Part Number: OPA992
Other Parts Discussed in Thread: OPA2992, OPA4992, , OPA2196, TLV2197-Q1
Tool/software:
HI , see the sch , is it OPA2992 (OPA4992 ) can work at vcc=5v and vee=-35v , input -15v , and want to get -15v*2=-30V , is it ok ? thanks.
and we can find the output is pulled down at the power up ( the RED wire , 0us~0.5us) , why ? What methods can be used to avoid it?
Hi Jianjian,
is it OPA2992 (OPA4992 ) can work at vcc=5v and vee=-35v , input -15v , and want to get -15v*2=-30V , is it ok
Yes. Your Vcc = V+ = 5Vdc and Vee = V- = -35Vdc, Vs = 5 - (-35) = 40Vdc, which is operating within the supply range. Your input signal is operating within the recommended Vcm range or linear operating region. So unsymmetrical supply rails are ok to use.
Here is the Tina simulation:
OPA2992 Asymmetrical Rails 07212025.TSC
we can find the output is pulled down at the power up ( the RED wire , 0us~0.5us) , why ? What methods can be used to avoid it?
In the simulated example (Gain=2V/V), your Vcc and Vee are different from the initial proposal. OPA992's slew rate is a function of input signals (small vs. large signals). The delay is due to the variation in slew rate, where 30V/32V/usec = 0.94 usec is based on the large input signal. For small input input signal, the slew rate is lower, and delay may be longer, see the attached figure below.
If you want to have consistent slew rate, you may have use other op amps such as OPA2196, OPA21974 or TLV2197-Q1 or similar ones etc.. Please provide us with your design and cost requirements for your application.
Best,
Raymond
hi, Raymond , thanks . about the question NO.2 ,please focus on the time(0us~0.5us) , after that , the op work as I want . that mean , when the power on (or when the simulate start ) , the output is pulled down to about -15v . and then it gradually rises back to 0v 。 I check the current of R54 (The inrush current upon power-on) almost to be 65
mA , look like the OPA2992 can not provide ,so ,the output is pulled down . now , I don't know ,why has so high inrush current .
Hi Jianjian,
(The inrush current upon power-on) almost to be 65
Why are you driving 1ohm load? Isc refers to short circuit current. 1ohm and short circuit are basic equivalent in this case. Since you shorted the op amp's output, then it is sourcing the max. current, close to ±65mA typical.
This is op amp, it can not drive such low load, -30V/1Ω = Short Circuit, that is why the op amp is sinking 65mA. The op amp is only able to source or sink approx. +/-30mA range, which means at -30V/30mA ≈ 1kΩ (sinking). For normal operation, your Rload should be greater than 2kΩ.
If you touch the surface of OPA992 momentarily, the IC will get very hot and 0us~0.5us change are likely having something to do with the IC's thermal shutdown and thermal recovery, where Tj is close to 150C.
Can you tell me what you are trying to do?
Best,
Raymond
Hi，Raymond 。 thanks。I know OPA2992 is a OP amp . but
Our usage is a bit unique. Judging from the temperature changes caused by power consumption, there seems to be no problem. It's just that when it's first powered on, the waveform seems incorrect.
Here is my usage, power consumption calculation. 
There should be no mistakes, right? Can it be used like this?
Hi JianJian,
OPA2992 is dual channel op amps. The power dissipation and Tj temperature rise may be estimated based on the table and equations below. Most of heat is dissipated through the lead frames and PCB surface. So the op amp's surface temperature does not really indicates how hot is the Tj of the internal Si-die.
https://www.ti.com/lit/an/slua844b/slua844b.pdf?ts=1753146317805
look like the OPA2992 can not provide ,so ,the output is pulled down . now , I don't know ,why has so high inrush current .
My guess is that it is likely that the load is triggered the IC's thermal protection circuitry during power on instant. If you remove the 1Ω and capacitive load at output, for instance, you should not see the (0us~0.5us) events. Or you can increase the load resistance to >2kohm.
FYI. Because of Tj surface area is small, temperature rise inside of the IC will be fast.
You mentioned that the following current profile is measured across R54 or 1Ω resistor during power on events. Are you sure that the op amp has not been damaged for some unknown reasons? If this is correct, then the OPA2992's output stage is shorted to GND through the 1Ω resistor.
No op amp is able to drive capacitive load without oscillation resulted from poor phase margin. 250pF may be too large, and OPA2992 is unable to drive 250pF without additional loop compensation. My suggestion is to increase 1Ω to 50-100Ω and see if the output will be stabilized. This could be the other reasons. The system is required to have at least >45 degree phase margin to be stable. At 250pF capacitive load, the phase margin is approx. ~23 degrees.
after that , the op work as I want .
If I understand your description correctly, the 0-5usec is only occurred at OPA2992's power on events. Say after 10usec, the output returns to normal at -15Vdc and the output voltage is stabilized. Please clarify.
Best,
Raymond
Hi, Raymond , thanks .In my application, I can't accept a power jump when the power is on because it's not good for my load. Here, I changed the load to a 10K resistor, but still when powering on, I saw that the output of OPA2992 was pulled down a little (as indicated by the yellow circle in the picture). Could you tell me why this is?
