A couple of simple questions about a simple op amp circuit

[Bob Engelhardt]

I'm a new poster here. I've posted a couple of times to SED, but always
felt that the general level of SED was _way_ above me. SE_Basics_ is
more my level.

My questions are about this battery tester (an "electronic load"):
http://home.comcast.net/~bobengelhardt/eLoad.jpg

1. What is the purpose of R5? My guess is that the ckt would work
without it, but it's in some way better with it.

2. The 2nd op amp: "The voltage reference, U2, provides a stable
2.5-volt reference voltage ...". How does it do that? It's not
connected, so it must be internal, but ...??

Thanks,
Bob

 

[Michael Black]

On Sun, 1 Apr 2012, Bob Engelhardt wrote:

I'm a new poster here. I've posted a couple of times to SED, but always felt
that the general level of SED was _way_ above me. SE_Basics_ is more my
level.

If you aren't designing, it doesn't belong there. You are trying to

interpret an existing schematic.

My questions are about this battery tester (an "electronic load"):
http://home.comcast.net/~bobengelhardt/eLoad.jpg

This is all about loading the battery so you can see what the voltage is

when under load. A battery may show decent voltage when unloaded, but
then drop significantly when under load. This sort of thing means when you
measure the voltage, it's more like when the battery is in the circuit.

You could just load the battery with a big resistor, but that has
problems, so the fancier circuit.

U1 and Q1, the fet, form a constant current sink, meaning no matter what
the voltage the FET sees, it keeps drawing the same current (a resistor
across the battery would draw current varying with the voltage on the
battery).

Since the circuit turns voltage into a current draw, if there was no
voltage regulator, the drain would vary according to whatever is powering
this circuit, the 9v battery. I'm not sure that the voltage regulator
isn't overkill, but it doesn't add much in cost or size to the circuit,
that sort of voltage regulator can be found in plenty of computer
switching supplies, though maybe not that specific device.

1. What is the purpose of R5? My guess is that the ckt would work without
it, but it's in some way better with it.

U1 and the FET form a constant voltage sink.

The resistor is in the feedback circuit to the op-amp. No, it probably
isn't needed, it's there for isolation (note that c1, the .01uf capacitor,
is probably there to limit frequency response, and without R5, it wouldn't
work as well).

2. The 2nd op amp: "The voltage reference, U2, provides a stable 2.5-volt
reference voltage ...". How does it do that? It's not connected, so it must
be internal, but ...??

U2 is the TL531 in the upper left corner, a precision voltage regulator

(kind of like a fancy zener, where you can actually control the point at
which the regulator regulates). It supplies a stable voltage to U1.

The other half of the 358, is simply labelled "U" and is completely out of
the circuit (it's not used, so likely a dual op-amp was specified because
the 358 has certain characteristics that means it works better in a single
voltage power supply)

Michael

 

[Michael Black]

On Sun, 1 Apr 2012, Jamie wrote:

Bob Engelhardt wrote:

I'm a new poster here. I've posted a couple of times to SED, but always
felt that the general level of SED was _way_ above me. SE_Basics_ is more
my level.

My questions are about this battery tester (an "electronic load"):
http://home.comcast.net/~bobengelhardt/eLoad.jpg

1. What is the purpose of R5? My guess is that the ckt would work without
it, but it's in some way better with it.

No, you need R5 there. It is the feed back sense required to keep
U1 operating as a voltage comparator. The + input of U1 is the reference
voltage required and the (-)input would be the comparing point, in this case,
it is comparing the SOURCE (S) side of the mosfet transistor and will make
what ever needed adjustment output on U1 to bias Q1 to get there.
If the voltage exceeds at (S) of Q1, compared to what is sitting at (+)
input of U1, U1 output will then drop in bias voltage on the gate (G) of Q1.
This of course, will cause the Q1 to not conduct as much and lower
the voltage at (S) of Q1 to satisfy the voltage comparator circuit of U1.

But technically, if not for the capacitor from the output of the opamp to

the inverting input of the op-amp, the resistor is not needed.

It's no different from a voltage follower (or indeed, the other half of
the op-amp that lies unused at the lower left, the output connected to the
inverting input), though in this case, the FET is in that feedback loop.

But the opamp wants to see voltage, which it's already seeing on the
non-inverting input. Since no voltage amplification is done in that
stage, there's no absolute reason for the resistor, the voltage at the
FET is the same voltage as at the inverting input of the op-amp.

But of course, it isolates the capacitor that's going from the op-amp
output to the inverting input of that op-amp.

