Inclusion of a power resistor. - Page 1
Forum section: electronics / beginners

author: davelectronic
author_rank: Frequent Contributor
timestamp: December 13, 2016, 06:39:27 pm
content: I know this subject comes up a lot, but i can't find the answer. Being linear regulators are similar, but may have slightly different characteristics, I'm unsure of the power resistor.In the 7812 regulator circuit with a single series pass transistor, a 1R 7 watt power resistor is used. After searching a lot and reading lots of forum threads, one from here as well. With the LM317 no power resistor is needed, I'm wandering why this is ? I know with multiple series pass transistor in parallel it balances current sharing. But what purpose does it serve, firstly in thr 7812 regulator circuit, and why is it not needed with the LM317 single series pass transistor circuit ? I do understand the pitfalls of current limiting and over voltage protection. But im at a loss as to why the LM317 circuit doesn't need a power resistor. Any explanation really appreciated.The two circuits below, the LM317 i found on a thread here in the forum. I really want to know why it doesn't need a power resistor.

author: davelectronic
author_rank: Frequent Contributor
timestamp: December 13, 2016, 06:40:59 pm
content: The second schematic of the boost regulator circuit with power resistor.

author: Kleinstein
author_rank: Super Contributor
timestamp: December 13, 2016, 06:48:01 pm
content: The resistor sets how much of the current is flowing through the regulator only. With an 1 Ohms resistor, something like the first 600 mA flows through the resistor and the regulator only. Thus would give a power in the 0.4 W range. So a little larger resistor (e.g. 1-2 W) would be nice, but no need for a really high power one. Using a 7812 or lm317 does not make a big difference, except for the output voltage of cause.Using more current through the regulator might be a slight advantage, as the thermal protection of the regulator might engage before the external PNP is melting.

author: davelectronic
author_rank: Frequent Contributor
timestamp: December 13, 2016, 07:12:11 pm
content: Thanks for helping, so do you need a ceramic power resistor for the LM317 circuit ? I did read that in that schematic the LM317 does the first 60 mA of the current carrying, and above that the transistor does the work load. Does that sound accurate ? I would want to keep the regulator LM317 doing no more than about the first 100mA or so. I've seen a similar circuit where a 22R resistor is used on the input to the LM317 circuit diagram.So would i need a 1R power resistor in the LM317 circuit, or is a 10R 2 watt resistor ok for the first 60mA of current for the regulator. I would have thought the schematic where a 22R resistor is used, that would limit the regulators work to less than 60mA ? Thanks again.

author: davelectronic
author_rank: Frequent Contributor
timestamp: December 13, 2016, 07:58:43 pm
content: I guess no power resistor (ceramic 1R) would lower the LM317 workload, but still using that 10 ohm 2 watt resistor on the input.

author: Zero999
author_rank: Super Contributor
timestamp: December 13, 2016, 08:40:55 pm
content: The resistor doesn't dissipate much power. P = I2R so if the base-emitter voltage is 1V at high currents and the resistor is 1 Ohm, it will dissipate 1W at most.The circuit works by sensing the current going into the LM317 or LM78xx and turning on a transistor to bypass the regulator IC, when the current exceeds a certain threshold. Looking at the internal schematic of the LM317 with the external pass transistor may make it easier to understand. Assume the value of the unlabelled resistor is 1 Ohm and the transistor and the external PNP transsitor starts to turn on when the base emitter voltage exceeds 0.6V.The current at which the bypass transistor is required to be switched on, depends on the power dissipated by the LM317. When the difference between the input and output voltage is under 1V or so, the external transistor can be turned on when the current exceeds 600mA or so. If the input-output differential is higher than 15V, the LM317 will lower the maximum output current, so 100mA is a more reasonable threshold for activating the bypass transistor.

