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'[OT] Solenoid Driver - Need a few good ideas...'
2000\02\03@122106 by Jay

flavicon
face
Now that the list has slowed down discussion on relay drivers with low
voltage supplies, I need some ideas on the opposite problem, too much
voltage on my supply.

I need to drive a solenoid on and off from a logic level output.  The
solenoid has 28ohms of coil resistance, and a 500mA operating spec.  The
only available power I have to drive it must come from the 80VDC supply.  If
I had a lower voltage on the power supply, say 60V or less I could use an
off the shelf regulator design to give me a reasonable 14V output in which
to drive the solenoid.  Of course the simple solution is to use a power
resistor to 80V, in series with the solenoid, but this wastes way to much
power.

 I am thinking about using a hexfet as the switching device and of course
apply diode protection across the coil, and the needed current limiting
resistors to obtain 500mA.

The LM317HV looks like it could provide some solutions, but I don't think I
can exceed 60V from input to output.  Is there a way to cascade this device?
Here is a rough idea of what I had in mind.



                         80VDC Supply
                              |
                              |
                      Regulator Solution
                              |
                              |
                            +14VDC
                              |
                              |
                 Solenoid with diode protection
                              |
                              |
                 50ohm   ||---.
     Logic Out---\/\/\---||<--,  N Channel HEXFET
                         ||---|
                              |
                              |
                             Gnd

Priority of tradeoffs:
1) Reliability (room temp. operation only)
2) Needless Power loss
3) Number of components (complexity)
4) Price (don't want to spend more than $30)
5) Size (not much of a factor)

Thank you in advance for your help,
Jay

2000\02\03@123812 by Thomas C. Sefranek

face picon face
"Crosby, Jay" wrote:

> Now that the list has slowed down discussion on relay drivers with low
> voltage supplies, I need some ideas on the opposite problem, too much
> voltage on my supply.
>
> I need to drive a solenoid on and off from a logic level output.  The
> solenoid has 28ohms of coil resistance, and a 500mA operating spec.  The
> only available power I have to drive it must come from the 80VDC supply.  If
> I had a lower voltage on the power supply, say 60V or less I could use an
> off the shelf regulator design to give me a reasonable 14V output in which
> to drive the solenoid.  Of course the simple solution is to use a power
> resistor to 80V, in series with the solenoid, but this wastes way to much
> power.

ANY linear solution will waste just as much power!!!
(Including the one listed below!)
Skip the regulator, you can get cheap 100 volt FETs and a 50 ohm 15 watt
resistor.
(Believing 28 ohms and .5 amp is 56 volts to the coil!)

{Quote hidden}

WATTS are WATTS, you won't save any heat.

> 2) Needless Power loss

Use a switching converter! (Expensive!)

> 3) Number of components (complexity)

1 resistor, 1 FET

> 4) Price (don't want to spend more than $30)

$5 for the resistor, $1 for the FET

>
> 5) Size (not much of a factor)
>
> Thank you in advance for your help,
> Jay

--
Thomas C. Sefranek  WA1RHP
ARRL Instructor, Technical Specialist, VE Contact.
http://www.harvardrepeater.org
http://hamradio.cmcorp.com/inventory/Inventory.html

2000\02\03@134317 by wagner

flavicon
face
Ballast resistor will make your circuit's productivity less than 21%, so
terrible waste of energy, even being one of the most safe and cheap
solution.

With a productivity around 70 to 80% you could build a voltage inverter
with discrete components (since it is difficult to find step-down chips
with input voltage higher than 50Vdc).  Perhaps a simple 200Hz
oscillator made with logic gates (74HCT00) and a FET transistor driving
a small power transformer (110 x 24) would generate around 16 Vdc.

Your control circuit would simply turn on/off the oscillator.

                        +80Vdc        Diode
                             | .------->|---o---------.
                      Diode? 3:C            |         |
                             3:C            |     .-------.
__--                          3:C     100µF ===    | Relay |
          .------------.     | '--.        ---    '-------'
Control----| Oscillator |--FET     |         |         |
          |   200Hz    |     |    |         |         |
          '------------'     |    '---------o---------'
                             |              |        
                            Gnd            Gnd      

I guess this solution can cost you less than $15.  For sure better
designs can be made but this is the idea.

Oscillator:  $1
FET:         $1
Transformer: $7  XFR1635Q (Jameco)
Diodes+Cap : $2

Other idea is just use an adjustable pulse simetry oscillator, and drive
directly the solenoid coil with a FET, you can call it a PWM driver.
           
                       +80Vdc
                         |
                         3  Relay Coil
                         3  & Diode
                         |
           Assym         |
Control---Oscillator---FET
On/Off      15%ON         |
                         |
                        Gnd


You should find the best oscillator frequency and simetry to keep the
current curve never going above 700mA.
I guess this would cost you less than $3.