Michael

The output of U1 will not be following the + input reference voltage, it
will be higher than the 2.5v max you would get with that circuit.

Looking up the transistor, the sheet tells me you'll need ~3 Volts Plus
what is sitting at the Source of Q1 to get it to come on.



2. The 2nd op amp: "The voltage reference, U2, provides a stable 2.5-volt
reference voltage ...". How does it do that? It's not connected, so it
must be internal, but ...??

Thanks,
Bob

That device is like a programmable zener diode. Internally it has a
2.5v fixed reference (Band Gap), which is very stable. You can think of
that as a zener diode. That is used as a internal voltage comparator against
the control pin voltage.

Since the internal is set for 2.5 volts, the component stops clamping
the load when it hits 2.5 volts, because the internal comparator as hit the
balance point of the internal fixed voltage reference.

Now, if you were to apply a scaled reference of what is appearing at the
top side to the control pin, you can then force it to elevate its clamping
voltage at a higher point. A scale reference would be like a
voltage divider network where it derives its source from the top side of
the programmable zener here.


Btw, U2 is the 3 terminal voltage reference, not the OP-AMP.

The 358 is a dual unit, it looks like they are simply terminating the
leads due to lack of use so it won't damage the chip.


Hope that shed some light on the subject.


Jamie

 

[Jamie]

Bob Engelhardt wrote:

I'm a new poster here. I've posted a couple of times to SED, but always
felt that the general level of SED was _way_ above me. SE_Basics_ is
more my level.

My questions are about this battery tester (an "electronic load"):
http://home.comcast.net/~bobengelhardt/eLoad.jpg

1. What is the purpose of R5? My guess is that the ckt would work
without it, but it's in some way better with it.

No, you need R5 there. It is the feed back sense required to keep
U1 operating as a voltage comparator. The + input of U1 is the reference
voltage required and the (-)input would be the comparing point, in this
case, it is comparing the SOURCE (S) side of the mosfet transistor and
will make what ever needed adjustment output on U1 to bias Q1 to get there.
If the voltage exceeds at (S) of Q1, compared to what is sitting at (+)
input of U1, U1 output will then drop in bias voltage on the gate (G) of
Q1. This of course, will cause the Q1 to not conduct as much and lower
the voltage at (S) of Q1 to satisfy the voltage comparator circuit of U1.

The output of U1 will not be following the + input reference voltage,
it will be higher than the 2.5v max you would get with that circuit.

Looking up the transistor, the sheet tells me you'll need ~3 Volts
Plus what is sitting at the Source of Q1 to get it to come on.

2. The 2nd op amp: "The voltage reference, U2, provides a stable
2.5-volt reference voltage ...". How does it do that? It's not
connected, so it must be internal, but ...??

Thanks,
Bob

That device is like a programmable zener diode. Internally it has a
2.5v fixed reference (Band Gap), which is very stable. You can think of
that as a zener diode. That is used as a internal voltage comparator
against the control pin voltage.

Since the internal is set for 2.5 volts, the component stops clamping
the load when it hits 2.5 volts, because the internal comparator as hit
the balance point of the internal fixed voltage reference.

Now, if you were to apply a scaled reference of what is appearing at
the top side to the control pin, you can then force it to elevate its
clamping voltage at a higher point. A scale reference would be like a
voltage divider network where it derives its source from the top side of
the programmable zener here.


Btw, U2 is the 3 terminal voltage reference, not the OP-AMP.

The 358 is a dual unit, it looks like they are simply terminating the
leads due to lack of use so it won't damage the chip.


Hope that shed some light on the subject.


Jamie

 

[Jamie]

Michael Black wrote:

On Sun, 1 Apr 2012, Bob Engelhardt wrote:

I'm a new poster here. I've posted a couple of times to SED, but
always felt that the general level of SED was _way_ above me.
SE_Basics_ is more my level.

If you aren't designing, it doesn't belong there. You are trying to
interpret an existing schematic.

My questions are about this battery tester (an "electronic load"):
http://home.comcast.net/~bobengelhardt/eLoad.jpg

This is all about loading the battery so you can see what the voltage is
when under load. A battery may show decent voltage when unloaded, but
then drop significantly when under load. This sort of thing means when
you measure the voltage, it's more like when the battery is in the circuit.

You could just load the battery with a big resistor, but that has
problems, so the fancier circuit.

U1 and Q1, the fet, form a constant current sink, meaning no matter what
the voltage the FET sees, it keeps drawing the same current (a resistor
across the battery would draw current varying with the voltage on the
battery).

