Junction temperature of operational amplifier  - Page 1
Forum section: electronics / general

author: joniengr081
author_rank: Regular Contributor
timestamp: August 17, 2024, 09:27:15 am
content: Hello,I am using operational amplifier in inverting configuration. Let's have input 100 mV and the output is -500 mV. The gain is 5.Here is the formula for inverting configuration. The input has to be applied at -IN. The +IN is connected to ground. The value of R1 is 1 k ohm and the value of R2 is 5 k ohm.Vout = -(R2/R1) VinHow do we perform thermal analysis ? How can we calculate the junction temperature ?The datasheet of operational amplifiers have a table for thermal resistances in thermal section. In order to calculate the junction temperature we need to know ambient temperature which can be considered as 40 degC. The thermal resistance which is given in the datasheet. And also the power dissipation. The only thing which is unknown to me is how to calculate the power dissipation ? I am looking for the formula to calculate the power dissipation for inverting input operational amplifier.

author: magic
author_rank: Super Contributor
timestamp: August 17, 2024, 10:03:26 am
content: Power dissipation is current times voltage drop experienced by the current. You have two components to worry about inside the chip:1. Internal quiescent current, flowing from Vcc to Vee. Multiply Iq given in the datasheet by (Vcc - Vee) and that's it.2. Load current, flowing in/out of the output pin. Multiply Iout by (Vs - Vout), where Vs is whichever supply that provides the output current.Load current will be at least the current flowing through feedback resistors, and all other things connected to the output pin.

author: joniengr081
author_rank: Regular Contributor
timestamp: August 17, 2024, 11:30:07 am
content: From the datasheet of the operational amplifier I can find the quiescent current. I also know the load current which is far less then the maximum output current of the operational amplifier.The power supply for the operational amplifier is +12V and -12V and the load voltage is -0.5V.Which voltage do I need to use in the following calculations.1- Internal quiescent current x (VCC - VEE) = Internal quiescent current x (12V+12V) ?2- Load current x (Vs - Vout) = Load current x (12V - 0.5V) ? which Vs should I use 12 V or 24 V ? The Vout is -ve 500 mV. Do I need to add (Vs + 0.5V) ?

author: magic
author_rank: Super Contributor
timestamp: August 17, 2024, 12:44:46 pm
content: Current times voltage drop experienced by the current.1. Correct, 24V.2. If load current is sunk by the chip, then it enters the output pin at -0.5V and exits through Vee at -12V, so voltage drop inside the chip is 11.5V.

author: joniengr081
author_rank: Regular Contributor
timestamp: August 17, 2024, 01:52:56 pm
content: Yes the load is sink. The other end of the load is connected to ground.

author: joniengr081
author_rank: Regular Contributor
timestamp: August 17, 2024, 01:54:48 pm
content: How about the third factor which is power due to input bias current. I understand that the input bias current is very small as found in datasheets but if we need to use that in formula theoretically then which voltage to use to calculate power ?

author: Doctorandus_P
author_rank: Super Contributor
timestamp: August 17, 2024, 02:18:30 pm
content: Quote from: joniengr081 on August 17, 2024, 09:27:15 amThey may put some number for that in the datasheet, but I would not trust it. Thermal resistance is highly dependent on the amount of copper that the pins of the IC are soldered to on the PCB. This can easily lower the thermal resistance by a factor of 3 (assuming a small signal opamp).For power opamps with heatsinks... Heatsinks also have a thermal resistance in their datasheet, Usually it's not just a number, but a graph that shows how thermal distance changes with airspeed. A little bit of air movement has a huge impact, while a huge airspeed only has a small incremental impact. It does not help if air molecules get blown past the heatsink before they have time to absorb some heat. The main goal of the air movement is to remove the "bubble" of hot air close to the heatsink and in between the heatsink fins.It's good to know some things about thermal calculations, but it is of no use to be very accurate here, because there is just too much variation in the variables involved. In a real (production) environment, the design should always be verified with a thermomenter (thermal camera) and under realistic circumstances (closed housing, possibly under a hot car bonnet etc).

author: joniengr081
author_rank: Regular Contributor
timestamp: August 17, 2024, 06:57:15 pm
content: There are operational amplifiers offering more then 100 mA output current. I found AD8397. There are two operational amplifers in the package. Let's say I have a power supply +12V and -12V for each operational amplifier. And I am having inverting configuration which basically invert the polarity with a gain set by the resistors R2/R1. Consider if the load on each operational amplifier draw continuous current 75 mA. The other end of the load is connected to ground.Calculating the power dissipation for Vout = -3V.P = V x I = 75/1000 x (12-3) = 675 mWThe same is true for the other operational amplifier.Would that be a problem ? What consideration do I need to consider in PCB layout.The total power dissipation would be 1.35 W for two operational amplifiers.According to datasheet: the thermal resistances are given below for two packages with and with exposed pad.8-lead SOIC_N: θJA = 157.6°C/W8-Lead SOIC_N_EP: θJA = 47.2°C/WLet's say I use the one with exposed pad so the thermal resistance would be 47.2 °C/W. Let's consider the ambient temperature 40 °C.Junction temperature = ambient_temperature + thermal_resistance x power-dissipationJunction temperature = (40 + 47.2 x 1.35) °CJunction temperature = (40 + 63.72) °CThis is clearly above 100 °C.Question: How can I use AD8397 for 2 x 75 mA continuous current through two operational amplifiers connected to two loads ? Is that impossible or still can be done by some PCB design consideration ?