author: davelectronic
author_rank: Frequent Contributor
timestamp: December 13, 2016, 09:48:56 pm
content: Its the maths i struggle to get my head round. I'm trying to work out how to adjust the formula in the relationship between the transistor base resistor, and the 1 ohm ceramic power resistor. In all honesty my maths is not a strong point.I can work out from some formulas and equations, but others like this have me stumped. I'd like to be able to understand how to arrive at a resistor value to increase or decrease the current the workload the LM317 or 7812 does. And understand how i got at that figure or value. Don't know if you could help me with that ? Sure would be handy, rather than breadboard and multimeter to find the current values measuring the resistor in series to find the answer.So its me kind of not understanding what way to go with the transistor base resistor, to turn the transistor on early at low current levels. Say about 100mA then the transistor takes up above that level to deliver the current. But I've no idea mathematicaly how to calculate that resistor value. I could do it on a breadbreadboard and use a multimeter. But it would be cool to know how to work it out on paper. Thank you for your help, that's to all.

author: davelectronic
author_rank: Frequent Contributor
timestamp: December 13, 2016, 09:59:42 pm
content: Its also confusing because there are scematics like below, there is no 1 ohm ceramic resistor, and again I've no idea how the transistor knows when to turn on in those circuits, let alone what ratio currents are between regulator and transistor.

author: davelectronic
author_rank: Frequent Contributor
timestamp: December 13, 2016, 10:01:16 pm
content: Sorry same scematic above. Othe non ceramic power resistor schematic below.

author: Zero999
author_rank: Super Contributor
timestamp: December 13, 2016, 10:57:37 pm
content: Quote from: davelectronic on December 13, 2016, 09:59:42 pmThe transistor simply turns on when the voltage on the base drops to 0.6V below the emitter. The voltage across the emitter resistor can be calculated using Ohm's law. V = IR. Rearranging the formula to make R the subject: R = V/I. The calculation for the resistor value is simply R = 0.6/I, where I is the desired current in Amps and R the resistor value in OhmsNote that the base-emitter voltage will be much higher than 0.6V, at high currents (the transistor data sheet will give more information) so the actual current flowing through the regulator, at full load, will be higher than that required to turn the transistor on. If the transistor turns on at 0.6V but the base is 1.2V below the emitter at full load, then the regulator should have a good enough heat sink and be capable of passing double the transistor's switch on current.The final circuit you posted uses two PNP transistors, a configuration known as the Darlington pair (look it up using a search engine for more information). Because the base-emitter junctions in a Darlington pair are connected in series, it takes double the voltage to turn it on, so the base voltage needs to be 1.2V below the emitter to turn it on.Quote from: davelectronic on December 13, 2016, 09:48:56 pmThe base resistor is not essential. It can be added to ensure that the maximum base current is not exceeded when a very large current is draw, for a short period of time. It will also reduce the gain and make oscillation less likely. The circuit will work without it, but a good rule of thumb is 10 times the value of the emitter resistor. Because only a small current flows through the base of the transistor, there will only be a small voltage drop across the base resistor, so it can be ignored, for the purposes of the calculation.

author: davelectronic
author_rank: Frequent Contributor
timestamp: December 13, 2016, 11:51:29 pm
content: Thank you for taking the time to explain this for me. So 0.6 volts / the current i want flowing through the regulator. Would that be like this 0.6/0.100mA = 6 ohms for the ceramic power resistor, and 10 ohms for the transistor base resistor. Not sure that's right.

author: davelectronic
author_rank: Frequent Contributor
timestamp: December 13, 2016, 11:57:40 pm
content: So its the ceramic power resistor that controls current through the regulator, and the 10 ohm transistor base resistor just limits base current. So i need a larger resistor than 1 ohm to get the regulators workload down to 100mA. Would a 6 ohm 10 watt ceramic power resistor be the right choice for this current ?

author: davelectronic
author_rank: Frequent Contributor
timestamp: December 14, 2016, 12:11:08 am
content: I'm sure I've got it wrong, a 6 ohm emitter resistor will burn up i think, why do different scematics have different transistor base resistor values. For these circuits I've seen 10 ohms 2.2 ohms 22 ohms 3 ohms. I've built both the single transistor boost circuit, at 40 watts load the emitter resistor is hot, but with in safe limits. As are the other components. I've built a 4 x transistor circuit using 0.1R emitter resistors and 100R base resistor, both circuits worked well. I just wanted to find out which value resistor controls the regulators workload.If i increase the emitter resistor to say 6 ohms ( not sure its the right thing to do) it will probably have to be 20 watts as not to burn up.