Wagner

2000\02\03@145311 by l.allen

picon face
> I need to drive a solenoid on and off from a logic level output.  The
> solenoid has 28ohms of coil resistance, and a 500mA operating spec.  The
> only available power I have to drive it must come from the 80VDC supply.

> Priority of tradeoffs:
> 1) Reliability (room temp. operation only)
> 2) Needless Power loss
> 3) Number of components (complexity)
> 4) Price (don't want to spend more than $30)
> 5) Size (not much of a factor)
>
> Thank you in advance for your help,
> Jay

Wagner has already said that PWM is the answer.. he is right but I
would add that if this is being driven by a PIC then the PIC should
have little difficulty providing the PWM.
I would add that there is a gotcha with this technique..... the PWM
duty cycle should be calculated to the power NOT the voltage. As
in 80V down to 14V at 500mA is derived by V div by A... so at 14
volts the solenoid uses 7 watts but at 80 volts the solenoid uses
(instantaneously) 228 watts, so the ratio for the PWM is 3%!!!!
(NOT 17.5% as in voltage only).
Been there ... blown up that by not thinking of this.

_____________________________

Lance Allen
Technical Officer
Uni of Auckland
Psych Dept
New Zealand

http://www.psych.auckland.ac.nz

_____________________________

2000\02\03@152619 by Thomas McGahee

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face
I would not use any form of linear regulator, as the
power dissipation would be about 33 Watts.

Wagner had the right idea: use the same FET that
switches the relay current ON as a switchmode
regulator. Since the load is pre-defined as being
14 V at .5 Amps we have all the info we need.

The FET will be held OFF whenever the relay coil
is to be un-energized. When the relay coil is to be
energized, the duty cycle will be based on 14/80
which is 17.5% ON time.

If you want, you can use the PIC to provide the
17.5% PWM signal. When the output is LOW, the
FET and relay are OFF. When the PIC provides a
17.5% PWM output, then the FET will chop current
into the relay and create an average of 14 V across
the coil.

If you don't want the PIC to generate the PWM
signal, then you can use a low power
555 timer, 2 resistors, one diode and two capacitors
to generate the 17.5% duty cycle rectangle wave.

You then need some way for the PIC to "gate" the
oscillator so that it can unconditionally turn
the oscillator off. A simple way to do this is to
directly pull the gate of the FET LOW.



              +------*---*----o +5
              |      |   |
              r      |8  | 4
              r R1 ---------
              r   7|       |
       -------*----|       |
       |      |    | Low   |
       V     ---   | Power |
      ---     A    |       |
       |      |    | 555   |3      17.5% Duty Cycle
       |      r    | Timer |------> Osc. Out
       |      r  R2|       |
       |      r    |       |
       |      |   6|       |
       *------*----|       |-----+
       |      |   2|       | 5   |
      ---     +----|       |    ---
      --- C1       |       |    --- .01 uf
       |           ---------     |
       |               | 1       |
      Gnd             Gnd       Gnd

Equations: R1/(R1+R2) = 17.5/100

Suggest 18k for R1, and 82k for R2.
This will give 18% which is close enough.

The value of C1 will not affect the duty cycle,
but it does determine the operating frequency.
Something around .01 to .1 should work fine.
You may have to play with the resistor/capacitor
values to find out which what duty cycle and
frequency actually works best.

Frequency must be high enough to prevent
"chattering" of the relay when it closes.
Also, too low of a frequency can allow
excessive current flow. I suggest starting with
a frequency of at least a few kilocycles.
If you select the right frequency you can
reduce power dissipation and maybe not
even need a heatsing on the power FET.
But ALWAYS use a good heavy duty heatsink when
testing, since "accidents do happen"

The nice thing about the 555 circuit is you can test
the technique without actually having to connect
it up to a PIC. Where we show the PIC I/O line
below you can substitute a pushbutton to ground
for testing purposes.


                         80VDC Supply
                              |
                              |
                              *-----+
                              |     | cathode
                              c    ---
                  relay coil  c     A anode
                              c     |
                              |     |
                              *-----+
                              |
                  1k     ||---.
      Osc. Out>--rrrr--*-||<--,  N Channel HEXFET
                       | ||---|
                       |      |
      PIC I/O>---------+      |
                             Gnd

The PIC should tristate the line when it want the
relay ON, and pull the line LOW when it wants the relay OFF.

If you experiment with this method, post your
experiences (and values used) to the PIC list.

