PWM circuit...

Hello,

I was searching the web for easy-to-build DC motor PWM control
circuits, and I came across this one:

http://www.cpemma.co.uk/555pwm.html

and I am wondering what the 100 uF capacitor C4 is for. It seems a bit
strange to me that it is wired between the +12V and ground, and not in
parallel with the motor... then again, I do not have a great deal of
experience with these types of circuits.

Thanks in advance,

Mike Darrett
www.darrettenterprises.com

 

Thanks for the kind replies.

When substituting an IRF530 for the TIP31, how do the base, collector,
and emitter for the TIP31 correspond to the gate, source and drain for
the IRF530? (This will be my first time using a MOSFET in a
circuit...)

Mike

 

Ah, so R2 can be as low as 47 ohms with the IRF530? The article said
"... the same value or a higher (up to a 1k) gate resistor R2 can be
used." so now I'm a bit confused if I should go higher or lower than
100 ohms...

Would an NTE587 (UF4001 equivalent) be as good as an 1N5817/5818?

Thanks for the kind replies,

Mike Darrett

 

[Andrew Wagstaff]

http://www.cpemma.co.uk/555pwm.html

Could this circuit be modified to run on 6V (for a 6V motor) ?

Regards

Andrew

 

Could you recommend some larger FETs, say 30 amps?

Could I simply hook up a second FET in parallel (gate-to-gate,
source-to-source, drain-to-drain)?

(I am kind of new to these kind of circuits

Thanks,

Mike

 

[Jamie]

mike-nospam@darrettenterprises.com wrote:

Could you recommend some larger FETs, say 30 amps?

Could I simply hook up a second FET in parallel (gate-to-gate,
source-to-source, drain-to-drain)?

(I am kind of new to these kind of circuits

Thanks,

Mike

Mosfets are resistor type devices via the drain and source

you can go to digikey and look up Hexfets
how about an
IRF2804-ND
40 Vds, 280 Amps cont! $3.33 each from an older cat.

anyways, i am sure that gives you some idea's

 

jpopelish@rica.net wrote:

mike-nospam@darrettenterprises.com wrote:
Ah, so R2 can be as low as 47 ohms with the IRF530? The article
said
"... the same value or a higher (up to a 1k) gate resistor R2 can
be
used." so now I'm a bit confused if I should go higher or lower
than
100 ohms...

Lower gives faster (higher efficiency, cooler transistor temperature)
switching, but more radio waves. Higher resistance gives slower
switching (quieter, but hotter transistor, lower overall efficiency).
I can see why they want the switching to be slow, with that 1N4001
diode struggling to turn off as the transistor turns on.

Would an NTE587 (UF4001 equivalent) be as good as an 1N5817/5818?

Not as good, because it is a junction diode (I think) so will drop
about twice as much voltage in the forward direction as the schottky
(will run hotter and lower overall efficiency a little). But it will
certainly switch off faster than the 1N4001.

--
John Popelish

To increase the maximum rated current from 15A to 30A, would it be a
good idea to simply hook up a second IRF530 in parallel (gate-to-gate,
source-to-source, drain-to-drain)?

If so, would the resistor R2 need to be changed? I'm currently using
50 ohms (two 100-ohm resistors in parallel).

Mike

 

[John Popelish]

mike-nospam@darrettenterprises.com wrote:

To increase the maximum rated current from 15A to 30A, would it be a
good idea to simply hook up a second IRF530 in parallel (gate-to-gate,
source-to-source, drain-to-drain)?

"good idea" is probably overstating it, but it might work. A bigger
device would be more reliable, but for a one off, the important thing
is the tab temperature. It shouldn't burn your knuckle. Remember
that every watt of heat showing up in the parts, is one less watt
getting ot the motor, but still draining out of the battery.

If so, would the resistor R2 need to be changed? I'm currently using
50 ohms (two 100-ohm resistors in parallel).

I would separate the gates, and run one resistor to each.

You will also need a diode rated for a fair fraction of peak motor
current. The big schottky I listed a while back might be close. But
you can get bigger ones in the same tab package as the fets which are
easier to heat sink.

--
John Popelish

 

[John Popelish]

mike-nospam@darrettenterprises.com wrote:

John Popelish wrote:
mike-nospam@darrettenterprises.com wrote:

To increase the maximum rated current from 15A to 30A, would it be
a
good idea to simply hook up a second IRF530 in parallel
(gate-to-gate,
source-to-source, drain-to-drain)?