author: Audioguru
author_rank: Super Contributor
timestamp: December 14, 2016, 12:12:57 am
content: An LM317 or 78xx regulator limits the maximum current to about 2.2A. When a current-boosting transistor is used then it does not limit the current and damage will occur if the current gets too high.Another transistor can add current-limiting.0.6V is the voltage when the transistor begins to turn on. The voltage across the current-sensing resistor might reach 0.8V when the current in the regulator will be 133mA.Limiting the current in the regulator to only 100mA limits the base current of the transistor then its max output current might not be high enough. If the hFE of the transistor is only 25 times then with a base current of 100mA the transistor current is only 2.5A which the regulator can almost do all by itself.The power in the 6 ohms resistor will be only (0.1A squared) x 6= 0.06W but 133mA produces heating of 0.11W. Any little resistor (not a power resistor) will be fine.

author: davelectronic
author_rank: Frequent Contributor
timestamp: December 14, 2016, 12:36:04 am
content: Quote from: Audioguru on December 14, 2016, 12:12:57 amThe resistor i was going to have at 6 ohms was the emitter ceramic power resistor, keeping the transistor base resistor as 10 ohms 0.5 watts. So if i have tried a single transistor boost regulator circuit with known values of 10 ohm 0.5w transistor base resistor, and emitter resistor at 1 ohm 7 watts. At a 40 watt load the 1 ohm power resistor was hot, but in safe limits. Where would a 6 ohm resistor be with the same load on the output. Please don't think I'm disagreeing, I'm trying to understand it. So would a 6 ohm emitter resistor, and a 10 ohm base resistor be ok for 100mA through the regulator. And the rest of the work done by the transistor. If I've got that right, then i can go down to say 3 ohms as an emitter resistor, and the regulator do more workload.In the write up on the boost regulator circuit it says 3 watts ceramic power resistor for currents up to 3 Amps, and 7 watts ceramic power resistor for currents up to 5 Amps. The base resistor is 10 ohms 0.5 watts. Its the ceramic power resistor i planned to alter to make the regulator do less work. Is that correct ?

author: Ian.M
author_rank: Super Contributor
timestamp: December 14, 2016, 12:51:33 am
content: Quote from: davelectronic on December 13, 2016, 11:57:40 pmNo. I reconfigured my LTSPICE sim (I linked to your previous topic) to match your PNP + base resistor schematic. Assuming a max total output current of 5A, try 15R 1/4W. It does however depend on the gain of the pass transistor @5A Ic and the value of its base resistor as high current PNP transistors usually have fairly low gain so the base current is a significant contribution the the regulator IC input current.Incidentally, using a PNP Darlington pair pass transistor is really undesirable because it doubles the Vbe drop and thus the headroom required and more than doubles the dissipation in the regulator feed resistor. If you use aSziklai pairwith a PNP driver and a NPN power transistor, it gives Darlington-like gain and performance with half the Vbe drop.If you are going to use a Darlington or Sziklai pair, its a good idea to include an emitter resistor to better control the ratio of the current through the regulator and the pass transistor. Done right this can also protect the transistor using the regulator's short-circuit protection. It also slugs the overall gain without increasing the base current which goes a long way towards preventing oscillation with difficult loads. As the emitter resistor has to pass the bulk of the load current, and increases the overall headroom required, it should be kept as small as possible while still doing its job - start with 0.1 Ohms.

