[Mcgyver]

I have question re the use of voltage regulator ICs like an lm7812. while from the datasheet it says its virtually indestructible, I couldn't figure out what input voltages it can handle or if it matters

for the input for these ICs, is it good practice to bring the voltage down to something closer (say by resistor dividing) or can you just hook it up to rectify mains @ 170V DC. That would seem a bit extreme, otherwise why bother are there transformers, but I can't seem to figure what is a reasonable input??

[ESjaavik]

From Datasheet:
For Vo 5-18V: Max 35V
For Vo 20,24V: Max 40V

Also consider the max power dissipation. (Vin-Vo)*I that's often the limiting factor.

When the input voltage is high, I usually go for a switching regulator.
http://www.elfa.se/pdf/73/732/07327711.pdf
Easy to use, and quite efficient / cool running. But it won't do 170V.

[Mcgyver]

thx, didn't think it would do the 170 either, but the datasheet i googled either didn't say or i didnt' see it. thanks for the pointing out the power dissipation, i'm new at this so sometimes the simple points elude me

[dgolka]

Great info on the switching regulator. I have tried a number of things to reduce the in put voltage for a 7805 but it would continue to cut in and out. thx

[philba]

What kind of current you planning on drawing?

To your original question - no, don't use resistors. Use a transformer. I would target the rectified output of the transformer to be about 15V (3 above the output voltage of the regulator). That way you aren't heating up the vreg with wasted voltage. A 12V transformer will have, after rectification, about 15.5 V (12*1.414 - 1.4 from 2 silicon rect drops) as input to the 7812 which is pretty reasonable.

If you are going to be pulling more than a couple hundred mA at 12V, you will need a heatsink. (12V * 200mA is 1.2 W)

[Mcgyver]

its a pwm motor controller, motor running at rectified mains voltage. i need to step down for the astable 555 part that is going to drive the fet. the reason why i was thinking a voltage divider was that i thought the current is going to be low and it would be simplier than a transformer. my thought was to use 1 watt resistors. Also, the use will be infrequent and with a low duty cycle.

having said that, i am inexperienced so with this description, am i ok with a resistor divider or should i still go the transformer route?

thanks for the advise

[pminmo]

What is the specific PN of the 555? If that is all your going to be running from the 12V, you can use a resistor. We just need to know the 555 circuit. Your not going to be pulling enough current that it should be a problem. I suspect a diode, resistor an a small electrolitic cap infront of the regulator is all you need.

[Mcgyver]

sorry for the delay, its two 555's + the gate on the irf740's (two). Reading a spec sheet has the 555 @ 10mA - the irf seems to draw next to nothing. I know you asked for specific part numbers, but i still have to acquire the parts. I wanted to check in and see if I'm at all understanding the theory here.

to go from 160, to say 12 Volts, I need a voltage divider (as per illustration) or resistors in the ratio in a ratio of 148:12, right? next I need to figure out how how many ohms to give me the amps I need, right (?)

ie if R1 was 7.4k and R2 was 600, I'd drop 148 across R1 and if my circuit consumed 20mA, I'd need R1 to be rated greater than 3Watts

Am I close or lost in left field?

[smarbaga]

you can also use a 12v zener doide and resistor,
or 1 per 555 if u wish

[gar]

051219-0853 EST USA

philba:

The power dissipated in the series drop regulator is as ESjaavik stated the voltage drop across the regulator times the current thru the regulatior. So your example should read source voltage = 15.5 so the regulator drop is 3.5 v and the regulator dissipation is 3.5 * 0.2 = 0.7 Watts. But also note that the load power is 12 * 0.2 = 2.4 Watts not 1.2 .

In the design of an series pass regulator you must design for min and max input supply voltage. A not uncommon industrial design criteria range is 95 to 135 VAC input. This puts severe limits on the design. If you design for a minimum source DC voltage of 12 + 2 at maximum load and 95 VAC in, then at 135 VAC the source voltage will be about 14 * (135/95) = 14 * 1.42 = 19.9 V. This maximum source voltage is where you need to calculate the maximum power dissipation in the regulator. In this example it is (19.9 - 12) * 0.2 = 7.9 * 0.2 = 1.58 Watts.

In your actual design determine the minimum source voltage for the regulator. This is regulated output plus the regulator dropout voltage. Then select your transformer-rectifier-filter to provide this voltage at your criteria for min input AC voltage. Next determine the maximum source voltage to the regulator at your max AC input voltage. This source voltage minus the regulated output times the load current is the maximum power you will dissipate in the regulator at maximum line voltage, and must be used for your heat sink design at maximum ambient temperature.


Mcgyver:

If you go to www.national.com and to the data sheet on LM78XX and to the page on "Absolute Maximum Ratings" Input Voltage is listed as 35 V, and at the bottom of this page is dropout voltage. Page 5 has useful curves on power dissipation, and page 6 has dropout voltage vs load current and temperature (Dropout Voltage curve set).

On your question about a resistor to drop voltage from 170 to 12 V vs a transformer it is a question of power dissipation, and how much load current variation you have and what means you want to compensate for load current variations. But here is a simple example:

Load is 12 V @ a constant 0.01 A. Load power is 12 * 0.01 = 0.12 W. If the source voltage is 170 V, then the series dropping resistor is (170-12)/0.01 = 15,800 ohms, and power dissipation is 158 * 0.01 = 1.58 W. Therefore you are dissipating 1.58/0.12 = 13.2 times as much power in the series dropping resistor as in the load. At this current level this may be OK. But at 10 to 20 times the current it probably is not.

Now consider what happens to load voltage if your load current varies from 0.01 to 0.002 A. The load voltage will go to 170 - (15,800 * 0.002) = 170 - 31.6 = 138.4 V. This is where smarbaga's suggestion of a Zener diode comes into play.

.