"good idea" is probably overstating it, but it might work. A bigger
device would be more reliable, but for a one off, the important thing
is the tab temperature. It shouldn't burn your knuckle. Remember
that every watt of heat showing up in the parts, is one less watt
getting ot the motor, but still draining out of the battery.

If so, would the resistor R2 need to be changed? I'm currently
using
50 ohms (two 100-ohm resistors in parallel).

I would separate the gates, and run one resistor to each.

You will also need a diode rated for a fair fraction of peak motor
current. The big schottky I listed a while back might be close. But
you can get bigger ones in the same tab package as the fets which are
easier to heat sink.

--
John Popelish

My setup with one IRF530 gets uncomfortably hot at only 7 amps (12VDC).
Then again, it's not connected to a heatsink. I'm afraid to connect
the load (large DC motor) for more than 2 seconds at a time.

Is this rate of heat generation at the MOSFET normal?

Mik

It is expected. There are two parts to the loss (heating) problem.
There is the simple ohmic loss when the transistor is on that is
calculated by load current squared times the on resistance. And there
are switching losses that can approach the supply voltage times the
load current. fortunately, these come in short spikes and are
minimized by getting through the switching process quickly.

But with a 7 amp load and a 6 volt supply, you are controlling about
35 watts of output power. A single watt will heat an un heat sunk
TO220 tab enough to burn your finger. But that watt is only
1/(1+35)=2.8% of the total power from the source, so you are operating
a quite efficient regulator.

How hot does the diode get when you run a 50% duty cycle (diode
carrying the load half the time)?

A 30 amp regulator will probably need at least a small heat sink,
because even if you find a huge, low resistance device to keep the DC
losses down, its high gate capacitance will slow the switching process
and generate heat there. At lease it will unless you also beef up the
gate drive to something that can deliver more current than a C555 and
a 47 ohm resistor.
--
John Popelish

 

[John Popelish]

mike-nospam@darrettenterprises.com wrote:

John Popelish wrote:
(snip)
But with a 7 amp load and a 6 volt supply, you are controlling about
35 watts of output power. A single watt will heat an un heat sunk
TO220 tab enough to burn your finger. But that watt is only
1/(1+35)=2.8% of the total power from the source, so you are
operating
a quite efficient regulator.

Ok... but, how do you calculate the 35 watts? IIRC, the calculation
goes 7A x 6V = 42 watts...

That is the answer you get if you can multiply correctly. I hadn't
had my tea yet when I wrote that. Sorry.

even if I divide this by sqrt(2) this is
still only about 30 watts (rms)... or is there something here I'm not
quite understanding...?

How hot does the diode get when you run a 50% duty cycle (diode
carrying the load half the time)?

I'm estimating between 60 and 80 degrees Centigrade, after running for
about 2-3 seconds. (I was being cautious with the run times, because
if I made a mistake in the wiring, I didn't want the electronic
components to explode in my face.) I was using the 12V battery in my
car, and a 12VDC surplus kiddie car motor which draws 7A on startup
(about 1/2 second), about 4A continuous, with no load.

I would want a bigger diode than that, if it is getting that hot.

A 30 amp regulator will probably need at least a small heat sink,
because even if you find a huge, low resistance device to keep the DC
losses down, its high gate capacitance will slow the switching
process
and generate heat there. At lease it will unless you also beef up
the
gate drive to something that can deliver more current than a C555 and
a 47 ohm resistor.

So, if the gate drive delivers more current, the MOSFET operates more
efficiently? (This implies, remove the resistor?)

Or reduce it. But you will quickly run into the limited output
current capability of the 555. Switching a really big fet very
quickly takes an amp or so for a fraction of a microsecond.

--
John Popelish

 

John Popelish wrote:

mike-nospam@darrettenterprises.com wrote:

John Popelish wrote:
(snip)
But with a 7 amp load and a 6 volt supply, you are controlling
about
35 watts of output power. A single watt will heat an un heat
sunk
TO220 tab enough to burn your finger. But that watt is only
1/(1+35)=2.8% of the total power from the source, so you are
operating
a quite efficient regulator.

Ok... but, how do you calculate the 35 watts? IIRC, the
calculation
goes 7A x 6V = 42 watts...

