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Motor Control Circuit Problem

S

Steve

Jan 1, 1970
0
Hello,
My group and I built the following dc motor control circut:
http://forum.webzila.com/index.php?act=Attach&type=post&id=2441

I replaced the SK100 transistor with NTE262 and the SL100 with TIP110

We have a high speed 9-18V DC Motor (18,000 RPM / 1.98A max).
When we plug a 12 V DC power supply directly to the motor it spins very
fast but when we use the circuit to drive the motor then it seems to
spin much slower.

With power supply set to 12V, we just measured to voltage going to the
motor and it says 1.5V .... what is wrong?

We just now set the power supply to 20V and the leads that connect to
the motor still only output 1.5V
 
T

Terry Given

Jan 1, 1970
0
Steve said:
Hello,
My group and I built the following dc motor control circut:
http://forum.webzila.com/index.php?act=Attach&type=post&id=2441

I replaced the SK100 transistor with NTE262 and the SL100 with TIP110

We have a high speed 9-18V DC Motor (18,000 RPM / 1.98A max).
When we plug a 12 V DC power supply directly to the motor it spins very
fast but when we use the circuit to drive the motor then it seems to
spin much slower.

With power supply set to 12V, we just measured to voltage going to the
motor and it says 1.5V .... what is wrong?

We just now set the power supply to 20V and the leads that connect to
the motor still only output 1.5V

Its because you have a built-in "will not work" feature. The npn
transistors Q3,Q5 in your H-bridge have no series resistors. Consider
only Q1,Q3,Q4 (the same is true for Q2,Q5,Q6) we see that:

with Q1 on, its collector sits at Vcesat = 0.2V (say), and pnp Q4 is on
(base current flows thru the 1k into Q1). Q3 is off. When Q1 turns off,
its collector 1k resistor pulls the collector high - at least it tries
to, but when Q1 collector rises to 0.7V (say) then Q3 base-emitter
junction is forward-biased, clamping Q1 collector to 0.7V - thereby
ensuring BOTH Q3 and Q4 are on. oops.

If you stick a resistor in series with Q3 base, it will work a bit
better - but the value of base resistor is important to ensure Q3,Q4
both turn off. use say 22k resistors for the base of Q3 and Q4, and also
slap 10k across the b-e junctions of Q3,Q4. This ensures that Q1
collector will pull up to 22k/23k*9V = 8.6V, ensuring Q4 will be off.

You will also have cross-conduction at every switching edge - at some
point both Q3 and Q4 will be on simultaneously, attempting to short out
the supply rail. A careful choice of base resistor (and perhaps a
speedup cap) is required. or a better circuit.

Cheers
Terry
 
C

ChrisGibboGibson

Jan 1, 1970
0
Terry Given wrote:

[snip]
Its because you have a built-in "will not work" feature.

Cracking phrase Grommit.

Never stopped Sinclair, Amstrad or Tandy.

Gibbo
 
S

Steve

Jan 1, 1970
0
john said:
The circuit is not right.
When Q1 is switched ON then the circuit works OK with Q3 OFF, (.7V at base),
and Q4 ON (11.3V at base).
When Q1 is switched OFF its collector voltage will rise to switch ON Q3
*but* the voltage at the base of Q4 cannot rise to 12.0V to allow it to
switch off. That's because there maybe a couple of ma's being taken by Q3
base when it's ON and this current is supplied by the 1k Q2 resistor and the
1k Q4 base resistor. This causes a few volts to be dropped and Q4 is still
switched ON.
The same thinking applies to both halves of the circuit so at any given time
one half of the circuit must be passing a big current through 2 of the
transistors (hot!) and not the motor. The effect is to bring the motor
voltage nearly to nothing
A redesign is needed. Not brilliant but maybe make Q1 Q2 resistors 220ohm.
Stick a 2k2 resistor in Q3 Q5 base. Change Q4 Q6 resistor to 2k2.
Where did'y'get the circuit from originally?.
regards
jhon

Originally we got the circuit from here:
http://www.electronic-circuits-diagrams.com/motorimages/motorckt4.shtml

Then someone told us that there is no current limiting between the
bases of the transistors. So we modified the circuit like this:
http://forum.webzila.com/index.php?act=Attach&type=post&id=2441

Would the original design work better?

Would the advise of modifying the various resistors that you and Terry
provided fix this issue? At this point we are not asking for the motor
to receive the full voltage that is supplied to the circuit but we need
as much as possible so that it can pull a 2lbs door.

We cant really afford to build a new circuit unless it can use
identical parts as the current one.

Can you again explain the changes that we should make or show a simple
circuit diagram of the way we should modify the circuit? When you say
to use a 22K resistor for Q3/Q5 base, do you mean put a resistor going
from the base of Q3 to the base of Q5? Also when you talk about the
Q4/Q6 resistor, are you refering to the 1K resistor that is between Q4
and Q3 & Q5 and Q6?

