Linear regulator boost with NPN transistor - Page 1
Forum section: electronics / beginners

author: sureshot
author_rank: Regular Contributor
timestamp: July 09, 2018, 08:42:23 am
content: I was looking at this simple circuit, what are the negatives of using an NPN transistor to boost current of a linear voltage regulator ? I think there would be a forward volts drop. Would a diode in the ground path resolve that problem, i don't know the name for it but it increases the output by 0.7 Volts per diode. And finally what are the implications of there being no current limiting base resistor. Thanks for reading and any advise appreciated.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 09, 2018, 10:34:44 am
content: Yah, you've hit all the main points.Dropout voltage (minimum input = output + droppout) is ~0.7V higher.Regulation is worse by the "softness" of the diode curve. It's not inaccurate by a diode drop, because of the diode compensating it. But the diode's current is ~constant (whatever the regulator GND current is), while the transistor's Vbe depends on current draw. This, plus transistor resistance, gives maybe 100mV of variation. Worse than a 7812 -- but not bad for most purposes.No current limit, or rather it's hFE limited, which is a bad idea all the same. If you need that capability, expect to add more components to implement it. Note that it's not as simple as pulling down the base with a current sense resistor and transistor, because the 7812's current is what has to be sunk.Nor can you quite pull it off with a LM317, which has 1.25V output, more than the Vbe drop so you can't quite completely turn it off by pulling ADJ down to GND (with a few transistors wrapping around to do that).Tim

author: Zero999
author_rank: Super Contributor
timestamp: July 09, 2018, 10:43:42 am
content: Yes, the base emitter voltage increases the drop-out voltage and worsens the regulation.No base resistor is required, because the base current is self-limiting. When the transistor turns on, the voltage at the emitter rises, reducing the potential difference between the base-emitter junction, therefore the base current, until an equilibrium is reached. This is an example of negative feedback. Look up emitter follower.An NPN transistor can be used, with a PNP transistor, as per the schematic I posted in the other thread.https://www.eevblog.com/forum/beginners/voltage-multiplier-120443/msg1648847/#msg1648847

author: sureshot
author_rank: Regular Contributor
timestamp: July 09, 2018, 10:51:46 am
content: Thank you for the help, i think i I've got most of what you've explained there. I did think about the LM317 in that configuration. Just a question about this circuit using the 7812 voltage regulator.What would be the implications on the transistor at say 3 Amps, would it thermal runaway ? And where would you see the voltage drop under that load ? With a zener diode and fuse for protection, would it be capable of supplying a cb radio with an RX current of 250mA, and a TX current of 1.2 Amps ? I know the 7812 is almost capable of that. But thought it might be suitable for that application. I do have some thyristors if a crowbar circuit would be better than just the zener and fuse protection. Thanks again for your help.

author: sureshot
author_rank: Regular Contributor
timestamp: July 09, 2018, 10:57:44 am
content: Quote from: Hero999 on July 09, 2018, 10:43:42 amThank you for the schematic, i could put that circuit together easily enough. I just like the idea of a circuit that needs little parts count, works reasonably efficient and gets the intened item powered up. I was just curious of the behaviour of the transistor with no base resistor. And the overall performance of the circuit up to 3 Amps maximum. Thanks again for the help with this.

author: sureshot
author_rank: Regular Contributor
timestamp: July 09, 2018, 11:47:40 am
content: I have a couple of darlington TIP142 transistors. Would the use of one them help with the voltage drop underload ? I'm thinking this as I've used darlington transistors in pnp regulator boost current circuits. And under load they do quite well with the TIP147 I've used before.

author: sureshot
author_rank: Regular Contributor
timestamp: July 09, 2018, 01:15:51 pm
content: Just on a final question, might be obvious to most, but i can't see why its there in the image below. Is the 1 ohm 5 watt resistor beneficial, believe its the same circuit but drawn differently. Does that 1 ohm 5 watt resistor limit the output current in anyway ? Thanks for reading.

