than 60 s, (iii) exceeding a selected depth, or (iv) exceeding a selected duration." LINK
 
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DickCappels

Joined Aug 21, 2008
6,011
Your burn wire is a postitive temperature coefficient resistor. When it is cold the resistance is much lower than when hot.

First choice would be to use an independent battery for the burn wire.

Reading literature on Ultracapacitors makes me think that its very low impedance would prevent much of a voltage dip, so maybe placing capacitors across the batteries will solve the problem. If Ultracapacitors can be used to start diesel train engines in freezing weather they should be able to power burn wires.
 

Thread Starter

Zurn

Joined Mar 4, 2019
73
You may be seeing the gate charge current transient of the MOSFETS sucking a lot
of current out of uC output pin. Use a low value R between uC output pin and gate
of MOSFET to see if that helps. Use a scope to help in evaluation monitoring supply.

The gate R will slow turn on of MOSFET so there is a tradeoff. Also MOSFET will spend
more time not hard on hence dissipating more power. So try 50 ohms, maybe then 100
ohms to see if that helps.


Regards, Dana.
I'll give this a shot. I actually already had 10k resistors from the gate to the source on the MOSFETs, I just forgot to update my graphic (which I just did).

First choice would be to use an independent battery for the burn wire.
This would be great but it's (currently) not an option due to weight/size restrictions... may go this route for future prototypes.

Reading literature on Ultracapacitors makes me think that its very low impedance would prevent much of a voltage dip, so maybe placing capacitors across the batteries will solve the problem. If Ultracapacitors can be used to start diesel train engines in freezing weather they should be able to power burn wires.
Interesting, I'm not yet familiar with ultra/supercapacitors, so I'll have to look into that. Thanks for the tip @DickCappels and @Danko.
 

danadak

Joined Mar 10, 2018
3,716
Interface wants to look something like this -

1575036957869.png

R1 you adjust to see if that helps with transient problem. I am thinking 50 - 100 ohms.

Ignore "electric strike", thats just your wire load. Also I do not think you need diode as
thats for inductive load turnoff transient suppression, and your wire probably not much
L.


Regards, Dana.
 

crutschow

Joined Mar 14, 2008
23,798
Adding a large capacitor from the supply to ground should help.

You could also add a Schottky diode in series with the power to the electronics that's being affected, also with a large capacitor to ground (1mF or so).
That capacitor will then provide the power for the electronics during the 1-2ms time of the momentary battery voltage dip with the diode isolating the drop from the electronics during that time.
 

Thread Starter

Zurn

Joined Mar 4, 2019
73
Interface wants to look something like this -

View attachment 193399

R1 you adjust to see if that helps with transient problem. I am thinking 50 - 100 ohms.

Ignore "electric strike", thats just your wire load. Also I do not think you need diode as
thats for inductive load turnoff transient suppression, and your wire probably not much
L.


Regards, Dana.
This didn't have any noticeable effect though I appreciate the suggestion.

Adding a large capacitor from the supply to ground should help.

You could also add a Schottky diode in series with the power to the electronics that's being affected, also with a large capacitor to ground (1mF or so).
That capacitor will then provide the power for the electronics during the 1-2ms time of the momentary battery voltage dip with the diode isolating the drop from the electronics during that time.
This sounds reasonable. There must be a 1mF cap laying around here somewhere...

No takers on the NTC thermistor route?
 

Thread Starter

Zurn

Joined Mar 4, 2019
73
Adding a 50 ohm resistor to the anode of my burn wire is currently doing the trick (after five or so tests). I figure replacing it with a similarly resistant thermistor would bypass the extra power consumption (it typically remains active for 5 minutes).

I wonder if anyone with experience sees any risks in this that I'm not aware of.
 