Since the circuit turns voltage into a current draw, if there was no
voltage regulator, the drain would vary according to whatever is
powering this circuit, the 9v battery. I'm not sure that the voltage
regulator isn't overkill, but it doesn't add much in cost or size to the
circuit, that sort of voltage regulator can be found in plenty of
computer switching supplies, though maybe not that specific device.

1. What is the purpose of R5? My guess is that the ckt would work
without it, but it's in some way better with it.

U1 and the FET form a constant voltage sink.

The resistor is in the feedback circuit to the op-amp. No, it probably
isn't needed, it's there for isolation (note that c1, the .01uf
capacitor, is probably there to limit frequency response, and without
R5, it wouldn't work as well).
Ha? I think you better look at that again, it is very needed..unless you

want a run away system.

2. The 2nd op amp: "The voltage reference, U2, provides a stable
2.5-volt reference voltage ...". How does it do that? It's not
connected, so it must be internal, but ...??

U2 is the TL531 in the upper left corner, a precision voltage regulator
(kind of like a fancy zener, where you can actually control the point at
which the regulator regulates). It supplies a stable voltage to U1.

The other half of the 358, is simply labelled "U" and is completely out
of the circuit (it's not used, so likely a dual op-amp was specified
because the 358 has certain characteristics that means it works better
in a single voltage power supply)

Michael
They terminate un-used op-amps in a package like that to insure no damage

comes to them, which would most likely propagate over to the used side.
Yes, it is a dual op-amp, to be exact.


Jamie

 

[Jamie]

Michael Black wrote:

On Sun, 1 Apr 2012, Jamie wrote:

Bob Engelhardt wrote:

I'm a new poster here. I've posted a couple of times to SED, but
always felt that the general level of SED was _way_ above me.
SE_Basics_ is more my level.

My questions are about this battery tester (an "electronic load"):
http://home.comcast.net/~bobengelhardt/eLoad.jpg

1. What is the purpose of R5? My guess is that the ckt would work
without it, but it's in some way better with it.


No, you need R5 there. It is the feed back sense required to keep
U1 operating as a voltage comparator. The + input of U1 is the reference
voltage required and the (-)input would be the comparing point, in
this case, it is comparing the SOURCE (S) side of the mosfet
transistor and will make what ever needed adjustment output on U1 to
bias Q1 to get there.
If the voltage exceeds at (S) of Q1, compared to what is sitting at (+)
input of U1, U1 output will then drop in bias voltage on the gate (G)
of Q1. This of course, will cause the Q1 to not conduct as much and lower
the voltage at (S) of Q1 to satisfy the voltage comparator circuit of U1.

But technically, if not for the capacitor from the output of the opamp
to the inverting input of the op-amp, the resistor is not needed.

No, the R5 is needed. The cap there serves to pad the feed back down so
it won't oscillate. Just think of adding more miller effect on top of
what is already there inside of the op-amp.

In order to have a constant current you need the circuit to monitor
the actual current and make calibrations at the gate drive. You can only
do this if you have a way to monitor the current and R5 is that. It
simply is monitoring the voltage which would be a function of current on
those high power R's there.

Jamie

 

[John Larkin]

On Sun, 01 Apr 2012 11:34:26 -0400, Bob Engelhardt
<bobengelhardt@comcast.net> wrote:

I'm a new poster here. I've posted a couple of times to SED, but always
felt that the general level of SED was _way_ above me. SE_Basics_ is
more my level.

My questions are about this battery tester (an "electronic load"):
http://home.comcast.net/~bobengelhardt/eLoad.jpg

1. What is the purpose of R5? My guess is that the ckt would work
without it, but it's in some way better with it.

2. The 2nd op amp: "The voltage reference, U2, provides a stable
2.5-volt reference voltage ...". How does it do that? It's not
connected, so it must be internal, but ...??

Thanks,
Bob

R5/R4/C1 keep both Q1 and U1 from oscillating. There are probably
three oscillation modes: Q1 RF oscillating on its own, U1 oscillating
from being loaded by the Q1 gate capacitance, and an overall loop
oscillation.

U2A appears to do nothing. It looks like the designer attempted to
keep it running closed-loop to keep it from affecting the other
section, but he didn't do it right.

It would be more sensible to use U2A to buffer the reference, increase
R1, and save a bunch of 9V supply current.


--

John Larkin Highland Technology Inc
www.highlandtechnology.com jlarkin at highlandtechnology dot com

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom timing and laser controllers
Photonics and fiberoptic TTL data links
VME analog, thermocouple, LVDT, synchro, tachometer
Multichannel arbitrary waveform generators

 

[Jim Thompson]

On Sun, 01 Apr 2012 11:34:26 -0400, Bob Engelhardt
<bobengelhardt@comcast.net> wrote:

I'm a new poster here. I've posted a couple of times to SED, but always
felt that the general level of SED was _way_ above me. SE_Basics_ is
more my level.