Also, when you say make Q1 and Q2 resistors 220 ohm...are you refering
to the 10K resistors that are going to the base or the 1K resistors
that are connected to the emitter?

Thanks
 
T

Terry Given

Jan 1, 1970
0
Steve said:
by Q3


and the


given time



Originally we got the circuit from here:
http://www.electronic-circuits-diagrams.com/motorimages/motorckt4.shtml

Then someone told us that there is no current limiting between the
bases of the transistors. So we modified the circuit like this:
http://forum.webzila.com/index.php?act=Attach&type=post&id=2441

Would the original design work better?

Would the advise of modifying the various resistors that you and Terry
provided fix this issue? At this point we are not asking for the motor
to receive the full voltage that is supplied to the circuit but we need
as much as possible so that it can pull a 2lbs door.

We cant really afford to build a new circuit unless it can use
identical parts as the current one.

Can you again explain the changes that we should make or show a simple
circuit diagram of the way we should modify the circuit? When you say
to use a 22K resistor for Q3/Q5 base, do you mean put a resistor going
from the base of Q3 to the base of Q5? Also when you talk about the
Q4/Q6 resistor, are you refering to the 1K resistor that is between Q4
and Q3 & Q5 and Q6?

Also, when you say make Q1 and Q2 resistors 220 ohm...are you refering
to the 10K resistors that are going to the base or the 1K resistors
that are connected to the emitter?

Thanks

the original circuit was even worse, for the same reasons I outlined -
except with no base resistors at all (as opposed to half the amount you
require) its a recipe for frying transistors whilst maintaining a
stationary shaft on the dc machine.

John's fix is essentially the same as that I proposed - base resistors
about, but choose values wisely; some semblence of understanding is
required.

Cheers
Terry
 
N

nospam

Jan 1, 1970
0
Steve said:
My group and I built the following dc motor control circut:
http://forum.webzila.com/index.php?act=Attach&type=post&id=2441
With power supply set to 12V, we just measured to voltage going to the
motor and it says 1.5V .... what is wrong?

The circuit diagram is wrong, and so is what you built if you followed it.

Swap Q3 with Q4 and Q5 with Q6. The emitters of Q3 and Q4 should be
connected and drive one side of the motor, Q5 and Q6 similarly driving the
other.
 
S

Steve

Jan 1, 1970
0
john said:
The circuit is not right.
When Q1 is switched ON then the circuit works OK with Q3 OFF, (.7V at base),
and Q4 ON (11.3V at base).
When Q1 is switched OFF its collector voltage will rise to switch ON Q3
*but* the voltage at the base of Q4 cannot rise to 12.0V to allow it to
switch off. That's because there maybe a couple of ma's being taken by Q3
base when it's ON and this current is supplied by the 1k Q2 resistor and the
1k Q4 base resistor. This causes a few volts to be dropped and Q4 is still
switched ON.
The same thinking applies to both halves of the circuit so at any given time
one half of the circuit must be passing a big current through 2 of the
transistors (hot!) and not the motor. The effect is to bring the motor
voltage nearly to nothing
A redesign is needed. Not brilliant but maybe make Q1 Q2 resistors 220ohm.
Stick a 2k2 resistor in Q3 Q5 base. Change Q4 Q6 resistor to 2k2.
Where did'y'get the circuit from originally?.
regards
jhon

I changed the Q1 and Q2 resistors to 220 ohms
Based on this circuit diagram can you show me what you mean by the
following:
Stick a 2k2 resistor in Q3 Q5 base. Change Q4 Q6 resistor to 2k2.
http://users3.ev1.net/~srudenko/img/hbridge.gif

I also have the EWB file (maybe if you have EWB you can show me what
you mean by making the changes in there).

Or just explain exactly where to plug in the resistors you are refering
to.

Thanks
 
J

john jardine

Jan 1, 1970
0
Steve said:
Hello,
My group and I built the following dc motor control circut:
http://forum.webzila.com/index.php?act=Attach&type=post&id=2441

I replaced the SK100 transistor with NTE262 and the SL100 with TIP110

We have a high speed 9-18V DC Motor (18,000 RPM / 1.98A max).
When we plug a 12 V DC power supply directly to the motor it spins very
fast but when we use the circuit to drive the motor then it seems to
spin much slower.

With power supply set to 12V, we just measured to voltage going to the
motor and it says 1.5V .... what is wrong?