Fr. Tom McGahee




{Original Message removed}

2000\02\03@160941 by paulb

flavicon
face
Lance Allen wrote:

> I would add that there is a gotcha with this technique..... the PWM
> duty cycle should be calculated to the power NOT the voltage. As
> in 80V down to 14V at 500mA is derived by V div by A... so at 14
> volts the solenoid uses 7 watts but at 80 volts the solenoid uses
> (instantaneously) 228 watts, so the ratio for the PWM is 3%!!!!
> (NOT 17.5% as in voltage only).

 No, the gotcha bit *you* on this one, Lance.  Your calculations apply
for resistive loads, this one is inductive.  He needs (as Thomas notes)
to keep the chopper frequency high enough so that the coil doesn't
saturate.

 The duty cycle will correspond to keeping the specified *current*, and
it will be closer to 17%.  It will certainly need to be "tuned" though.
And a suitable diode across the coil, plus a 100V capacitor on the
supply are essential.  Maybe a small decoupling resistance before the
capacitor also.

 It would be far more elegant to get the PIC to do this than invoke the
trusty 555.  In addition, you can use a higher initial duty cycle to
make the solenoid pull in and then when it is either detected, or
expected after a certain time to have done so, drop the duty cycle
(current) to half or less.
--
 Cheers,
       Paul B.

2000\02\03@161346 by Thomas McGahee

flavicon
face
Lance Allen is right, you have to base the duty cycle on the
ratio of the desired power consumption and the max power
consumption, not just the voltages. I was so intent on
demonstrating a working circuit with values that I failed
to notice that I was using a voltage instead of power ratio.
Actually, though, the duty cycle has to be higher than the
3% that Lance mentioned, due to the fact that the current
in the coil will be ramping up, not going up instantaneously.
In fact, if the frequency is made sufficiently high, it is the
inductive reactance of the coil that will dominate, not the
DC resistance. So, the value of the PWM is between 3% and
17.5% for the example we have been illustrating.

There are also differences when driving an inductive, resistive,
or capacitive load using a fixed PWM. Purely capacitive loads
charge up, and the current decreases while the voltage tends towards max.
Inductive loads will have ramping currents, and if a shunting
diode or snubber is not used, the coil can generate some really
nasty spikes when the FET turns OFF. Purely resistive loads
are the tamest of all loads

Also be careful with regards to the actual switching FREQUENCY
used. When the frequency is too low and you have an inductive
load, then you will have moments when the current will exceed
the maximum you want and may do permanent damage. You are safer
starting with higher frequencies and lower duty cycles and
experimenting until you find the best combination. Or eliminate
all the guesswork and use a technique that includes negative
feedback. That will cost you some more pennies, though...

Even a load like a light bulb can be burned out if a voltage
in excess of the lamps max voltage is applied for too long.
The actual operating frequency IS important!

Lance Allen said:

>I would add that there is a gotcha with this technique..... the PWM
>duty cycle should be calculated to the power NOT the voltage. As
>in 80V down to 14V at 500mA is derived by V div by A... so at 14
>volts the solenoid uses 7 watts but at 80 volts the solenoid uses
>(instantaneously) 228 watts, so the ratio for the PWM is 3%!!!!
>(NOT 17.5% as in voltage only).
>Been there ... blown up that by not thinking of this.

{Original Message removed}

2000\02\03@181445 by hmiller

picon face
Jay

There is a voltage regulator, TL783C that handles up to 125V input and
variably adjustable outputs 1.5-125V up to 700 ma. It looks like and
works exactly like the LM317T, minimal parts required. I got mine from
TechAmerica, and they were US$2.75 from the C898 catalog.

http://www.techam.com

Harley L. Miller     spam_OUThmillerTakeThisOuTspamsound.net

2000\02\03@182733 by Harold M Hallikainen

picon face
On Thu, 3 Feb 2000 15:33:46 -0500 Thomas McGahee
<.....tom_mcgaheeKILLspamspam@spam@SIGMAIS.COM> writes:
> I would not use any form of linear regulator, as the
> power dissipation would be about 33 Watts.
>
> Wagner had the right idea: use the same FET that
> switches the relay current ON as a switchmode
> regulator. Since the load is pre-defined as being
> 14 V at .5 Amps we have all the info we need.
>
> The FET will be held OFF whenever the relay coil
> is to be un-energized. When the relay coil is to be
> energized, the duty cycle will be based on 14/80
> which is 17.5% ON time.
>

       This does sound like a good approach to the problem. Remember to put a
diode across the relay coil, or you'll fry the FET! You can treat the
relay coil as an inductor, so when you turn the FET on and place a
voltage across it, the current linearly ramps up until you either hit
core saturation (making the inductance drop and the current ramp up
faster), or the current limit imposed by the resistance of the coil. When
the FET is opened, the coil current will ramp down, through the inductor.
I think it's important that the FET be off long enough to allow the
current to ramp all the way down. If it's not off that long, the coil
current will start ramping up from the ending current when the FET comes
back on. Eventually the current will ramp up to some dangerous value. You
could probably experimentally watch the FET current waveform (perhaps
with a small resistor in the source) to insure that it ramps up from zero
each time the FET comes on, then throw a little extra off-time in there
just for sure.
       Also, have a look at the Burr-Brown DRV102. This solenoid/valve driver
chip can handle 60V and 2.7A. It has an internal adjustable PWM generator
running at 24 kHz. It also has an adjustable "kick start" where the duty
cycle is 100% for some period so the solenoid is engaged with high power,
then backs off to hold it with PWM. Interesting chip...


