Hi everyone, I need to use linear voltage regulator for RF part. I decided to go with 7809. I have 24V power. I need to convert it from 24V to 9V so I checked the data sheet and it says input voltage max 35V. (My 24V is 220 AC to 24V DC 5A). My RF part requires max 120mA.
120mAx15V=1.8W
What is the critical value of power dissipation? Is 1.8 W too hot ?
Now, I am using 24V to 9V with 7809 without a problem. What happen if I made a PCB and 7809 work like 1 month ?
If there is no problem when using heat sink, what is the critical value for using heat sink ? 600 mW?
Everyone suggesting me I need to use switching regulator for 24V to 12v the use linear voltage regulator but why? I don't need 12V, 7809's data sheet says voltage input max 35V. Why just I can't use 24V to 9V with 7809 ?
Thank you.
1.8W is not a temperature, it is a a power dissipation. Rule of thumb is that if you can keep your finger (or thumb) on it then it's not too hot. That said electronic components can operate perfectly well at temperatures too hot to touch, but it's best if they don't. Use a heat-sink. Heat-sinks are rated in degrees C per Watt, so a heat-sink rated at 10 degrees per Watt will be 18 degrees above ambient temperature with what you have, which is fine. If in doubt use a bigger one.
The formula is Tj= Ta + (Rja x power) datasheet
tj=125 , ta=25, power= 1.8W , rja= 50 , Ta + (Rja x power) = 115.
How can I keep it around 50 Celcius. Is it like this :
50/1.8=27.77 , so I need to buy 28°C/W is it correct ?
Thank you i will read all of them. Which way do you think is the best ? 24V to 12V switching regulator then 12V to 9V linear or 24V to 9V linear regulator? I am only using 120mA max. 24V to 9V is cheap. 24V to 12V regulator 9,99 $ , heat sink 1$.
9 V @ 120 mA is 1.1 W
24->9 V burning 1.8 W in heat
This 1.8w will be reflected on the electricity bill and is tihs the problem with 24V to 9V linear regulator ?
If I go with 24V to 12V switching regulator then 12V to 9V 3VX120A=360 mW. If I use switching and linear regulator , only 360 mW will be reflected on the electricity bill . Am I thinking correct? Sorry but I just want to learn (I know this is easy question )
1.8 W will take 555 hours to consume 1 kW-hr of energy, or in my area, $0.25 to operate for 555 hrs
.36 W takes 2777 hrs for the same cost
Assuming $8 price difference in components between the two, it will take ~22000 hrs of operation to realize the $8 savings from going to the switching regulator vs going for the linear regulator
Okay, thank you. I just wanted the know why everybody suggest me the switching regulator. Now I got it. Probably RF part won't draw 120 mA all the time. So it is okay to use 7809 with heat sink I guess. Thank you.
You may consider adding i.e. 50R resistor between 24 V and 7809 input. It will drop about 6 V, leaving enough voltage for the regulator but reducing its dissipation. With the input cap for 7809 it will also act as a low pass filter reducing noise from the power supply.
You now need an output bypass on the 7809, since the 1 mF is isolated from it by the 1 Ohm(!) resistor.
33 uF electrolyte capacitor okay for output bypass ? I read someone suggest it In data sheet, they re suggesting 0.1 uF ceramic.
Now I am thinking of use this schematic.
In a circuit with a "normal" sensitivity to electrical noise a buck regulator (switching) will handle your voltage drop with ease. However they do generate significant noise.
Linear regulators are relatively low noise but burn the lost energy in heat.
Designing heat dissipation systems are not so cut and dry. I'll try to go through one option here, I hope it helps.
The 7809 regulator in a TO220 case in (very) free air will rise (per their spec) about 65 °C/watt. This regulator is internally limited for power and temperature so the regulator when it gets hot will drop the output voltage.
It states at two watts the rise will be 65°C + the 5°C from junction to case + another 5°C for the TO220 case to heatsink results in a 75 °C rise.
With a 25°C ambient temp the junction of the device will be at approximately 100°C
But remember this assumes free airflow to the heatsink. So no or big enclosures.
The is not great but doable.
You could add a power resistor before the regulator.... say large enough to drop 10 Volts.
R = E/I = 10 / .120 = 82 ohms. If you choose to do this be sure you add extra capacitance before the regulator.
I have one more question. For example this özdisan.com
This one is cheap and It states at two watts the rise will be 20°C . Isn't it ? This one is working good for me. 20+5+5 = 30°C.
If the ambient temp goes up even 60°C ( I want to run in 60 degree environment in summer ) , It will be okay because 90°C under the 150°C.
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In data sheet, they re suggesting 0.1 uF ceramic.
