Author Topic: LT3083 current boost  (Read 640 times)

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Offline dweeb99

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LT3083 current boost
« on: February 09, 2020, 09:56:22 pm »
Can someone explain the workings of this circuit from the LT3083 datasheet


This gives 4.5A total  - LT1963 (Max 1.5A) boosted by LT3083 (max 3A)

I don't understand the 33K to ground on the SET pin - can anyone explain this?

The use of LT3083 as a current source shows SET pin connected to Vout, not ground
« Last Edit: February 10, 2020, 06:07:27 pm by dweeb99 »
 

Offline David Hess

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Re: LT3083 current boost
« Reply #1 on: February 09, 2020, 11:14:10 pm »
The divider sets the LT3083 output low by about 4 millivolts (3.3V - 4mV) so with no load or a light load, the LT3083's output current falls to zero.  The LT3083 does not start supplying output current until the voltage across the 20 milliohm resistor reaches 4 millivolts or about 200 milliamps.
 
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Offline dweeb99

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Re: LT3083 current boost
« Reply #2 on: February 10, 2020, 05:54:32 pm »
The divider sets the LT3083 output low by about 4 millivolts (3.3V - 4mV) so with no load or a light load, the LT3083's output current falls to zero.  The LT3083 does not start supplying output current until the voltage across the 20 milliohm resistor reaches 4 millivolts or about 200 milliamps.
Thanks
Ok, here are my problems in understanding - I thought the LT3083 was operating in Iout mode (SET pin tied to OUT pin through Rs)?
If it operates in Vout mode the SET pin is tied to GND through an R but for 3.3V output this R should be 66K, not 33K?
So is the LT3083 operating in Vout mode or Iout mode?
 
Just so I understand it as I want to use it for other voltages - Is the voltage divider comprised of R1 20mOhm & R2 to ground of 42ohm - which would give Vout drop of 2mV?

This would be correct if If 42ohm was to ground but there is also 3300ohm to ground - so what part does this R play & does it not change the voltage divider. For 3.3V output from LT3083

If I understood the function of 3300ohm to ground I might be able to simplify this?

Maybe an example - If I was using this with an adjustable LT1963 which has a voltage divider to give approx 8V output (voltage divider R2 of 1K & R1 of 178 to gnd) how would I calc the various Rs needed?

PS: I looked at the LT3081 datasheet & see the same current booster diagram but component values are changed
- 42ohm is now 8.2ohm
- 33K ohm is now 6.2Kohm
- 10mohm on OUT is now 20mohm

Sorry for all the questions - I know I'm making hard work of this - maybe I'm being dumb but I need to understand the role of the components in order to scale up for different voltage outputs?
« Last Edit: February 10, 2020, 07:02:07 pm by dweeb99 »
 

Offline dweeb99

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Re: LT3083 current boost
« Reply #3 on: February 10, 2020, 07:08:46 pm »
According to LT3083 datasheet "The LT3083 sources a 50μA reference current that flows out of the SET pin"
Which would make sense as the datasheet specifies Rset  of 66K for 3.3V, not 33K?
This is what threw me first - I just thought the 33K was a typo & it should be 66K but the LT308X are operating in Iout mode, not Vout mode so I'm missing something

In Vout mode the same Rset of 66K specified on the LT3081 for output of 3.3V
And yet the same current boost diagram on LT3081 has 6.2K (instead of 33K) to ground so I doubt this is a operating to set the Vout of the LT3081?
   
« Last Edit: February 10, 2020, 07:29:40 pm by dweeb99 »
 

Offline dweeb99

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Re: LT3083 current boost
« Reply #4 on: February 10, 2020, 08:01:35 pm »
This might throw some light on it? Found this on the forum here https://www.eevblog.com/forum/beginners/lt3083-current-limiting-how-does-it-work/msg623800/#msg623800

Quote
"LT3080/3083 internally assures that voltage supplied to the pin SET is identical to the voltage on the pin OUT (it's called error amplifier, if OUT pin goes higher, that causes internal circuits to "increase resistance" of the regulating transistor, and in turn the OUT pin's voltage goes down. In the case of the Voltage Regulation circuit, potential (voltage) of the SET pin is defined (for LT3083) as 50uA (provided by the IC) times resistor you call Vset. Note that one side of that resistor is connected to the GND!

