Unsure about a circuit. - Page 1
Forum section: electronics / beginners

author: davelectronic
author_rank: Frequent Contributor
timestamp: November 06, 2016, 11:43:38 am
content: Just found this circuit, but not sure it could deliver the extra current given the transistors location. I know the would be a slight voltage drop across the transistor. Can anyone tell me if this circuit would work ? Thanks for reading, any help appreciated.

author: pyroesp
author_rank: Regular Contributor
timestamp: November 06, 2016, 12:20:56 pm
content: Ve = Vb - 0.6VIc = Ib * hFEIe = Ic + IbYour LM317 sets the Vb voltage.Output current depends on hFE and base current.Don't know the schematic of the LM317 off the top of my head, but it should work.You might want to change the transformer to a 30VAC one. I don't think you'll get 30VDC under full load.

author: davelectronic
author_rank: Frequent Contributor
timestamp: November 06, 2016, 12:38:20 pm
content: Thanks for the help here, I just wandering about the output open emItter and the 220 ohm voltage set resistor doubling up as a based limit resistor to the 2N3055 transistor. My use, if this circuit was viable was 13.80 Volts fixed, and 3 to 5 Amps. But closer to 3 Amps continuous.

author: pyroesp
author_rank: Regular Contributor
timestamp: November 06, 2016, 12:57:56 pm
content: Be aware that there's no constant current circuitry in that schematic.

author: SeanB
author_rank: Super Contributor
timestamp: November 06, 2016, 02:03:18 pm
content: Yes it can deliver the current, at the expense of having some truly horrid voltage regulation, along with the output voltage now being dependant on both the current and the temperature of the output device.However, if you just want something with some regulation, no real current limit other than the transformer or bridge rectifier burning out, and just to emulate a simple lead acid battery as a supply then this will do, just set it to 13V. Just do not try to charge batteries with it, or to power anything that expects a well regulated supply, and beware the leakage current of the transistor can raise the output voltage, with no ability to control it, if your heatsink is poor and the pass transistor gets way too hot.If you use a transformer of 18VAC and 5-10A rating, a 25A bridge rectifier ( you may think it is overkill, but it is cheap, has a metal case to attach to a heatsink to cool it, and has large connector lugs), some bigger capacitors ( 4 2200uF 35V capacitors in parallel at a minimum) and a decent heatsink it will do 13V8 at 3A fine, and higher depending on the transformer.

author: davelectronic
author_rank: Frequent Contributor
timestamp: November 06, 2016, 06:57:23 pm
content: Thanks for your reply and help. Very interesting, what, or how bad a regulation would I be looking at ? Using heavy components is not a problem as I have those already. As for voltage climb, what wold be the expected, given an output set at 13.80 Volts ? Meaning any climb in voltage under lodds up to 3 Amps. Thanks again for everyone' help.

author: davelectronic
author_rank: Frequent Contributor
timestamp: November 06, 2016, 07:16:39 pm
content: In this circuit schematic does the incuded 10 ohm 2 watt resistor improve the circuit ? Or is that resistor of not much benefit at all ? Thanks again for reading and your help.

author: SeanB
author_rank: Super Contributor
timestamp: November 06, 2016, 07:22:46 pm
content: Does help the power transistir leakage, but otherwise only means the regulator supplies the first 600mA of current before being assisted. Will still droop slightly, but still the same regulation figures. You would need to use the datasheet recommended high side boost to improve things, but for an application like you want to use, as a battery replacement, either will work for up to 5A and will show roughly the same voltage drop with load as a real battery.

author: davelectronic
author_rank: Frequent Contributor
timestamp: November 06, 2016, 07:37:24 pm
content: Ok understood, thank you again for your help. And thanks to others that replied, much appreciated.

author: Cliff Matthews
author_rank: Unknown
timestamp: November 06, 2016, 08:27:30 pm
content: Quote from: SeanB on November 06, 2016, 07:22:46 pmWouldn't that be 60ma before the assist? 600ma on 10ohm resistor = 6v (and 3.6 watts..)

author: davelectronic
author_rank: Frequent Contributor
timestamp: November 06, 2016, 09:17:36 pm
content: Quote from: Cliff Matthews on November 06, 2016, 08:27:30 pmSo i guess it acts like a series load resistor ? And possibly burn out ? Maybe..

