Driver to drive?

[Robert Monsen]

Jonathan Kirwan wrote:

On 24 Oct 2004 09:24:22 -0700, Winfield Hill <Winfield_member@newsguy.com
wrote:


. +Vs o------+----------------+-------,
. 40 - 55V | | |
. R4 R5 BFC
. 1k00 249 |
. | npn | gnd
. Q2 e\| |/e
. pnp |----+-------| Q3
. c/| | |\c 100V pnp
. | | | small heat sink
. +------' |
. | |
. |/c Q1 | bipolar current source
. +15V --| npn | 20mA fs, 30V compliance
. |\e |
. | | R4 R2 1
. R1 ,----|--R2--, | Io = - Vin -- ( ---- + -- )
. 10k0 | _| 10k0 | | R5 R1 R3 R1
. Vin --/\/\--+--|- \ | x |
. -10V pk in | >----+ +---- Io out
. = +20mA out ,--|+_/ | |
. | | R3 2k49 |
. gnd | | |
. | gnd |
. | |
. and etc for neg portion


I ran some simulations on the extended version of this schematic and output
current isn't particularly flat relative to the driving voltage as one might
hope for. (I selected the OPA227 as the opamp, the 2N3906/2N3904 for the Q1
equivalents, and the 2SB709A and the 2SD601A for the (+) and (-) side pairs.)

Firstly, accuracy is off for reasons I cannot completely fathom. I'm getting an
Iout that is about 12% lower than what the equation predicts. I can see where
the equation arises, I think, but my understanding doesn't appear to change the
simulation results (of course):

[1] The current from Vin to the (-) input node is Vin/R1.
[2] The current flowing from the (-) input node to the opamp output
is then -Vin/R1.
[3] This current, [2], must flow through R2, thus the voltage at
the opamp output must be -R2*(Vin/R1).
[4] The current flowing through R3 is then -R2*(Vin/R1)/R3.
[5] These currents sum, [2] and [4], so the total current arriving at
(or leaving) is -Vin/R1 + -R2*(Vin/R1)/R3 or -Vin*(1/R1+R2/(R1*R3)).

The output must either source or sink this, so this is what either draws an
extra from the V+ or the V- of the opamp, yielding the behavior. As I see it,
anyway.

[6] So, this is then applied to one side of the current mirror and is
gained up by the ratio of R4/R5.

But regardless of this logic above, I still get 9.293mA at -5V. The calculation
suggests:

-5V*1k/249*(1/10k+10k/(10k*2.49k)) = 10.072mA

That's about 92% of the predicted value. Of course, this is simulation. But
I'm curious if you think this is a simulation problem or if you think it's
probably real.

Aside from this accuracy difference from the equation, I'm curious about another
behavior I noticed in simulation. The relative gain varies rather widely as the
controlling input varies from -5V to +5V. (I used a .dc sweep.) As I sweep the
DC controlling voltage from -5V towards 0V, the relative change (using -5V as
the standard of 1.000) is:

Vin Iout gain Rel %
------ ------- ---------- -------
-5V 9.293mA 1.859mMhos 1.000
-1 1.947 1.947 1.047
-.5 1.026 2.052 1.104
-.4 842.1uA 2.105 1.132
-.3 657.8 2.193 1.180
-.2 473.4 2.367 1.273
-.1 289.0 2.890 1.555
-.05 196.7 3.934 2.116

etc.

Does this mean it needs a micro to support a correction table? Or is this more
likely to be some problem in my simulation?

Jon

Add the traditional PNP transistor between base and collector of Q2, so
its emitter is at the base of Q2, base is at the collector of Q2, and
its collector is grounded. That will minimize the current swiped from
the collector, and make the mirror more accurate.

[R4] [R5]
| |
| |
| |
| |

| |
|-------o-------------|

/| | |\
| | |
| |< Iout
o-------|
| |\
| |
| GND
|
o
Iin

created by Andy´s ASCII-Circuit v1.25.250804 www.tech-chat.de

Also, the current ratio isn't really R4/R5, because there is a small
correction factor. It can easily be derived from ebers-moll. Assuming a
desired ratio of x:1, a control current of I, and a resistance on the
control transistor of R, the emitter resistance on mirror transistor is

Rm = (R*I - 25.3e-3 * ln(x))/(x*I)

So, for the example above, you have

Rm = (1k * 4m - 25.3m * ln(4))/(4 * 4m) = 247.8

Not much of a difference unless the ratio is large. For a 10x mirror
with a 1mA steering current, for example, the resistance Rm will be 94.2
rather than 100. TC will also adversely affect the linearity.

A more obvious way (for me...) to generate the steering current would be

Iout+
|
15V |/
--------------o--------|
| |>
| |
| |
| |/
| .----|
| | |>
|\| | |
.-----|-\ | |
| | >--o o------.
Vin----------|+/ | | |
| |/| | | |
| | | | |
'------------------o |
| | | |
| | |< \
| '----| / R=2.5k
| |\ \
| | /
| | |
| |< ===
--------------o--------| GND
-15V |\
|
Iout-

created by Andy´s ASCII-Circuit v1.25.250804 www.tech-chat.de

It's basically the same as Win's solution, but the driver for his is
internal to the opamp package. With this one, however, you can easily
make the steering current equal to the output current, even up to fairly
large currents, and thus eliminate one source of non-linearity in the
mirrors (ln(1) = 0). This might also help with any tempco problem.

