Need help with values on adj. regulator (warning: it's an oldie from the datasheet)

Thread Starter

Hamlet

Joined Jun 10, 2015
252
I want 12v 1A from a 90v source,
what might be some appropriate values
to start with for the power bjt, resistor, &
zener to the left of the regulator in the
schematic below?



LM7812, TI, Page 15:

upload_2018-7-24_20-46-52.png
 

dl324

Joined Mar 30, 2015
8,862
You need the zener to drop the regulator input to around 16V and select a series resistance to limit power dissipation in the zener to something reasonable. The transistor needs to be able to handle the current desired; and be able to dissipate the power.

A switching regulator would be more appropriate.
 

crutschow

Joined Mar 14, 2008
23,297
So for a 90V source, the zener should be about 75V, with a base-emitter resistor of about 10kΩ.
The transistor should be rated for a least 100V and will dissipate about 7.5W so should be a power transistor on a heatsink
For a 0.1A load, the zener current should be no more than about 1mA, giving a zener dissipation of 75mW.

As dl324 noted, a switching regulator (or preregulator) is probably more appropriate as you don't have all that power to dissipate, requiring a high voltage, power transistor.
 

Kjeldgaard

Joined Apr 7, 2016
377
At such high input voltage, a series of Zener diode is not easy to get optimal.
With 90 Volt Input Voltage and A 75 Volt, 5% Zener, the 7812 could get an input voltage between 11 Volt and almost 19 Volt.

If there is to be linear regulation, I would suggest re-arranging the circuit to a series controller as the pre-regulator:
PreReg_1.jpg
Z1 can be either a 16 or 18 volt type.

Also, take a look at @crutschow thoughts about power and voltage.
 

Thread Starter

Hamlet

Joined Jun 10, 2015
252
That's a different take... thanks, I'll transcribe your schematic.

I wanted to add a digital voltmeter to a remote end of a 90v DC supply.

Maybe I should just buy a switcher, if I can find one cheap enough for a 90v
input, or look for a small panel meter.
 

Thread Starter

Hamlet

Joined Jun 10, 2015
252
I get that.
I'm still not happy.
I wanted a simple solution that I could understand. I thought I found it in the datasheet.
I don't understand switchers enough to design my own, so I can't really own it,
I can't catalog it, & thus I can't refer back to it as a building block in later projects,
except as a curiosity (jewel thief).
 

dl324

Joined Mar 30, 2015
8,862
Can you understand how this switching regulator (from a Signetics applications manual) works?
upload_2018-7-25_20-47-47.png
 

Thread Starter

Hamlet

Joined Jun 10, 2015
252
Oh, it's an old datasheet too!
I found some on ebay, but nothing on Digikey.
 

WBahn

Joined Mar 31, 2012
24,677
At such high input voltage, a series of Zener diode is not easy to get optimal.
With 90 Volt Input Voltage and A 75 Volt, 5% Zener, the 7812 could get an input voltage between 11 Volt and almost 19 Volt.

If there is to be linear regulation, I would suggest re-arranging the circuit to a series controller as the pre-regulator:
View attachment 156870
Z1 can be either a 16 or 18 volt type.

Also, take a look at @crutschow thoughts about power and voltage.
You can also put a resistor in series with the collector of the transistor to shift the power dissipation from your active transistor to a passive component that is less picky about operating hot. It can also provide current limiting.

Right now you are asking the transistor to drop something around (90 V - 15 V) or 75 V at 100 mA, so about 7.5 W of power. That's manageable, but it would be nice to reduce it. Let's have the transistor saturate at about 120 mA (or pick whatever value suits you) and call Vcesat about 1 V. That means we want 74 V across our resistor at 120 mA, so we need 617 Ω. We might find a 620 Ω resistor, but that will reduce our current limit (which might be fine). However, the next lower standard size is 560 Ω and it is quite a bit more common, so let's use that. That means that our transistor will now saturate at about 135 mA.

So what does this do to the power dissipation (of these two components), particularly in the transistor?

The total power dissipation of the two doesn't change. It's still

P_total = (75 V)·I

but now it is split between the resistor

P_resistor = I²·R

and the transistor

P_transistor = P_total - P_resistor = (75 V)·I - I²·R = (75 V - I·R)·I

At low currents, there will be little power dissipated in either.

At high currents, nearly all the power is dissipated in the resistor and virtually none in the transistor. The peak power dissipation in the transistor will be something a bit over 1/4 of the original peak power (half the current and half the voltage drop). It would be exactly 25% if the resistor were chosen to bring the voltage across the transistor to zero at the full current of 100 mA, but since that won't work, we can't achieve that. In this case, the peak power in the transistor would be about 2.5 W, which will allow it to run a lot cooler with the same heat sinking or, alternatively, operate at the same temperature with less heatsinking (the former is the better choice, if practical).
 
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