Can Op Amp provide control voltage near to ground?

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
**broken link removed**

Yes, the data sheet does say this opamp can go down to 20mV, but only if the load resistance is less than 10K. If the input to the PWM module does not present this kind of load to ground, or worse yet, provides some pull-up, then this low output voltage cannot be guaranteed. You might try adding some load resistance to ground, about 5K.

Thanks Tunelab Guy,

There s no pullup resistance at the LD input pin....but i was going to add one there to Vdd so that it tends to pull up to the maximum current ene voltage unless the opamp pulls it down to some level.

But you are saying that the pullup cannot be done?.....surely the opamps output resistance is only about 80 Ohms, and so a 10K pullup i going to put very little voltage at the opamps output?

treez said:

Thanks Tunelab Guy,

There s no pullup resistance at the LD input pin....but i was going to add one there to Vdd so that it tends to pull up to the maximum current ene voltage unless the opamp pulls it down to some level.

But you are saying that the pullup cannot be done?.....surely the opamps output resistance is only about 80 Ohms, and so a 10K pullup i going to put very little voltage at the opamps output?

Click to expand...

Read your datasheet. The line where it says "Output Low Voltge" of 5mV nominal, 20mV max has a condition mentioned: RL < 10K. That's what Texas Instrument said, not me. You can believe them or not. It is your choice.

Thanks,

The short coircuit current at 5V is 0.06A, so that implies that the output resistance in the opamp is 83R.

...from that i would presume that if i place a 10K pullup resistor from output to a 10V rail, then the opamps output can only swing to a minimum of 80mV?......i presume this is correct?

There's no reasonable purpose of a pull-up resistor at LD. I guess, you didn't think about it thoroughly. A pull-down resistor without concurrent current sources will allow an LM358 output voltage effectively down to zero (< 1 mV).

P.S.: Your output resistance calculations are missing the point. The datasheet clearly shows about 800 mV residual voltage with 1 mA sink current. If you want to understand why, you should review the (simplified) transistor level circuit in the datasheet. Above 50 µA sink current, the output voltage rises to one diode voltage drop. Below 50 µA, it's essentially the saturation voltage of a current mirror. With a pull-down resistor, it's zero plus leakage currents.

I think the erroneous pull-up resistor concept has been elaborated sufficiently now...

By the way, the pull-up resistor won't ever serve the intended purpose, it's always overriden by the LM358.

The pullup resistor would be to ensure that LD is high at startup...hence allowing the SMPS that is being controlled here to start (sorry i failed to mention this)

But in any case, 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/en/AppNotes/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 there really is something in this....how can i manage this?....i guess i must just put a potential divider to the "LD" pin of the HV9910B, and then drive that potential divider with the output of the opamp, so that i don't need the opamps output to go near to ground......ie, i should keep the opamp's output say at around >1V above ground......

...is this the best thing to do?

I'm always skeptical on the output range of R-R opamps. I've used many kinds that claim their outputs can go down to <10mV, but most of them cannot actually do this in practice, even when I put a pulldown resistor on the output to ground.

Using a divider on the op amp's output sounds like a good idea. It will provide some pull down current as well.

LM358 isn't R-R, but claiming output swing down to zero. As with R-R, you can ask "how many mV is considered rail voltage". From LM358 datasheet figures and internal schematic, it's quite clear, that <20 mV as asked in the initial post isn't a problem even with open output. I can also confirm the fact from previous designs.

P.S.: To refer to the circuit details, I already mentioned that the output voltage is essentially the saturation voltage of a (bipolar) current mirror. As any BJT transistor saturated in forward operation, it can be expected to show a few mV residual voltage. If it matters, a voltage divider can help.

thanks,
FvM:

As any BJT transistor saturated in forward operation, it can be expected to show a few mV residual voltage

Click to expand...

...i thought Vce for a BJT in saturation was around 0.2V?