One can use a cheapo opamp like an LM324 if crossover distortion or slew
rate isn't an issue. It sounds like the OP will want to set the current,
then either shunt it to ground using his optoisolator control, or let it
rip into the poor torture victim/grad student. If so, opamp slew rate
won't be an issue.

--
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.

 

[Fred Bloggs]

Tom Seim wrote:

Fred Bloggs <nospam@nospam.com> wrote in message news:<417DC399.8050607@nospam.com>...

Tom Seim wrote:

Rich Grise <rich@example.net> wrote in message news:<pan.2004.10.25.20.41.16.115088@example.net>...


On Mon, 25 Oct 2004 13:04:25 +0000, John S. Dyson wrote:



In article <ckspf8$rbo$5@blue.rahul.net>,
kensmith@green.rahul.net (Ken Smith) writes:


In article <8oicd.14564$nj.6932@newssvr13.news.prodigy.com>,
Clarence <no@No.com> wrote:


"Robert Monsen" <rcsurname@comcast.net> wrote in message
news:Xzecd.376828$mD.361109@attbi_s02...


Jim Yanik wrote:
Saddam was NO THREAT to the United States, or anybody else.

Especially now, sitting in jail!

Actually now he is a bigger threat.


Remember: the intelligence agencies from ALL OVER THE WORLD were

Remember: John S. Dyson is a Nazi Sympathizer!

Cheers!
Rich


John F. Kennedy is a Nazi by today's Lib standards.

Not really- JFK is revered by most hard core liberals. He was a man for
the people. What do you call a joke for /profits/ at any cost? GW Bush
comes to mind- the most incompetent joke to ever sit in the Oval Office.


Hell no. The "real" JFK was tough on defense and cut taxes. I admired
him and mourned his loss. If JFK showed up today with those ideals he
would be ridiculed by the likes of Dean, Kerry, and Edwards.

Everyone was tough on defense back then- it was expected. Joseph Sr was
most likely responsible for the tax cuts.

 

[Mac]

On Tue, 26 Oct 2004 17:42:39 -0700, Dave wrote:
[snip]

Sorry I forgot, I am trying to obtain 2-4ms bi-phasic square pulses from
the current source (isoalted events). The resistance (500ohm to 2kohms)
values are guesstimates based on experiments done in the past (the
resident expert on this topic is
on vacation at the moment). Also, I saw some of the other suggestions
with supply voltages around 40-55V, I dont think that I will be able to
obtain sources above +-15V for the supply.

You also stipulated that the muscle be grounded, and that you might need
to drive 10 mA into 2k. This is impossible, I believe, since 10mA * 2k =
20 Volts.

Anyway, If you are only going to have +/- 15 Volt supplies, and you only
need 10 mA, I would think you could work something out with op-amps.

For example, I still think you could use a current sensing resistor with a
difference amplifier across it, and take the output of the difference
amplifier back to the driving op-amp. You may need to use rail-to-rail IO
op-amps all around.

I don't know, maybe the offset voltages and stuff would add up and kill
you. I'm not going to do the analysis. ;-)

But, just to be more concrete, you could have an op-amp driving the
load through a 10-Ohm current sensing resistor. Across this resistor
would be a difference amplifier with a gain of, say, 10, whose output is
then fed back to the inverting input of the driving op-amp.

So, when you apply 1 V to the op-amp, the output current would increase
until you had 0.1 V across the CS resistor, which would imply 10 mA into
the load. So the slope is 10 mA per volt, but it could be reduced to
whatever you want with a voltage divider. Maybe you need to offset it,
too. It's not clear what the spec is from your description. But that
should be easy with one more op-amp.

At the maximum current of 10 mA, you would only be dropping 0.01 * 10 =
0.1 Volts in the current sense resistor. You could probably make it even
bigger. Maybe 49.9 Ohms.

Anyway, good luck!

--Mac

 

[Jonathan Kirwan]

On Wed, 27 Oct 2004 03:22:07 GMT, Robert Monsen <rcsurname@comcast.net> wrote:

Jonathan Kirwan wrote:
On 24 Oct 2004 09:24:22 -0700, Winfield Hill <Winfield_member@newsguy.com
wrote:


. +Vs o------+----------------+-------,
. 40 - 55V | | |
. R4 R5 BFC
. 1k00 249 |
. | npn | gnd
. Q2 e\| |/e
. pnp |----+-------| Q3
. c/| | |\c 100V pnp
. | | | small heat sink
. +------' |
. | |
. |/c Q1 | bipolar current source
. +15V --| npn | 20mA fs, 30V compliance
. |\e |
. | | R4 R2 1
. R1 ,----|--R2--, | Io = - Vin -- ( ---- + -- )
. 10k0 | _| 10k0 | | R5 R1 R3 R1
. Vin --/\/\--+--|- \ | x |
. -10V pk in | >----+ +---- Io out
. = +20mA out ,--|+_/ | |
. | | R3 2k49 |
. gnd | | |
. | gnd |
. | |
. and etc for neg portion


I ran some simulations on the extended version of this schematic and output
current isn't particularly flat relative to the driving voltage as one might
hope for. (I selected the OPA227 as the opamp, the 2N3906/2N3904 for the Q1
equivalents, and the 2SB709A and the 2SD601A for the (+) and (-) side pairs.)

