Maker Pro
Maker Pro

Need insight how to use 4017 counter with MOSfet

V

Vencislav

Jan 1, 1970
0
Do you have any insight on how to use the decade counter with
MOSfets to pulse power through an array of coils?

I've heard from some that a diode needs to be connected in reverse across
the MOSfet to prevent damaged from the Lenz Law Effect (or high voltage self
induction due to the varying magnetic field of the pulse). The back
EMF could otherwise destroy the MOSfet.


Thank you!
 
T

Tim Wescott

Jan 1, 1970
0
Vencislav said:
Do you have any insight on how to use the decade counter with
MOSfets to pulse power through an array of coils?

I've heard from some that a diode needs to be connected in reverse across
the MOSfet to prevent damaged from the Lenz Law Effect (or high voltage self
induction due to the varying magnetic field of the pulse). The back
EMF could otherwise destroy the MOSfet.


Thank you!

To drive it from the 4017 you just need "logic level" n-channel MOSFETs.
You want one of two diode protection circuits. The most commonly used
one is:


VCC
+
|
o-----o
| |
C| |
C| -
C| ^
| |
o-----o
|
||-+
||<-
----||-+
|
|
===
GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

But your relay will close quicker if you use:




VCC
+
|
|
|
C|
C|
C|
|
|
o------o
| |
||-+ \-\
||<- ^
------||-+ |
| |
| |
=== ===
GND GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

with the zener diode voltage set to prevent the MOSFET drain-source
voltage from being exceeded.
 
J

John Popelish

Jan 1, 1970
0
Vencislav said:
Do you have any insight on how to use the decade counter with
MOSfets to pulse power through an array of coils?

I've heard from some that a diode needs to be connected in reverse across
the MOSfet to prevent damaged from the Lenz Law Effect (or high voltage self
induction due to the varying magnetic field of the pulse). The back
EMF could otherwise destroy the MOSfet.

Thank you!

Since the gates of mosfets appear to the outside world much like small
capacitors, they can be directly connected to CMOS outputs, as long as
the frequency is not so high that the gate capacitance can not be
charged up and down in the duration of the output states. If you use
N channel mosfets with their sources connected to the negative supply
rail used by the cmos, then a log ic high output will turn the mosfet
on, and a load connected between any more positive supply rail
(assuming that supply shares a negative side with the CMOS supply) and
the drain will get connected when the mosfet turns on.

The diode protection you mention involves inductive loads. An
inductance produces voltage in proportion to the rate of change of
current, in a direction that tends to resist such change. This makes
it easy to switch a voltage into an inductor, because as the inductor
current begins to rise, the inductor produces a voltage that cancels
the applied voltage and slows that current rise. So there is no turn
on current surge.

But when you try ot open a switch that is carrying inductor current,
the inductor produces voltage that adds to the power supply, to keep
the current from instantaneously falling to zero. This extra voltage
must be interrupted by the switch. If the switch is mechanical, this
effect causes arching between the separating contacts. If the switch
is a solid state device, like a mosfet, this effect can cause over
voltage failure of the device.

A common solution to this problem is to parallel the inductive load
(not the switch) with a diode, connected so that the diode is reverse
biased by the normal on state voltage. This prevents the inductor
from producing more than a forward biased diode drop when the current
to the inductor is interrupted and its voltage reverses. Instead of
the current falling quickly (which is what causes the inductor to make
all that high voltage), the current simply detours to the parallel
diode till the resistance of the coil and the voltage drop of the
diode slowly use up the stored magnetic energy. The most extra
voltage the switch can be exposed to is 1 diode drop above the
positive supply.

Of course, this protective method has the draw back that it makes it
impossible to turn the inductive device off very fast.
 
K

Ken Smith

Jan 1, 1970
0
Tim Wescott said:
To drive it from the 4017 you just need "logic level" n-channel MOSFETs.

This is not true if this is a CD4017 which can be run from +15V and drive
the gate high enough for non-logic level MOSFETs. The down side is that
CD4XXX parts are slow and can't output much current.


[.. deleted suggested good spike catchers ..]

