NejTech Evolution - a new BOOST driver - Nearly finished!

[Wolfman29]

Re: NejTech Evolution - a new BOOST driver - preparing to build prototypes

So I did a lot of studying last night... I think I got the basic understanding of switching supplies that are in a constant VOLTAGE topology.

However, I don't understand how to go from CV to CC! I have looked everywhere and I am not able to find anything describing how to get CC out of switching supplies. Is it as simple as placing a set resistor in series with the driver which is in turn in series with the load? If so, couldn't I just take a CV driver and stick a resistor in series with the output to get a CC driver?

Anyway, thanks for all the help.

 

[random person]

Re: NejTech Evolution - a new BOOST driver - preparing to build prototypes

The feedback pin always wants to see 1.2V (or Vref) regardless of whether its a switching supply or a linear supply. Its all about how you go about giving the feedback pin its voltage. Do you give it a voltage proportional to the output voltage or a voltage proportional to the output current?

 

[Wolfman29]

Re: NejTech Evolution - a new BOOST driver - preparing to build prototypes

So how does the first one give it voltage proportional to the output voltage? And if that's true, why is it that some feedback pins on those switching power supplies are like 50mV?

And in your diagram, where is the output "current" or whatever? It's not labeled, and it's rather confusing -.- Is the single wire coming out the feedback pin? Or is it the output?

 

[random person]

Re: NejTech Evolution - a new BOOST driver - preparing to build prototypes

You know what a voltage divider is right? http://en.wikipedia.org/wiki/Voltage_divider

In the first diagram, Vout is ratio based resistance, not current based. Those resistors could be 100 and 200, or 1M ohm and 2Mohm... the voltage will always be some ratio of the input voltage to the bridge regardless of the current.

Anyhow, heres another attempt at explaining it. Remember that the FB for switching and for linear works the same way.

So heres a linear supply with a standard resistor bridge feedback pin.

Here is a linear regulator. First off, we need some fixed variable called a reference voltage. Now, the voltage cross across a diode is always fixed within reason (as opposed to LDs) so they are used as a reference voltage. (most commonly zeners are used) So the voltage drop across those diodes in the circuit is 1.2V. Now, via op amp theory the + terminal and the - terminal of the op amp must produce the same voltage. Therefore the opamp will keep driving more and more current into the base of the transformer until it sees that they are the same. So in this case were driving that current into a bjt that turns it more and more on. If that bjt were fully on the output would be 20V. To prevent the output from flying away, once the - pin exceeds 1.2 volts it will start scaling back until its in equilibrium... Thus regulated.

Again watch this: ‪EEVblog #90 - Linear and LDO regulators and Switch Mode Power Supply Tutorial‬‏ - YouTube





Now if we take that FB pin and put it somewhere in the circuit where the voltage of node is proportional to current again, it makes our supply proportional to current as opposed to proportional to voltage. So in this case the op amp will keep turning on until it sees 1.2V on the - pin. It will see 1.2V on the - pin once there is 1.2V/ 2 ohms = 600mA. So, regardless of the load (provided its a small load), it will always drive 600mA.

 

[Wolfman29]

Re: NejTech Evolution - a new BOOST driver - preparing to build prototypes

So current flows into the FB pin.

So I drew up this schematic, and it doesn't seem to want to work.

With a 1 Ohm resistor stuck right there, it should be giving me 1.2A, right? It doesn't.

And it seems that the FB pin is seeing like 3.6V. What did I do wrong here? I feel helpless -.-

EDIT: I watched the video - I will watch it again tonight to see if I can make more sense of it. But again, it was talking about voltage regulators. Essentially, a current regulator is a voltage regulator minus a resistor (between Iout and FB?).

And in that circuit, it was giving me nearly 900mA... shouldn't it have been giving me 1.2A?

 

[random person]

Re: NejTech Evolution - a new BOOST driver - preparing to build prototypes

Nope, very little current generally flows into a FB pin...Its a high impedance input to an internal op amp. That would be the equivalent of a very very tiny capacitor. It is merely a voltage sense. maybe 1 nA flows through the FB pin. Another way to think of it is like multi meter probes. Basically the FB pin is just probes a certain point on your circuit and will keep pumping juice until it sees 1.2V.

For your example... no, that would not give you 1.2A over the diode. All your current needs to go through Rsense resistor!!! IN your case 1.2A would go over the current sense resistor but an unkown amount would go through the diode.

Now for your circuit.... first off SHDN should be pulled high. As for calculating a current output, I looked in the datasheet and the reference voltage of the LT1308A was 1.22V. So 1.22V / 3 ohms = 0.406666mA driver. Used diodes because they don't have a good LED sample in LT spice. output =between 405mA and 409.2 mA... Thats pretty darn good regulation Note green graph is the current through those diodes.

Heres how it should be wired: (look LPF, heres how you make a flex drive or something close =P )

 

[qumefox]

Re: NejTech Evolution - a new BOOST driver - preparing to build prototypes

No offense to either involved party, but if Wolfman doesn't understand how/why a voltage divider works, much less a simple op-amp regulator, shouldn't you rewind back to those first RP? Otherwise I think you'll probably find your having to explain the same basics over and over again in every post.

The java circuit simulator might not be the best thing around, but it's really good for basic learning.

 

[Wolfman29]

Re: NejTech Evolution - a new BOOST driver - preparing to build prototypes

@random person: OH. I think I got you. So basically, the idea in the circuit that you gave as an example is essentially that an "unknown" amount of current passes through those diodes, but then, because the resistor is there, it forces the 1.2V that the feedback sees to also pass through that 3 Ohm resistor, meaning that it makes it so it acts a current sink, shoving around 406mA through the diodes?

