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Hi Thomas Kuehl,
I take my previous subject (which was closed) "OPA541: Paralleled Operation for the OPA541"; additional data that can change a lot of things: I told you that the load was "The load: It's a coil of 1.9H with a resistance of 11Ω which we can add cable resistance, at the maximum 40Ω.". There is a modification, here is the real charge "The load: It's a coil of 0,46H (about) with a resistance of 1.4Ω (about) which we can add the cable resistance, at the maximum 4Ω." It is totally different.
I did the test with my previous schematic, in different steps :
- First, I simulate my load with a resistance on the output of 180Ω if I remember. My signals were OK.
- After, I plugged the real load : a coil of 0,46H with a resistance of 1,4Ω. It wasn't worked (my amplifiers were heating up and my signal was distorted). I can join you pictures if you want.
Thank you for your answer.
Regards.
Hi Thomas,
Please find in attached :
Regards
Hi Florian,
The OPA541 TINA simulation circuit you provided indicates that the input signal is a 4.4 Vpk, 1.2 kHz sine wave. I believe the waveform presented on DSO channel 1 confirms that is the approximate input condition.
When I analyze the 0.46 H load, it has an inductive reactance of almost 3.5 k-Ohms (j3500) at 1.2 kHz. This is accompanied by a small series resistance of 1.4 Ohms. Therefore, the output load is almost entirely reactive. The circuit is set up for a voltage gain of 8.2 V/V so for a 4.4 Vpk input the expected peak output voltage would be about 36.1 Vpk. That would be a problem.
If I lower the peak input voltage such that the output is within its linear range the current is surprisingly low. For example, if the output is kept to 30 Vpk, the current is only about 30 Vpk/ 3500 Ohms, or 8.5 mA peak, for the reactive load. That is actually a much lighter load than the 180 Ohm resistor poses, which would draw about 170mA.
When a TINA simulation is attempted with the 0.46 H reactive load it indicates that the output is hitting the rails which causes the simulation to stop. It appears that the gain is too high and also there may be some gain peaking occuring because of the circuit Q. You mention the twisted pair cable. That is adding some capacitance across the inductor and adds another reactance to the load. If I reduce the input voltage down to 4 Vpk, then the output doesn't hit the rail and simulation finished. Try reducing the input voltage and see if the distortion decreases.
You may find it necessary to reduce the drive slightly, or decrease the gain slightly, and add some resistance across the inductor to lower the Q so that the output voltage remains within the linear range of the OPA541.
Regards, Thomas
Precision Amplifiers Applications Engineering
Thomas,
After checking, this is the 2 signals (input and output). Output signal looks a liitle bit better than last time, but he is still distorted. And my input signal hasn't the same amplitude with and without load. I don't understand.
Regards, Florian
Hi Florian,
It appears that the upper NPN output transistor is turning off momentarily. You are driving the ouput swing to the maximum and beyond and there may not be enough voltage remaining to keep the current source and output Darlington turned on during that portion of the waveform. Remember you are driving a complex impedance and the peak current and peak voltage are out of phase. Try reducing the drive and see if the waveform returns to a sine wave, or if you have the option of increasing the supplies try doing that.
Regards, Thomas
Precision Amplifiers Applications Engineering
Thomas,
This is how the resistor is connected. I need many amperes, that's why I use two amplifiers in paralell.
Regards, Florian
Hi Thomas,
As you said, I removed the TVS and I increased supply voltage. I tried with +/-37,3V, +/-38,3V and +/-38,8V.
I still have the same problem :
I don't want to increase the supplies too close to the maximum power supply voltages.
Regards, Florian
I did two others tests. For the two, supplies is always the same, +/-38,8V.
For the first one, I decrease the gain to 6,55 (I used 10k and 1,8k). The curves looks better, but not stil ok :
For the second one, I reduced the gain to 5,87 (I used 5.36k and 1,1k). This is the curves :
The curves looks ok but :
- The second heats up a lot and very quickly, so it's not normal. Maybe the power supply voltage (38,8V) is too high?
- When I measure the current with my system, I've about 4,7A. The aim was to use two amplifiers in parallel to have more current than one. With this voltage and current, it's not necessary to use 2. How can I increase the power (voltage and current) in the coil?
Regards, Florian
Hello Florian,
Regarding your questions:
- When I changed the load (see the beginning of the discussion), the excitation signal is not the same (about 10V instead of 8,8V) with a loaded output. How can it affect the input signal?
One would not expect the input signal level to change, with a change in the op amp output load conditions. That is not something that I have ever observed with a properly functioning operational amplifier. The input stage of the OPA541 is a JFET differential amplifier and its impedance is very high. It offers a very light load to the source when the non-inverting input is driven as is the case with your circuit. Could you try driving the OPA541 input with an independent low-impedance signal generator and see if the input and output signal levels are different without, and with, the output load attached? If the input and output levels remain about the same, then that might tell us something about the on-board signal source.
- How can I calculate the real value of the output current?
That is an involved question, because once you move away from a purely resistive load to a complex load that includes reactance you have to consider the real power (Watts), the reactive power (Volt-Amperes reactive) and Apparent Power (Volt-Amperes). It is a pretty straightforward subject that incorporates the power triangle relationship between these three different types of power. Rather than repeating what is covered nicely on line, see the following tutorial provided by Arrow.com, and copyrighted by Aspencore Inc. Here's the link:
- Maybe this amplifier is not suitable for my application?
