simple switcher constant current supply with high side sense

[Tim Mitchell]

I'm trying to design a simple switcher constant current circuit (using
the LM2574-ADJ). It's easy enough if you put the sense resistor between
the load and ground. However the snag is that the current sense resistor
has to be on the high side of the load (there are 3 current supplies
driving 3 devices with a commoned ground).

Does anyone know of a clever way of doing this without getting into
differential amplifiers?
--
Tim Mitchell

 

[budgie]

On Thu, 3 Feb 2005 09:07:02 +0000, Tim Mitchell <timng@sabretechnology.co.uk>
wrote:

I'm trying to design a simple switcher constant current circuit (using
the LM2574-ADJ). It's easy enough if you put the sense resistor between
the load and ground. However the snag is that the current sense resistor
has to be on the high side of the load (there are 3 current supplies
driving 3 devices with a commoned ground).

Does anyone know of a clever way of doing this without getting into
differential amplifiers?

Are you also using the switcher to buck or boost the input voltage
substantially? If not, I wonder why you are approaching a constant current task
that way.

 

[Terry Given]

Tim Mitchell wrote:

I'm trying to design a simple switcher constant current circuit (using
the LM2574-ADJ). It's easy enough if you put the sense resistor between
the load and ground. However the snag is that the current sense resistor
has to be on the high side of the load (there are 3 current supplies
driving 3 devices with a commoned ground).

Does anyone know of a clever way of doing this without getting into
differential amplifiers?

www.zetex.com

they make clever high-side current sense thingies that level-shift the
signal down to the low side. digikey even sell them.

And if they do, others will too. Bear in mind that a differential
amplifier is really a resistor mismatch detector.

Cheers
Terry

 

[Tim Mitchell]

In article <au1401tuerp1nl40jpb9nnenqq13iqloua@4ax.com>, budgie
<me@privacy.net> writes

On Thu, 3 Feb 2005 09:07:02 +0000, Tim Mitchell <timng@sabretechnology.co.uk
wrote:

I'm trying to design a simple switcher constant current circuit (using
the LM2574-ADJ). It's easy enough if you put the sense resistor between
the load and ground. However the snag is that the current sense resistor
has to be on the high side of the load (there are 3 current supplies
driving 3 devices with a commoned ground).

Does anyone know of a clever way of doing this without getting into
differential amplifiers?

Are you also using the switcher to buck or boost the input voltage
substantially? If not, I wonder why you are approaching a constant
current task
that way.

No (sorry should have given more details), it's a supply for luxeon
LEDs, but there can be a variable number connected in series. There's a
48V supply on one end and from 1 to 12 LEDs in series on the output. So
a switcher is the only way to go otherwise too much power has to be
dissipated if low numbers of LEDs are connected.
--
Tim Mitchell

 

[Fred Bloggs]

Tim Mitchell wrote:

No (sorry should have given more details), it's a supply for luxeon
LEDs, but there can be a variable number connected in series. There's a
48V supply on one end and from 1 to 12 LEDs in series on the output. So
a switcher is the only way to go otherwise too much power has to be
dissipated if low numbers of LEDs are connected.

Mind telling why the LEDs *have* to be grounded?

 

[Ol' Duffer]

In article <uZRS$5B2meACFAm8@tega.co.uk>, timng@sabretechnology.co.uk
says...

I'm trying to design a simple switcher constant current circuit (using
the LM2574-ADJ). It's easy enough if you put the sense resistor between
the load and ground. However the snag is that the current sense resistor
has to be on the high side of the load (there are 3 current supplies
driving 3 devices with a commoned ground).

Does anyone know of a clever way of doing this without getting into
differential amplifiers?

Doesn't sound like you need much accuracy, so how about
a pnp transistor and about 3 resistors to form a crude
current mirror?

 

[Tim Mitchell]

In article <420226DA.7030902@nospam.com>, Fred Bloggs
<nospam@nospam.com> writes

Tim Mitchell wrote:

No (sorry should have given more details), it's a supply for luxeon
LEDs, but there can be a variable number connected in series. There's
a 48V supply on one end and from 1 to 12 LEDs in series on the
output. So a switcher is the only way to go otherwise too much power
has to be dissipated if low numbers of LEDs are connected.

Mind telling why the LEDs *have* to be grounded?

They don't, but it's an rgb triad with the cathodes commoned, so I can't

measure current on the cathode side.

These are luxeon LEDs which have a drive current of 350mA, not yer
normal 20mA. The blues have a Vf of just below 4 volts, pretty much the
whole 48V input is needed to drive 12 of them.

Thanks to everyone for the ideas.
--
Tim Mitchell

 

[Tim Mitchell]

In article <4202EE40.1040007@nospam.com>, Fred Bloggs
<nospam@nospam.com> writes

Tim Mitchell wrote:

They don't, but it's an rgb triad with the cathodes commoned, so I
can't measure current on the cathode side.

And you also cannot stack these triads in series because you have to
connect each anode to the common cathode of the triad above it which
places all LEDs in the triad in parallel.

These are luxeon LEDs which have a drive current of 350mA, not yer
normal 20mA. The blues have a Vf of just below 4 volts, pretty much
the whole 48V input is needed to drive 12 of them.

You seem to be having trouble with basic circuits.

No, I'm just having trouble getting all the necessary information

across...

The www.romanblack.com website Mark Jones suggested has got some
amazingly simple switcher circuits which make me realise I don't need to
use a "simple switcher" at all. Impressive stuff.
--
Tim Mitchell

 

[Tony Williams]

In article <uZRS$5B2meACFAm8@tega.co.uk>,
Tim Mitchell <timng@sabretechnology.co.uk> wrote:

I'm trying to design a simple switcher constant current circuit
(using the LM2574-ADJ). It's easy enough if you put the sense
resistor between the load and ground. However the snag is that
the current sense resistor has to be on the high side of the
load (there are 3 current supplies driving 3 devices with a
commoned ground).

Sideways thought.....

If a flyback converter is arranged to dump a fixed
Joules/cycle, then making the Frequency proportional
to Vout should result in a constant Iout.

--
Tony Williams.