My questions are about this battery tester (an "electronic load"):
http://home.comcast.net/~bobengelhardt/eLoad.jpg

1. What is the purpose of R5? My guess is that the ckt would work
without it, but it's in some way better with it.

Both R5 _and_ R4 are to isolate the OpAmp from a capacitive load, to
prevent instability.

2. The 2nd op amp: "The voltage reference, U2, provides a stable
2.5-volt reference voltage ...". How does it do that? It's not
connected, so it must be internal, but ...??

Thanks,
Bob

The second OpAmp is not U2, it's the second section of U1... it's not
used and is "dummy" connected" to prevent undefined currents.

U2 is the TL431, an adjustable shunt regulator, connected here simply
as a 2.5V _zener_.

...Jim Thompson
--
| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.

 

[Tim Wescott]

On Sun, 01 Apr 2012 14:25:26 -0400, Michael Black wrote:

On Sun, 1 Apr 2012, Jamie wrote:

Bob Engelhardt wrote:

I'm a new poster here. I've posted a couple of times to SED, but
always felt that the general level of SED was _way_ above me.
SE_Basics_ is more my level.

My questions are about this battery tester (an "electronic load"):
http://home.comcast.net/~bobengelhardt/eLoad.jpg

1. What is the purpose of R5? My guess is that the ckt would work
without it, but it's in some way better with it.

No, you need R5 there. It is the feed back sense required to keep U1
operating as a voltage comparator. The + input of U1 is the reference
voltage required and the (-)input would be the comparing point, in this
case, it is comparing the SOURCE (S) side of the mosfet transistor and
will make what ever needed adjustment output on U1 to bias Q1 to get
there. If the voltage exceeds at (S) of Q1, compared to what is sitting
at (+) input of U1, U1 output will then drop in bias voltage on the
gate (G) of Q1. This of course, will cause the Q1 to not conduct as
much and lower the voltage at (S) of Q1 to satisfy the voltage
comparator circuit of U1.

But technically, if not for the capacitor from the output of the opamp
to the inverting input of the op-amp, the resistor is not needed.

But quite practically, if not for C1 and its associated resistors then
the circuit will oscillate strongly. That circuit is a classic for
driving a MOSFET gate without oscillations.

It's no different from a voltage follower (or indeed, the other half of
the op-amp that lies unused at the lower left, the output connected to
the inverting input), though in this case, the FET is in that feedback
loop.

It's considerably different from a voltage follower into a resistive
load, though, because the MOSFET gate is very capacitive, and the LM358
(like lots of op-amps) Really Doesn't like driving capacitive loads, and
will oscillate like mad if you try.

But the opamp wants to see voltage, which it's already seeing on the
non-inverting input. Since no voltage amplification is done in that
stage, there's no absolute reason for the resistor, the voltage at the
FET is the same voltage as at the inverting input of the op-amp.

Unless, of course, you want the circuit to actually work correctly.

--
Tim Wescott
Control system and signal processing consulting
www.wescottdesign.com

 

[Tim Wescott]

On Sun, 01 Apr 2012 11:34:26 -0400, Bob Engelhardt wrote:

I'm a new poster here. I've posted a couple of times to SED, but always
felt that the general level of SED was _way_ above me. SE_Basics_ is
more my level.

My questions are about this battery tester (an "electronic load"):
http://home.comcast.net/~bobengelhardt/eLoad.jpg

1. What is the purpose of R5? My guess is that the ckt would work
without it, but it's in some way better with it.

First, you need _some_ connection between the MOSFET source and the
negative feedback to the op-amp, because it's by servoing that source
voltage to your set voltage that you're controlling the current.

The reason for R5 is because you need R4 and C1 in there for stability:
the MOSFET gate is highly capacitive, which slows the response of U1 down
considerably. If you just connected U1 straight to Q1, then there would
be enough added lag in the loop formed by U1 and Q1 that the circuit
would oscillate. R4 isolates U1's output from the MOSFET gate (but
leaves in the lag). C1 lets U1's output 'talk' directly to U1's input
(which introduces lead, which counteracts the lag). R5 gives C1 a chance
to work, while still providing a DC path from the MOSFET source voltage
(which is what you want to set) to U1's negative input.

2. The 2nd op amp: "The voltage reference, U2, provides a stable
2.5-volt reference voltage ...". How does it do that? It's not
connected, so it must be internal, but ...??

