Request for Dummy Load Project Critique - Page 1
Forum section: electronics / beginners

author: PSanf
author_rank: Contributor
timestamp: April 06, 2016, 01:48:19 pm
content: I've decided to build a Dummy Load based off Dave's design in EEVBlog #102 on YouTube. I've also watched a series of videos by another YouTuber, mjlorton, who did a similar project to make sure I understood how things worked. His project was also based on Dave's design, and his forum has a lot of useful information about similar projects other people have done. I'm going to throw in a temperature controlled fan based off circuits I've found in a couple of places, and modified/combined. My basic idea is to have the fan turn on when the temperature rises above a set point, and then stay on for a set period of time after it's fallen back below the set point. I've included links to all of these different videos below.Would anyone like to critique my circuit? I've included an image below. I haven't determined the final values for all of the components yet, but have the general idea laid out in KiKad.I'd like my final product to handle a load of >2A, but plan on limiting it to 2A by altering the values of the voltage divider resistors to limit the non-inverting input to U4.I'm using LM741 Op Amps instead of one IC with multiple Op Amps just for my own thinking. I may alter this before I build the final version.I have a DC-DC switching (buck) converter on hand that I'm planning on using to power everything. At this time, I think I'll be using 12V for the power supply. I'd eventually like to get a wall-wart to provide power, but will be using my bench top power supply for prototyping.I do have some questions that I haven't been able to find an answer for, and hope people will be friendly enough to offer their insights.Should I add a voltage regulator to the VCC line on the dummy load to make sure it stays stable when the fan turns on? I plan on using a small computer fan that shouldn't pull much current, but don't want my test load to wobble when it starts or stops.Since I'll be using a common ground, I'm concerned about where some of the electricity will be going. Should the 1 Ohm resistor across the test loop connect to the common ground, or only connect to the negative test point, P2? Everything I've seen shows it connecting to the common ground, but that's a question I haven't been able to find an answer for.Should I have a pull-down resistor on the non-inverting input to U4 to make sure off is really off?mjlorton YouTube Playlist:https://www.youtube.com/playlist?list=PLDBuVMDVJaX29FdhwEDVX5TS01ianiKtoTemperature Controlled Fan:http://www.electroschematics.com/4788/temperature-controlled-dc-fan/Dave's EEVBlod #102:Op Amp Timer Video:May 21 UpdateFinally finished my dummy load project! O0Here's an Imgur album of the final product.

author: djQUAN
author_rank: Frequent Contributor
timestamp: April 06, 2016, 03:07:13 pm
content: First thing that pops to mind is that the voltmeter should be in parallel with the input voltage not in parallel with the ammeter.Another is that the fan control can be done with one op amp. or you can use this:http://sound.westhost.com/project42.htm

author: mswhin63
author_rank: Frequent Contributor
timestamp: April 06, 2016, 03:12:37 pm
content: One thing that is missing is the drive Op-Amp should be configured as an integrator RC or C to prevent instant changes from overloading before the current settles.I have been building one myself with 2 MOSFETs with separate drives and I wanted to see the results with and without the Cap and as predicted switching the source from one voltage to another caused a significant surge of current to blow the MOSFET. The other worked fine. I have still more testing before I submit a further video. Currently bit busy with other things before continuing.Mal

author: langwadt
author_rank: Super Contributor
timestamp: April 06, 2016, 04:17:49 pm
content: you'd probably need atleast a small cap between pin 2 and 6 on U2 or it will oscillate.safe operating area for the transistor is barely 20V at 2A

author: mij59
author_rank: Frequent Contributor
timestamp: April 06, 2016, 04:38:30 pm
content: The common mode voltage of the 741 doesn’t go down to zero volt, it will not function below about 2V, use e.g. LM324 or LM358 or some rail to rail type.U1 is not needed, use a simple voltage divider with resistors.Place a resistor of 100 Ohm between the output of U2 and the gate of Q1.Place a resistor between the source of Q1 and the inverting input of U2, try 10k.Place a capacitor as indicated by langwadt, try 1nF.Q2 not needed, should be NPN not PNP.Q3 should be NPN not PNP.U4 is configured as a comparator , will be more like an on/off function.

author: PSanf
author_rank: Contributor
timestamp: April 07, 2016, 01:39:40 am
content: Thanks to everybody for the suggestions. I've made some changes to my design. A new schematic is below. I did read everyone's post, and consider each suggestion. I really appreciate everyone's attention to the details. Further suggestions are welcome.I placed an order with Mouser that should be arriving tomorrow. Once I'm able to start prototyping I'll be able to experiment with the design. I intend to examine the use of pull up/down resistors at some selected places to make op amp output rail one direction or the other as appropriate. I'll be reading this thread again as I'm prototyping to look at all of my options.QuoteCan you elaborate on this?QuoteThat's the intent.