author: davelectronic
author_rank: Frequent Contributor
timestamp: December 14, 2016, 01:14:09 am
content: Quote from: Ian.M on December 14, 2016, 12:51:33 amThanks for replying, my confusion is in a schematic below, i thought i could follow that and remove the ceramic power resistor ? But I'm not sure. Do i need a ceramic power resistor in the emitter line ? I think i do, but to obtain lets say half the 600mA load on the regulator, then I've no idea of its value for this. Then i know i need a base resistor to limit current at the transistors base. As these make up a divider ive no idea no its value either. To use a 10 ohm resistor as per picture below would be ok, but there is no emitter resistor in that circuit.In another forum a similar discussion clocked up 14 plus pages....I ended up using the known values in the 7812 regulator and 6 x TIP2955 psu schematic. 6 x 0.1R emitter resistors, 100R 0.5 watt input to the regulator.Back then i was trying to work out the resistance values, and possibly altering them. But other transistors where discussed there. Yes a high powered darlington that did oscillate. So i won't use one again for now. So, simply, do i need an emitter resistor in the schematic below ? And if not is a 2 watt 10R resistor ok for the base value ? Thanks for all the help.

author: davelectronic
author_rank: Frequent Contributor
timestamp: December 14, 2016, 01:23:06 am
content: Yes Ian I'm certain i need an emitter resistor, so starting at 0.1 ohms say 10 watts ( already have them) then to control the regulators workload i need to play a bit with the 0.5 watt bsse resistor, and or both. So try 15 ohms 0.5 watt base resistor, and 0.1 Ohms 10 watt emitter resistor. For a start to obtain a base line. Think that's right...

author: Zero999
author_rank: Super Contributor
timestamp: December 14, 2016, 06:52:14 pm
content: You already do have an emitter resistor. It's R3 in the previously posted schematic. It doesn't need to be a high power resistor. P = I2R, where I is the current in Amps and R is the resistance in Ohms.

author: davelectronic
author_rank: Frequent Contributor
timestamp: December 14, 2016, 10:26:03 pm
content: The two resistors i plan to use are an emitter resistor, and a base limiting resistor. The emitter resistor will sence when to turn on the transistor, I'm sticking with 1 ohm 10 watts. The base resistor I'm going to play with that value to see what current under load conditions i get. And select a value that gives me 300mA workload for the regulator, and above this the transistor takes up the current needed for the load. I might play with lower regulator currents say 100mA see how it behaves. Save soldering resistors in and out i will mount a couple of pcb pins so i can swap resistors in and out. Also measure current the regulator is doing. Then solder final resistor to pcb pins. Because there are differences between load and no load, i think its best to play with resistor values.

author: davelectronic
author_rank: Frequent Contributor
timestamp: December 14, 2016, 10:30:49 pm
content: I think the power emitter resistor is preference, but I'm sure this simple circuit works better with both resistors.

author: Zero999
author_rank: Super Contributor
timestamp: December 14, 2016, 10:49:31 pm
content: Quote from: davelectronic on December 14, 2016, 10:26:03 pmGo with that, by all means, but the power rating is overkill. The resistor will dissipate 1.2 Watts at most. A 2W metal film resistor is more than enough.QuoteA 1R emitter resistor will make the transistor turn on when the current gets to about 600mA. The value of the base resistor will make very little difference to this, as long as it isn't excessively high.What are your requirements? How much current do you need to draw? What's are the input and output voltages?

author: davelectronic
author_rank: Frequent Contributor
timestamp: December 15, 2016, 01:20:35 am
content: Linear transformer with a secodary of 15 volts so about 21 volts after rectifier and filter. Dc output of about 13.80 volts at 3 Amps continuous 5 Amps 50% duty cycle. Which resistor value single or combination would you suggest so the regulator only does about 300mA of the total full load of 5 Amps ? I was going to keep the 1 ohm emitter resistor, and increase the base resistor until i reach 300mA. Or is that the wrong way to go ? With only two resistors, one or both must be able to limit the regulators workload to about 300mA. I was going to start at the base resistor and increase its value to 22 ohms. Then upwards if its still drawing more than 300mA.

author: Ian.M
author_rank: Super Contributor
timestamp: December 15, 2016, 01:49:55 am
content: That almost certainly wont work. The transformer and reservoir cap would need to be massively over-sized to deliver 5A DC after the rectifier without the troughs of the ripple on the output voltage dropping below 16.8V (as boosted 78xx or LM317 regulators generally require >3V headroom).