Firstly, accuracy is off for reasons I cannot completely fathom. I'm getting an
Iout that is about 12% lower than what the equation predicts. I can see where
the equation arises, I think, but my understanding doesn't appear to change the
simulation results (of course):

[1] The current from Vin to the (-) input node is Vin/R1.
[2] The current flowing from the (-) input node to the opamp output
is then -Vin/R1.
[3] This current, [2], must flow through R2, thus the voltage at
the opamp output must be -R2*(Vin/R1).
[4] The current flowing through R3 is then -R2*(Vin/R1)/R3.
[5] These currents sum, [2] and [4], so the total current arriving at
(or leaving) is -Vin/R1 + -R2*(Vin/R1)/R3 or -Vin*(1/R1+R2/(R1*R3)).

The output must either source or sink this, so this is what either draws an
extra from the V+ or the V- of the opamp, yielding the behavior. As I see it,
anyway.

[6] So, this is then applied to one side of the current mirror and is
gained up by the ratio of R4/R5.

But regardless of this logic above, I still get 9.293mA at -5V. The calculation
suggests:

-5V*1k/249*(1/10k+10k/(10k*2.49k)) = 10.072mA

That's about 92% of the predicted value. Of course, this is simulation. But
I'm curious if you think this is a simulation problem or if you think it's
probably real.

Aside from this accuracy difference from the equation, I'm curious about another
behavior I noticed in simulation. The relative gain varies rather widely as the
controlling input varies from -5V to +5V. (I used a .dc sweep.) As I sweep the
DC controlling voltage from -5V towards 0V, the relative change (using -5V as
the standard of 1.000) is:

Vin Iout gain Rel %
------ ------- ---------- -------
-5V 9.293mA 1.859mMhos 1.000
-1 1.947 1.947 1.047
-.5 1.026 2.052 1.104
-.4 842.1uA 2.105 1.132
-.3 657.8 2.193 1.180
-.2 473.4 2.367 1.273
-.1 289.0 2.890 1.555
-.05 196.7 3.934 2.116

etc.

Does this mean it needs a micro to support a correction table? Or is this more
likely to be some problem in my simulation?

Jon

Add the traditional PNP transistor between base and collector of Q2, so
its emitter is at the base of Q2, base is at the collector of Q2, and
its collector is grounded. That will minimize the current swiped from
the collector, and make the mirror more accurate.

[R4] [R5]
| |
| |
| |
| |
| |
|-------o-------------|
/| | |\
| | |
| |< Iout
o-------|
| |\
| |
| GND
|
o
Iin

created by Andy´s ASCII-Circuit v1.25.250804 www.tech-chat.de

I had already done that, but didn't decide to post the results until comments
started coming in. The new table looks like:

Vin Iout gain Rel %
------ ------- ---------- -------
-5V 9.463mA 1.8926mMhos 1.000
-1 1.961 1.961 1.036
-.5 1.023 2.046 1.081
-.4 835.0uA 2.088 1.103
-.3 647.2 2.157 1.140
-.2 459.8 2.299 1.215
-.1 271.9 2.719 1.437
-.05 177.9 3.558 1.880

A little better, but the trend is still there. I'm imagining that this is due
to the re in the current mirror transistors, but I need to study more to see if
this is true and to what degree.

(By the way, I was simulating with +/-60V rails.)

Also, the current ratio isn't really R4/R5, because there is a small
correction factor. It can easily be derived from ebers-moll. Assuming a
desired ratio of x:1, a control current of I, and a resistance on the
control transistor of R, the emitter resistance on mirror transistor is

Rm = (R*I - 25.3e-3 * ln(x))/(x*I)

So, for the example above, you have

Rm = (1k * 4m - 25.3m * ln(4))/(4 * 4m) = 247.8

Not much of a difference unless the ratio is large. For a 10x mirror
with a 1mA steering current, for example, the resistance Rm will be 94.2
rather than 100. TC will also adversely affect the linearity.

Okay. I generally know what you are referring to and I knew I'd need to account
for it, but I decided to just go with the equation Win posted up for now. I'll
look at exactly what you are saying here and make sure I follow.