author: Audioguru
author_rank: Super Contributor
timestamp: April 07, 2016, 04:10:10 am
content: Your new schematic still has many errors:

author: mij59
author_rank: Frequent Contributor
timestamp: April 07, 2016, 11:13:10 am
content: Quote from: PSanf on April 07, 2016, 01:39:40 amThe polarity of Q2 is wrong, the transistor will not function.The schematic vaguely suggest that Q2 is an emitter follower, but it the emitter resistor is missing.Why do you need an emitter follower for ?

author: PSanf
author_rank: Contributor
timestamp: April 07, 2016, 12:34:38 pm
content: Quote from: Audioguru on April 07, 2016, 04:10:10 amThanks for pointing those places out so well! I plan on testing everything on a breadboard in the near future.Why attenuate? That's supposed to be a voltage divider, and is something I've seen used successfully elsewhere. It's used to adjust the input voltage to U2, and limit the output. That's done so U1 can be fully open, but still limit the overall current that will flow across Q1.I guess I don't fully understand where the voltmeter should be. Should it be between SW2 and Gnd? If need be, I'll figure that out on the breadboard.Quote from: mij59 on April 07, 2016, 11:13:10 amI'll keep what you're saying in mind, but I believe that what I have will work. The breadboard will decide.The basic idea is that the resistance of TH1 decreases as the temperature increases. That will cause U3 to output a logical low when the temperature increases to the set point determined by RV3. (I may need to put a pull down resistor on that line.) Q2 will then allow current to flow, and that will go to C1, and pin 2 of U4. At that point, the output of U4 will also go low, and that will allow Q3 to let current flow, and turn on the fan. When U3 returns to a high output Q2 will stop letting current flow. When that happens, C1 should have a charge on it that will hold the output of U4 low for a time. The amount of time can be adjusted with RV4. The end result will be that the fan will turn on when the temperature rises above a set point, then stay on until the temperature drops plus some additional time. The reasoning is that I don't want the fan rapidly kicking on and off as the temperature fluctuates around the set point. I want to run the fan until the temperature drops well below the set point.

author: mij59
author_rank: Frequent Contributor
timestamp: April 07, 2016, 12:55:28 pm
content: Quote from: PSanf on April 07, 2016, 12:34:38 pmThat's called hysteresis, a resistor between pin 3 and 6 of U4 is all you need.

author: Audioguru
author_rank: Super Contributor
timestamp: April 07, 2016, 02:59:40 pm
content: Quote from: PSanf on April 07, 2016, 12:34:38 pmBut the input voltage range from the pots is much too wide so that with a 12V power supply the first opamp U1 will not work when its input is below 3V or above 9V anyway. The voltages from the pots should be attenuated to a range that the old 741 opamps can handle. Also, your current sensing resistor R3 is measuring voltages that are out of range for the inputs on the old 741 opamp U2.QuoteIt should be at the output and gnd. The output fuse will also cause a small voltage drop.QuoteYou do not understand transistor basics. For Q2 and Q3 you show PNP transistors but the polarities are wrong for PNP transistors. You do not need Q2 anyway since R5 by itself can charge C1. If you add hysteresis then C1 is also not needed.

author: PSanf
author_rank: Contributor
timestamp: April 09, 2016, 01:32:29 am
content: Quick update for everybody.Using an LM741 op amp to control a BD140 transistor works as intended. I did a small test of U3 and Q2 using a potentiometer to stand in for TH1. I can get the transistor to output a steady and consistent 0 or 1v using the op amp as a comparator. I can now see how this may not meet my needs, however. If the 1v output off the resistor isn't enough to sufficiently charge the capacitor before it goes back to 0v then there may be an issue. I'll just have to see!

author: Audioguru
author_rank: Super Contributor
timestamp: April 09, 2016, 04:04:59 pm
content: Quote from: PSanf on April 09, 2016, 01:32:29 amThere is no way your circuit will work with the BD140 PNP transistors as your schematic shows.

author: PSanf
author_rank: Contributor
timestamp: April 09, 2016, 06:29:25 pm
content: Another quick update! The fan control is done, and working. I did need to switch the second BD140 for an MTP3055.I've uploaded a video of the circuit functioning on my breadboard to YouTube.https://youtu.be/nLGuLAJNGjMI've updated my schematic to reflect what's on my breadboard.

