Electronic – Difference between input current and supply current

First, the TLC59116 should be able to drive 120mA Per channel, ideally with a small VLED - VF voltage of VOL, to minimize heat issues. The higher the "unused" voltage, the less current can be driven per channel before thermal shutdown happens.

A TI employee responded to a user with the TLC59116. More info at the link

Question:

1. The TLC59116 (TSSOP package) gets hot and nearly goes into thermal shutdown when we run 16 strings of LEDs at 50 mA. Is that to be expected? We think we can get away with using the IC because we will be using the quad package which has higher dissipation.

Answer:

1. This depends on the voltage you are using on the LEDs. If the LED anode voltage is 10V and the string consists of 2 LEDs with a forward voltage of 2V each, the IC has to dissipate 16* 6V * 50mA = 4.8W. The power dissipation capability of RHB package is better than PW package because RHB has a thermal pad that has to be soldered to the board. Please make sure that you have a good thermal connection to the GND plane on your board.

So you need to match your LED VCC to the Forward Voltage of each led on that channel. Don't have too much left over voltage, as the internal transistor will need to dissipate that into heat.

From the Datasheet:

Figure 9 shows the output voltage versus the output current with several different resistor values on REXT. This shows the minimum voltage required at the device to have full VF across the LED. The VLED voltage must be higher than the VF plus the VOL of the driver. If the VLED is too high, more power will be dissipated in the driver. If this is the case, a resistor can be inserted in series with the LED to dissipate the excess power and reduce the thermal conditions on the driver.

As for more current, you can tie multiple channels together.

An IC is a load, it's just more complex than a simple resisitve or capacitive load. There are a few things to note about op-amps, though. They will consume some current to keep their circuitry biased properly, but they will also source current from the power supply pins to whatever is connected to the output. This makes it a little difficult to characterize an op-amp's current requirements in isolation as it really depends on what ciruit is built up around the op-amp. Looks like the power consumption of that amplifier is max 85 mW, so you can work out the current draw from that.

However, you probably don't want to use a 741 for your project if your only power supply is 9.6 volts. Here's why. The output voltage swing with +/- 15 volt rails is only guaranteed to +/- 12 volts into a 10k load and +/- 10 volts into a 2k load. We'll call 9.6 volts 10 volts here, so a split rail supply would be +/- 5 volts. If you assume the margin is independent of voltage, powering the amp on +/- 5 volts will provide minimum 0 volts of output swing. In other words, nada. You either need to use a higher voltage power supply, or what is called a 'rail to rail' op amp. A rail to rail op amp can drive its output to within less than a volt of the power supply rails. A rail to rail op amp powered on +/- 5 volts should easily provide a +/- 4 volt swing into a reasonable load.

For input offset current and other op-amp parameters, take a look at chaper 11 of http://www.ti.com/lit/an/slod006b/slod006b.pdf . This guide discusses many of these parameters. Also, take a look at the rest of the guide - it's a good reference for anyone using op-amps.