Series Pass Transistor for TL-783

[sonata149]

I have designed a high voltage regulator to supply 185V as shown below. I’m using a 200-0-200V @ 50mA trans which after a CLC stage gives me around 260V unregulated d.c. Calculating the divider network for 185V isn’t difficult – it’s given in the datasheet. I’m using a TIP50 as series transistor, however I’m not sure how to calculate the bias resistor.

I have done very little work with transistors and the books I consulted weren’t helpful in this area. The regulator will power a Steve Bench RIAA preamp discussed in another thread and according to the schematic it draws 10mA. From my understanding of the circuit the output voltage and current will be supplied by the regulator which is rated at 700mA max.

Again as I understand it, the bias resistor has to drop the voltage difference between the unregulated supply (260V) at the collector and the zener diode (120V) at the base (i.e = 140V). Is this correct or should it be the output voltage (i.e 185V) subtracted from the raw dc (260V) ??

Now, for what current should this resistor be tailored? One article says it’s a small current (2 - 3mA) to bias the zener, while from other sources I got the impression (may be wrong) that it’s the load plus the regulator currents.

Can someone please explain what goes on in this area of HV regulators.

Thanks and regards.
Joe A

[Stixx]

Sonata,
your schematic is waay too small...

[sonata149]

Sorry Stixx and thanks for your fast reply. Hope I get it right this time.

[sonata149]

No image there!

[kenpeter]

I hope thats a 12V Zenier, and not 120V as drawn.

[sonata149]

Yes, it's 120V zener. It's from the datasheet application information here: http://focus.ti.com/lit/ds/symlink/tl783.pdf

[DougL]

Looking at the Application note, The Zener does not normally draw current. In your example 185V+ 120V = 305V Which is higher than the raw supply. This means that no current is flowing through the Zener.
The Current the base needs is determined by Ic = Beta * Ib.
Ic is the current draw of your circuit plus the draw of the regulator. If you have a circuit that draws 100 ma, the transistor beta is 100, 1 mA is required to flow through the base.

I have a feeling that the resistor is not critical.
If the resistor is 10K (pulled out of the air), the current is 1 mA (from previous discussion) the drop will be 10 V.
That will place 280V - 10V - 1.4V (darlington drop) = 270V at the input of the regulator. That will allow 270 V - 185V = 85V across the regulator.

Basically, the Pass transistor and the 120V zener are there to protect the regulator from voltages higher than 120V.

Hope this helps.

Doug

[sonata149]

Thanks DougL.

If the resistor is 10K (pulled out of the air), the current is 1 mA (from previous discussion) the drop will be 10 V. That will place 280V - 10V - 1.4V (darlington drop) = 270V at the input of the regulator. That will allow 270 V - 185V = 85V across the regulator.

From the example you gave, does it mean that for such a small drop across the transistor (10V approx.) I could feed the unregulated supply directly to the input of the regulator? If so there would be 280V - 185V = 95V across the regulator, which is well within the 125V maximum spec - correct?

Excuse my newbie questions.

Thanks and regards.

[DougL]

Looking at the app note, the reason for the transistor and zener was for short circuit protection.
If the output of the regulator was held at zero volts, the pass transistor limited the current and the voltage across the regulator.
During normal operation, the additional parts do not really do much.

From the example you gave, does it mean that for such a small drop across the transistor (10V approx.) I could feed the unregulated supply directly to the input of the regulator? If so there would be 280V - 185V = 95V across the regulator, which is well within the 125V maximum spec - correct?

That's true. At this point its your choice whether to keep the short circuit protection.

Doug

[sonata149]

Thanks DougL for your explanation.

The most important thing I learned from this is that

a) For a series pass transistor in a voltage rectifier, Ic = beta (hfe) * Ib. In choosing the biasing resistor for the base one divides Ic (consisting of the the regulator and load currents) by beta.

b) The voltage specified for the regulator is the difference between the input and output pins and not the voltage seen at the regulator input. This means that, for example, in order to obtain a 24V reg. dc from a 317 one can feed it with say 50V raw dc, the difference being 26V, which falls within the 317 spec. which is around 30V-35V.

I am just thinking aloud. Hope I am correct.

Thanks and regards,
Joe A