Does the OpAmp heat if the slew rate of an OpAmp is less than the input signal ?
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Hello Preetham,
Please provide the circuit parameters and the OpAmp you are using.
Thank you,
~Leonard
Hello Leonard,
Thanks for the reply
Part number used is OPA2130UA/2K5. Supplies provided are +25V and -5V
Given Input waveform specs:
Frequency : 100KHz
Rise/Fall time : 2us
High / low time : 3us
Amplitude : 0 to +20V
Gain : 1
Hello Leonard,
Thanks for the reply
Part number used is OPA2130UA/2K5. Supplies provided are +25V and -5V
Given Input waveform specs:
Frequency : 100KHz
Rise/Fall time : 2us
High / low time : 3us
Amplitude : 0 to +20V
Gain : 1
Hi Preetham,
OPA2130UA/2K5 is unable to support 100kHz at the 10V amplitude due to the low slew rate of 2V/usec.
The slew rate 2V/usec (dv/dt) is equal to Vo*2πf_max, where Vo, the output amplitude is 10V and f_max is the max. output frequency.
f_max = 2V/usec/(10*2π) = 31.8kHz, which is <100kHz per your configuration and requirements.
Regarding heating issue: I do not know what load the application is driving. Isc is specified at +/-18mA at room temperature. If the op amp is driving in high current application, perhaps the part is heating up due to high source/sink current. I do not have parameters to explain the operating states.
If you have additional questions, please let us know.
Best,
Raymond
Hello Preetham,
Regarding your question - Does the OpAmp heat if the slew rate of an OpAmp is less than the input signal ?
If the input signal slew rate exceeds the slew rate of the op amp, the op amp's slew will be the limiting factor and the maximum slew that can be attained. Internal to the op amp the current required to charge the compensation capacitor will be maximum when the op amp is in slew rate limiting. This will cause a slight increase in power dissipation and self heating.
However, if you are applying a 0 to 20 V input having a rise time of 2 us the edge rate would be the equivalent of a 10 V/us slew rate. This far exceeds the 1 V/us SR of the OPA130. When you attempt to drive an op amp with a much higher slew rate than their inherent slew rate some unexpected behaviors could occur depending on its design.
Some op amps will go into a shoot-through condition when the output driver and/or output stage cannot keep up with applied signal. Transistors inside those stages continue to sink/source high current when they should have idled back to their quiescent state. That results in extra current running through from the op amp V+ supply path to the V- path. The current can be significant and will increase power dissipation and heating.
It is best to avoid this abnormal operating condition because what long term effect it will have on the op amp is an unknown. It is better to use an op amp that has a slew rate more in line with the circuit operating conditions.
Regards, Thomas
Precision Amplifiers Applications Engineering
Hi Raymond,
The load connected to the OPA2130UA/2K5 is one more OPA2130UA/2K5, which is acting as an amplifier. So the current drawn at the output stage of the voltage follower is less which should not cause heating.
I understood that the slew rate of this OpAmp is much lower than the slew rate of the input and because of which the output of voltage follower is distorting.
But I need to know whether the limited slew rate of this OpAmp is responsible for its heating.
Thanks and regards,
Preetham
Hi Preetham,
The load connected to the OPA2130UA/2K5 is one more OPA2130UA/2K5, which is acting as an amplifier.
Could you send me the above drawing or schematic?
If you are talking about the parallel op amp configuration, as shown below, please see the attached article.
https://e2e.ti.com/blogs_/archives/b/thesignal/posts/paralleling-op-amps-is-it-possible
But I need to know whether the limited slew rate of this OpAmp is responsible for its heating.
When you talking about heating in this part: are you talking about the op amp is hot to a touch or you may place a finger on the surface of op amp indefinitely. Please be more specific. Temperature reading from op amp's surface will be good.
I do not believe that the limited slew rate in a single op amp configuration will cause heating (if the heating is referenced to significant temperature rise in the IC, excluding the R load's effects). However, if you configure op amps that marked in "X" as shown above, you may experience heating issues that output of A1 and A2 may fight each other due to imbalanced output voltage.
Please provide me the actual op amp configuration and load in your setup, I will take a look at the root causes of heating issues.
Best,
Raymond
Hi Raymond,
Thanks for the reply.
I have attached the circuit diagram of OpAmp configuration being used. Here U1 which is used as voltage follower is heating too much which can be felt by touching IC surface.
The signal specs flowing through the current sense resistor is given below:
Frequency : 100KHz
Rise/Fall time : 2us
High / low time : 3us
Amplitude : 0 to +20V
Gain : 1
As of now no load has been connected in the line after current sense resitor.
Please have a look at the circuit and let me know the reason behind heating.
Thanks and regards,
Preetham
Hi Preetham,
Thanks for sharing the circuit drawing with us. I have some questions, and will reply the inquiry by tomorrow.