A more obvious way (for me...) to generate the steering current would be

Iout+
|
15V |/
--------------o--------|
| |
| |
| |
| |/
| .----|
| | |
|\| | |
.-----|-\ | |
| | >--o o------.
Vin----------|+/ | | |
| |/| | | |
| | | | |
'------------------o |
| | | |
| | |< \
| '----| / R=2.5k
| |\ \
| | /
| | |
| |< ===
--------------o--------| GND
-15V |\
|
Iout-

created by Andy´s ASCII-Circuit v1.25.250804 www.tech-chat.de

It's basically the same as Win's solution, but the driver for his is
internal to the opamp package. With this one, however, you can easily
make the steering current equal to the output current, even up to fairly
large currents, and thus eliminate one source of non-linearity in the
mirrors (ln(1) = 0). This might also help with any tempco problem.

Yes, but isn't it the case that the output drive section of an OPA227 is pretty
darned good, already? I need to look more closely at what you've posted here,
though, as I'm still getting more familiar with these arrangements and don't
just instinctively *know* their behavior. Thanks!

Also, your diagram here makes it much clearer why that resistor was needed.
Should have seen it beforehand, but this clarifies it.

One can use a cheapo opamp like an LM324 if crossover distortion or slew
rate isn't an issue. It sounds like the OP will want to set the current,
then either shunt it to ground using his optoisolator control, or let it
rip into the poor torture victim/grad student. If so, opamp slew rate
won't be an issue.

Thanks much for the comments. I'm still seeing the non-linearity of >=4% in the
range of +/-1V and that bothers me to see so much when a full volt is at the
input as compared to when 5V is there. But I think I'll find it in the BJT's re
(which is really a 25.3mV thing, at 25C), though that's a rather uneducated
guess. But if I were building one of these, I'd be wanting to use it to put
precision 0.1uA currents up through say 1mA, in steps, through diodes, for
example, and then measuring the V across them and plotting the results. I'd
want a precise, wide dynamic range without gross non-linearities in the output
at small value currents and still able to deliver 'high currents' in the 1mA to
even maybe 10mA area. I don't think I'd ever care to go higher than 10mA,
except for VERY SHORT times, I suppose.

I'm thinking automated BJT tester, eh?

Jon

 

[Clifford Heath]

Jonathan Kirwan wrote:

I'm thinking automated BJT tester, eh?

Why not make it three-way, with a uC driving them, and you have
a curve-tracer that can handle any type of 2 or 3 terminal device
in any orientation? I want one!

 

[Robert Monsen]

Dave wrote:

Winfield Hill <Winfield_member@newsguy.com> wrote in message news:<clgkvm03l1@drn.newsguy.com>...

Winfield Hill wrote...

Dave wrote...

I have a particular application that I'm looking for some advice on.
I am trying to build a 32 electrode array of bipolar stimulators for
electrophysiology related experiments. The specs that I have right
now are the load varies from 500 ohms to 2Kohms, the current range
needed is -10mA to +10mA (+-15mA would be even better) and the muscle
needs to be grounded. The array is used to fire a pulse into the
muscle at sucessive distances and times. The current needs to be
voltage controlled from 0 to 10V and current pulse width is between
2 and 5ms, I hope to achieve switching ON & OFF of the electrodes by
digital switches in a pre-optoisolated stage. I have tried the
Improved Howland and found the variation in load to be a problem,
and since I will need 32 of these using potentiometers to match
resistances seemed complicated to me.

[ snip ]

I'd appreciate any suggestions on which path to pursue or on the
making the Howland work, or any other ideas.

I'll type up a bipolar power current-mirror circuit candidate.

. +Vs o------+----------------+-------,
. 40 - 55V | | |
. R4 R5 BFC
. 1k00 249 |
. | npn | gnd
. Q2 e\| |/e
. pnp |----+-------| Q3
. c/| | |\c 100V pnp
. | | | small heat sink
. +------' |
. | |
. |/c Q1 | bipolar current source
. +15V --| npn | 20mA fs, 30V compliance
. |\e |
. | | R4 R2 1
. R1 ,----|--R2--, | Io = - Vin -- ( ---- + -- )
. 10k0 | _| 10k0 | | R5 R1 R3 R1
. Vin --/\/\--+--|- \ | x |
. -10V pk in | >----+ +---- Io out
. = +20mA out ,--|+_/ | |
. | | R3 2k49 |
. gnd | | |
. | gnd |
. | |
. and etc for neg portion

This circuit is based on a pair of current mirrors, made from the pnp
positive mirrors Q2 - Q3 and their negative-side npn mates, not shown.
The gain of 4x mirrors have an emitter resistors that drop 5V at full
scale, eliminating the need for matched transistors. In this circuit
one wants high-beta Q3 parts, although the base-current error can be
reduced by adding an extra transistor (see AoE figure 2.50, page 90).

The opamp voltage-current converter in this circuit is unusual. First
note that any opamp quiescent supply current creates identical currents
that cancel at the Io output, even as it also creates a class-A output
transistor bias current. This bias will be 4mA if an opamp with 1mA of
quiescent current is used, and will create 0.2W of class-A heating in
each output transistor, which is fine. Second, realize that when the
voltage at the opamp output (point x) is not zero volts, a predictable
unbalanced programming current is created for the output mirrors, via
the opamp's output transistors, and through Q1 and its opposite mate.
The formula shows how to select resistor values to precisely set the
circuit's transconductance gain with 1% resistors, without requiring
any trimpots.