Danko

Joined Nov 22, 2017
843
Adding a 50 ohm resistor to the anode of my burn wire is currently doing the trick (after five or so tests). I figure replacing it with a similarly resistant thermistor would bypass the extra power consumption (it typically remains active for 5 minutes).
I wonder if anyone with experience sees any risks in this that I'm not aware of.
Nothing bad.
Only time between activation burn wire and anchor releasing will increased.
This time is inversely proportional to current.
EDIT:
Adding a 50 ohm resistor to the anode of my burn wire
Well, one lead of 50 Ohm resistor is connected to anode of burn wire,
so where is connected second lead of resistor 50 Ohm???
 
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Thread Starter

Zurn

Joined Mar 4, 2019
73
Doesn't the burn-wire go open circuit after it is activated?
I think of the burn wire as a node on the anchor release circuit that's only really "open" when it's out of the water (no sea water = no resistive/capacitive load therefore an incomplete circuit). Even when the burn is complete and the anchor drops, there's still a complete circuit between the power supply, the sea water, and ground until the circuit is switched off by the MOFSET. Unless I'm mistaken, of course...

Well, one lead of 50 Ohm resistor is connected to anode of burn wire,
so where is connected second lead of resistor 50 Ohm???
To the positive lead of my power supply (after the respective diodes of course). The resistor is inline with my burn wire circuit, simply acting as a current limiter.

Your burn wire is a postitive temperature coefficient resistor. When it is cold the resistance is much lower than when hot.
Just thinking about this point, which could lead to problems further down the road. I'll have to be sure to test in lowest temperatures.

Thanks for the responses everyone.
 

Thread Starter

Zurn

Joined Mar 4, 2019
73
So it turns out it wasn't the burn wire inrush current after all... in fact the problem has nothing to do with this part of the system!

It's due to the MOSFET I'm using to turn the system power on/off. My batteries are controlled by a uC that sends a shifting TTL state the gate of a MOSFET when a reed switch is activated. This MOSFET connects/disconnects the entire system from the negative rail of the batteries.

I still don't understand why it was acting up, its rating is for well above the entire system current. (@danadak's first reply might be a good place to start). I've since removed the MOSFET altogether though and everything's working fine. Quickly learning to never assume anything in electronics!

Now to redesig my power switching circuit, which calls for another thread....
 

Thread Starter

Zurn

Joined Mar 4, 2019
73
Vgs(th) is min. 2V, max 4V... I remember thinking at the time (way back in July, fresh out of technician school) that that meant a gate signal between those values would turn the MOSFET on. Have I misunderstood this?

The MOSFET in question in an IRF510, N-Channel.
 

Chris65536

Joined Nov 11, 2019
139
Vgs(th) is min. 2V, max 4V... I remember thinking at the time (way back in July, fresh out of technician school) that that meant a gate signal between those values would turn the MOSFET on. Have I misunderstood this?
That Vgs is not a voltage window you need to be within. It is the minimum voltage to turn on the MOSFET. I believe the min and max values are just the manufacturing tolerance for Vgs threshold. You can use 5v to be sure it's turned on. The absolute maximum Vgs is +-20v.
 

crutschow

Joined Mar 14, 2008
23,798
that meant a gate signal between those values would turn the MOSFET on. Have I misunderstood this?
Yes, completely. :)
The Vgs(th) is where the MOSFET just starts to turn on (usually 0.1-1mA, see below for the IRF510), not fully on.
And if the Vgs(ths) for the particular MOSFET you buy is 4V it will be totally off for a Vgs less than that.

1575658579365.png

So the usual Vgs to fully turn on a standard MOSFET is ≥10V, and for a logic-level MOSFET is ≥3-5V depending on its design.
 
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Thread Starter

Zurn

Joined Mar 4, 2019
73
AHA! Very good to know... so if my Vgs(th) max. was 4V, it would explain why a 3.3V TTL signal would be unreliable in keeping it on, correct? What it really wants is something more akin to 10V.

Well, I could track down a logic level MOSFET... though I'm currently looking into the world of SSRs...
 

bertus

Joined Apr 5, 2008
20,185
Hello,

The IRF series mosfets are standard mosfets.
The IRL series mosfets are logic gate mosfets.
To find the Vth, have a look at the various datasheets of the mosfets.

Bertus
 
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