In the case of the current limiter, that "same" resistor (now you call it Iset) is connected "after" the current sense resistor (Rlimit).  At this point in diagram, potential is OUT-Rlimit*Ilimit. To have OUT and SET at the same potential, voltage on Iset has to be 50uA*Iset, or simply said Rlimit*Ilimit = 50uA*Iset => I=50uA*Iset/Rlimit (note: Iset is actually resistor, but that's how it's marked in your diagram  :) ).

Unless it reaches Ilimit, voltage drop on Rlimit is lower then votage drop on Iset resistor, so error amplifier sees that OUT pin is not "high" enough, and internal circuitry opens the transistor that regulates the output to the max, effectively not impacting the next stage. If the current gets to Ilimit value and attempts to go higher, error amplifier will see that OUT has higher potential then SET, and will take actions to "reduce" the Vout as much as needed to get current under control. That will in turn lower the supply voltage of the next block (voltage Regulator) and in turn lower the output to your load"

In the circuit being talked about Iset (Resistor on SET pin) is 20K & Rlimit (R on OUT pin) is 0.5ohm so Iout is 50uA*20000/0.5 = 2A
But I still can't figure out the role the Rs perform around the SET pin in the datasheet current booster?
« Last Edit: February 10, 2020, 08:51:17 pm by dweeb99 »
 

Offline dweeb99

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Re: LT3083 current boost
« Reply #5 on: February 10, 2020, 10:54:20 pm »
Ok, getting closer now, I believe

3.3V through a voltage divider of 8.2ohm & 6.2K ohms gives a voltage of 3.296V i.e drop of 4mV on LT3081
3.3V through voltage divider of 42ohm & 33K ohm gives a voltage of 3.296V i.e drop of 4mV on LT3083

So this 4mV is on the SET pin in both cases


Not sure why the same voltage divider isn't used on both configs to give the same 4mV drop?
Is it because of the different max current outputs of LT3081 (1A) Vs LT3083 (3A)? But how, exactly?
What are the compliances that needs to be adhered to in choosing these Rs?
« Last Edit: February 10, 2020, 11:02:42 pm by dweeb99 »
 

Offline dweeb99

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Re: LT3083 current boost
« Reply #6 on: February 12, 2020, 06:34:27 pm »
Still working my way through this but NOT making progress!

I'm surprised nobody already has a handle on this or used it in this way (or wants to offer advice/help)?

Here's the datasheet text on using LT3081 as a current source (not in other LT308X datasheets, AFAIK)

Quote
The 50µA reference current from
the SET pin is used with one resistor to generate a small
voltage, usually in the range of 100mV to 1V (200mV is a
level that rejects offset voltage, line regulation, and other
errors without being excessively large). This voltage is
then applied across a second resistor that connect from
OUT to the first resistor

So, apart from the 4mV reduction in the above voltage dividers, there's also the text above to work through.
To achieve 200mV across Rset, it needs to be 4.0Kohm,  1V use 20Kohm - so Rset between 4K & 20K are the recommended min/max values

So for the 8.2ohm/6.2Kohm voltage divider, I'm not sure how to treat this? Does the 50uA still apply on Rset or is it a higher current based on the 3.3V through the 6.2Kohm to ground? I reckon this would give 530uA added to 50uA for 580uA.

Now running this through 8.2 would NOT give anything between 200mV & 1V

The 3.3V isn't going through a voltage divider of any note to reduce the 3.3V to between 200mV & 1V

What is Vset in the scenario of the current booster schematic?