Well, I am not sure if this is really an issue with the OPA541 power op amp or something else inthe system. I think the complexities introduced by highly reactive load creates much more different and complex conditions than those encountered when driving a purely resistive load. The fact that the output is rising to levels that are very near, or exceed, the power supply rails indicates that there are things are happening in the circuit that fall outside of the normal operating range.
If you are not obtaining satisfactory results with the OPA541 and OPA549 you may want to consider an even higher power op amp. The OPA541 and OPA549 are TI's highest power op amps and I don't have anything else I can suggest from our line-up. When we run into that situation, then we often suggest having a look at the high power op amps offered by Apex Microtechnology. They offer a wide variety of power op amps that exceed the output voltage and current capabilites of ours. TI has no affiliation with Apex Microtechnology, but mention them as a courtesy.
Regards, Thomas
Precision Amplifiers Applications Engineering
Hello Florian,
I've spent a lot of time analyzing the OPA541, dual amplifier circuit we have been deiscussing. However, before I did that I analyed the 0.46 H + 1.4 Ohm series load that you need to drive. The results are may be surprising to you.
When the impedance of the series 100 uF output capacitor, 0.46 H inductor and 1.4 Ohm resistance are considered the calaculated impedance at 1.2 kHz is 1.4+ j3465.6 Ohms, or 3466∠89.97° Ohms.
That is a high, reactive imepdance and the associated ac current will be low even at moderately high output voltage levels. For example, if the required output is 36 Vpk, the peak current is:
ipk = Vpk / ZT = (36 Vpk ∠ 0°) / (3466∠ 89.97°) = 10.39 mA∠ -89.97°
That is much lower current than expected, and shouldn't require a high power op amp but instead one that can provide the required high output voltage swing and moderate output current.
When I simulated and analyzed the two OPA541 circuit, I found that there was a high circulating current flowing between the two outputs; upwards to 300 mA peak. That may explain why the amplifiers were getting hot when you drove the circuit with the sine wave generator. The circulating current comes about from the master/slave tied output configuation and the voltage offset of the slave amplifier.
I propose that a different approach using a high voltage, much lower output current op amp be tried. Since all that needs to be provided is an output current of about 10 mA, a HV op amp such as the OPA452 should work, without the need for the master/slave circuit. The OPA452 is usable with supplies up to +/-40 V and is capable of a 50 mA output. If the output current is kept to the 10 to 15 mA peak level, the output swing to the supply rails is about 1.5 V from them. That is much closer than what can be achieved with the high power op amps.
I've began simulating the circuit below, but am finding something that may or may not be real. The series output load has a resonant frequency of 23.5 Hz and there appears to be a low-level amplitude component in the peak-to-peak output current. That is why the waveform isn't symetrical about zero. That ripple is right around 23 Hz. If this is real, that is an issue that will have to be dealt with.
I don't know if you want to pursue another path, but I wanted to present you my latest findings.
Regards, Thomas
Precision Amplifiers Applications Engineering
Hello Florian,
That changes everything in terms of the analysis, and drives the design back to the two OPA541 power op amps. I ran a quick simulation of the circuit with the load inductor chnaged from 0.46 H to 0.46 mH. The result is shown below.
The peak current is now in excess of 6 Amperes and the peak to peak voltage at the load is very high, ~+/-50 Vpk. It appears there is a near resonance between the inductor and the output coupling capacitor. This would need to be examined more.
Note that I will soon be taking a break for the holidays. I will not be answering anymore E2E posts after tomorrow, and not until the second week of January 2018. Have good holidays!
Regards, Thomas
Precision Amplifiers Applications
If the current through the second OPA541 is about equal to that of the first, and it has the same thermal envirornment as the first, it should not be getting any hotter than the first. They should share the output power load and heat nearly equally.
The two series 100 uF output capacitors (50 uF total) are in conjunction with the 430 uH inductor create a resonance at 1.1 kHz in the output circuit. That is very close to the 1.2 kHz input frequency. The result of the resonance is output peaking which is casuing the output voltage to be higher than the predicted closed loop gain of 8.5 V/V (18.6 dB). According to the simulation the overall gain is 20.6 V/V (26.3 dB), and reduces a little at 1.2 kHz to 19.5 V/V (25.8 dB). The apparent gain is much higher at 1.2 kHz than the 8.5 V/V and that is why the output can exceed the supply rails. Operational amplifiers are not normally operated under such conditions.
If the 50 uF series output capacitor is removed only a small dc level will be present at the output load. This is due to the op amps offset. Removing the capacitor eliminates the resonant circuit setup and the gain becomes exactly 8.5 V/V in the pass band. You can see this result comparing the output voltage vs frequency response with, and without the capacitor. See the plot shown below. I recommend not using the capacitor. Try doing so and see how the circuit behaves.
The OPA541 is rated for continuous current of 5 Amperes, and a peak current of 10 Amperes. The limiting factor is the power dissipation which must keep the transistor junction temperature below 150 C. Use the Figure 11 Safe Operating Area graph as a guideline.
I will be on holiday the next 3 weeks. I hope you make progress with your circuit application.
Regards, Thomas
Precision Amplifiers Applications Engineering
Hi Florian,
We use the AAVID 530002B02500G heat sink on our OPA547/OPA548 EVM. It has about the lowest heatsink-to-air thermal resistance for the TO-220 package that we could find. The datasheet lists a natural thermal resistance of 2.60°C/W, but that can be reduced to 2.0°C/W with a 300 LFM forced air flow. You can find the catalog information on-line about this heatsink, at companies such as DigiKey:
Also, you can see how we install it on our OPA547/OPA548 EVM:
Regards, Thomas
Precision Amplifiers Applications Engineering