author: Zero999
author_rank: Super Contributor
timestamp: July 09, 2018, 01:29:39 pm
content: Quote from: sureshot on July 09, 2018, 11:47:40 amA Darlington transistor would be fine. If VDoublerand VRectifierwere connected together, it wouldn't make any difference to the drop-out voltage, because Tr1 can pull its base nearly all the way up to the supply voltage and the LM7805 will drop more voltage, than the Darlington. If If VDoubleris much higher than VRectifierand the latter is providing most of the current, then the drop-out voltage will be slightly higher, since the minimum voltage drop on a Darlington pair is 0.6V, even if the base voltage is above the emitter voltage.Quote from: sureshot on July 09, 2018, 01:15:51 pmThe 1R resistor won't do much. The voltage drop across it will be low, since the base current is low and it won't do anything to limit the current.

author: schmitt trigger
author_rank: Super Contributor
timestamp: July 09, 2018, 01:35:04 pm
content: Like other posters have mentioned, I also recommend to use a LM317, because then you can include the booster transistor inside the feedback loop. and thus have a much tighter regulation than your original circuit.

author: David Hess
author_rank: Super Contributor
timestamp: July 09, 2018, 04:23:18 pm
content: Do it the way National showed so that the regulator's current limit can be applied to the pass transistor. If the regulator is mounted to the same heat sink and the thermal resistance is adjusted properly, then its thermal protection will extend to the pass transistor as well.

author: sureshot
author_rank: Regular Contributor
timestamp: July 09, 2018, 06:51:26 pm
content: Thank you again for the help with this. So figure 7.4 is the way to go, if it's an npn transistor as the final power element ? And the pnp transistor drives it. Are R2 and R3 replaced with the 50 ohm and 75 ohm resistors ? And is the diode on the input for an intended voltage drop, as I can't see it's for protection, being it's forward biased. Thanks again for the help, and circuit schematics.

author: David Hess
author_rank: Super Contributor
timestamp: July 09, 2018, 10:43:51 pm
content: Quote from: sureshot on July 09, 2018, 06:51:26 pmR1 and R2 have nothing to do with the 50 and 75 ohm base-emitter shunt resistors. The base-emitter shunt resistors are needed to remove charge from the base-emitter junctions and bypass Icbo leakage and R3 corresponds with the 50 ohm resistor shown and fulfills the same purpose in the power PNP version of the circuit. R1 and R2 control the ratio of current between the output transistor and regulator so that when the regulator goes into current limit, the transistor current is well defined.The diode roughly compensates for the base-emitter voltage drop of the PNP transistor. Without it, the current limit will not be as accurate.

author: sureshot
author_rank: Regular Contributor
timestamp: July 10, 2018, 12:32:09 am
content: I think i get that. Thank you for taking time to explain that, that's where i struggle a bit with the maths. i need to find a decent book for these equations that doesn't go to far down the complex physics route. I will probably give both the first circuit I posted a go, and the national semicoductor schematic you kindly posted. And possibly try these with an LM317 just to see if and adjustment for voltage drop under load is needed. But I would prefer to stick with a 7812 for this first npn transistor current boost circuit.Thank you to all that helped me out, appreciated.

author: sureshot
author_rank: Regular Contributor
timestamp: July 10, 2018, 12:45:06 am
content: Quote from: T3sl4co1l on July 09, 2018, 10:34:44 amI often go back over posts in case I've missed something, and I think I have, form Tim's post to David's post. The mention of the work the regulator does in the national semicoductor schematic. I see it uses resistors to adjust shared parameters between transistor and regulator. Yet the schematic I posted in my opening at the top, what stops the 7812 from not drawing over 1 Amp of load current. Especially closer to 3 Amps, or is it's internal circuitry of the 7812 that current limits the regulator ? Sorry if it's a lot of questions, that one is not immediately obvious to me. Shouldn't guess really. But I think it's the 7812's circuitry, please correct me if I'm wrong with that thought.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 10, 2018, 02:48:36 am
content: Yes, it's only limited by the 7812's internal circuitry. That current is amplified by the following transistor, so the current limit at the output terminal will be vastly higher.Tim