author: davelectronic
author_rank: Frequent Contributor
timestamp: November 06, 2016, 09:20:17 pm
content: So would i be right in thinking a 5 watt or higher at 10 ohms would be more appropriate ? Not sure.

author: davelectronic
author_rank: Frequent Contributor
timestamp: November 06, 2016, 09:27:23 pm
content: So just thinking about before assistance from the transistor, 0.600 mA at 13.80 Volts would be 8.2 watts, and 0.060 mA would be 0.828 watts, if i have got that right.

author: davelectronic
author_rank: Frequent Contributor
timestamp: November 06, 2016, 09:33:37 pm
content: Would the LM317 be drawing a max current of 0.060 mA and the transistor does all the work above that. I would have thought the LM317 would be doing more work than that.

author: davelectronic
author_rank: Frequent Contributor
timestamp: November 06, 2016, 10:02:54 pm
content: So i have built a pnp pass transistor with 78xx regulators. This gave me fairly good results. This time i was looking at trying the 2N3055 in this circuit with the LM317. But as the circuits are an unknown source i'm not sure about it. I wouldn't know how to achieve feedback from the emitter, and would an added base resistor of 1R be like the 1R 10 ohm ballast resistor i used in the pnp 78xx circuit i built.

author: davelectronic
author_rank: Frequent Contributor
timestamp: November 06, 2016, 10:12:34 pm
content: My humble knowledge says the 2 watt 10 ohm resistor in the second schematic i posted will be ok..

author: Cliff Matthews
author_rank: Unknown
timestamp: November 07, 2016, 12:16:25 am
content: Quote from: davelectronic on November 06, 2016, 09:27:23 pmWhy would that resistor drop 13.8v? (It only takes 0.6v to get the transistor to start conduction..)600ma through a 10 ohm resistor is a high 6 volt Vbe (I think SeanB typo'ed an extra zero). 6v X 0.6A would have been 3.6watts

author: davelectronic
author_rank: Frequent Contributor
timestamp: November 07, 2016, 01:20:39 am
content: Yes my mistake, it won't drop 13.80 Volts across it. I'm going to try and build it with the addition of the 2 watt 10R resistor. I'm interested to see if it can match similar performance to the pnp 78xx psu circuits I've built. I will post back results in a while. Its powering cb radio gear, and the pnp versions with 7812 regulators worked great with 4 x transistors up to 10 Amps. So going to try this.

author: David Hess
author_rank: Super Contributor
timestamp: November 07, 2016, 03:22:55 am
content: Quote from: xavier60 on November 06, 2016, 09:40:46 pmThis will not help at all. The LM317 wants to maintain 1.25 volts between its output pin and adjustment pin no matter where the feedback is taken from. Adding resistance or voltage drop between the output and feedback point makes regulation worse because it alters the 1.25 volt reference voltage which gets multiplied by the resistor divider.The original circuit will deliver regulation within 100s of millivolts as the Transistor Vbe changes which is completely adequate for many applications. No short circuit current protection to the pass transistor is provided however.The PNP version of this circuit using for instance a 2N2955 will deliver much better regulation and is short circuit protected as well.

author: xavier60
author_rank: Super Contributor
timestamp: November 07, 2016, 03:25:42 am
content: Quote from: David Hess on November 07, 2016, 03:22:55 amYes, you are right.

author: David Hess
author_rank: Super Contributor
timestamp: November 07, 2016, 03:56:33 am
content: Quote from: xavier60 on November 07, 2016, 03:25:42 amCheck out page 12 of theLinear Technology LT1084 datasheetfor a discussion of this issue.

author: Zero999
author_rank: Super Contributor
timestamp: November 07, 2016, 10:41:49 am
content: Adding an emitter follower ruins the voltage regulation and isn't the right way to do this.The proper method is to add a PNP pass transistor, which preserves the voltage regulation and increases the current capability.https://www.eevblog.com/forum/beginners/linear-regulated-power-supply-with-lm317t-mosfet-50n06/msg656109/#msg656109

author: davelectronic
author_rank: Frequent Contributor
timestamp: November 07, 2016, 10:44:31 am
content: Thank you for the information, i would add a zener and a fuse to the output. In the second diagram i posted, will the 2 watt 10R resistor be of benefit, or is it pointless in adding this ? Thank you to you all again for your help here with this.