In some cases a series resistor + capacitor combination across the coil
is a good way to go too. Depending on the relay, it may actually open
faster if the current rings than if it stops suddenly.
 
S

Stefan Heinzmann

Jan 1, 1970
0
Tim said:
To drive it from the 4017 you just need "logic level" n-channel MOSFETs.

That depends on the supply voltage of the 4017. If you power it from 10V
or more any old MOSFET will do. That's the idiocy with this marketing
term "logic level": There are about 5 different definitions of logic
levels in common use and umpteen more in not so common use. It just
confuses people.
You want one of two diode protection circuits. The most commonly used
one is:


VCC
+
|
o-----o
| |
C| |
C| -
C| ^
| |
o-----o
|
||-+
||<-
----||-+
|
|
===
GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

But your relay will close quicker if you use:

Surely you meant to say: "will open quicker"?
 
T

Tim Wescott

Jan 1, 1970
0
Stefan said:
That depends on the supply voltage of the 4017. If you power it from 10V
or more any old MOSFET will do. That's the idiocy with this marketing
term "logic level": There are about 5 different definitions of logic
levels in common use and umpteen more in not so common use. It just
confuses people.

OK, I forgot this one, and I've been tagged for it twice.

The "logic level" MOSFETs seem to be specified for both 5V and 3.3V
these days, which is nice. I haven't seen any with specs for 1.8V yet,
though :(.
Surely you meant to say: "will open quicker"?

What, you mean you use normally open relays? :)
 
S

Stefan Heinzmann

Jan 1, 1970
0
Tim said:
OK, I forgot this one, and I've been tagged for it twice.

The "logic level" MOSFETs seem to be specified for both 5V and 3.3V
these days, which is nice. I haven't seen any with specs for 1.8V yet,
though :(.


What, you mean you use normally open relays? :)
 
S

Stefan Heinzmann

Jan 1, 1970
0
Tim said:
Stefan said:
Tim Wescott wrote: [...]
To drive it from the 4017 you just need "logic level" n-channel MOSFETs.

That depends on the supply voltage of the 4017. If you power it from
10V or more any old MOSFET will do. That's the idiocy with this
marketing term "logic level": There are about 5 different definitions
of logic levels in common use and umpteen more in not so common use.
It just confuses people.


OK, I forgot this one, and I've been tagged for it twice.

We're cruel, aren't we?
The "logic level" MOSFETs seem to be specified for both 5V and 3.3V
these days, which is nice. I haven't seen any with specs for 1.8V yet,
though :(.

I've seen 2.5V specs. That's a start, at least.
 
W

Winfield Hill

Jan 1, 1970
0
Spehro Pefhany wrote...

Yes, there are a host of 1.8V rated power mosfets. Furthermore,
NEC has 1.5V parts, like the uPA675T and 2sk2159, etc., as does
Sanyo, like the mch6602, mch6615, etc. Toshiba's HN1K05FU is
rated at 1.2V, and there are a few others rated even lower.

And don't forget bipolar transistors, rated at under 0.65 volts!

Thanks,
- Win

(email: use hill_at_rowland-dot-org for now)
 
K

Ken Smith

Jan 1, 1970
0
Tim Wescott said:
I haven't seen any with specs for 1.8V yet,
though :(.

Try Supertex, if you don't need too much power.

TN2501 -> 1.0V
VN4012 -> 1.8V
 
K

Ken Smith

Jan 1, 1970
0
Winfield Hill said:
And don't forget bipolar transistors, rated at under 0.65 volts!

Germanium is better.

JFETs have a Vgsth less than zero.
 
W

Winfield Hill

Jan 1, 1970
0
Ken Smith wrote...
Tim Wescott wrote:
[.. logic voltage MOSFETs ..]
I haven't seen any with specs for 1.8V yet, though :(.

Try Supertex, if you don't need too much power.

