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Can opamp provide low voltage output?

eem2am

Aug 3, 2009
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Hello,

I wish to use the LM358 opamp as an error amplifier.

The LM358's output will be providing a voltage into the "LD" pin of the HV9910B PWM controller, in order to control the peak current level of the HV9910B.

The problem is that the control voltage will need to be in the range of around 30mV to 250mV.

The supply of the opamp will be 10VDC (single supply).

Do you think this opamp will be able to produce a voltage from its output that's this low?

(pg 6 of the opamp datasheet appears to say that LM358 can give an output low voltage of 20mV, but i think that the opamp is not working properly whenever its output is below about 0.5V?.....is this true?

LM358 DATASHEET
http://www.ti.com/lit/ds/symlink/lm358.pdf

HV9910B DATASHEET
http://www.supertex.com/pdf/datasheets/HV9910B.pdf
 

Harald Kapp

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i think that the opamp is not working properly whenever its output is below about 0.5V
Why do you think so? If the datasheet states Vol=20mV, the device has to satisfy this requirement. Otherwise it wouldn't pass the tests at the factory.
 

eem2am

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Well.... "why" is a good question........ my fears are confirmed, as i have read point 3 at the very bottom of page 8 of the following....

http://ww1.microchip.com/downloads/e...tes/00682c.pdf

......it basically says that you can't rely on a single supply opamp to drive to within 200mV of either rail.

So now i really am worried....

...and i am more worried still, because i have just read the right hand side of page 9 of the LT1006 opamp datasheet........

http://cds.linear.com/docs/Datasheet/1006fa.pdf

....there it clearly states that the OP20 opamp cannot go to less than 600mV above ground....it also states that the LM158 & LM124 opamps cannot sink more than a few uA when swinging their output to ground.

...so the LM158 can drive near ground but only when sinking a few uA and no more......thats useless to me.

How can i find an opamp that can drive to ground?
 

Harald Kapp

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The link to microchip doesn't work.
Looking into datasheets of other OpAmps is of no use if you want to know the specs of a specific device.
so the LM158 can drive near ground but only when sinking a few uA and no more
that is only partly true. I admit that the datsheet is a bit difficult to read.
There is a field stating 12µA at 200mV over the full range.
There is also another field saying -10mA for Vo=0V.
I'm always having a bit of trouble with the bad habit of people (don't feel offended, I refer to the datasheet's authors ;) ) not to denote voltage and current directions with clear arrows. Therefore I am unable to resolve the mystery of these two contradictory data fields.
However, there is a way around the problem: The test setup (here: figure 1) shows a load resistor RL to GND. You can use this resistor to deliver current in the low state by making it smaller than 10kOhm (that's the test condition for Vol from the datasheet). You can go as low as 2kOhm (that's the test condition for Voh). That gives you another 100µA.

If that is not sufficient, please tell us:
- how much current is needed (why is it more than a few µA)?
- why has the output voltage to be under 250mV?
- is the signal analog or digital (a simple NPN driver could solve your problem in the latter case)
- give us an idea of the application (extract from the schematic plus some explanation)

We need more information to be able to identify possible solutions to your problem (which may, as I have stated above, look rather different than just using another type of OpAmp).

Regards,
Harald
 

CDRIVE

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I read the HV9910B data sheet and I see what you're referring to. You want to use the LD (Linear Dimming) input pin that's spec'd at 0 to 250mV. If you're going to stick with the LM358 you may want to consider doing something like this. The Diode in the output should provide the zero or near zero voltage you need.

Chris
 

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KrisBlueNZ

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I second CDRIVE's suggestion. If the op-amp's output won't go all the way down to 0V you can just connect a diode in series with it, with its anode to the output pin, and a resistor from its cathode to ground to ensure some current flows in it.

In this case, the LD input of the HV9910B just feeds a comparator, so it should have a relatively high impedance, so there shouldn't be a problem.

Chris's diagram also shows a two-resistor voltage divider to the LD pin, which limits the maximum voltage that the op-amp can feed into the LD pin, and also means that the op-amp's output is operating over a moderate voltage range, rather than a range only 250 mV wide, which is probably also a good idea.

I would suggest taking the feedback from the cathode of the diode, rather than the anode, for several reasons.

If you have a provisional circuit diagram for the error amplifier, you should be able to make those changes fairly easily. Otherwise post the diagram and we can make suggestions.
 

CDRIVE

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Kris, that's a very good point and an brain fart oversight on my part. I moved the feedback node to the cathode of the diode.

Here's the new plot results. No knee!!! ;)

Chris
 

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eem2am

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Hello,

Cdrive, Harald and KriBlueNZ……Many thanks, these are excellent words.

I kind of get it about the diode……though the diode will tend to have a varying Vf at the very small current that it passes….and I wonder if this varying Vf will upset the feedback loop?

….if not, then I will add the diode where you have done so, ….inside the feedback loop.

