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I would like to use an LT1210 op-amp with a +/-12 VDC supply.

enter image description here

The input is from a signal generator (10 Vpp, 400 Hz sine wave). The required circuit output is 5 Vrms (14.01 Vpp), 400 Hz and drives a 600 mA load. The load is around 8.33 Ω.

I use the example calculation from the datasheet:

enter image description here

My result is negative 1.21 W (-1.21 W). The load/feedback dissipation is too large at around 3.04 W, while the op-amp device is at only 1.824 W.

What am I missing here?

UPDATE: Thanks everyone for the tips! Using your inputs and LTSpice, I will proceed with this design and I will use TO-220 package with heatsink:

enter image description here

Assuming total TO-220 thermal resistance at 15C/W, max Tj is 123.2C at 70C ambient temp. Max device Tj is 150C so this should work!

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4 Answers 4

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Your voltage swings from +7 to -7 volts. When the output voltage is +7 volts, the current from the positive rail to power the load is 7/8.33 or 840 mA. At this point you are dropping 5 (12-7) volts between the rail and the output, so the peak power inside the op amp is 5 x 0.84 or 4.2W. Similarly, the negative swing will also have 4.2W instantaneous peak power, this time from the negative rail. The power will drop to zero as the output passes through zero volts.

To use the method described, you would need to measure the actual current in the op amp rails. The author is just taking the total power and subtracting the load power. The rest is lost in the op amp.

To reduce power dissipated in the op amp, keep your rails as close as practical to the peak output voltage.

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  • \$\begingroup\$ if I lower voltage supply to +/-10VDC, how do I know opamp will not blow? Any idea on how to estimate current in the rails? looking at datasheet, it seems author just multiplied it by 2 (graph in page 7). \$\endgroup\$ Commented May 8 at 4:18
  • \$\begingroup\$ i think datasheet example is incorrect. how can supply current be just 76mA when load current is 140mA? \$\endgroup\$ Commented May 8 at 5:19
  • \$\begingroup\$ thanks for your input! i will proceed with the design (see updated post) using TO-220 package. It should work, fingers crossed. \$\endgroup\$ Commented May 8 at 9:43
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  • Pd = 0.28A x 24V - (5V)² / 8.33 Ohm = 3.7W

enter image description here

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  • \$\begingroup\$ how do you get Idc? \$\endgroup\$ Commented May 8 at 6:25
  • \$\begingroup\$ thanks for your input! i will proceed with the design (see updated post) using TO-220 package. It should work, fingers crossed. \$\endgroup\$ Commented May 8 at 9:43
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    \$\begingroup\$ @Ariel * Take the peak-peak current (1.7A) through the load and divide by 2*Pi. On top of that add the DC supply current (35mA) of the CFA when unloaded. To be on the safe side calculate with a total of 0.305A instead of 0.28A . \$\endgroup\$
    – Raonoke
    Commented May 8 at 14:08
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The problem is you used the supply current from the datasheet schematic 76mA, but you need to use the real supply current from your circuit instead.

With a supply Vs, if the opamp is outputting Vo into load resistor RL, then:

  • Output current is Io=Vo/RL

  • Voltage drop across the output transistor is (Vs-Vo)*Io

  • Dissipated power is (Vs-Vo)*(Vo/RL)

enter image description here

Thus maximum power dissipation is 4.3W at Vo=6V.

This calculation works for DC, or for instantaneous power. When reproducing a signal, what counts for thermal modeling is average power, which will be lower. However, using the maximum dissipated power for DC means the amp will also be safe if it is used for DC signals or very low frequency AC.

A correction can be made for the idle current of the opamp, but this includes the idle bias current of the output stage, which is class-AB. The datasheet mentions 50mA, so let's go with 10mA for the small signal stages, and 40mA for output stage idle bias. When output current is high enough to matter for thermal modeling (say 500mA) then the output stage will already be in class B, so we don't have to add dissipated power due to output stage bias current, but we do have to add 0.24W for idle current in the small signal stages.

With an AC signal, average power will be lower than this maximum. Here the output voltage is stepped:

enter image description here

If the load has series inductance, then current will not be in phase with voltage, which increases peak instantaneous power dissipation, but does not change the average much. Load inductance matters more for checking if output transistor SOA is not violated.

Lowering the supply voltage will reduce dissipation, but if it is too low the opamp will clip or distort. According to the datasheet, the chip has a maximum output voltage of +/-10V into 10 ohm load with +/-15V supplies, so with +/-12V supplies it's already close to your desired peak output voltage of 7V. You can test if it'll work with +/-10V supplies but it's not guaranteed.

So let's go with 5W max. The datasheet mentions RthJC=5°C/W for the TO-220 package, let's add 1°C/W for the thermal pad and a 4°C/W heat sink, we get a total of 11°C/W. Thus the chip will be at most 55°C above ambient, which means 95°C in 40°C ambient, this is fine.

For the DPAK, the datasheet gives RthJA in the 25-35°C/W range which is inadequate, but most of this comes from the thermal resistance of the board to ambient. The package RthJC is probably the same as TO-220 which means the DPAK would probably work, with thermal vias to carry the heat to the back of the PCB, then the PCB pressed against a heat sink with a thermal pad. This is a lot more cumbersome to build by hand than bolting a TO220 to a heat sink though.

For the SO package, the datasheet mentions RthJA in the 40-50°C/W range which is way too high. The chip doesn't have a thermal pad, so it will only really work at low supply voltage if high output currents are needed.

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  • \$\begingroup\$ thanks for your input! i will proceed with the design (see updated post) using TO-220 package. It should work, fingers crossed. \$\endgroup\$ Commented May 8 at 9:44
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You're asking the opamp to supply 7.07V peak on 8.33Ω which requires 848mA (I = 7.07V / 8.33Ω). At that voltage, 5V will be dropped across the op-amp output stage (12V - 7V = 5V), so the instantaneous power dissipation in the op-amp will be:
Pd = 5V * 0.848A = 4.24W.

Note that this is the instantaneous power dissipation; if the input is a sine wave then the voltage, current, and power dissipation will vary in time with the input signal, and the average power dissipation will be lower than this. But if the input is set to a constant 5V (ie: dc), then the opamp will suffer this 4.2W of power loss constantly. In both cases, I suspect the op-amp will burn-up.

There are two ways to reduce the power loss in, and hence heating of, the opamp:

  1. Reduce the supply rails. If the highest output voltage is +/-7V, then why do you need +/-12V supply rails? Reducing them to, say, +/-9V will reduce power loss by more than 50%.
  2. Reduce the current flowing out of the op-amp output pin. You can do this by putting a voltage-follower buffer at the op-amp output, and let this buffer supply most of the current to the load. This buffer can be made from transistors of suitable power rating so they don't get hot themselves. Here is one design that may be suitable:

https://electronics.stackexchange.com/a/689296/341959

Here is an explanation of a "voltage follower", aka "unity-gain buffer":
Unity Gain Amplifier Purpose

Just a minor typo in your math: The signal generator output is a sinewave with 10Vp-p (5V peak), 400Hz. Ignoring frequency for now, you require the output to be 5V RMS, which is 7.07V peak, or 14.14V p-p. R3 (910Ω) and R1 (2k2) set the gain to be 1.414, so for Vin=5.00V peak, Vout=7.068V peak, or 14.14V peak-to-peak (not 14.01V).

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    \$\begingroup\$ thanks for your input! i will proceed with the design (see updated post) using TO-220 package. It should work, fingers crossed. \$\endgroup\$ Commented May 8 at 9:43

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