Can anyone steer me in the right direction? 
« Last Edit: February 13, 2020, 12:37:57 am by dweeb99 »
 

Offline dweeb99

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Re: LT3083 current boost
« Reply #7 on: February 16, 2020, 10:59:10 pm »
Well, it's working when powering a circuit that draws current but when charging battery or supercapacitors it does not regulate even when I have resistor from Out to GND drawing 5mA current.

I'm pretty much ready to give up on these parts.
 

Offline dweeb99

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Re: LT3083 current boost
« Reply #8 on: February 25, 2020, 11:43:34 pm »
Can anyone answer why LT3081 & LT3083 use two different voltage dividers when operating as current boosters on 3.3V regulator - the voltage divider in both cases reduces the 3.3V by 4mV to the SET pin?

I want to use a similar circuit on 5V, 7V 9V voltage regulators with LT3083 as current booster
 

Online macboy

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Re: LT3083 current boost
« Reply #9 on: February 26, 2020, 03:11:57 pm »
Looking at the simplified schematic (as depicted inside the IC symbol), one should be able to see that the LT308x device maintains the voltage at the OUT pin to be equal to the voltage at the SET pin. In the simplest form of normal operation, a resistor from SET to ground is all that is needed to provide this reference voltage, due to the current source which sources 50 micro Amp of current out of the SET pin. Another way to program an output voltage is to drive the SET pin to a specific desired voltage (ensuring that the circuit driving it can sink that 50 uA).

In the booster designs, the voltage at the set pin is not determined by that simple V=IR resistor arrangement. Instead, it is explicitly set to track the output voltage of the other regulator (LT1963-3.3). The output of that regulator has a voltage divider (8.2 & 6200 ohms or 42 & 33000 ohms) which provides the reference voltage to the SET pin of the LT308x. The actual value of the lower leg resistor does not directly determine output voltage, since the equivalent impedance at this node is dominated by the smaller value (upper leg) resistor going to the regulator output, rather than the lower leg resistor to ground. The 50 microAmp current flows through this resistor into the load. The LT308x maintains its output to be  at that voltage which is slightly lower (due to the voltage divider) than the output of the LT1963. It will only source current when the voltage drop across the 20 mOhm resistor in series with the the LT1963 output exceeds 4 mV, and at that point, the two regulators will share current based on the ratio of their output resistors as well as their respective output voltages.

Analyzing the resistor network (in post #1) deeper, assume that the output of the LT1963 is fixed at exactly 3.3 V.  The current through the 33 k resistor is the sum of the current through the 42 Ohm resistor plus 50 uA (from the LT3081 booster SET pin). Assume that the voltage at this node is approximately 3.3 V (due to the relatively low impedance to the 3.3 V node, the output voltage). Then the current flowing through the 33000 ohm resistor is ~ 100 uA (remember, 50 uA of that comes from the SET pin). Then, about 50 uA needs to flow through the 42 Ohm resistor. This makes the voltage at that node approximately (42 * 50e-6) 2.1 mV lower than the output node.  You can see that altering the value of the 33 k resistor will change the (mV level) offset of the booster compared to the LT1963, but will not directly set its output voltage in the "normal" way with V=(50uA*R).
 
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Offline dweeb99

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Re: LT3083 current boost
« Reply #10 on: February 26, 2020, 06:54:07 pm »
Looking at the simplified schematic (as depicted inside the IC symbol), one should be able to see that the LT308x device maintains the voltage at the OUT pin to be equal to the voltage at the SET pin. In the simplest form of normal operation, a resistor from SET to ground is all that is needed to provide this reference voltage, due to the current source which sources 50 micro Amp of current out of the SET pin. Another way to program an output voltage is to drive the SET pin to a specific desired voltage (ensuring that the circuit driving it can sink that 50 uA).