I cant yet show a schematic of the circuit with the HV9910B, but the following schematic is a like-for-like circuit, but with the constant off time control provided by a 555 IC………..
(I didn’t have a .model for the HV9910B)

Schematic:
http://i50.tinypic.com/nd3gqx.jpg


…Should you wish to run the above in the free download LTspice simulator, then here is the .txt file which you can convert to .asc file, and then run it……

http://www.2shared.com/document/Wncy-saV/555_monostable_boost.html


….The LTC1841 comparator used is very slow, with a Propagation delay of 4us to 12us…….this slowness makes it less-than-ideal in this application….and I hope that the HV9910B’s internal comparator is much faster than this.

The HV9910B datasheet, on page 3, appears to show that the comparator delay is just 80ns, …..I hope so.

LTC1841 DATASHEET:
http://cds.linear.com/docs/Datasheet/184123f.pdf

HV9910B datasheet:
http://www.supertex.com/pdf/datasheets/HV9910B.pdf



In the above circuit…..the comparator is so slow that the control voltage (on the non-inverting input of the comparator) is actually *below* the ramp on the inverting input of the comparator….this cannot be a good thing, and I am amazed that the simulation actually works as perfectly as it does.

…here is waveforms showing the (filtered) current sense ramp signal and the control voltage at the comparator input terminals……

http://i47.tinypic.com/dfftax.jpg

…..(please note the unfortunate situation of the control voltage being below any part of the actual ramp….this occurs with the lower output voltages……also note that this SMPS regulates its own input current, instead of its output voltage or current)

Do you believe that the internal comparator inside the “LD” pin of the HV9910B is a lot faster than the LTC1841 comparator? (I hope it is…)
 

CDRIVE

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Hello,

Cdrive, Harald and KriBlueNZ……Many thanks, these are excellent words.

I kind of get it about the diode……though the diode will tend to have a varying Vf at the very small current that it passes….and I wonder if this varying Vf will upset the feedback loop?

Did you see any in my plot? After moving the feedback node, as Kris suggested, It's a straight line.

Chris
 

eem2am

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OK thanks,

But can i actually use 2 diodes in series?, .....because i think that at these low currrents , the diode Vf for a small signal diode will be only about 0.2V.
 

eem2am

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Hi,

Moving the diode inside the feedback loop of the error amplifier seems a good idea......but why do they not do it with the UC3843 PWM control chip?

UC3843 DATASHEET:
http://www.ti.com/lit/ds/symlink/uc3843.pdf

..The block diagram on page 1 of this datasheet shows the diodes in the orignal place that CDrive had it.....i.e. outside the feedback loop.
 

KrisBlueNZ

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I kind of get it about the diode……though the diode will tend to have a varying Vf at the very small current that it passes….and I wonder if this varying Vf will upset the feedback loop?
No, because the feedback is taken after the diode, so the voltage after the diode is what is being controlled by the op-amp.
The HV9910B datasheet, on page 3, appears to show that the comparator delay is just 80ns, …..I hope so.
Yes, that specification seems pretty clear.

But can i actually use 2 diodes in series?, .....because i think that at these low currrents , the diode Vf for a small signal diode will be only about 0.2V.
It won't be THAT low! I think one diode is enough, but there's no problem using more than one, since the feedback is taken after the diode.
Do make sure that the feedback loop itself has a reasonably high impedance. The output post-diode won't have the low impedance that the op-amp output has - at least, not when the voltage is falling, because that node is only pulled to ground by the voltage divider resistors.

Moving the diode inside the feedback loop of the error amplifier seems a good idea......but why do they not do it with the UC3843 PWM control chip?
I think because the purpose of the diodes in the UC384x is different. See on page 7 under Shutdown Techniques: "Shutdown of the UC1842 can be accomplished by two methods; either raise pin 3 above 1 V or pull pin 1 below a voltage two diode drops above ground." In your case, the diodes are only there because the op-amp output won't go all the way to ground.
 

eem2am

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Thanks Kris

The output post-diode won't have the low impedance that the op-amp output has - at least, not when the voltage is falling, because that node is only pulled to ground by the voltage divider resistors.

..This is an interesting point, i am going to go back and review if i need the diodes inside the loop now, because i want the output to be able to fall rapidly, as i want current to be able to decrease pronto when necessary.



[Also, by the way, Thanks Harald above, here is the link to Microchip......

http://ww1.microchip.com/downloads/en/appnotes/00682c.pdf

...its point 3 at the very bottom of page 8 of the microchip link which sends fear through my bones about opamps that need to drive their output near to ground...such as in my case.]
 
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KrisBlueNZ

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(re higher impedance after the diode)
This is an interesting point, i am going to go back and review if i need the diodes inside the loop now, because i want the output to be able to fall rapidly, as i want current to be able to decrease pronto when necessary.
I think you're overestimating the problem. Unless you have a lot of capacitance in the feedback loop, you should be fine. This is why I asked you to post your proposed error amplifier circuit. I doubt there will be a problem.
 
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