Go Down

Topic: Perboard Brownout but Works with Wires (Read 1 time) previous topic - next topic

TomGeorge

Hi,
I have left HOPE with a feedback note, enquiring about the DC specs.
Its poor when they provide a 9Vdc option for "battery" operation yet do not provide any of the ESSENTIAL power consumption figures to back it up. 

Tom.... :)
PS, If they still balk at the request, I'll try with my business email and letter head.
Everything runs on smoke, let the smoke out, it stops running....

larryd

9V (no resistance nor current specs).

Do you have a DMM, measure the resistance.


No technical PMs.
The last thing you did is where you should start looking.

larryd

#32
Jul 20, 2018, 05:55 am Last Edit: Jul 20, 2018, 05:55 am by larryd
PS, If they still balk at the request, I'll try with my business email and letter head.
+1  for being a good guy!



No technical PMs.
The last thing you did is where you should start looking.

czu001

#33
Jul 20, 2018, 07:41 pm Last Edit: Jul 20, 2018, 07:49 pm by czu001
+100 from me.  Ok.  I used Sketch (illustration app for Mac) to try and draw this diagram.  I know it's probably horrible, but I'm trying to communicate the concept and how I've wired it into the rails on the perfboard.  I definitely need to learn how to use a real schematic application, but I really hope it's enough to show how I've got everything wired up.

I've also included a close up of the board itself so that it shows how things are actually soldered.  I wouldn't be surprised if the diagram concept doesn't actually match the perfboard soldering.  I really appreciate the help, and just let me know if this diagram makes no sense. 

Also, where did you leave feedback at?

czu001

#34
Jul 20, 2018, 08:17 pm Last Edit: Jul 20, 2018, 08:18 pm by czu001
9V (no resistance nor current specs).

Do you have a DMM, measure the resistance.
I took my DMM, and I measured about 4ohms.  Here are the specs of the motor driver and voltage regulator.

Motor Driver
Output current: up to 1.2 A continuous (5 A peak for a few milliseconds)

Voltage Regulator
Typical maximum continuous output current: 1.5 A (when input voltage is close to the output; the Typical efficiency and output current section below shows how the achievable continuous output current depends on the input and output voltages)

Perhaps I need a bigger resistor?  I'm wondering if I am shorting out one of the components with the 10ms pulse.

TomGeorge

#35
Jul 21, 2018, 02:06 am Last Edit: Jul 21, 2018, 02:08 am by TomGeorge
Hi,
I got a response but not enough info.
Tom,


Quote
The specs you will need are the following:
DC voltage
50 millisecond pulse
 6-9 volts
I have replied, saying it is for a battery application with a microcontroller and need to know current to see how to drive it.
Its the start of the  weekend so may need to wait for a reply.

Can you please post a picture of the valve/solenoid and its terminals?

Tom... :)
PS. What device that is designed for battery application does not have a power or current spec?????
PPS. I suspect the guy I'm talking to is a salesmen, not tech guy.
Everything runs on smoke, let the smoke out, it stops running....

wvmarle

4 Ohm resistance that would suggest a peak current of 3A at 12V. That's a lot.

That photo also shows a solderless breadboard, not a soldered perfboard. Those breadboards may not be able to carry that much current.
Quality of answers is related to the quality of questions. Good questions will get good answers. Useless answers are a sign of a poor question.

TomGeorge

Everything runs on smoke, let the smoke out, it stops running....

TomGeorge

#38
Jul 21, 2018, 02:39 am Last Edit: Jul 21, 2018, 02:41 am by TomGeorge
Hi,
In your diagram, what are the boxes?
Where are you getting your power from?

Are you boosting low voltage to a higher voltage?
If so, do you realize that you need MORE input current for any output current that is needed.
POWER IN = POWER OUT + LOSSES.

POWER IN  >> POWER OUT.

Are you trying to drive the DC latching valve directly from the controller?

Tom.... :)
Everything runs on smoke, let the smoke out, it stops running....

czu001

#39
Jul 21, 2018, 03:35 am Last Edit: Jul 21, 2018, 03:36 am by czu001
That photo also shows a solderless breadboard, not a soldered perfboard. Those breadboards may not be able to carry that much current.
It's a perma-protoboard from Adafruit.  It is soldered, but the pads look just like a breadboard.  https://www.adafruit.com/product/1609

"50 millisecond pulse"

It's definitely a 10ms pulse.  I can't remember where I read that from, but I know it used to be somewhere on Hunter's website.  I'll post a picture of the solenoid in the next post.

In your diagram, what are the boxes?
Where are you getting your power from?

Are you boosting low voltage to a higher voltage?

Are you trying to drive the DC latching valve directly from the controller?
Starting at the top:

The box on the right is the Adafruit ESP32 HUZZAH.

The box next to it on the left is a pololu voltage regulator that will boost the 3.3V to 9V. (https://www.pololu.com/product/2869)

The box underneath is a pololu brushed dc motor controller. (https://www.pololu.com/product/2960)

I am trying to boost voltage from 3.3V to 9V and pulse the solenoid open via the capacitor+resistor.

The logic flow is as follows:

When it's time to open the valve, I ENABLE the voltage regulator, let the regulator fill the capacitor up via resistor, then after some time (470ms), I pulse the solenoid open (a 10ms pulse to either INPUT 1 to open or INPUT 2 to close on the motor controller).

