High voltage pre-regulator circuit

Would some one mind explainging how these circuit work and the difference between them.

1.

The left-hand transistor network acts just like a power zener. It subtracts the zener value from the incoming supply. (Drop an additional 0.6V because of the transistor bias.)

Suppose the incoming is 40V. And the zener value is 15 V. This produces 24.3 V output.

Then it goes to the 78xx. Suppose it's a 7809 regulator.

If it gets an incoming supply 2V above nominal, then it will output 9 V.

It only produces output when the incoming supply is greater than than 27.6 V (the sum of 15 + .6 + 9 + 2.)



2.

The left-hand transistor network is a voltage regulator. It will limit the incoming supply to the zener value (minus 0.6V).

It works very much like a 3-pin regulator. It's handy for taking some of the burden off the 78xx, by dropping the incoming supply to a middle level.

BradtheRad said:

1.

The left-hand transistor network acts just like a power zener. It subtracts the zener value from the incoming supply. (Drop an additional 0.6V because of the transistor bias.)

Suppose the incoming is 40V. And the zener value is 15 V. This produces 24.3 V output.

Then it goes to the 78xx. Suppose it's a 7809 regulator.

If it gets an incoming supply 2V above nominal, then it will output 9 V.

It only produces output when the incoming supply is greater than than 27.6 V (the sum of 15 + .6 + 9 + 2.)



2.

The left-hand transistor network is a voltage regulator. It will limit the incoming supply to the zener value (minus 0.6V).

It works very much like a 3-pin regulator. It's handy for taking some of the burden off the 78xx, by dropping the incoming supply to a middle level.

Click to expand...

Does the equation in figure 28 also apply to figure 29?

In first picture, output Vin will follow variation in Vi. It will be,
Vin= V1-(Vz+0.7)
. So the ripple will apear on Vin point as it is present on V1.

In second diagram, the transistor circuit regulates to,
Vin= Vz-0.7.
It is a voltage regulator circuit.
The Ic regulator is the same in both circuit.
0.7V denotes Vbe

ALERTLINKS said:

In first picture, output Vin will follow variation in Vi. It will be,
Vin= V1-(Vz+0.7)
. So the ripple will apear on Vin point as it is present on V1.

In second diagram, the transistor circuit regulates to,
Vin= Vz-0.7.
It is a voltage regulator circuit.
The Ic regulator is the same in both circuit.
0.7V denotes Vbe

Click to expand...

Is this correct for the second version?

The zener on the transistor base fixes the voltage across bE at the zener nominal voltage, say 16V.
If BE is 16V then CE is 16V and the power dissipated by the zener varies with the input voltage.


16V / 5W zener with 36V in.

P=VI
5=(36-16)I
5=20I
I=5/20=0.25A max

V=IR
36=0.25R
R=36/0.25=144Ohms to limit the current through the zener to 250mA for 36V in.

So then I need a transistor which is biased to up to 10A or so with 250mA of base current, and the BU913P darlington I have laying around seems to fit the bill because, according to the data sheet, 250mA of base current will give me about 10A CE current.

I actually tried this circuit based on the above calculations and it failed spectacularly - I set the darlington on fire.

The zener and the resistor appear untouched and I did not even get to connect the darlington emitter to anything. I would guess that means the BE current or voltage limit was exceeded

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Found this website regarding these pre-regulator circuits: **broken link removed**

This is got the necessary equation for determining the value of R.

Tried another transistor with hFE 8 and input of 24V. The result was that a piece blew off the transistor.

Then I took more notice of Vceo <= Vin

I re-called that 2n6569 had a Vceo of 40V compared to the 60V input of this particular spec.

Can some one explain to me the signficance of Vceo <= Vin.

I thought Vceo was the max reverse voltage that the junction can withstand. So why do you want this less than the input voltage?

I did not even get to connect the darlington emitter to anything. I would guess that means the BE current or voltage limit was exceeded

Click to expand...

Base current sinks through emitter. If emitter is not connected, there is no base current. Probably you connected the transistor wrong way.


In this circuit, voltage between collector and emitter iof transistor is 30.6V.

I thought Vceo was the max reverse voltage that the junction can withstand.

Click to expand...

You are right. VCE of the transistor must be at least equal or higher the operating voltage.
Cant you see this circuit posted next on the website.

Most of components are in wrong position and circuit wont work. Search for a better page.
**broken link removed**

ALERTLINKS said:

Base current sinks through emitter. If emitter is not connected, there is no base current. Probably you connected the transistor wrong way.


In this circuit, voltage between collector and emitter iof transistor is 30.6V.

You are right. VCE of the transistor must be at least equal or higher the operating voltage.
Cant you see this circuit posted next on the website.

Most of components are in wrong position and circuit wont work. Search for a better page.
**broken link removed**

Click to expand...

I am in no way knowledgeable enough about this stuff yet to spot website mistakes, as you have just pointed out, with any confidence or reliability.

In both cases I was pretty careful about which transistor pins I connected where - I consulted the datasheet to make sure I got it right.

The fact that I have blown two transistor trying to implement the same circuit would suggest I am doing something else wrong.

Even if I connected the C and E pins around the wrong way, the second transistor I used had a Vceo of 1500V, or something like that, so I would expect that it could withstand a mere 36V or 24V.

Can you suggest anything else I might have done wrong here, because it has me a little stumped - I mean it is not as if this is a complicated circuit.

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I will add that the first transistor I used was an automotive darlington and the second was a HR output transistor from an old tv that has a built in damper diode......if that helps in suggesting what my error might be.

Put a series resistor (220ohm 1 watt or similar) in +ve lead of your supply. This is to limit current so that any mistake in connections wont burn any component. Test voltages and current at different points without load with multimeter eg voltage on zener diode, on emitter of transistor, drop on resistor etc.
Have a look on these pages.



http://www.circuitstoday.com/zener-controlled-transistor-voltage-regulators

**broken link removed**

http://brettcave.net/howto/circuits/Transistor_VoltageRegulators-BJT-Simple.pdf

Well this circuit has me absolutely stumped despite its simplicity.

As far as I can tell I have wired the positive part up on my bread board correctly.

16V zener, R1 = 1k, Vin=24V

I measure 24V in with my multimeter and 24V out. And R5 10R heats up massively such that I can't leave the circuit connected for more than a second or so - I have the multimeter probes connected via alligator clips.

Clearly I am interpreting something wrongly on this schematic while transposing to the bread board but I am damned if I can see what.

Don't suppose some one could wire it up on their bread board, photograph it and post the photo so I can see what the hell I am doing wrong?
**broken link removed**

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This is so *** damned frustrating that such a trivial circuit can absolutely stump me!