1 KHz PWM FET driver

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geojoe1
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1 KHz PWM FET driver

Post by geojoe1 » Wed Sep 13, 2006 6:44 am

I am running a brushed 12VDC motor with an IRF 1010 N chanel FET and 3705 FET driver. I am using a PIC 16F876 (Way more than needed but I had one in a drawer and so used it) to produce a PWM pulse train to the FET driver. The positive duty cycle of the PWM is based on a POT input for speed control of the motor. The period of the PWM is 1 KHz. Everything is working great!

My question is: Should I be concerned with the 1 KHz period? Is this going to produce ugly RF interference when running on a 12V car battery. I measured the Source voltage and can see a 1 Khz ripple at about 200 mV p-p. Should I up my period or filter the input better?

Thanks for the help.

Joe

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philba
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Post by philba » Wed Sep 13, 2006 2:20 pm

that's a lot of ripple but I assumeyou are using a vreg to get +5 for the PIC and other electronics so it shouldn't be an issue. also, anything else in the car draawing 12V that can't handle the ripple is probably not going to last anyway.

I don't know but I'd guess that would have some problems at very low duty cycles. can you hear a 1khz noise when running the motor? I'd push the pwm frequency up to 20K.

rshayes
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Post by rshayes » Wed Sep 13, 2006 6:47 pm

The potential for radio frequency interference is not dependent on the repetition rate of 1 kilohertz but rather on the rate of rise of the ripple waveform. A little filtering of the ripple to slow down the edges should reduce the radio frequency interference to a tolerable level.

geojoe1
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Post by geojoe1 » Thu Sep 14, 2006 3:19 pm

Yes I can hear the 1k hz tone on the motor. And in my current sense resistor. I thought that was kinda cool. I put a 12uH inductor on the positive input line. This sharpend up the input greatly. there is only a very small ripple current now. However, I just guessed at what value to use. How would I calculate what value inductor to use. Or could I use caps too?

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MrAl
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Post by MrAl » Fri Oct 13, 2006 9:02 pm

Hi there,

I didnt see any mention of a catch diode? Do you have one installed?

You can approximate the ripple of an inductor/cap with load
approximately like this if you assume a 50% duty cycle:

dt=50e-6
L=100e-6
v=5
C=100e-6
di=v*dt/L
i=di
dv=i*dt/C/2
vpp=dv/2

where
dt is the pulse width (assumed half the period here)
L is the inductance in Henries
v is the pulse voltage above that of the output
C is the capacitance in farads
di is the calculated change of current in the inductor
dv is the voltage that would appear across the cap with a square pulse
vpp is the ripple voltage, which is dv compensated for the fact that
the inductor current is a triangle and not a square pulse.

Example:
10kHz pulse source, 50% duty cycle pulsing 0v to 10v
100uH inductor
100uf cap
50 ohm load across cap

dt is one-half of the 10kHz wave, or dt=50us=0.000050
di=5*0.000050/0.000100=2.5 amps
dv=2.5*0.000050/0.000100/2=0.625 volts
vpp=0.625/2=0.3125 volts peak to peak ripple approximately

Note that if we increased the cap to 200uf (twice the original value)
we would reduce the ripple to half that of what it was, which means
now it would be roughly 150mv instead of around 300mv. This of
course means that you can decrease the ripple by simply increasing
the value of the cap. Also, you can reduce the value of the ripple
by increasing the value of the inductor too...if we increase the inductor
value to 200uH instead of the cap we would again get half the ripple.

The inductor value isnt the only criterion however, we also need to select
the inductor with an appropriate sat current as well as the correct value,
and also make sure it has a tolerable equivalent series resistance (ESR).
When possible it's nice to select an inductor with an ESR of 1/50 of that
of the load resistance. Thus, if we have a load of 1 amp at 5 volts
(that would be 5 ohms) we would want an ESR not more than 0.1 ohms.
Less is better, but it leads to a physically larger inductor size which takes
up more space.
As for the sat current rating, we need at least the peak current plus
some safety margin. Selecting this usually means looking at the
inductor curve for L vs Idc, but for an approximation select twice
the expected peak current. The peak current Ipk is:
Ipk=iLoad+di/2
where
iLoad is the load current and
di was calculated above.
Thus, the min sat rating should be
IsatMin=Ipk*2
but to be even safer select IsatMin=Ipk*4.
This kind of selection depends on how much you want to spend
for the inductor as well as how much space you have for mounting.
If it's not a critical space situation, select a nice size toroid inductor
with a low ESR and a high sat rating.
LEDs vs Bulbs, LEDs are winning.

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