The power supply comes off of the Arduino's voltage regulator either via USB or 3.7V Lipoly battery.  I'm using USB right now so that I can see the serial output; however, it will use the 3.7V Lipoly battery in the field.

If so, do you realize that you need MORE input current for any output current that is needed.
POWER IN = POWER OUT + LOSSES.

POWER IN  >> POWER OUT.
My hope that this circuit would behave like a Capacitor Discharge Unit.  Since the pulse is so short, I was hoping the capacitor would build up enough energy to "kick" it open.  It works on breadboard and on the protoboard via wires from the Arduino, but it fails when going directly from the soldered headers.

Thank you so much for the help!

TomGeorge

Hi,


Quote
The logic flow is as follows:

When it's time to open the valve, I ENABLE the voltage regulator, let the regulator fill the capacitor up via resistor, then after some time (470ms), I pulse the solenoid open (a 10ms pulse to either INPUT 1 to open or INPUT 2 to close on the motor controller).
Its probably when you enable and have to fill the cap AND pulse the valve that the converter is consuming peak current and causing power drop outs.

What happens if you leave the DC-DC enabled ALL the time and just pulse the controller PCB????


Do you have the motor controller disabled until you need to pulse?

Thanks.. Tom... :)
Everything runs on smoke, let the smoke out, it stops running....

wvmarle

The box next to it on the left is a pololu voltage regulator that will boost the 3.3V to 9V. (https://www.pololu.com/product/2869)

The box underneath is a pololu brushed dc motor controller. (https://www.pololu.com/product/2960)

I am trying to boost voltage from 3.3V to 9V and pulse the solenoid open via the capacitor+resistor.
To pulse your 4Ω solenoid for 10 ms, while the voltage drops from 9V to 6V, you need a 2,200 µF capacitor. For a 50 ms pulse you need over 11 mF of capacitance. This means that your boost converter will have to supply the full 2.25A (9V/4Ω) for at least a few ms as the voltage drops, and that means it will try to draw nearly 7A on the input side. I was using this calculator for the capacitor.

As it's an inductive load the current is not instantaneous, that will help lessen the load and you to get away with a somewhat smaller cap than calculated, still it won't be long before it does reach full current.

You say you try to charge the capacitor via a resistor (I suppose to limit the current), but I don't see that in your circuit, only a 470Ω one in parallel to the capacitor (doing nothing more than wasting power that way).
Quality of answers is related to the quality of questions. Good questions will get good answers. Useless answers are a sign of a poor question.

czu001

#42
Jul 23, 2018, 02:07 am Last Edit: Jul 23, 2018, 02:17 am by czu001
1) Thank you for the great feedback.

2) I think I figured out what was causing the brownout.  The motor controller has high impedance on the inputs when both are LOW.  I didn't have a pull-up resistor on the pins from the Arduino driving it, so when the Arduino would go LOW on both inputs all of the current was running through the solenoid.

3)  The information from wvmarle is great!  I now realize I need to break this circuit down into 2 different parts and get each working.  Now that I have the resistance of the solenoid, I need to calculate the capacitance required to boost the circuit (and keep the voltage from dipping too low), the amount of time to charge the capacitor, and the resistance required to keep the voltage regulator from shorting (current limiter).

I think the values are 2,200 uF, 10ohm current limiting resistor, and 20ms to charge the capacitor (9V @ 10ohm @ 2,200 uF).  Seems like I tried to charge the cap with 20ms and it wasn't enough, but I'm willing to bet it was how I had it wired.  I used this calculator (https://www.allaboutcircuits.com/tools/capacitor-Charge-and-time-constant-calculator/).

Also, the resistor is in parallel, but it needs to be in series as mentioned.

I'm also going to play with the regulator to see if I can lower the voltage to 7 volts since the salesman mentioned 6-9 volts (which is probably right since it runs off of a 9V battery).

Also, I learned that I guess a breadboard doesn't have the same current as a perfboard. It seems like more current can run through a perfboard (is that right?).

I'll post my updates as I move further along.  Hopefully, I'll get it working this time.  I could have a separate 9V power supply for the solenoid, but I really want a single power supply so that I can more easily replinish/recharge the battery.

czu001

I added the truth table from the motor controller.  I hope I'm correct on this stuff.  Thanks so much for the help.

wvmarle

I think the values are 2,200 uF, 10ohm current limiting resistor, and 20ms to charge the capacitor (9V @ 10ohm @ 2,200 uF).  Seems like I tried to charge the cap with 20ms and it wasn't enough, but I'm willing to bet it was how I had it wired.  I used this calculator (https://www.allaboutcircuits.com/tools/capacitor-Charge-and-time-constant-calculator/).
2200 uF * 10 Ohm = 22 ms for RC time constant.

To fully charge a capacitor you need 5xRC so in this case 110 ms.

You have to do this charging shortly before you need the pulse as capacitors do self-discharge. Also you really should charge it to 9V, as the moment you use the solenoid the voltage starts to drop, fast. The 9V supply will also start to help out a bit, so it doesn't drop as fast as when it would be just the capacitor (I'm sure it can be calculated), but still, it'll drop fast.

And indeed solderless breadboards can't carry much current (typically around 0.5A), a perfboard can handle a lot more.
Quality of answers is related to the quality of questions. Good questions will get good answers. Useless answers are a sign of a poor question.

Go Up