author: davelectronic
author_rank: Frequent Contributor
timestamp: December 15, 2016, 02:29:54 am
content: Well it does, in an Alti i restored after i found the internals wrecked. But that one is doing exactly as per the circuit resistors, that heatsink can handle 600mA for the regulator, at the back of the unit.The smaller one has a much smaller TO3 heatsink, so i will reduce thst 600mA down to about 300mA.

author: davelectronic
author_rank: Frequent Contributor
timestamp: December 15, 2016, 02:39:55 am
content: The psu i plan to repair, with the LM317K doing 300mA of the workload. Note the smaller heatsink on the back.

author: davelectronic
author_rank: Frequent Contributor
timestamp: December 15, 2016, 03:03:45 am
content: So if i want the LM317K to do 300mA of the psu workload, what resistor value would be appropriate. In heros schematic it says a single 10 ohm resistor input to the regulator and transistor and emitter base junction would allow the regulator to do 66mA after that the transistor does the work. Its posted on page 1.Is it really that easy, and if 10 ohms restricts the regulators workload to 600mA in one schematic, how can the same single value resistor of 10 ohms restrict it to 66mA in another schematic, but less the 1 ohm emitter bias resistor ? All i know is it makes no sense.

author: davelectronic
author_rank: Frequent Contributor
timestamp: December 15, 2016, 11:25:36 am
content: I'm going to try ditching the 1 ohm ceramic power resistor, and on the input to the LM317K use a 10 ohm 2 watt resistor. See how that goes, the schematic on page i says anything above 66mA is handled by the transistor. I would have gone for 6 ohms 2 watts, but none to hand. I know there's little output protection, so i will add a simple crowbar to protect the output.

author: davelectronic
author_rank: Frequent Contributor
timestamp: December 15, 2016, 06:23:07 pm
content: Thanks for the help from everyone I'm going with the schematic below, but ordered some 4.7R 3 watt resistors. I was thinking a bit more current through the regulator than 66mA as it would be with 10 ohms. Hopefully it will work ok.

author: Zero999
author_rank: Super Contributor
timestamp: December 15, 2016, 11:02:13 pm
content: Quote from: davelectronic on December 15, 2016, 02:29:54 amHow much smaller is the heat sink?The one in the picture should be enough for the full output current.One thing to bear in mind is the output ripple. If the voltage across the regulator IC is too low, then the output may measure close to the correct voltage but it might have ripple on it, which can only be seen with an oscilloscope. The ripple will only be there at high load currents and when the mains voltage is on the low side.

author: davelectronic
author_rank: Frequent Contributor
timestamp: December 16, 2016, 12:26:15 am
content: The big heatsink on top has the MJ2955 on it, the rear heatsink is, i would say a quarter the size of the top heatsink. This circuit from what i can make out, and varieties of it use different resistor combination. You have some schematic with single base emitter resistor, and some with two resistors, the base and emitter resistor. Values vary from 0.3 to 0.7 to 1 ohm for the emitter resistor, and the base resistors range from 2.2 ohm up to 22 ohm and a far few values in between.I figured leave out the two resistor schematic and use a single resistor. 3 watts was the cheapest i could find. A 3 watt 4.7 ohm resistor would give the regulator about 127mA if the maths is correct. It seemed a good compromise between 2.2 - 22 ohms and every value in between. If i use that 1 ohm emitter resistor then I'm still passing 600mA through the regulator.

author: davelectronic
author_rank: Frequent Contributor
timestamp: December 16, 2016, 01:21:34 am
content: Im not a million miles away from this Fairchild schematic below. I think 3 watt 4.7 ohms for an LM317K and MJ2955 will be fine. I've trawled through many forum threads. It should be fine.

author: Zero999
author_rank: Super Contributor
timestamp: December 16, 2016, 09:09:51 am
content: That should be fine but I advise moving the 0.33µF capacitor to the input side of R1. Otherwise, when the power is first applied, Q1 will turn hard on, passing the full input voltage directly to the output.

author: davelectronic
author_rank: Frequent Contributor
timestamp: December 17, 2016, 10:13:13 am
content: Thanks again all.