We just now set the power supply to 20V and the leads that connect to
the motor still only output 1.5V

The circuit is not right.
When Q1 is switched ON then the circuit works OK with Q3 OFF, (.7V at base),
and Q4 ON (11.3V at base).
When Q1 is switched OFF its collector voltage will rise to switch ON Q3
*but* the voltage at the base of Q4 cannot rise to 12.0V to allow it to
switch off. That's because there maybe a couple of ma's being taken by Q3
base when it's ON and this current is supplied by the 1k Q2 resistor and the
1k Q4 base resistor. This causes a few volts to be dropped and Q4 is still
switched ON.
The same thinking applies to both halves of the circuit so at any given time
one half of the circuit must be passing a big current through 2 of the
transistors (hot!) and not the motor. The effect is to bring the motor
voltage nearly to nothing
A redesign is needed. Not brilliant but maybe make Q1 Q2 resistors 220ohm.
Stick a 2k2 resistor in Q3 Q5 base. Change Q4 Q6 resistor to 2k2.
Where did'y'get the circuit from originally?.
regards
jhon
 
T

Terry Pinnell

Jan 1, 1970
0
john jardine said:
The circuit is not right.
When Q1 is switched ON then the circuit works OK with Q3 OFF, (.7V at base),
and Q4 ON (11.3V at base).
When Q1 is switched OFF its collector voltage will rise to switch ON Q3
*but* the voltage at the base of Q4 cannot rise to 12.0V to allow it to
switch off. That's because there maybe a couple of ma's being taken by Q3
base when it's ON and this current is supplied by the 1k Q2 resistor and the
1k Q4 base resistor. This causes a few volts to be dropped and Q4 is still
switched ON.
The same thinking applies to both halves of the circuit so at any given time
one half of the circuit must be passing a big current through 2 of the
transistors (hot!) and not the motor. The effect is to bring the motor
voltage nearly to nothing
A redesign is needed. Not brilliant but maybe make Q1 Q2 resistors 220ohm.
Stick a 2k2 resistor in Q3 Q5 base. Change Q4 Q6 resistor to 2k2.
Where did'y'get the circuit from originally?.
regards
jhon

Out of interest I tried simulating the original (symmetrical) circuit
http://www.electronic-circuits-diagrams.com/motorimages/3.gif
with CircuitMaker. But after an hour or so I'm darned if I can get it
working. Typical error messages include:
Warning: Gmin step failed
Warning: source stepping failed
doAnalyses: Iteration limit reached'

Anyone care to try it in some other package please, such as LT
SWCADIII? I'm assuming there's some nifty trick for getting an
H-Bridge to converge in Spice...
 
N

nospam

Jan 1, 1970
0
Terry Pinnell said:
Out of interest I tried simulating the original (symmetrical) circuit
http://www.electronic-circuits-diagrams.com/motorimages/3.gif
with CircuitMaker. But after an hour or so I'm darned if I can get it
working.

Geees the original circuit shorts out the power supply with the power
transistor base emitter diodes - what did you expect to simulate?

As you have gone to the trouble of drawing it simply drag Q4 to where Q3 is
and Q3 to where Q4 is, the same for Q5 and Q6.

You then have a pair of simple class B complimentary amplifiers driving the
motor in a bridge. Why no one else can see this is how the original circuit
was supposed to be is beyond me.
 
R

Rich Grise

Jan 1, 1970
0
Out of interest I tried simulating the original (symmetrical) circuit
http://www.electronic-circuits-diagrams.com/motorimages/3.gif

This is the circuit configuration known as:
"Blow up four transistors simultaneously." ;-)
(note current path from +9v-Q4E-Q4B-Q3B-Q3E-GND, Q5,6 same)
with CircuitMaker. But after an hour or so I'm darned if I can get it
working. Typical error messages include:
Warning: Gmin step failed
Warning: source stepping failed
doAnalyses: Iteration limit reached'

Anyone care to try it in some other package please, such as LT
SWCADIII? I'm assuming there's some nifty trick for getting an
H-Bridge to converge in Spice...

Cheers!
Rich
 
T

Terry Pinnell

Jan 1, 1970
0
nospam said:
Geees the original circuit shorts out the power supply with the power
transistor base emitter diodes - what did you expect to simulate?

As you have gone to the trouble of drawing it simply drag Q4 to where Q3 is
and Q3 to where Q4 is, the same for Q5 and Q6.

You then have a pair of simple class B complimentary amplifiers driving the
motor in a bridge. Why no one else can see this is how the original circuit
was supposed to be is beyond me.
I understand that, but that wasn't the point. The circuit should still
simulate for each of the 4 input conditions.
 
T

Terry Pinnell

Jan 1, 1970
0
Rich Grise said:
This is the circuit configuration known as:
"Blow up four transistors simultaneously." ;-)
(note current path from +9v-Q4E-Q4B-Q3B-Q3E-GND, Q5,6 same)

See my reply to nospam.
 