The second op-amp is mislabeled (as is the first). The first one should
be U1A, and the second one (which is in the same package) should be
labeled U1B.

U2 is the thing that looks like a zener with a wire sticking out of it,
hooked back to the cathode -- it's a shunt voltage reference: stick a
reasonable current into it, and it'll hold a defined voltage.

--
Tim Wescott
Control system and signal processing consulting
www.wescottdesign.com

 

[Jamie]

Tim Wescott wrote:

On Sun, 01 Apr 2012 14:25:26 -0400, Michael Black wrote:


On Sun, 1 Apr 2012, Jamie wrote:


Bob Engelhardt wrote:


I'm a new poster here. I've posted a couple of times to SED, but
always felt that the general level of SED was _way_ above me.
SE_Basics_ is more my level.

My questions are about this battery tester (an "electronic load"):
http://home.comcast.net/~bobengelhardt/eLoad.jpg

1. What is the purpose of R5? My guess is that the ckt would work
without it, but it's in some way better with it.

No, you need R5 there. It is the feed back sense required to keep U1
operating as a voltage comparator. The + input of U1 is the reference
voltage required and the (-)input would be the comparing point, in this
case, it is comparing the SOURCE (S) side of the mosfet transistor and
will make what ever needed adjustment output on U1 to bias Q1 to get
there. If the voltage exceeds at (S) of Q1, compared to what is sitting
at (+) input of U1, U1 output will then drop in bias voltage on the
gate (G) of Q1. This of course, will cause the Q1 to not conduct as
much and lower the voltage at (S) of Q1 to satisfy the voltage
comparator circuit of U1.


But technically, if not for the capacitor from the output of the opamp
to the inverting input of the op-amp, the resistor is not needed.


But quite practically, if not for C1 and its associated resistors then
the circuit will oscillate strongly. That circuit is a classic for
driving a MOSFET gate without oscillations.


It's no different from a voltage follower (or indeed, the other half of
the op-amp that lies unused at the lower left, the output connected to
the inverting input), though in this case, the FET is in that feedback
loop.


It's considerably different from a voltage follower into a resistive
load, though, because the MOSFET gate is very capacitive, and the LM358
(like lots of op-amps) Really Doesn't like driving capacitive loads, and
will oscillate like mad if you try.


But the opamp wants to see voltage, which it's already seeing on the
non-inverting input. Since no voltage amplification is done in that
stage, there's no absolute reason for the resistor, the voltage at the
FET is the same voltage as at the inverting input of the op-amp.


Unless, of course, you want the circuit to actually work correctly.

Well, at least it gives people something to talk about. But I still
can't see how he expects the circuit to be a constant current source for
the test load if you don't bother to account for the sharp knee on the
gate turn on voltage point and the load varying due to a battery
discharging while under test.

I almost get the impression that maybe he thinks that is a jfet or
non-enhanced, when in fact, it's not. Maybe using the correct foot print
may have removed the mystery behind that.

Jamie

 

[Tim Wescott]

On Sun, 01 Apr 2012 17:52:50 -0400, Jamie wrote:

Tim Wescott wrote:

On Sun, 01 Apr 2012 14:25:26 -0400, Michael Black wrote:


On Sun, 1 Apr 2012, Jamie wrote:


Bob Engelhardt wrote:


I'm a new poster here. I've posted a couple of times to SED, but
always felt that the general level of SED was _way_ above me.
SE_Basics_ is more my level.

My questions are about this battery tester (an "electronic load"):
http://home.comcast.net/~bobengelhardt/eLoad.jpg

1. What is the purpose of R5? My guess is that the ckt would work
without it, but it's in some way better with it.

No, you need R5 there. It is the feed back sense required to keep U1
operating as a voltage comparator. The + input of U1 is the reference
voltage required and the (-)input would be the comparing point, in
this case, it is comparing the SOURCE (S) side of the mosfet
transistor and will make what ever needed adjustment output on U1 to
bias Q1 to get there. If the voltage exceeds at (S) of Q1, compared to
what is sitting at (+) input of U1, U1 output will then drop in bias
voltage on the gate (G) of Q1. This of course, will cause the Q1 to
not conduct as much and lower the voltage at (S) of Q1 to satisfy the
voltage comparator circuit of U1.


But technically, if not for the capacitor from the output of the opamp
to the inverting input of the op-amp, the resistor is not needed.


But quite practically, if not for C1 and its associated resistors then
the circuit will oscillate strongly. That circuit is a classic for
driving a MOSFET gate without oscillations.


