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Electronics => Projects, Designs, and Technical Stuff => Topic started by: void_error on February 23, 2015, 03:49:51 pm

Title: UC3843 questions
Post by: void_error on February 23, 2015, 03:49:51 pm
I'm designing a dimmable boost converter for a 10W LED spotlight to be used as a bench lamp. Already ripped out the current regulator it came with as it was way too bright for bench use - 250mA CC output, LED voltage around 35V and its only use would have been frying ants with a magnifying glass during the winter, assuming one could find any.

I chose the UC3843 (http://www.ti.com/lit/ds/symlink/uc3842.pdf) as I'm going to power it from a 12-15V source and I want to keep the parts count and total cost relatively low while using locally available components (what my local electronics shop has in stock) and what I have lying around (lots of stuff).

For LED voltage/current limiting I'll be using a LM358 with the outputs OR'd like the error amps of the TL494 (http://www.ti.com/lit/ds/symlink/tl494.pdf). The half that handles the current will have adjustable gain.

One nice feature of this IC is the MOSFET driver with current limiting. Unfortunately it requires a relatively high voltage across the sense resistor - 1V - which means that for a current limit of 1A it'll be wasting towards 1W depending on the duty cycle and I'd be better off with something lower, preferably.

Any ideas on how I can bias the current sense pin to have a lower sense voltage?
Title: Re: UC3843 questions
Post by: T3sl4co1l on February 24, 2015, 12:31:50 am
You could clamp the COMP pin to a lower voltage (say 2 or 1V; it's internally clamped to around 3V, at least equivalent as the current comparator sees it), which saves you a little range, but maybe doesn't save enough.  The methodology is bad; it's burning free dynamic range, making the rest of the circuit worse (less input voltage for the current comparator to trip on, internal noise is a larger percentage of full scale).  So maybe you just turn up the shunt itself.

You can use an op-amp to do this, but it has to be pretty damn fast (>20MHz).  A much more effective alternative is using a current transformer, especially if you need more than a few amperes.  This essentially eliminates shunt resistor losses, as well as the issue of spikes due to stray inductance.  If you move it to the drain terminal, you also eliminate the error of gate current pulses (though diode capacitance / recovery pulses remain).

As it's seeing pulsed DC, it needs a diode in series with the secondary to do its job.  It's wired like this;

(http://seventransistorlabs.com/Images/Mag_Amp_PSU.png)

There are a couple things that you'll want to change from this rather old drawing, but it shows the general idea.

As for current feedback, you'll still need a shunt to sense DC, which will spend power.  This can be amplified, for which a slow LM358 is just fine.  Rather than diode-ORing the outputs, you can run a zener from +Vout to the current sense / summing node.  No need for it to be precise or anything.  The 3843 can be wired as error amplifier, with your dimming signal being some VREF bled into the summing node through a pot and series resistor.

Tim
Title: Re: UC3843 questions
Post by: nickm on February 24, 2015, 02:45:11 am
This is nothing new.  You bias up the CS pin with a resistor to a stable voltage reference such as pin 7 or 8.  You are shifting up where zero current is so you will lose information but that's a tradeoff.  You should leave yourself adequate headroom to keep any fuzz on the rising edge from tripping the current sense prematurely. Using the symbols in the attached picture the threshold becomes (I think):

I = [1 + R(1 - Vbias) / Rb] / Rs

At zero current the voltage on the CS pin is:

Vcs = R*Vbias / (R + Rb)

http://imgur.com/4rAeJUw (http://imgur.com/4rAeJUw)
Title: Re: UC3843 questions
Post by: void_error on February 24, 2015, 09:15:58 am
I'm going with nickm's suggestion, the current limit will be quite low at about 1A so biasing the sense pin at around 0.7V with 0.33V across the sense resistor should limit the power dissipation enough for using 3 1R 0805 resistors in parallel. Calculations say the maximum duty cycle will be under 80% for 12V input and 33V output @ 200mA.

As for current feedback, you'll still need a shunt to sense DC, which will spend power.  This can be amplified, for which a slow LM358 is just fine.  Rather than diode-ORing the outputs, you can run a zener from +Vout to the current sense / summing node.  No need for it to be precise or anything.  The 3843 can be wired as error amplifier, with your dimming signal being some VREF bled into the summing node through a pot and series resistor.

Tim
The dimming signal will be coming from a 40192 or 40193 with an R-2R DAC to generate a control voltage. It'll be limited to 3 bits but it should be enough. I'm planning to run it off the UC3843's VREF pin as it can source enough current. The zener is a good idea but it leaves me with an unused half of the dual opamp.

Thanks for the replies

Time to breadboard the whole thing