As for the internals: the feedback doesn't draw much of that 406mA off because it's essentially just a voltage sensor, so it just demands that 1.2V is at that? So we're actually passing 1.2V *over* the diodes Vf's through those diodes, but the current is regulated anyway, so those diodes aren't effected by the 1.2V, right?

I think I am starting to get it (+1 for RP, by the way, for all the help).

@qumefox: Yeah, that *may* have been helpful. Hopefully, I will get a chance to take an electronics class once classes start back up so I can get some of the theory behind this more rigorously. But when it comes to a voltage divider, I think I got that (not intuitively yet, but I think I understand it intellectually). And actually, the op-amp regulators are making more and more sense the more I think about them. I think I am making progress!

 

[random person]

Re: NejTech Evolution - a new BOOST driver - preparing to build prototypes

@random person: OH. I think I got you. So basically, the idea in the circuit that you gave as an example is essentially that an "unknown" amount of current passes through those diodes, but then, because the resistor is there, it forces the 1.2V that the feedback sees to also pass through that 3 Ohm resistor, meaning that it makes it so it acts a current sink, shoving around 406mA through the diodes?

Your getting there. However, your forgetting that all loads in series share the same current. So if we know the current through one element in the series loop (the Rfeedback resistor) we know the current through all the other components in the loop(aka the diodes). Now its a series loop because the FB pin doesn't count since its a capacitive load that only draws a few micro amps of current at most. Picture to illustrate

So remember, the voltage driver is going to keep pumping out more and more voltage until it sees 1.2V at the FB pin.

So if I already know the voltage drop across the resistor attached to the feedback pin is going to be 1.22V, I can pick any resistor accordingly such that all other elements in series with that experience the same amount of current as the current that goes through that resistor. Say I want to build a 2A driver. I would then pick a ..R=1.22V / 2amps = .61 ohm resistor for a feedback resistor. (if such a value existed).


Now acourse... theres practical limitations. I can't pick a .001 ohm resistor FB resistor and expect the current driver to pump out 1220amps. Nor can I exspect it to pump out a super high voltage then the IC is rated for...ect. Just trying to explain the basics right now

 

[Wolfman29]

Re: NejTech Evolution - a new BOOST driver - preparing to build prototypes

I think I am getting it!

I was doing some testing randomly with some ICs (trying out the new knowledge I have), and sometimes, the Vfb would drop lower when I try to use higher currents. Like, I would stick in a .75Ohm resistor to attempt to get 1.62A or so, but then when I ran it, I measured the Vfb to be around .9V or so, therefore only giving me 1.2A.

Is this due to the IC not being able to pump out enough voltage to cover the Vd of the resistor, the 1.22V of the Vfb, AND the Vf of the load? I thought that most ICs could handle fairly large amounts of voltage (I chose something with a max output voltage of 40V) just by drawing larger currents?

 

[Johnyz]

Re: NejTech Evolution - a new BOOST driver - preparing to build prototypes

$6.85 for 6 boards incl shipping? Not bad at all!! Boards will be fabbed at 25th July.

 

[random person]

Re: NejTech Evolution - a new BOOST driver - preparing to build prototypes

$6.85 for 6 boards incl shipping? Not bad at all!! Boards will be fabbed at 25th July.

Welldone! Was that with makepcb.com? Best of luck to you with your driver!.

I think I am getting it!

I was doing some testing randomly with some ICs (trying out the new knowledge I have), and sometimes, the Vfb would drop lower when I try to use higher currents. Like, I would stick in a .75Ohm resistor to attempt to get 1.62A or so, but then when I ran it, I measured the Vfb to be around .9V or so, therefore only giving me 1.2A.

Is this due to the IC not being able to pump out enough voltage to cover the Vd of the resistor, the 1.22V of the Vfb, AND the Vf of the load? I thought that most ICs could handle fairly large amounts of voltage (I chose something with a max output voltage of 40V) just by drawing larger currents?

That is a very specific examples where you would need to provide datasheets for all parts involved as well as a diagram. Can't say for certain.

As far as "IC"s go(99.9%), they can't produce a real[/b[] voltage gain. Like an op amp ...ect. If you put 5V in, you can't get more then 5V out. Unless your using a boost convertor like ours you can't really get true voltage gain. In addition, read the datasheets. If I put 3.7V into a boost convertor I can't expect to get 12V out at 1A. The higher above the source you go, the less potential you have for current. In addition, remember, voltage is used up each component in a series string including the current sense. So say I want to drive a 445nm at 1A. My supply will need 5V for the 445nm then an extra voltage to go over the current sense resistor. So if that exceeds the output ability of the IC, it won't work.

 

[Kmor2004]

Re: NejTech Evolution - a new BOOST driver - preparing to build prototypes

Wolf, not sure what IC's you were testing with, but if you were trying to use the IC shown in Randoms post, "it" doesn't pump out a voltage, it creates pulses for the inductor to generate negative spikes.

If you were using some other ic, then it's not a question of the amount of voltage it "pumps out" but ratehr the amount of current the output pin can pull in or push out most IC's are <1A sink or source unless its a specialty one.

 

[qumefox]

Re: NejTech Evolution - a new BOOST driver - preparing to build prototypes

Those look like dorkbot pbx prices.. Plus july 25th is when the next batch through them goes in according to their website. Look forward to seeing those pretty purple boards.

 

[Wolfman29]

Re: NejTech Evolution - a new BOOST driver - preparing to build prototypes

So the IC I was using was the LT1946, supposedly a boost IC capable of boosting to 2.1A, typical. And it should be able to boost up 3.3V-4.2V to around 5V, no? That doesn't seem out of the ordinary.

 

[Johnyz]

Re: NejTech Evolution - a new BOOST driver - preparing to build prototypes

Aaargh! Apparently, when the forum was down, we lost one page!