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'[EE] PSU current output measurement'
follow-up from the diode question
I'd like to add a PIC and 8 x 2 LCD to this PSU to display Vout
and Iout. Measuring Vout is easy enough, but what about I out ?
The two simplest options appear to be a sense resistor in either
the (a) 0V or (b) Vout line
(a) would be the easier, using a PIC analogue pin, with probably
a DC amp to improve the resolution. But the load will not see a
true ground, which could be a problem if it shares a common chassis/
mains ground with the PSU. Mains earth will go to the PSU chassis
but I don't necessarily have to connect 0V to it as well
(b) is said to be more accurate but needs more circuitry, being a
high-side differential. It does measure the real Iout however
If this is a variable 0-35V 5A supply, what value would you make
the sense resistor ? For example to get 5mA resolution (1000th the
full Iout, or ~1 bit of ADC)
If I were to put the sense resistor in the Vout line (my academic
preference) could I have a pointer to a differential circuit that's
compatible with a PIC ? I note that there are many high-side battery
monitor ICs, none of which I have. If there's something that could
be knocked up with a reasonable-quality op-amp that'd be peachy
There are lots of ways to do this and you've hit upon the two that are
probably most common.
A shunt resistor in the gnd line is easy because it's the same gnd as the
rest of the sensing circuit - you don't have to level shift the signal
(which can be a pain).
The downside is... well there's several, but the one that always bugged me
is now the negative output terminal, your load connection point, is now some
value above ground - a variable value dependent on current at that. Lots of
loads don't mind, some do. It's a recipe to create weird ground loop
problems when you hook up a scope probe or serial port to a PC.
If you want to go this way, calculating the resistor is easy. You have an
analog input that can measure 0 to 5 volts, so you want a current signal
that is 5 volts at full load. 5V/5A = 1 ohm. Granted that's a _functional_
solution, it's not one I would recommend. First you need a huge resistor for
that (remember power is I^2*R) and second you loose as much as 5 volts from
your power supply's capability. (it has some other unpleasant effects as
well, but we'll skip them at the moment). The better solution is to use a
smaller resistor and an amplifier. You could reduce the resistor to 0.1 ohm
and add a 10x amp to achieve the same goal, but now your resistor is
only dissipating 2.5 W instead of 25, much better. This can be extended to
even better systems, but the complications start to rise significantly.
High side monitoring is basically the same process, but now you need level
shifting. There's probably a dozen ways to achieve this, but I've seen some
nice parts from Maxim that can handle this quite easily. Poke around on
their website, your looking for "high-side current monitors".
But in your case, I would see about some parts from Allegro Micro. They have
some really nice current sensors that use Hall-Effect sensors to measure the
magnetic field in a conductor. The real upside is effectively zero
resistance for the sensor, several kV of galvanic isolation, and the output
is internally amplified to a useful range for the A/D converter. Take a look
at the ACS714 for starters.
> Take a look at the ACS714 for starters
Thanks Denny. I was not aware of that part
I've used their UGN type sensors (isolated clamp meters for
example for large currents) and seen all the Maxim battery
charger monitors you mention
Digikey sell the ACS714 at a tolerable 1-off price. Could add
one to my next order
There's also this no-fuss product
On Sun, May 29, 2011 at 6:41 PM, Denny Esterline <gmail.com> wrote: desterline
> A shunt resistor in the gnd line is easy because it's the same gnd as the
> rest of the sensing circuit - you don't have to level shift the signal
> (which can be a pain).
> The downside is... well there's several, but the one that always bugged me
> is now the negative output terminal, your load connection point, is now some
> value above ground - a variable value dependent on current at that. Lots of
> loads don't mind, some do. It's a recipe to create weird ground loop
> problems when you hook up a scope probe or serial port to a PC.
I don't see why putting a shunt resistor in the negative line is a
problem at all. Most power supplies (at least of the lab or bench
variety) have ground-isolated outputs. I would just take care to make
sure that no points in my circuit on the isolated side of the input
transformer were connected to earth or chassis ground. Then, if I am
careful to run the reference for my regulator right to the negative
output terminal, the regulation should be fine (despite the shunt) and
also the user is free to connect (or not) the negative terminal (or
the positive terminal) to earth/chassis ground without affecting
On 5/29/2011 4:54 PM, IVP wrote:
>> Take a look at the ACS714 for starters
> Thanks Denny. I was not aware of that part
> I've used their UGN type sensors (isolated clamp meters for
> example for large currents) and seen all the Maxim battery
> charger monitors you mention
> Digikey sell the ACS714 at a tolerable 1-off price. Could add
> one to my next order
> There's also this no-fuss product
Two devices for high side I have used successfully are:
Zetex ZT1010 (cheap)
- and -
Linear Tech LTC6101
I like the LTC6101 as it has a separate gain resistor to set the gain if you care about exact results.
