Does voltage drop resistor affect power dissipation?

Question

I need to power an RFID from 24 V DC voltage. The RFID requires max. 9 V and max. 120 mA. The RFID suggests a 7809 linear voltage regulator. I thought too much and I want to use a 7809 because if I use a switching regulator (10$) for 24 V to 12 V and then 12 V to 9 V with 7809, it costs a lot.

If I go with 24 V to 9 V, (24 V - 9 V) * 120 mA = 1.8W,

If I go with 24 V to 12 V and then 12 V to 9 V ,(12 V - 9 V) * 120 mA = 0.36 W.

It will take ~22000 hrs of operation to realize the $8 savings from going to the switching regulator vs going for the linear regulator and that's okay for me.

• I made this circuit; someone suggested to me to use a 82 Ω, 2 W resistor. If I use that resistor, would I still need to use a heat sink? If I add that resistor (82 Ω), is power dissipation = (10.2 V - 9 V) * 120 mA = 0.24 W ? (I used a 75 Ω resistor for the RFID part)
• I read somewhere (I couldn't find the link) that says you need to put in 3 different output capacitors for clear power, is that correct? Are these capacitors okay? And all these capacitors are ceramic, should I use electrolytic ones?
• I found this heatsink özdisan.comdatasheet, this heatsink has only graphic and it says that at 2 W the rise in temperature will be 20°C, 5°C from junction to case, plus another 5°C for the TO220 case to heatsink, resulting in 30°C isn't it wonderful?

I have another heat sink farnell.com, it doesn't have a graphic but it says Thermal Resistance is 14°C/W , how can I compare which is better? I couldn't get its Thermal Resistance: 14°C/W . Thank you.

Answer 1

Got a bunch of switchers on Farnell, starting at 1.69€ plus tax.

It's cheaper than a 7809 plus heat sink.

someone suggested to me to use a 82 Ω, 2 W resistor. If I use that resistor, would I still need to use a heat sink?

Plot the regulator dissipation (24-9-i*R)*i with i from 0 to your max current and check. With the resistor, the regulator maximum dissipation will not be a t maximum current but a bit below.

If the load draws 120mA maximum during pulses, but has much lower average current, then perhaps you can get away with a smaller heat sink.

If load current is constant, then make sure to use a 3W resistor if you intend it to dissipate 2W, and solder it with long legs. Otherwise it'll barbecue your board.

Also 7809 needs more than 1.2V drop.

this heatsink has only graphic and it says that at 2 W the rise in temperature will be 20°C, 5°C from junction to case, plus another 5°C for the TO220 case to heatsink, resulting in 30°C isn't it wonderful?

20°C/2W = 10°C/W, That's good.

I have another heat sink farnell.com, it doesn't have a graphic but it says Thermal Resistance is 14°C/W , how can I compare which is better? I couldn't get its Thermal Resistance: 14°C/W . Thank you.

Thermal resistance seems worse, you need a powerful soldering iron to solder it because of the thick feet, and it's heavier.