In the booster designs, the voltage at the set pin is not determined by that simple V=IR resistor arrangement. Instead, it is explicitly set to track the output voltage of the other regulator (LT1963-3.3). The output of that regulator has a voltage divider (8.2 & 6200 ohms or 42 & 33000 ohms) which provides the reference voltage to the SET pin of the LT308x. The actual value of the lower leg resistor does not directly determine output voltage, since the equivalent impedance at this node is dominated by the smaller value (upper leg) resistor going to the regulator output, rather than the lower leg resistor to ground. The 50 microAmp current flows through this resistor into the load. The LT308x maintains its output to be  at that voltage which is slightly lower (due to the voltage divider) than the output of the LT1963. It will only source current when the voltage drop across the 20 mOhm resistor in series with the the LT1963 output exceeds 4 mV, and at that point, the two regulators will share current based on the ratio of their output resistors as well as their respective output voltages.

Analyzing the resistor network (in post #1) deeper, assume that the output of the LT1963 is fixed at exactly 3.3 V.  The current through the 33 k resistor is the sum of the current through the 42 Ohm resistor plus 50 uA (from the LT3081 booster SET pin). Assume that the voltage at this node is approximately 3.3 V (due to the relatively low impedance to the 3.3 V node, the output voltage). Then the current flowing through the 33000 ohm resistor is ~ 100 uA (remember, 50 uA of that comes from the SET pin). Then, about 50 uA needs to flow through the 42 Ohm resistor. This makes the voltage at that node approximately (42 * 50e-6) 2.1 mV lower than the output node.  You can see that altering the value of the 33 k resistor will change the (mV level) offset of the booster compared to the LT1963, but will not directly set its output voltage in the "normal" way with V=(50uA*R).
Many thanks for this reply, macboy

I have delved into it some more & there are two factors that I believe are in operation:
- the parts must have a minimum load to stay in regulation - 1mA load for LT3083 & 5mA for LT3081 which begins to explain the differences seen on the schematic between the voltage dividers used on these parts in their role as current booster to the same 3.3V LT1963
- the other factor is that it tries to keep the voltage on the SET pin lower by 4mV than the OUT pin - this seems to be to turn off the current output from LT308X when this difference falls below 4mV differential? Is this correct? I'm not sure how this works, though i.e the how is there a different voltage differential between SET & OUT when load or no-load?

So for LT3083 acting as current booster to 3.3V would require R to gnd of 3.3K (for 1mA load) & hence 4.2ohm as the second R in the voltage divider (to give this 4mV drop) - the datasheet shows X 10 of these values which reduces this load by 10 to 0.1mA so what gives???? There's also a differential between the R connecting the SET pin to load (20mohm) & OUT pin to load R (10mohm) but I doubt this comes into play?

Same applies to LT3081 - as current booster for 3.3V, R to gnd of 660ohms (for 5mA load)  & hence 800ohm for other resistor in voltage divider - using X10 would give 6.6K (6.2K closest shown in the datasheet which only gives a load of 0.5mA - so again what gives?). here, there's NO differential between the R connecting the SET pin to load (20mohm) & OUT pin to load R (20mohm)

Is my analysis correct? If so what's up with the datasheet values??

I guess whether the SET pin is 2.1mV or 4mV below OUT pin is not of much consequence but the min load requirement of 1mA seems to be important & the datasheet R value of 33K only gives 0.1mA load - hence no regulation would result?     
« Last Edit: February 26, 2020, 07:16:11 pm by dweeb99 »
 

Offline dweeb99

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Re: LT3083 current boost
« Reply #11 on: February 27, 2020, 05:09:14 pm »
Just so I can be sure I understand :

- if I want to use LT3083 as current booster for regulated 5V, then a voltage divider of 50 & 33000 ohm will provide 5.1mV of bias between SET & OUT pins?

- I can ignore minimum load requirements?
 

Offline dweeb99

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Re: LT3083 current boost
« Reply #12 on: February 28, 2020, 09:16:58 pm »
I believe there might be another issue in using the LT308X as a current booster - the feed voltage regulator needs to be able to withstand reverse current as I seem to have damaged a number of TPS7A4700 regulators trying this.
 


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