Thanks for the circuit. I was wondering, is it possible to configure
something that would require less voltage, like from a +-15V supply.
I forgot to mention that I would be using the ciruit to fire 2 to 4ms
biphasic square pulses into the muscle in a non-repetitive fashion. I
have read in some data sheets that the opamps did not have the
required speed, what steps do you suggest?

If you want to fire 15mA pulses into a 2k load, you need 30V of what's
called 'compliance', which just means the voltage the current source can
vary up to. With a +-15V supply, your compliance will be more like
+-12V, min, so that means that for a fixed load of 2k, the most current
you can get through it is 12/2k = 6mA.

That's why Win's design went up to 40V.

--
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.

 

[Clifford Heath]

Dave wrote:

Thanks for the circuit. I was wondering, is it possible to configure
something that would require less voltage, like from a +-15V supply.

For +-15, just drop Q1 (join its C and E).

I forgot to mention that I would be using the ciruit to fire 2 to 4ms
biphasic square pulses into the muscle in a non-repetitive fashion. I
have read in some data sheets that the opamps did not have the
required speed, what steps do you suggest?

Some op-amps would have trouble reaching 2-4us, none will have any
difficulty with 2-4ms.

 

[Tony Williams]

In article <02mtn0hpvfpgnn6t4le1j7t9j03fptlcgj@4ax.com>,
Jonathan Kirwan <jkirwan@easystreet.com> wrote:

Vin Iout gain Rel %
------ ------- ---------- -------
-5V 9.293mA 1.859mMhos 1.000
-1 1.947 1.947 1.047
-.5 1.026 2.052 1.104
-.4 842.1uA 2.105 1.132
-.3 657.8 2.193 1.180
-.2 473.4 2.367 1.273
-.1 289.0 2.890 1.555
-.05 196.7 3.934 2.116

Does the simulation show an Iout offset of just over
+100uA at Vin= 0v?

--
Tony Williams.

 

[Paul Burke]

Bill Sloman wrote:

The case stops being obvious to those whose IQ is rather higher than
50 - not one of my real-world U.S. acquaintances is going to vote for
Bush, and the common factor here is that they have all earned
post-graduate degrees and work as academics, which means that their
IQ's are all going to be higher than 110.

Bill, it's no use arguing with Americans. They inhabit a different
world, where values are as different from ours as the mediaeval Ottoman
empire. Over there, even Daily Telegraph readers would count as commies.
There are only two probables:

(1) They will re-elect Bush.

(2) Six months later, nobody will admit that they voted for him.

Paul Burke

 

[Terry Given]

Alan Perry wrote:

Terry Given <my_name@ieee.org> wrote in message news:<ackfd.2693$lF1.69232@news.xtra.co.nz>...

Alan Perry wrote:

Having had experience of ground loops and general interference
I wanted to design a system that was as isolated as possible, but
still safe. To start with the mains power feed I researched around
and I believe it boils down to this...

Is this How you use an Isolation Transformer and RCD to protect equipment
and break ground loops?

Primary | Secondary

--------------
L -------- | --------| |------------- L2
( | ( | |
)| ) | |
( | ( | |
)| ) | RCD 30mA |
( | ( | |
N -------- | ----+---| |------------- N2
| | | |
| ----| |------------- E2
E ---------- --------------

^
|
Isolation transformer

Also can charge build up be solved with a 1M resister between the
two earths?

Cheers

Alan

Sorry Alan, but you use EITHER an RCD or an isolating transformer. As it
stands, your isolating transformer adds weight, cost and NO PROTECTION
WHATSOEVER to the user. Here is why:


who said I wanted to protect the user? oh yeah I did say safe....


An isolation transformer is used to minimise electric shock hazards in
MEN (Multiple Earthed Neutral) power systems. With the MEN system,
distribution transformer secondary star-points (ie neutral connections)
are connected to "the body of the earth" at multiple points. There are
thus 2 ways of getting an electric shock:
1) hook yourself up across Phase & Neutral
2) hook youself between phase and the planet (eg wet concrete floor, or
muddy paddock etc).

Obviously the latter is the most common technique for electrocution, as
in general we are always in good contact with the earth. MEN is used to
ensure a low impedance to fault currents (which are correspondingly
high) thereby ensuring protective devices quickly trip. This is a good
thing; the downside is the increased risk of electrocution.

The use of an isolation transformer removes the connection between the
neutral conductor and earth. This removes #2 as a means for getting
zapped - now you must hook up to both phase and neutral, which takes
considerably more effort - but is achievable, if you try hard enough

Alas in your schematic the second "earth" terminal is of course not
connected to earth, but is actually the neutral conductor. In the case
of metallic equipment, you have now ensured that the case is securely
connected to neutral - ie you just achieved half of the necessary
connections for electrocution to occur. Its worth noting that an
isolation transformer has no "phase" and "neutral" outputs - they are
interchangeable (hey, its AC...). Iso trannys generally have a fuse in
each output lead.