Questions I have are:
1. What is common mode voltage across the current sensing resistor?
2. What is the current through Current Sensing Resistor? Is the powerline through sensing resistor AC or DC? (You mentioned that frequency is 100kHz. I want to confirm where 100kHz is presented. Do you mean that there is 100kHz 0-20V signal through the sensing resistor?).
3. I assumed 2E means 2Ω. we do not use the convention in U.S.
4. what is 0.014F, which is 14000uF? Is this a typo?
5. Are there any oscillation at the output U1A & U1B?
6. Do you have any scope shots you may be able to share with us?
Best,
Raymond
It is not enough to say that the differential input voltage across the shunt resistor is 0 to 20V - we also need to know what is the input common-mode voltage, Vcm, with respect to system ground as shown below? If it is above +25V or below -5V, you will forward-bias input ESD protection diodes (shown below) drawing HIGH current, which may not only heat up the part but also damage the input stage of OPA2130.
Hi Raymond,
Thanks for the reply.
1. What is common mode voltage across the current sensing resistor?
It will be 20V
2. What is the current through Current Sensing Resistor? Is the powerline through sensing resistor AC or DC? (You mentioned that frequency is 100kHz. I want to confirm where 100kHz is presented. Do you mean that there is 100kHz 0-20V signal through the sensing resistor?).
As of now no load has been connected on the line after Current Sensing Resistor and therefore current will be 0. When constant voltage of 20V is given the OpAmp is not heating but when a input signal changing from 0 to 20V with the slew rate more than the OpAmp can support is given then OpAmp starts heating.
3. I assumed 2E means 2Ω. we do not use the convention in U.S.
Yes it is 2 Ohm.
4. what is 0.014F, which is 14000uF? Is this a typo?
It is 0.01uF
5. Are there any oscillation at the output U1A & U1B?
I don't see any oscillation at both the outputs.
6. Do you have any scope shots you may be able to share with us?
I have attached below
Hi Preetham,
Per the circuit description, I duplicated the circuit in Tina. Something is missing in the configuration, and I am unable to duplicate the output.
1. The 2nd OPA2130 stage in difference amplifier should have gain in 10V/V (first stage is just a buffer). According to the description, the output should be in saturation, even at Vin = 10Vdc.
2. Your input scope shot looks ok (0-20V with fin = 83.76kHz), but you output screen shot is biased at 5V, the waveform may be centered at approx. 10V. In addition, output voltage amplitude from the scope shot is between ~12.6V to 5V approximately (7.6Vpp at output). Per the schematic configuration, your output signal should be saturated, but the scope shot is not.
If Vin =20V at the input, output will be saturated as it is configured. Even at 10V, it will be saturated, see the image below. Please ignore Vout waveform, the output should not be in sinewave and should be distorted totally (see its vertical axis from 23.61V to 23.62V). Vp node is slewing, which it looks correct before the second stage; and the second stage has a gain = 10V/V, which means that the output will be saturated from 41.2V to 181.8V ideally, if it is not limited and capped by the power rail (25Vdc max).
Enclosed is the Tina simulation. Let us get the setup consistent first, then I will explain what it is going on.
Best,
Raymond
Preetham,
You have several issues with your application:
1. The input differential signal developed across the Rshunt is different than the input common-mode voltage, Vcm. Applying even 0V differential signal while floating the Vcm will result in the output collapsing on one of its rails - see below.
Therefore, in order for the circuit to work you MUST reference the Vcm with respect to the system ground - see below
2. In order to accommodate +/-10V (20Vpp) input signal you would have to lift the Vcm by 10V so the output of the first stage is within 5V of either rail - see below. But in the gain of G=-10, the output would have to go to +/-100V, which is not possible of -5V/25V supply - I used below ideal op amps to show what it would take in terms of the output voltage range to accommodate the 20V input signal.
3. Using actual OPA130 with 20V input signal referenced to the system ground will result in the Vout to saturate against its negative rail causing shoot through current in the output stage. In believe this is the reason your part heats up.
4. To fix the issue you would have to lower the Rshunt so the input signal is 2Vpp. You would also need to reference the output with Vref of 10V so Vout is between 0V and 20V - see below.
In terms of current measurement of 0 to 8A, resulting in 0 to 2V voltage across Rshunt of 0.249ohm, you would need to reference the output to 20V so the output is between 0 to 20V - see below.
However, as you may see above the sinusoidal input signal becomes triangular (slewing) at the output because f=83.76kHz frequency exceeds the full power bandwidth of OPA2130 - see below. The maximum frequency OPA2130 may accommodate for Vout=20Vpp is around 30kHz - see graph below.
Lowering the frequency to 30kHz results in undistorted output waveform - see below.
I have attached below Tina-TI schematic so you may use for your own simulation.