J

Jim Meyer

Jan 1, 1970
0
nospam said:
The circuit diagram is wrong, and so is what you built if you followed it.

Swap Q3 with Q4 and Q5 with Q6. The emitters of Q3 and Q4 should be
connected and drive one side of the motor, Q5 and Q6 similarly driving the
other.


You are of course 100% right. That's the only way the original
circuit could have worked. The schematic got screwed up. No current
limit resistors needed at all.


Jim
 
R

Robert Monsen

Jan 1, 1970
0
Terry said:
I understand that, but that wasn't the point. The circuit should still
simulate for each of the 4 input conditions.

What happens is that the current goes too high on one of the top
transistors, causing the computation to fail. Try putting a 1 ohm
resistor into the base of both Q4 and Q6. That will enable the
simulation to complete without failing. Then, decrease the value of that
resistor, and see what happens. When you get down to some very small
value, it'll start to fail again.

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.
 
T

Terry Given

Jan 1, 1970
0
john said:
Steve wrote:
[clip]

the original circuit was even worse, for the same reasons I outlined -
except with no base resistors at all (as opposed to half the amount you
require) its a recipe for frying transistors whilst maintaining a
stationary shaft on the dc machine.

John's fix is essentially the same as that I proposed - base resistors
about, but choose values wisely; some semblence of understanding is
required.

Cheers
Terry


I've wondered if newcomers are ever put off the subject, after spending
hours trying to wrestle some sense out of useless designs.
I take a mechanical engineering mag' that recently did a short series on
'electronics in the workshop'. The author showed an H bridge circuit with
just the same problem as Steves. Poor proof reading or whatever, the author
should have known better. Shame on him :).
(ps, If I'd seen the group posts 'updating' I wouldn't have bothered with my
post. I was wasting bandwidth just repeating what you had said)
regards
john

Still, their revised circuit is an improvement on the original rubbish,
even if it doesnt yet work right. its getting closer.....

Cheers
Terry
 
T

Terry Pinnell

Jan 1, 1970
0
What happens is that the current goes too high on one of the top
transistors, causing the computation to fail. Try putting a 1 ohm
resistor into the base of both Q4 and Q6. That will enable the
simulation to complete without failing. Then, decrease the value of that
resistor, and see what happens. When you get down to some very small
value, it'll start to fail again.

Thanks, Robert, that sorted it immediately!
 
J

john jardine

Jan 1, 1970
0
Steve said:
clip

I changed the Q1 and Q2 resistors to 220 ohms
Based on this circuit diagram can you show me what you mean by the
following:
http://users3.ev1.net/~srudenko/img/hbridge.gif

I also have the EWB file (maybe if you have EWB you can show me what
you mean by making the changes in there).

Or just explain exactly where to plug in the resistors you are refering
to.

Thanks

Q1Q2 keep as 220ohms, then bridge resistors like this ...

-----+---------------------+-------- +V (eg 12V)
| |
| |
| Q4 Q6 |
| NTE262 NTE262 |
___ |< >| ___
,-|___|--| |-|___|--,
| 2k2 |\ /| 2k2 |
| | | |
| | | |
| | | |
| | Motor | |
| | ___ ___ | |
to Q1-----+ +-----o-|___|--UUU-o--+ +--to Q2
| | | |
| | 9ohm 30mH | |
| | | |
| | | |
| | | |
| ___ |/ \| ___ |
'--|___|-| |-|___|--'
2k2 |> <| 2k2
| |
| Q3 Q5 |
| TIP110 TIP110|
| |
---------+---------------------+-----0V
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

This setup's not optimum as it's a wee bit reliant on the actual base
current pulled by Q3 or Q5.

A nicer and simpler design (by Antonio Raposo) can be found at ...
http://www.cs.uiowa.edu/~jones/step/circuits.html

Antonio's circuit design uses a total of only 4 transistors and all could
be say TIP110 Darlingtons.
regards
john
 
J

john jardine

Jan 1, 1970
0
Terry Given said:
[clip]

the original circuit was even worse, for the same reasons I outlined -
except with no base resistors at all (as opposed to half the amount you
require) its a recipe for frying transistors whilst maintaining a
stationary shaft on the dc machine.

John's fix is essentially the same as that I proposed - base resistors
about, but choose values wisely; some semblence of understanding is
required.

Cheers
Terry

I've wondered if newcomers are ever put off the subject, after spending
hours trying to wrestle some sense out of useless designs.
I take a mechanical engineering mag' that recently did a short series on
'electronics in the workshop'. The author showed an H bridge circuit with
just the same problem as Steves. Poor proof reading or whatever, the author
should have known better. Shame on him :).
(ps, If I'd seen the group posts 'updating' I wouldn't have bothered with my
post. I was wasting bandwidth just repeating what you had said)
regards
john
 
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