author: davelectronic
author_rank: Frequent Contributor
timestamp: November 07, 2016, 10:47:53 am
content: I have built this with 7812 regulators and it worked very well. I was going to do the same with the LM317. But wanted to try the 2N3055 version of this similar circuit.

author: Zero999
author_rank: Super Contributor
timestamp: November 07, 2016, 12:04:33 pm
content: Quote from: davelectronic on November 07, 2016, 10:44:31 amAdding a resistor between the base and emitter of the emitter follower will only make the regulation worse, rather than better.At low currents, then the emitter follower won't conduct, so the output voltage will be close to the value determined by R1 & R2 in the usual LM317 circuit. At higher currents, the emitter follower will start to conduct and the voltage will drop to between 0.6V and 1.5V (depending on the saturation voltage of the transistor) less than the LM317's output voltage.With no emitter-base resistor, then the emitter follower will always drop near 0.6V, even at low currents.Example:Set the output to 5V on the LM317 circuit.With no base-emitter resistor:Vunloaded = 5 - 0.6 = 4.4VVfullyloaded = 5 - 1.5V = 3.5VAn output change of 0.9V from unloaded to fully loaded. If 5V is desired, then the LM317's output voltage can be set to 5.6V or more to compensate for the emitter follower voltage drop.With a base emitter resistorVunloaded = 5VVloaded = 3.5VAn output voltage drop of 1.5V from unloaded, to fully loaded.In any cause, just use the pass transistor method. The only advantage the emitter follower has is being able to go down to lower output voltages, but this is of limited use if the regulation is poor, especially at lower voltages and there are better ways to do that too.Quote from: davelectronic on November 07, 2016, 10:47:53 amIf you can't find a high power PNP transistor, then it's possible to use a small PNP transistor to drive the NPN transistor.

author: Cliff Matthews
author_rank: Unknown
timestamp: November 07, 2016, 12:55:52 pm
content: You have the 2n3055 so get a $0.25 PNP like the TIP42 you can still use it (eg:http://www.instructables.com/id/5A-Variable-Linear-Power-Supply/may be a quick example). Lot of other examples on Google images too:https://www.google.ca/search?biw=1600&bih=788&tbm=isch&sa=1&q=LM317+TIP42+2N3055.. just pick one from a reputable site with EE user feedback

author: davelectronic
author_rank: Frequent Contributor
timestamp: November 07, 2016, 01:08:39 pm
content: Yes i understand, a voltage drop of 1.5 Volts loaded is a little undesirable, i guess this is the poorer regulation by using the transistor after the regulator. I also understand using a modest pnp transistor to drive an npn transistor before the regulator. I might except the voltage drop under load. If not then use a MJ2955 before the LM317. Most of what i try is from searches of schematics. Thanks again for the help

author: Zero999
author_rank: Super Contributor
timestamp: November 07, 2016, 07:52:18 pm
content: One of the disadvantages of using a PNP transistor to drive an NPN is lack of short circuit protection.Another transistor and current limiting/sensing resistors can be added to give overcurrent protection. The attached schematic will limit the current to about 5A.

author: David Hess
author_rank: Super Contributor
timestamp: November 07, 2016, 09:56:30 pm
content: Here is an alternative way to implement the current limiting from the National 1980 Voltage Regulator Handbook although it increases the dropout voltage. The text also mentions how the regulator's thermal protection can protect the power transistor.

author: Audioguru
author_rank: Super Contributor
timestamp: November 08, 2016, 12:04:16 am
content: A 7812 regulator regulates as good as an LM317. In your circuit the output of either regulator has good voltage regulation but the emitter output of the 2N3055 emitter-follower is not regulated. The PNP transistor added to any voltage regulator allows a properly regulated output voltage.

author: davelectronic
author_rank: Frequent Contributor
timestamp: November 08, 2016, 01:16:32 am
content: Thanks for all your help, much appreciated. I will go with the LM317 and a single pnp for now and see how it goes. I will load it up gradually to see how it performs. Really i only need 3 Amps at around a 50% duty cycle. But a much bigger version might be worth a try some way down the road..

author: David Hess
author_rank: Super Contributor
timestamp: November 08, 2016, 01:23:40 am
content: Quote from: Audioguru on November 08, 2016, 12:04:16 amThe line and load regulation of a 317 is much better than on a 7812. Also, the adjustment node of the 317 which is not available on a 7812 can be bypassed with a capacitor for lower noise.