I have also used the ACS7xx series but they don't have very good specs compared to the above, I think.
Bob Blick has a circuit on an SX forum, I happened across it the other day.
7 resistors and one op-amp (rail to rail) high side sensing.
V+in - sense_R ---------- V+out
R_div_1a R_div_2a R_div_1b R_div_2b
Op-amp V+ connects to R_div_1 centre tap
Op-amp V- connects to R_div_2 centre tap
Op-amp feedback resistor from op_v_out to op_V-
Resistor from op_V+ to 0v
R1b,R2b 100K (input voltage dived by 11)
Feedback R and R to 0v 10M
Sense resistor 5v/current to be sensed.
I notice that TI and Linear use either FET's or BJT's in the feedback loop, byt they have converted from V to I and back to V.
cdb, btech-online.co.uk on 30/05/2011 colin
Web presence: http://www.btech-online.co.uk Hosted by: http://www.justhost.com.au
On 30/05/2011 08:08, cdb wrote:
People used to use LM3900 to do high side current sense and PSU input volts as the inputs are essentially current driven sinks there is apparently no + rail limit. It can exceed amp rail.
4 x "Norton" input opamps. AKA LM2900
> Bob Blick has a circuit on an SX forum, I happened across it the other day.
Presumably you mean this one
Info only: Comment on this by Roman Black indicates it is
'quite some years old'.
The arrangement can be useful, but circuits like this can give you
unexpectedly bad results if implemented without understanding their
It is important to note that circuits of this sort require
increasingly good "common mode rejection" by the opamp as the ratio of
sense voltage to high side voltage increases and that errors in the
divider components also become increasingly critical as the divide
For circuits with small sense voltages, either due to low sensed
currents or high division ratios the op amp choice may become
For accuracy achievable with given components.
Memory suggests (and it's not utterly intuitive) that a very rough
rule of thumb is that you can expect a error about N times worse than
the tolerance of your divider components.
"Select on test" can help this greatly. It's been a wee while since I
worked through this myself (so left as an exercise for the student) -
there may be eg a factor of 2 either way in that. Also, components
that cluster tightly around their nominal values (as often happens)
may yield better accuracies than expected.
If I'm right (ask the student), in Bob's circuit the divide ratio is
11:1 so 1% untrimmed resistors in the divider *may* typically produce
a 10% error.
Student please run figures and report back :-)
On 30/05/2011 08:47, RussellMc wrote:
> If I'm right (ask the student), in Bob's circuit the divide ratio is
> 11:1 so 1% untrimmed resistors in the divider*may* typically produce
> a 10% error.
Or you measure also with accurate bench meter and put a calibration factor in SW.
Then limitation original test gear, temperature and ageing drift
> Or you measure also with accurate bench meter and put a calibration
> factor in SW.
I think that comes under
>> ... understanding their limitations. :-)
> Then limitation original test gear, temperature and ageing drift.
Still some second order effects from eg temperature drift but much
less than the likely effect from initial component tolerance errors.
|This is the circuit I posted six or seven years ago for high side
current measurement. There is nothing novel or exciting about it:
/ 1M 1M /
| | 10M
| | |\ |
| *----|-\ |
| | | >-------*---- 5v=5A
| | | |/
\ \ \
/ 100K 100K/ / 10M
\ \ \
| | |
The main thing to remember in this and any other instrumentation
circuit, NO CARBON RESISTORS! Carbon resistors are crap. If you use 1%
metal film resistors that you've hand selected, the circuit is pretty
accurate. If you are mass-producing it, use 0.1% resistors so you don't
have to hand-select them. The shunt resistor (the low ohm one at the
top) doesn't need to be as fantastic as the others because its error is
not amplified disproportionately. The opamp needs to have output that is
valid to ground, and even so, if you are running single-supply, the
output can never actually be zero because of the 10M feedback resistor
working against the output impedance of the opamp at 0 volts. It amounts
to fewer than 5 millivolts on a crappy LM324 type opamp. Though I think
when I posted the circuit I was leaning toward an LMC6484 or something
On Mon, 30 May 2011 17:08 +1000, "cdb" wrote:
-- http://www.fastmail.fm - Send your email first class
> This is the circuit I posted six or seven years ago for high side
> current measurement
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