author: Audioguru
author_rank: Super Contributor
timestamp: April 09, 2016, 07:19:33 pm
content: Since you show the pin numbers WRONG on your schematic then maybe the BD140 PNP transistor is acting like a collector to base diode that keeps the capacitor charged and the fan running all the time.If you use a BD139 NPN transistor with correct pins instead then the circuit will do what you want, except there is no resistor to discharge the capacitor.

author: PSanf
author_rank: Contributor
timestamp: April 09, 2016, 08:25:05 pm
content: Quote from: Audioguru on April 09, 2016, 07:19:33 pmI don't have my pins wrong. We're looking at different datasheets.I've switched out my capacitors for 1uF to test that they're discharging properly. Everything works like I want it to. The capacitors take about 45 seconds to discharge when I have the potentiometer on the second op amp set for a long run time on the fan. If I want to run the fan for more than 45 seconds I can increase the capacitors. If I want it to run less then I can adjust the potentiometer. It turns on and off reliably.

author: Dave
author_rank: Super Contributor
timestamp: April 09, 2016, 10:30:07 pm
content: Guys, stop arguing, he clearly knows best and doesn't want any help with his project. Keep your opinions to yourself.The reason why this terrible circuit actually works is because the PNP is operating in reverse active mode. The current gain is awful, but it still works.The capacitor is discharging slowly thanks to the craptastic input bias current. Take a look at the internal diagram of the LM741, it becomes quite clear how it works.

author: PSanf
author_rank: Contributor
timestamp: April 10, 2016, 12:03:56 am
content: Quote from: Dave on April 09, 2016, 10:30:07 pmI've just read an article on Sparkfun that goes over different transistor modes.https://learn.sparkfun.com/tutorials/transistors/operation-modesThank you for posting the internals of the LM741. That will be helpful in the future, as well as now.My transistor does meet the qualifications of working in reverse active mode. I can see why this isn't ideal for the transistor based upon ideal current flow theory, but nonetheless it still works in this instance. Can you please elaborate on why this is bad? Having the current flow across the transistor from the VCC line to the capacitor and op amp was the intention. I'll need to see if I can dig up a PNP to do a comparison. When I think about the construction of a transistor, a sandwich of N and P materials, it doesn't seem like this would be harmful. Current should be able to flow in either direction between the collector and emitter without issue.

author: Audioguru
author_rank: Super Contributor
timestamp: April 10, 2016, 03:03:49 am
content: The pins in the datasheets are confusing because it is difficult on some datasheets to see if they are showing the metal or plastic side of the case. Most manufacturers call the emitter pin #1 except ST micro who calls the base pin #1.All manufacturers agree that the collector is pin 2 which is a P of a PNP when it should be the N of an NPN. If the emitter and collector of an NPN in this circuit are reversed then the transistor will barely work with low gain but here the base and emitter of a PNP might be reversed which is completely different. With the P collector at the positive supply then it is a conducting diode to the N base.

author: PSanf
author_rank: Contributor
timestamp: April 10, 2016, 03:30:40 am
content: Quote from: Audioguru on April 10, 2016, 03:03:49 amThat's good information. It explains some of the behavior I've observed with the transistor.The low gain is something I noted with my multimeter. I can rail U3 to high, but the voltage I'm getting off the transistor pin connected to the capacitors is 1V. Doesn't matter how high the op amp output is. The transistor will giver me either a solid 1V, or 0V. This is putting a limit on how fast the capacitors are charging, so there's actually a little delay between hitting the temperature set point, and turning on the fan. Once the transistor returns to 0V it takes the capacitors about 45 seconds to discharge low enough to turn the fan off. I take it we can now say that the capacitors are discharging thanks to the bias of U4. I'm basically thinking of that as "leakage current" inherent to the IC. Anyway, I need to adjust my schematic again.Update Schematic:

author: Dave
author_rank: Super Contributor
timestamp: April 10, 2016, 03:37:02 am
content: Quote from: PSanf on April 10, 2016, 12:03:56 amWatchthis video.

author: Audioguru
author_rank: Super Contributor
timestamp: April 10, 2016, 04:18:53 pm
content: Why do you have two 1uF capacitors in series? To make a 0.5uF capacitor?The PNP transistor isnotoperating in the "reverse active mode" because it is a PNP when it is supposed to be an NPN. Its P collector pin is connected to the positive supply so its N base will be one diode voltage drop (about 0.7V) lower ALWAYS! But since you probably have the transistor pins mixed up then the opamp feeds the emitter pin which is a P and the base which is an N is on the capacitor. Then the emitter-base junction is reverse-biased and has avalanche breakdown at about 9V (like a 9V zener diode) causing the capacitor to have +1V when the opamp output is high at about +10V (guessing that the supply is about +12V since the schematic does not show it). So the PNP transistor is not amplifying, it is simply a zener diode.When an NPN transistor is connected with its pins correctly then it is an emitter-follower and drives the capacitor quickly with amplified current to about +9.3V when the output of the opamp is high at about +10V.Why are you using a little BD140 POWER transistor anyway? We said earlier that Q2 is not needed. Why not use an ordinary diode instead? The 741 opamp has an output current of typically 25mA and can charge the capacitor fairly quickly.