It's no different from a voltage follower (or indeed, the other half of
the op-amp that lies unused at the lower left, the output connected to
the inverting input), though in this case, the FET is in that feedback
loop.


It's considerably different from a voltage follower into a resistive
load, though, because the MOSFET gate is very capacitive, and the LM358
(like lots of op-amps) Really Doesn't like driving capacitive loads,
and will oscillate like mad if you try.


But the opamp wants to see voltage, which it's already seeing on the
non-inverting input. Since no voltage amplification is done in that
stage, there's no absolute reason for the resistor, the voltage at the
FET is the same voltage as at the inverting input of the op-amp.


Unless, of course, you want the circuit to actually work correctly.


Well, at least it gives people something to talk about. But I still
can't see how he expects the circuit to be a constant current source for
the test load if you don't bother to account for the sharp knee on the
gate turn on voltage point and the load varying due to a battery
discharging while under test.

I almost get the impression that maybe he thinks that is a jfet or
non-enhanced, when in fact, it's not. Maybe using the correct foot print
may have removed the mystery behind that.

The op-amp accounts for that. If you put your thumb over C1, R4, and Q1,
then what's left is a voltage follower that drives the voltage at the top
of R13 to be equal to the voltage at pin 5 of U1. Since (at DC at least)
the MOSFET takes no gate current, that voltage is proportional to the
battery current, and hence it is the battery (or power supply) current
that is being servoed.

The circuit should do a pretty darned good job of holding the load
current steady, in fact.

--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com

 

[Jamie]

Tim Wescott wrote:

On Sun, 01 Apr 2012 17:52:50 -0400, Jamie wrote:


Tim Wescott wrote:


On Sun, 01 Apr 2012 14:25:26 -0400, Michael Black wrote:



On Sun, 1 Apr 2012, Jamie wrote:



Bob Engelhardt wrote:



I'm a new poster here. I've posted a couple of times to SED, but
always felt that the general level of SED was _way_ above me.
SE_Basics_ is more my level.

My questions are about this battery tester (an "electronic load"):
http://home.comcast.net/~bobengelhardt/eLoad.jpg

1. What is the purpose of R5? My guess is that the ckt would work
without it, but it's in some way better with it.

No, you need R5 there. It is the feed back sense required to keep U1
operating as a voltage comparator. The + input of U1 is the reference
voltage required and the (-)input would be the comparing point, in
this case, it is comparing the SOURCE (S) side of the mosfet
transistor and will make what ever needed adjustment output on U1 to
bias Q1 to get there. If the voltage exceeds at (S) of Q1, compared to
what is sitting at (+) input of U1, U1 output will then drop in bias
voltage on the gate (G) of Q1. This of course, will cause the Q1 to
not conduct as much and lower the voltage at (S) of Q1 to satisfy the
voltage comparator circuit of U1.



But technically, if not for the capacitor from the output of the opamp
to the inverting input of the op-amp, the resistor is not needed.


But quite practically, if not for C1 and its associated resistors then
the circuit will oscillate strongly. That circuit is a classic for
driving a MOSFET gate without oscillations.



It's no different from a voltage follower (or indeed, the other half of
the op-amp that lies unused at the lower left, the output connected to
the inverting input), though in this case, the FET is in that feedback
loop.


It's considerably different from a voltage follower into a resistive
load, though, because the MOSFET gate is very capacitive, and the LM358
(like lots of op-amps) Really Doesn't like driving capacitive loads,
and will oscillate like mad if you try.



But the opamp wants to see voltage, which it's already seeing on the
non-inverting input. Since no voltage amplification is done in that
stage, there's no absolute reason for the resistor, the voltage at the
FET is the same voltage as at the inverting input of the op-amp.



Unless, of course, you want the circuit to actually work correctly.



Well, at least it gives people something to talk about. But I still
can't see how he expects the circuit to be a constant current source for
the test load if you don't bother to account for the sharp knee on the
gate turn on voltage point and the load varying due to a battery
discharging while under test.

I almost get the impression that maybe he thinks that is a jfet or
non-enhanced, when in fact, it's not. Maybe using the correct foot print
may have removed the mystery behind that.


The op-amp accounts for that. If you put your thumb over C1, R4, and Q1,
then what's left is a voltage follower that drives the voltage at the top
of R13 to be equal to the voltage at pin 5 of U1. Since (at DC at least)
the MOSFET takes no gate current, that voltage is proportional to the
battery current, and hence it is the battery (or power supply) current
that is being servoed.

The circuit should do a pretty darned good job of holding the load
current steady, in fact.