The earth-neutral connection did bother me, hence the post....

the flip side to my argument is of course for equipment with metal
cases, any phase-to-case internal fault will immediately blow the iso
tranny fuses when connected as drawn. If you leave the earth pin
unconnected (as in a conventional iso tranny) then a phase-to-case fault
means the case is at the phase potential. Because phase is floating wrt
earth, this does precisely nothing. OTOH if you touch this appliances
case with one hand, and the neutral conductor with the other - zap,
you're dead (at least you probably are with 230Vac

An iso tranny with 1 outlet has NO CONNECTION to the earth pin at all.
An iso tranny with 2 or more outlets will have all of the earth pins
connected together, but going nowhere. The reason for this is as
follows: imagine two appliances, one having a phase-to-case fault, the
other with a neutral-to-case fault. If you plug the first one into a
dual-output iso tranny (with earth unconnected), the case now sits at
the phase conductor potential. The operator will not get a shock, and
the equipment will continue to operate. However if they plug the second
faulty device into the same iso tranny, one case is at phase potential,
the other at neutral potential. one hand on either appliance WILL KILL
YOU. OTOH if the two outlet earths are connected together, then as soon
as you plug in the 2nd appliance, one or both output fuses will rupture
(the first appliance means the phase conductor is at earth potential,
the 2nd shorts phase to neutral thru the earth terminal link).

an RCD is a controlled-electrocution device, and will trip an isolating
device very rapidly (< 300ms) when earth current (>= 30mA) is detected.
Of course this earth current must flow (for a little while). If faulty
equipment causes the earth leakage current, then the RCD will trip when
you plug in said equipment. If it is the operator getting a belt that
causes the earth leakage current, then the operator must get said belt
before the RCD will trip.

cheers
Terry


So I must connect my earths together (on my isolated side), but not to
'neutral' (which I am instinctively more happy with)

Bingo. And that connection is specifically for the two-faulty-appliance
case (pun intentional . Most countries have electrical regulations
governing the use of iso trannies, which specifically forbids connecting
the "earth" E2 to N2. NZ certainly does - I just passed my electrical
registration B exam a couple of weeks ago, allowing me to service
3-phase 400V fixed-wired toasters. We specifically studied the regs
concerning iso tranny's

If I include an RCD, this is another level of fault protection (in
addition to fuses in both output lines)

only if you connect your "earth" E2 to neutral N2 a-la the diagram. if
E2 is unconnected, then the RCD will never trip.

so you see that putting in the RCD requires the E2-N2 connection, which
you are not supposed to do with an iso tranny. normally one uses either
an rcd or an iso tranny, not both.

As I am expecting fault conditions to occur, any thing that helps
protect the equipment and user will be welcome (just as long as it's
not too expensive.... )

what equipment are you actually trying to protect?

Thanks

Alan

Cheers
Terry

 

[Terry Given]

Clifford Heath wrote:

Dave wrote:

Thanks for the circuit. I was wondering, is it possible to configure
something that would require less voltage, like from a +-15V supply.


For +-15, just drop Q1 (join its C and E).

I forgot to mention that I would be using the ciruit to fire 2 to 4ms
biphasic square pulses into the muscle in a non-repetitive fashion. I
have read in some data sheets that the opamps did not have the
required speed, what steps do you suggest?


Some op-amps would have trouble reaching 2-4us, none will have any
difficulty with 2-4ms.

and some opamps sneer at 2us. all depends on how much you feel like
paying. Shit, you can buy > 10GHz GBW opamps... but I'm pretty sure they
cost more than an LM324 *sigh*

Cheers
Terry

 

[Jonathan Kirwan]

On Wed, 27 Oct 2004 07:55:31 +0100, Tony Williams <tonyw@ledelec.demon.co.uk>
wrote:

Does the simulation show an Iout offset of just over
+100uA at Vin= 0v?

104.58uA in the first case I posted. 84uA in the second design case, with the
extra BJT.

Jon

 

[Robert Baer]

TomH wrote:

Greetings,

I need to use a large number of banana sockets for a patchbay. Rather
than buying these by the hundred, I find that I can buy hard brass
tube of the correct internal diameter for a few cents. Now collegues
have told me that I must have these plated with nickel or similar to
get a good contact, but I recall the patchbays used for low level
microphone signals used in recording studios were often unplated
brass, and no care was ever taken to clean them, in fact they were
often filthy with a black greasy deposit. Also scientific and
laboratory instruments up to the 20's were made with unplated brass
terminals.

What do people think, could I get away with unplated brass? My
application is for patching low level dc voltages in the range +/-10V.

Regards

TomH

As long as the atmospher there is not corrosive (sulphor from the salt
mines, or salt from the deep ocean), you should have no problems.