Harold



FCC Rules Online at http://hallikainen.com/FccRules
Lighting control for theatre and television at http://www.dovesystems.com

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2000\02\03@190616 by Wagner Lipnharski

flavicon
face
This is why I said to check the current ramp with scope and find out the
best pulse width "ON" so the current would never goes steady, what means
excessive time, current, and not positive use of it, as Thomas say
below. The wave form should represent the actual slope for magnetic flux
charging, nothing more. This is also the same idea to control stepper
motors pulse width.

The pulse width "OFF" would be the necessary timming to the magnetic
colapse enough and allow a new good "ON" slope again, and so on.

The relation here for ON/OFF can not be calculated so easily. It should
go for some practical ways, since there are several factors involved,
but one thing is for sure, it will not be less than 10% or higher than
90%.  I would start with 15% and adjust from there.  Frequency is just a
guess. I would say a 28 Ohms solenoid @ 500mA wire gage would respond to
a high frequency, lets guess 2kHz?

Current              xxxxxxxxxxxxxxxxx
 ^                  x
 |                  x
 |                 x:
 |                x :
 |              x : :
 |          x     : :
 |   x            : :
 +-------------------------------------> Time
     ^            ^ ^
    FET           | |
    Fires         | '---- Saturation
                  '-------Good Time to Cut FET
     |<---------->|
            |
            '-------------"ON" pulse period

If the FET is cut at the right point, the saturation would neve happens,
so current would be as low as possible, protecting FET and Solenoid
Coil.

There is another issue, the oscillator OFF control, or the PIC PWM OFF
situation to the FET, should allow the FET to stay in OFF even when PIC
is in reset.  So, I would suggest a very good fail safe design (I
already used it to control my Styrofoam machine PWM thermal wire output
drive):

                      Vcc
                       |
                      LOAD
                       |
PWM----||----o------|FET
            |          |
            R          |
            |          |
           Gnd        Gnd

If the oscillation stops or if the software hangs with the PWM output
pin in high level, the LOAD will fry. Using the above configuration, the
capacitor would allow only pulses to pass, not DC.

Capacitor and Resistor should be calculated to have a RC time at least 3
to 4 times bigger than the PWM pulse ON period, so it will deliver PWM
pulses nicely to the FET.

A simple example using square waves:
------------------------------------
If R=10k, and "ON" pulse period is 1ms (best found time before
saturation), then C ~ 100nF.  If this time is ok, probably you could
apply a square wave at the cap and the diferential CR would allow a
single 1ms pulse to the load for each raise time of the wave.  Starting
from 30Hz (3% load current time) this percentage increases as you
increase the square wave frequency. The top frequency in this example
would be 500Hz, because both ON time would be equal (1ms).

  .--------------.              .---------------.
  |              |              |               |
---'              '--------------'               '---------
10ms Freq = 100Hz
  .--.                          .--.
  |  |                          |  |
---'  '--------------------------'  '----------------
1ms Pulse*  = 10% current applied



  .--.  .--.
  |  |  |  |
---'  '--'  '---------
2ms Freq = 500Hz
  .--.  .--.
  |  |  |  |
---'  '--'  '---------
1ms Pulse  = 50% current applied

(*) Actual diferencial pulse is not a square, but:
  .---.
  |    \
  |     '.
  |       '.
---'         '---- Ground
   <----->
     1ms

Frequencies above 500Hz can cause unstable operation of the FET since
the the signal at the FET gate would not swing complete from + to zero
(in this particular exemple):

   ^      ^      ^
  '  '. .'  '. .'
       V      V
---------------------- Ground

Wagner

Thomas McGahee wrote:
[snip]
> Also be careful with regards to the actual switching FREQUENCY
> used. When the frequency is too low and you have an inductive
> load, then you will have moments when the current will exceed
> the maximum you want and may do permanent damage. You are safer
> starting with higher frequencies and lower duty cycles and
> experimenting until you find the best combination. Or eliminate
> all the guesswork and use a technique that includes negative
> feedback. That will cost you some more pennies, though...
[snip]

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