That is the answer you get if you can multiply correctly. I hadn't
had my tea yet when I wrote that. Sorry.

even if I divide this by sqrt(2) this is
still only about 30 watts (rms)... or is there something here I'm
not
quite understanding...?

How hot does the diode get when you run a 50% duty cycle (diode
carrying the load half the time)?

I'm estimating between 60 and 80 degrees Centigrade, after running
for
about 2-3 seconds. (I was being cautious with the run times,
because
if I made a mistake in the wiring, I didn't want the electronic
components to explode in my face.) I was using the 12V battery in
my
car, and a 12VDC surplus kiddie car motor which draws 7A on startup
(about 1/2 second), about 4A continuous, with no load.

I would want a bigger diode than that, if it is getting that hot.

A 30 amp regulator will probably need at least a small heat sink,
because even if you find a huge, low resistance device to keep
the DC
losses down, its high gate capacitance will slow the switching
process
and generate heat there. At lease it will unless you also beef
up
the
gate drive to something that can deliver more current than a C555
and
a 47 ohm resistor.

So, if the gate drive delivers more current, the MOSFET operates
more
efficiently? (This implies, remove the resistor?)

Or reduce it. But you will quickly run into the limited output
current capability of the 555. Switching a really big fet very
quickly takes an amp or so for a fraction of a microsecond.

--
John Popelish

Could I wire the output from the MOSFET to the control coil of a (20
amp or so) relay? Would the switching frequency need to be slowed down
a bit for the relay to respond?

Thanks,

Mike

 

[John Popelish]

mike-nospam@darrettenterprises.com wrote:

Could I wire the output from the MOSFET to the control coil of a (20
amp or so) relay? Would the switching frequency need to be slowed down
a bit for the relay to respond?

To quote an AOL commercial, "Slowed way down. Slowed down like holiday
traffic". Contacts can be the switching mechanism for a PWM output
but they are not very reliable. All the heat that would normally be
dumped into the mosfet during turn on and turn off will appear in the
tiny gas pocket between the contacts, converting it to little blasts
of plasma, which will quickly either heat the contacts till they melt
together, or will oxidize them till they no longer conduct. That 20
amp rating is based on the contacts cooling after each break before
the next make, and a PWM regulator does not meet that requirement.

You may cool a large mosfet quite a bit just by adding a complementary
emitter follower between the 555 and the mosfet as a current booster,
with something like a 10 ohm gate resistor. If this is for the 6 volt
version, you might try paralleling all 6 buffers in a 74HC4050 to act
as a high current gate driver. This will boost the drive
considerably, compared to the LMC555 output. But I would probably
still keep 10 ohms right at the gate or put a ferrite bead around the
source lead.

--
John Popelish

 

[John Popelish]

mike-nospam@darrettenterprises.com wrote:

John Popelish wrote:
mike-nospam@darrettenterprises.com wrote:

You may cool a large mosfet quite a bit just by adding a
complementary
emitter follower between the 555 and the mosfet as a current booster,
with something like a 10 ohm gate resistor. If this is for the 6
volt
version, you might try paralleling all 6 buffers in a 74HC4050 to act
as a high current gate driver. This will boost the drive
considerably, compared to the LMC555 output. But I would probably
still keep 10 ohms right at the gate or put a ferrite bead around the
source lead.

Ok; I think I follow this. (You mean, use the 74HC4050 as a
"pre-amplifier" for the mosfet gate, right?)

Right. Lots of other noninverting chips with high current capability
could work, too, including 3 state buffers (with the output strapped
to be constantly enabled. These come in 4s, 6s and 8s per pack.

But what exactly is a
"complementary emitter follower"?

It is a current amplifier made up of a pnp and npn transistor with
their bases connected together as the input and their emitters
connected as the output. The PNP collector ties to the negative rail,
and the NPN collector to the positive rail. Unfortunately, you lose
..6 volts in each direction to forward bias the base emitter junctions.
And with only 6 volts available, you cannot afford to lose any gate
drive voltage and expect to get minimum temperature out of the fet.

I will also investigate the possibility that the motor was originally
designed for 6V, as opposed to 12V. That would probably explain the
temperature rise, if I've been exceeding the design voltage for the
motor...!

That would do it. Of course, you could limit the duty cycle of the
PWM to no more than 50% and thus limit the average motor voltage to 6
volts, even though the circuit runs on 12 volts. Then you have lots
of gate voltage and can apply the complementary emitter follower.

--
John Popelish