 

[Robert Monsen]

John Larkin wrote:

On Wed, 27 Oct 2004 04:11:58 GMT, Robert Monsen
rcsurname@comcast.net> wrote:


John Larkin wrote:

On Tue, 26 Oct 2004 19:32:00 GMT, Robert Monsen
rcsurname@comcast.net> wrote:



By the way, if your scroll down to the bottom of the link that mr Yanik
has posted, which attempts to corroborate the iraqi terrorist story, you
find an add for an Ann Coulter book "How to talk to a Liberal (if you
must)". Coulter, of course, deserves a pie in the face:



What a disgusting statement.


A pie in the face isn't a bullet. It's not even a slap. It's a pie,
John. Whipped cream. It probably tasted good. She apparently laughed
about it.



So, audiences should maybe routinely pack pies to public forums, and
splatter anybody they don't agree with?

Crowds used to routinely bring vegetables to public performances, so
they could pelt people who sucked too badly. Maybe we should try to
bring that back. Unfortunately, given the loyalty oaths required to see
Bush, they would probably frisk people for tomatoes too.

http://www.thesmokinggun.com/archive/1022042coulter1.html

Unfortunately, they missed. (This is one of the attacks Dyson was using
to point out how republicans are under attack from smelly democrats.)


What's the point in having a democracy if all you care about is
winning, and will use violence to win? That's a dangerous position for
lefties to take, given that the Right is a bunch of tough hombres,
rather well armed. Seems like the left are today's facists, physically
attacking people they disagree with.


That's a stretch. A couple of kids throw a pie at ann coulter, and you
equate it to a left-wing fascist movement? Pretty pathetic reasoning.


Nobody's throwing pies at Ted Kennedy, or Hillary, or Dan Rather, or
even Al Sharpton. And it's not the Republicans who are shooting up and
trashing their opponents campaign headquarters. At least these creeps
were arrested for felonies.

I'm having a hard time finding any real evidence of democrats "shooting
up or trashing" RNC headquarters. I have seen an article about somebody
who shot out an RNC doorway in Knoxville, which I posted a link to
earlier in this thread. There isn't any evidence that that was done by a
mob of angry left wing brie eating elitists. Maybe you have more, but
those accusations sound like pre-election rabble-rousing to me.

Ann is a kick. People have been shooting up Republican campaign
offices; she cites, as proof that the shooters are Democrats, the fact
that they never hit anything.


Ann is a 'kick' alright. As a comedienne she is funny in a sleasy,
lying, shock-jock sort of way. As a political reporter and analyst,
which is what she claims to be, she is simply a dishonest shill for the
right.


Which opinion justifies pie-throwing? Why not paint? Or rocks?

Nobody is throwing paint or rocks. A cream pie, a few tomatoes, maybe...

Here are a few opinions from her (ironically titled) book "Slander:
Liberal Lies About the American Right":

"Liberals hate America"
"Liberals hate all religions except Islam"
"Democrats actually hate working-class people."
"Liberals hate society"
"Even Islamic terrorists don't hate Americans like liberals do"
"Democrats ... will destroy anyone who stands in their way. All that
matters to them is power"
"Liberals can't just come out and say they want to take more of our
money, kill babies, and distriminate on the basis of race."
"That's the whole point of being a liberal: to feel superior to people
with less money."
"Liberals are crazy"

Aren't these funny? She is such a 'kick'. They were collected by Al
Franken in his book "Lies, and the Lying Liars Who Tell Them: A Fair and
Balanced Look at the Right". Thankfully, I didn't have to read her book.
He suffered through it for me. Thanks, Al, I owe you one.

Democracy is a compact of mutual respect: we agree before we speak and
vote to respect people's opinions and respect the outcome. Without
that, only force matters.

Yes, it's important to have respect for your opponents, like the respect
our dear Ann Coulter showed when writing those ugly, hateful opinions above.

I know, sticks and stones... but a pie is just a funny political
statement, and doesn't hurt anybody (except the dumb al-pieda throwers,
who are now facing felony pie throwing charges for soiling a $3,000
backdrop).

--
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.

 

[nyffeler]

In article <c7b545b1.0410251953.48c0d2ab@posting.google.com>, fox@pub.xaonline.com (Fox zhou) wrote:

Dear all,

I'd like to charge a capacitor with a constant current source in short
period of time(1ns-10ns). At the end of the time, I want to stop the
charging. I propose to use a switch to control the constant current
source. Turn on at the start and turn off at the end.

But where can I find the corresponding switch? (1GHz)
Does any one know there are the controllable constant current source
chips can complete this task?

Are you looking for a time to amplitude converter?
If so, have a look at
http://ej.iop.org/links/q24/R3eIq,maJdzdv1f2yjwktw/e003n2.pdf
or a commercial one on
http://www.canberra.com/products/692.asp

 

[Hans-Bernhard Broeker]

[F'up2 cut down to one group --- should have been done by OP!]

In comp.arch.embedded yo <yotango@yahoo.com> wrote:

I understand it's complex to use SDRAMS. I have an idea that could
simplify the job. Instead of having my circuit do all the
initializations, etc. to the SDRAM, why not do it through software
on the PC?