author: David Hess
author_rank: Super Contributor
timestamp: July 10, 2018, 03:07:16 am
content: Quote from: sureshot on July 10, 2018, 12:32:09 amThe diode to ground in the output transistor version is an easy way to compensate for the base-emitter voltage drop but as T3sl4co1l pointed out, the regulation is "softer" but this is not normally an issue with a 12 volt output. The 317 can be used the same way by placing the diode at the bottom of the feedback divider which is essentially included inside the 7812. A resistor can be placed in series with the collector of the NPN transistor on the output to limit the current somewhat.

author: sureshot
author_rank: Regular Contributor
timestamp: July 10, 2018, 09:40:21 am
content: Thanks for your replys, i had a feeling it was the regulator self current limiting itself. Just because there are no resistors. It probably isn't the best of linear voltage regulator circuits, but I'm curious to see how well it does. The diode to lift the output voltage in the ground path i first discovered in a cb power supply some years ago. I had to do a bit of searching to find why it was in this retail power supply. Thanks you again to everyone for all the help with my questions.

author: David Hess
author_rank: Super Contributor
timestamp: July 10, 2018, 10:19:43 am
content: The regulator current limits just fine but the external transistor's current gain is high and by itself poorly controlled. Adding a resistor in series with the collector allows the collector voltage to drop at high currents limiting current; this is not ideal but it is better than nothing. The National circuit instead controls the current from the emitter side which works much better.

author: sureshot
author_rank: Regular Contributor
timestamp: July 10, 2018, 11:35:01 am
content: Yes i can see what you mean, up to 3 Amps current is it necessary to have a resistor in series with the collector ? I'm guessing if it is, something like a 10 ohm 10 watt ceramic resistor. Kind of the same value resistor if it was a pnp emitter series resistor. Not 100% sure that's the right value and power rating for a collector series resistor.

author: not1xor1
author_rank: Frequent Contributor
timestamp: July 10, 2018, 03:09:58 pm
content: Quote from: Hero999 on July 09, 2018, 10:43:42 amHi Hero999I do not understand how that circuit is supposed to work.If the purpose is to get a low dropout (through an higher voltage to the NPN base) then the input Voltage labels have been exchanged.The PNP + the IC should be supplied by the voltage doubler solving the problem of the higher dropout, while the NPN collector (or that of an array of power NPN transistors + ballast resistors) should be connected to the rectified voltage. Then the minimum dropout would be the saturation voltage of the NPN transistor(s + ballast resistors drop).

author: sureshot
author_rank: Regular Contributor
timestamp: July 10, 2018, 06:18:08 pm
content: I think the npn collector should be the regulated input potential, not sure how there's a third terminal. I didn't notice that until you mentioned it. My full rectifier voltage multiplier mock up had the two terminals from the capacitors + - one from each capacitors.

author: Zero999
author_rank: Super Contributor
timestamp: July 10, 2018, 07:06:12 pm
content: Quote from: not1xor1 on July 10, 2018, 03:09:58 pmYou're right. I got them the wrong way round. Here's the correct schematic.Quote from: sureshot on July 10, 2018, 06:18:08 pmThe VDoublerterminal is for the voltage doubler, which can be used to reduce the drop-out voltage to just the tranistor's saturation voltage. If you don't have a voltage doubler, just connect it to the VRectifierand it will work, but with a higher drop-out voltage of about 3V, at full load.

author: sureshot
author_rank: Regular Contributor
timestamp: July 10, 2018, 09:52:57 pm
content: Yes ok I get it, thanks for putting me straight on that, I might, only might, lol have been doing some head scratching somewhere down the the line.