author: Doctorandus_P
author_rank: Super Contributor
timestamp: August 17, 2024, 07:23:45 pm
content: AD8397 has an exposed heatsink pad on it's underside. Use it in the same way as is common for all IC's with such a think and which need a bit more heatsinking. Thermal performance is also often specified while maintaining a 25 degree centigrade for the "junction" or packaging. It is also common that thermal rating is a real limiting factor, and you have to derate the power due to temperature of the air, quality of the heatsink etc. His is a normal process for all power electronics.This opamp has a relatively high output current (for a SOIC-8 opamp) but also a wide power supply voltage (3 to 24V) Attempting to max out both these limits at the same time will exceed it's thermal rating. This is also normal beginners knowledge. For example have a look at the SOA (Safe Operating Area) of normal BJT's.

author: joniengr081
author_rank: Regular Contributor
timestamp: August 19, 2024, 08:01:37 am
content: I made some changes and added external transistors at the output of operational amplifier.Now I have prepared schematic available in attachment. The maximum load current requirement is 75 mA. I just put load as 1kohm but I the exact load impedance is still not known yet. The load voltage requirement is 0 to -3V adjustable which is set by the DAC not shown in the schematic. The DAC output will be connected in place of input voltage source Vin. The maximum voltage of the DAC is 2.5V. I just a use a sin wave in simulation -2.5 V to +2.5V to get a full swing output but in practice will be from 0V to 2.5V. The OpAmp and a push pull stage using external transistors are shown in the circuit. The gain of the inverted operational amplifier is set by R2/R1 which can be changed by resistor values.Question: The transistors in the attached circuit are able to drive load with continuous current 75 mA and voltage -3V ?. Back to the original question in this thread about power dissipation. How can I perform thermal analysis in the attached circuit. The output current of the operational amplifier is very small driving the base of the transistors. How to calculate the power dissipation on the transistors given the power supplies are +12V and -12V.

author: tszaboo
author_rank: Super Contributor
timestamp: August 19, 2024, 02:23:17 pm
content: Quote from: joniengr081 on August 19, 2024, 08:01:37 amVoltage drop on the transistor * current * thermal resistance + ambient temperature = max temperature.Same as for the opamp. If your output voltage is small, use a smaller voltage power supply. Either a voltage regulator, or just place a resistor in series with your opamp or transistor. So that the power is dissipated in those, not your active parts.

author: mawyatt
author_rank: Super Contributor
timestamp: August 19, 2024, 02:36:19 pm
content: Quote from: joniengr081 on August 19, 2024, 08:01:37 amIf you intend to just drive the load with low frequency/DC then you can forgo the added bias diodes and extra resistors. Just connect both the NPN and PNP bases together, add a ~1K from base to emitter (optional) and a 100~500 ohms from op amp output to bases. The op-amp feedback will "cover up" the cross over distortion at DC and lower frequencies.Best,

author: joniengr081
author_rank: Regular Contributor
timestamp: August 20, 2024, 07:25:31 am
content: Sorry for confusion. The input is from 0 to +2.5V. The output requirement is 0 to -3.0V. I have corrected in simulation. The output is adjustable from 0 to -3.0V as we change the input from 0 to +2.5V. The load current requirement is the same 75 mA. Kindly have a look again in attached circuit. Is there any component that run into thermal problem ? How can I calculate the power dissipation in the transistors ? I guess the last two transistors are push pull configuration.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: August 24, 2024, 07:27:05 am
content: You are shorting D1 and D2.Apart from that, it looks like the classical thermal runaway problem in (audio) amplifiers. As transistors get hotter, their Vbe drops, which results in more "shoot through" current through the two end transistors, which causes an even lower Vbe and this repeats until the magic smoke is released.There are two ways to prevent this. One is to add emitter resistors that also drop a bit of voltage when the current increases. The other is thermal coupling between the resistors and diodes, with the result that the base voltage for the BJT's also gets lower when the diodes get hot.Your BJT's are small SOT23 devices and you can't get a good thermal coupling with that. Those BJT's also have a maximum power rating of half a watt, so you just move the thermal problems from the opamp to the BJT's.