I fully understand how the circuit works, but my comment to the original

statement is, that R5 is needed when it was thought it could've been
removed because it looked like it wasn't needed. That is far from the
truth. That is all I was trying to convey, R5 is absolutely needed here.

This is a 101 constant current circuit using a differential circuit to
maintain it's current at R13 and R14.

Oh well, maybe I should pick a different brand beer. This Coors Light
is getting to me, then again, it could be the fact that I just got done
with my Taxes. And I hope all the free loaders enjoy getting my money
that I have to additionally pay on this year.


Jamie

 

[Michael A. Terrell]

Jamie wrote:

Oh well, maybe I should pick a different brand beer. This Coors Light
is getting to me, then again, it could be the fact that I just got done
with my Taxes. And I hope all the free loaders enjoy getting my money
that I have to additionally pay on this year.

I'm sure they'll enjoy the three bucks that you didn't blow on cheap
beer.


--
You can't have a sense of humor, if you have no sense.

 

[Tim Wescott]

On Sun, 01 Apr 2012 22:00:46 -0400, Jamie wrote:

Tim Wescott wrote:

On Sun, 01 Apr 2012 17:52:50 -0400, Jamie wrote:


Tim Wescott wrote:


On Sun, 01 Apr 2012 14:25:26 -0400, Michael Black wrote:



On Sun, 1 Apr 2012, Jamie wrote:



Bob Engelhardt wrote:



I'm a new poster here. I've posted a couple of times to SED, but
always felt that the general level of SED was _way_ above me.
SE_Basics_ is more my level.

My questions are about this battery tester (an "electronic load"):
http://home.comcast.net/~bobengelhardt/eLoad.jpg

1. What is the purpose of R5? My guess is that the ckt would work
without it, but it's in some way better with it.

No, you need R5 there. It is the feed back sense required to keep U1
operating as a voltage comparator. The + input of U1 is the
reference voltage required and the (-)input would be the comparing
point, in this case, it is comparing the SOURCE (S) side of the
mosfet transistor and will make what ever needed adjustment output
on U1 to bias Q1 to get there. If the voltage exceeds at (S) of Q1,
compared to what is sitting at (+) input of U1, U1 output will then
drop in bias voltage on the gate (G) of Q1. This of course, will
cause the Q1 to not conduct as much and lower the voltage at (S) of
Q1 to satisfy the voltage comparator circuit of U1.



But technically, if not for the capacitor from the output of the
opamp to the inverting input of the op-amp, the resistor is not
needed.


But quite practically, if not for C1 and its associated resistors then
the circuit will oscillate strongly. That circuit is a classic for
driving a MOSFET gate without oscillations.



It's no different from a voltage follower (or indeed, the other half
of the op-amp that lies unused at the lower left, the output
connected to the inverting input), though in this case, the FET is in
that feedback loop.


It's considerably different from a voltage follower into a resistive
load, though, because the MOSFET gate is very capacitive, and the
LM358 (like lots of op-amps) Really Doesn't like driving capacitive
loads, and will oscillate like mad if you try.



But the opamp wants to see voltage, which it's already seeing on the
non-inverting input. Since no voltage amplification is done in that
stage, there's no absolute reason for the resistor, the voltage at
the FET is the same voltage as at the inverting input of the op-amp.



Unless, of course, you want the circuit to actually work correctly.



Well, at least it gives people something to talk about. But I still
can't see how he expects the circuit to be a constant current source
for the test load if you don't bother to account for the sharp knee on
the gate turn on voltage point and the load varying due to a battery
discharging while under test.

I almost get the impression that maybe he thinks that is a jfet or
non-enhanced, when in fact, it's not. Maybe using the correct foot
print may have removed the mystery behind that.


The op-amp accounts for that. If you put your thumb over C1, R4, and
Q1, then what's left is a voltage follower that drives the voltage at
the top of R13 to be equal to the voltage at pin 5 of U1. Since (at DC
at least) the MOSFET takes no gate current, that voltage is
proportional to the battery current, and hence it is the battery (or
power supply) current that is being servoed.

The circuit should do a pretty darned good job of holding the load
current steady, in fact.

I fully understand how the circuit works, but my comment to the original
statement is, that R5 is needed when it was thought it could've been
removed because it looked like it wasn't needed. That is far from the
truth. That is all I was trying to convey, R5 is absolutely needed here.

Ah -- I thought you were commenting on the OP's post, not Mr. Black's
post. My error.

This is a 101 constant current circuit using a differential circuit
to
maintain it's current at R13 and R14.

Yes, and I was kind of surprised that you didn't realize it. But I've
seen some pretty astute analog circuit designers who've seen this circuit
(or similar ones) and not realized what was up.

