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Series Regulator Schematics

I'm designing a series regulator. I need 3.3V in my circuit, the circuit has 5V and 9V available in it.

What should be the minimum input DC voltage in a DC series regulator in general case?
In this particular case, is 5V enough, or should I supply it with 9V?

BD135 Datasheet


EDIT:

Below is an alternative circuit with opamp. You can also discuss on this one.

enter image description here

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    \$\begingroup\$ Circuitlab might be a great way to make your schematics and simulate them pretty easily online. If you need me to make an example of one of these just let me know and I will draw it up for you! \$\endgroup\$ – Kortuk Jun 14 '12 at 13:40
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    \$\begingroup\$ Your opamp circuit has some gotcha's: the opamp will require a minium supply voltage, and its output's range is limited. Check its datasheet for details. When you change the transistor to a PNP type you will in principle get a much lower minimum drop voltage (but the gotcha's will remain, and the extra gain might add some new problems). \$\endgroup\$ – Wouter van Ooijen Jun 14 '12 at 14:23
  • \$\begingroup\$ @Wouter - the limited output range is IMO not a problem, as it will always be around 4 V. He'll need 6 V in anyway for the transistor. \$\endgroup\$ – stevenvh Jun 15 '12 at 5:03
  • \$\begingroup\$ The opamp needs a series resistor between output and transistor. Otherwise the slightest input variation will cause large base current changes and it will become unstable. \$\endgroup\$ – stevenvh Jun 15 '12 at 5:05
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This is a very basic regulator, and it doesn't regulate well. The output voltage depends on how constant the zener's voltage and the base-emitter voltage. First the transistor:

enter image description here

This graph shows that base-emitter voltage can vary by more than 300 mV between 1 mA and 1 A load. Above 20 mA you have more than the 0.7 V drop you assumed.

Also a zener voltage is not as constant as you may think. If the load decreases zener current will rise, and with it zener voltage.

A better circuit uses feedback from the actual output voltage, so that B-E junction is compensated:

enter image description here

This is for a 13.6 V supply. D1, R1 and R2 determine output voltage:

\$ \mathrm{V_{OUT} = \dfrac{R1 + R2}{R2} (V_Z + 0.7 V)}\$

So this also depends on a B-E junction voltage, but the current variation in Q1 will be far less than in Q2.

edit
A few people commented on this schematic. I just wanted to show how you can create a better (though far from perfect yet) regulator if you take your feedback from the output voltage. I know you need other components for 3.3 V out. You can replace D1 by a 1N4148, or delete it altogether. You'll have 1.4 V resp. 0.7 V at the point marked "Feedback". Choose R1 and R2 accordingly.

Now your actual question. Let's start by decreasing R1 to 100 \$\Omega\$, 10 k\$\Omega\$ is too high. The datasheet specifies \$H_{FE}\$ for a minimum \$V_{CE}\$ of 2 V, so we'll need at least 5.3 V in if we want to rely on that \$H_{FE}\$. Then the current through R1 is 14 mA (with the 10 k\$\Omega\$ it would only have been 140 \$\mu\$A).
If 10 mA of that goes to the base then we can supply 250 mA to the load. That's not much, but like most power transistors the BD135 has a low \$H_{FE}\$, 25 typical at 500 mA collector current. Higher input voltage will give you a higher base + zener current.

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    \$\begingroup\$ OP wants 3.3V output. This looks more like 13.7V! Also, lose 1 Lathrop for the 4-way junction :) \$\endgroup\$ – MikeJ-UK Jun 14 '12 at 11:51
  • \$\begingroup\$ @Mike - It's a schematic I copied from somewhere else, it's about the working principle, but I'll clarify in my answer. Thanks for the feedback. \$\endgroup\$ – stevenvh Jun 14 '12 at 11:53
  • \$\begingroup\$ I do not see how 6.2V zener in your diagram is able to match 3.3v requirement without divider. I see a divider, but it seems to be on wrong side of comparator transistor. \$\endgroup\$ – user924 Jun 14 '12 at 12:18
  • \$\begingroup\$ @Rocket - use a 1N4148 instead of the zener, and the "feedback" point is at 1.4 V. Or delete D1 altogether and you have 0.7 V. \$\endgroup\$ – stevenvh Jun 14 '12 at 12:25
  • \$\begingroup\$ Can I make \$ R_1 = 0 \Omega \$ (feed back without scaling) in your circuit? \$\endgroup\$ – hkBattousai Jun 14 '12 at 12:29
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Calculate! Your picture indicates you need 1A. Divide that by the minimum Beta of the pass transistor, and you have the minimum base current. Add some minium current through the zener (enough that the voltage across it won't be too low for your taste) and you get the minimum required current through R1. Multiply this by R1 and you get the minimum BC voltage, from which you can calculate the minimum input voltage (you know the base voltage). Sanity check: the CE voltage at 1A must be lower, which will probably be the case for this configuration.

You might think you are done now, and you migt have lowered R1 to accomodate a lower input voltage. But now you must calculate again what the maximum current through R1 can be, using the highest input voltage that will evere arise. Both R1 and your zener must be able to cope with the dissipation caused by this current. And check whether the zender voltage won't be too high for your taste with this current!

Low-drop regulators do not use this configuration, instead they use a PNP pass transistor which can receive its base current from the gound, without the need for the voltage drop that occurs in your configuration due to R1.

If you are doing this for anything but the thrill, I would recommed that you get yourself a one-chip low-drop voltage regulator, either a 3.3V one or rig an LM317.

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