For the same reason you couldn't use the PC's own memory in the first
place: accesses through PCI bus to the SDRAM chips' pins will rather
likely be too slow to function correctly.

I don't think it's feasible for a one-off hobby project to implement
your own SDRAM controller logic.

--
Hans-Bernhard Broeker (broeker@physik.rwth-aachen.de)
Even if all the snow were burnt, ashes would remain.

 

[Winfield Hill]

Robert Monsen wrote...

Also, the current ratio isn't really R4/R5, because there is a small
correction factor. It can easily be derived from ebers-moll. Assuming a
desired ratio of x:1, a control current of I, and a resistance on the
control transistor of R, the emitter resistance on mirror transistor is

Rm = (R*I - 25.3e-3 * ln(x))/(x*I)

So, for the example above, you have

Rm = (1k * 4m - 25.3m * ln(4))/(4 * 4m) = 247.8

Not much of a difference unless the ratio is large. For a 10x mirror
with a 1mA steering current, for example, the resistance Rm will be 94.2
rather than 100. TC will also adversely affect the linearity.

This assumes Q2 and Q3 are identical, and would therefore be operating
at different current densities. But in practise a larger transistor
would be used for Q3, especially for high currents, considering that
the power dissipated by Q3 is so much greater than Q2. Now that the
subject has been raised, for high current ratios the large temperature
difference between Q2 and Q3 would substantially degrade the accuracy
of the current ratio. Under these conditions a different kind of
mirror would be appropriate.


--
Thanks,
- Win

(email: use hill_at_rowland-dotties-org for now)

 

[Paul Burridge]

On 26 Oct 2004 18:58:23 -0700, paulk880@yahoo.com (Paul K.) wrote:

Thanks everyone for your help.

JFETs are pretty interesting devices. In some ways the're much easier
to work with than BJTs.

They are conceptually easier to understand, but only suitable for a
small range of applications where they're indeed preferable to BJTs.
The biggest drawback with them is the large spread of parameters one
encounters with any given type number and the limited amount of gain
they can provide.

--

"What is now proved was once only imagin'd." - William Blake, 1793.

 

[Allan Herriman]

On 27 Oct 2004 10:26:04 GMT, Hans-Bernhard Broeker
<broeker@physik.rwth-aachen.de> wrote:

In comp.arch.embedded yo <yotango@yahoo.com> wrote:

I understand it's complex to use SDRAMS. I have an idea that could
simplify the job. Instead of having my circuit do all the
initializations, etc. to the SDRAM, why not do it through software
on the PC?

For the same reason you couldn't use the PC's own memory in the first
place: accesses through PCI bus to the SDRAM chips' pins will rather
likely be too slow to function correctly.

I don't think it's feasible for a one-off hobby project to implement
your own SDRAM controller logic.

You can get a cheap FPGA evaluation board with SDRAM, and use an open
source SDRAM controller core inside FPGA.
You can get an open source PCI target core for the FPGA as well.

This reduces the PC Oscilloscope to basically three chips: FPGA, SDRAM
and ADC.

http://www.fpga-faq.com/FPGA_Boards.shtml

Regards,
Allan

 

[Winfield Hill]

Jonathan Kirwan wrote...

Why use R3, at all? I'm struggling to understand that detail.

Adding R3 allows one to set (increase) the transconductance gain
independently from loading of the input programming signal. In
my example R3 could be eliminated and R1 = 2k00 used to achieve
the same gain. The benefits of R3 arise at higher currents, but
I admit one might then chose to keep the programming currents low,
increasing the Q3 Q2 current ratio instead; this would reduce the
appeal of adding R3 at high currents.

.. +Vs o------+---------------+-------,
.. 40 - 55V | | |
.. R4 R5 BFC
.. 1k00 249 |
.. | npn | gnd
.. Q2 e\| |/e
.. pnp |----+------| Q3
.. c/| | |\c 100V pnp
.. | | | small heat sink
.. +------' |
.. | |
.. |/c Q1 | bipolar current source
.. +15V --| npn | 20mA fs, 30V compliance
.. |\e |
.. | | R4 R2 1
.. R1 ,----|--R2--, | Io = - Vin -- ( ---- + -- )
.. 10k0 | _| 10k0 | | R5 R1 R3 R1
.. Vin --/\/\--+--|- \ | x |
.. -10V pk in | >----+ +---- Io out
.. = +20mA out ,--|+_/ | |
.. | | R3 2k49 | R5
.. gnd | | | Iq = Is ----
.. | gnd | R4
.. | |
.. and etc for neg portion

Another issue to consider is the setting of the output-transistor
class-A quiescent bias current. I've added an Iq equation showing
how the bias is set by the opamp's Is quiescent supply current.
The equation shows that a high R5/R4 ratio can lead to a possibly-
undesirable high class-A current. Adding R3 allows one to chose a
lower R5/R4 gain, while at the same time maintaining an attractive,
moderately high, Vin input impedance for the circuit.


--
Thanks,
- Win

(email: use hill_at_rowland-dotties-org for now)