Oh well, maybe I should pick a different brand beer. This Coors Light
is getting to me, then again, it could be the fact that I just got done
with my Taxes. And I hope all the free loaders enjoy getting my money
that I have to additionally pay on this year.

Coors Light. Ick. Surely in this day and age you can get some decent
craft beer with real taste and color!

--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com

 

[Jasen Betts]

On 2012-04-01, Jamie <jamie_ka1lpa_not_valid_after_ka1lpa_@charter.net> wrote:

Well, at least it gives people something to talk about. But I still
can't see how he expects the circuit to be a constant current source for
the test load if you don't bother to account for the sharp knee on the
gate turn on voltage point and the load varying due to a battery
discharging while under test.

???
the circuit compares the volage across in r13 + r14 with the preset
voltage in the non-inverting input and turns the mosfet up or down to
match the voltages. the mosfet response doesn't need to be linear, only
monotonic,

--
⚂⚃ 100% natural

--- Posted via news://freenews.netfront.net/ - Complaints to news@netfront.net ---

 

[Jasen Betts]

On 2012-04-01, Tim Wescott <tim@seemywebsite.please> wrote:

The reason for R5 is because you need R4 and C1 in there for stability:
the MOSFET gate is highly capacitive, which slows the response of U1 down
considerably. If you just connected U1 straight to Q1, then there would
be enough added lag in the loop formed by U1 and Q1 that the circuit
would oscillate. R4 isolates U1's output from the MOSFET gate (but
leaves in the lag).

C1 lets U1's output 'talk' directly to U1's input
(which introduces lead, which counteracts the lag).

nah, a capacitor from output to the inverting input reduces the AC gain
producing even more lag, this stops the op-amp from overshooting and
also ensure that it won't oscillate.


--
⚂⚃ 100% natural

--- Posted via news://freenews.netfront.net/ - Complaints to news@netfront.net ---

 

[Jim Thompson]

On 2 Apr 2012 07:13:33 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:

On 2012-04-01, Tim Wescott <tim@seemywebsite.please> wrote:

The reason for R5 is because you need R4 and C1 in there for stability:
the MOSFET gate is highly capacitive, which slows the response of U1 down
considerably. If you just connected U1 straight to Q1, then there would
be enough added lag in the loop formed by U1 and Q1 that the circuit
would oscillate. R4 isolates U1's output from the MOSFET gate (but
leaves in the lag).

C1 lets U1's output 'talk' directly to U1's input
(which introduces lead, which counteracts the lag).

nah, a capacitor from output to the inverting input reduces the AC gain
producing even more lag, this stops the op-amp from overshooting and
also ensure that it won't oscillate.

Jason, Could you elaborate on that? Like write us an equation that
shows how it reduces the AC gain?

...Jim Thompson
--
| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.

 

[Tim Wescott]

On Mon, 02 Apr 2012 07:13:33 +0000, Jasen Betts wrote:

On 2012-04-01, Tim Wescott <tim@seemywebsite.please> wrote:

The reason for R5 is because you need R4 and C1 in there for stability:
the MOSFET gate is highly capacitive, which slows the response of U1
down considerably. If you just connected U1 straight to Q1, then there
would be enough added lag in the loop formed by U1 and Q1 that the
circuit would oscillate. R4 isolates U1's output from the MOSFET gate
(but leaves in the lag).

C1 lets U1's output 'talk' directly to U1's input (which introduces
lead, which counteracts the lag).

nah, a capacitor from output to the inverting input reduces the AC gain
producing even more lag, this stops the op-amp from overshooting and
also ensure that it won't oscillate.

Reduces the AC gain of what? The loop from inverting input to inverting
input?

Keep the difference between your overall circuit response and the loop
gain straight, please.

--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com

 

[Jamie]

Jasen Betts wrote:

On 2012-04-01, Tim Wescott <tim@seemywebsite.please> wrote:

The reason for R5 is because you need R4 and C1 in there for stability:
the MOSFET gate is highly capacitive, which slows the response of U1 down
considerably. If you just connected U1 straight to Q1, then there would
be enough added lag in the loop formed by U1 and Q1 that the circuit
would oscillate. R4 isolates U1's output from the MOSFET gate (but
leaves in the lag).


C1 lets U1's output 'talk' directly to U1's input
(which introduces lead, which counteracts the lag).


nah, a capacitor from output to the inverting input reduces the AC gain
producing even more lag, this stops the op-amp from overshooting and
also ensure that it won't oscillate.


You mean it lowers the BW ?

Jamie