TN2501 -> 1.0V
VN4012 -> 1.8V

Actually, the TN2501 has its Ron spec'd at 1.2V, if anyone
is counting. That's not 1.0V, but it's still a pretty low
voltage. How low? It's so low this MOSFET's Ron suffers
substantially. Whereas it's Ron rating is 3.5 ohms max for
Vgs = 2.0V, which is moderately useful, it degrades by over
7x at 1.2V, all the way to 25 ohms. Whew, it's nearly off!

Even though the FDG327N that Spef suggested isn't spec'd at
1.2V (it's rated 0.14 ohms max at 1.8V), its curves (fig 4)
show it'll work quite well there, typically about 0.2 ohms.

The VN4012 is spec'd at 4.5V, and it has a fairly high
threshold voltage of 1.8V, so it's not a candidate.

Thanks,
- Win

(email: use hill_at_rowland-dot-org for now)
 
F

Fred Bloggs

Jan 1, 1970
0
Tim said:
To drive it from the 4017 you just need "logic level" n-channel MOSFETs.
You want one of two diode protection circuits. The most commonly used
one is:


VCC
+
|
o-----o
| |
C| |
C| -
C| ^
| |
o-----o
|
||-+
||<-
----||-+
|
|
===
GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

But your relay will close quicker if you use:




VCC
+
|
|
|
C|
C|
C|
|
|
o------o
| |
||-+ \-\
||<- ^
------||-+ |
| |
| |
=== ===
GND GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

with the zener diode voltage set to prevent the MOSFET drain-source
voltage from being exceeded.

The zener clamp sometimes requires a high peak power rating making for a
pricey component- the rectifier diode clamp can be assisted for very
fast turn-off like so:
View in a fixed-width font such as Courier.


VCC
+
|
o-----o
| |
| |
| -
| ^
| | | VD,max-VCC-Vdiode
IC | C| / Rc,MAX= ------------------
| C| Rc IC
| C| /
\ / | \ Rc=Rcoil gets VD,MAX ~ 2x VCC
| |
| |
o-----o <-- VD
|
||-+
||<-
-----+-----||-+
| |
/ |
10K |
/ |
\ |
+--------+
|
===
GND
 
T

Tim Wescott

Jan 1, 1970
0
Fred said:
Tim Wescott wrote:
-- snip --

The zener clamp sometimes requires a high peak power rating making for a
pricey component- the rectifier diode clamp can be assisted for very
fast turn-off like so:
View in a fixed-width font such as Courier.


VCC
+
|
o-----o
| |
| |
| -
| ^
| | | VD,max-VCC-Vdiode
IC | C| / Rc,MAX= ------------------
| C| Rc IC
| C| /
\ / | \ Rc=Rcoil gets VD,MAX ~ 2x VCC
| |
| |
o-----o <-- VD
|
||-+
||<-
-----+-----||-+
| |
/ |
10K |
/ |
\ |
+--------+
|
===
GND

If you're only energizing the relay occasionally you can also leave out
the diode and just use an R or RC snubber network, at the expense of
more current in the transistor.
 
S

Spehro Pefhany

Jan 1, 1970
0
If you're only energizing the relay occasionally you can also leave out
the diode and just use an R or RC snubber network, at the expense of
more current in the transistor.

Or put an LED in series with the resistor (with an antiparallel diode
or a parallel resistor) as an indicator.

Best regards,
Spehro Pefhany
 
T

Tom Del Rosso

Jan 1, 1970
0
In Tim Wescott typed:
But your relay will close quicker if you use:

VCC
+
|
|
|
C|
C|
C|
|
|
o------o
| |
||-+ \-\
||<- ^
------||-+ |
| |
| |
=== ===
GND GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

with the zener diode voltage set to prevent the MOSFET drain-source
voltage from being exceeded.


Now why would that de-energize faster? The stored current can still
flow after the transistor turns off, except that it flows to ground.
 
J

John Popelish

Jan 1, 1970
0
Tom said:
In Tim Wescott typed:

Now why would that de-energize faster? The stored current can still
flow after the transistor turns off, except that it flows to ground.

The current decays faster only if the zener voltage is higher than the
supply voltage, implying that there is a fixed voltage across the coil
as the current decays. If the zener voltage is less than the supply
voltage, the relay current never goes all the way to zero.
 
Top