author: PSanf
author_rank: Contributor
timestamp: April 10, 2016, 07:02:51 pm
content: Quote from: Audioguru on April 10, 2016, 04:18:53 pmI have two capacitors to increase the voltage being supplied to the op amp. I found that I wasn't able to get the second op amp to rail high using only +1V, and adding a second op amp was a quick way to bring that up to what I needed.I've uploaded a new video showing how I have the transistor connected. Your description of the +1V behavior is spot on. You're correct that I have the pins mixed up. I don't believe the issue is PNP over NPN. I should be able to use either, unless I'm missing something fundamental about the applications of each. To my understanding, PNP vs NPN is a matter of whether you drive the base high or low to control the collector to emitter voltage. A NPN will go to saturation mode when you drive the base high, and a PNP will go to to cut off mode when you drive the base high. They're a lot more than little switches. This conversation is proving to be educational. I have a smaller NPN transistor that I plugged into the circuit. I found that it allowed the capacitors to charge more quickly, but I also lost the time delay before the fan stopped running. Since that was just a drop in replacement I'm wanting to play with it some more. I'll need to redo a few of the wires on my breadboard.I'm using the BD140 because that's what I have that I feel comfortable pushing 12V through, and I'm using Q2 to learn how. I have some ideas on how I can eliminate the second op amp, but I'd like to make sure this is working properly, and I understand how it's working, before making major alterations it.https://youtu.be/f8lYc2Nmdts

author: Audioguru
author_rank: Super Contributor
timestamp: April 10, 2016, 10:36:43 pm
content: Quote from: PSanf on April 10, 2016, 07:02:51 pmThat makes no sense.QuoteIt is WRONG to use a PNP like you have.QuoteYou are using the transistor as an emitter follower. You wrongly have a PNP instead of an NPN and your PNP has the wrong polarity for its power supply voltage. Look at my attachment.Quote1) You lost the time delay because the capacitor was charged to about +9.3V which is the emitter of the transistor and when the base was driven low by the opamp then the reverse-biased emitter-base junction of the transistor had avalanche breakdown (like a zener diode) which quickly discharged the capacitor. Use a diode instead of a transistor.2) If the capacitor charges too slowly then its capacitance is too high. Reduce its capacitance so that the opamp can charge it quickly. But then it will discharge too soon so you must replace the lousy old 741 opamp with a modern opamp THAT HAS NO INPUT CURRENT and use a very high value resistor to discharge it in any amount of time that you want.QuoteThe BD140 is a very old POWER transistor and you are using it in a location where there is almost no power being used. It is in a stupid case where even datasheets do not know which pin is which. Use a diode instead.

author: Dave
author_rank: Super Contributor
timestamp: April 11, 2016, 05:19:04 am
content: Quote from: Audioguru on April 10, 2016, 10:36:43 pmHold on, who said he's using it as an emitter follower?Take a look at the thermistor arrangement. If he's using an NTC thermistor (which he likely is, because PTCs usually have very steep characteristics and are not suitable for this sort of application), then the non-inverting input of U3 dips below the inverting, resulting in the output pulling the base of Q2 low. The output current limit of LM741 kicks in to prevent Q2 from spilling the magic. It's a simple common emitter circuit, except his collector and emitter are flipped, resulting in the transistor operating in reverse active mode.The whole circuit is flawed, though. A single resistor is needed on U3 to give it some hysteresis and you can eliminate U4 and all of the capacitor charging shenanigans.U1 is redundant as well, merely adding an error to the current setting. It should be eliminated from the circuit.But most importantly, the ancient dinosaur that is LM741 should only be written in the history books. The design dates back to 1968. It's almosthalf a centuryold, for crying out loud. The specs are absolutely terrible and there is no damn reason why anyone should put it into their design.Quote from: Audioguru on April 10, 2016, 10:36:43 pmYup, absolutely zero freaking sense.

author: Audioguru
author_rank: Super Contributor
timestamp: April 11, 2016, 02:47:04 pm
content: I give up! The latest schematic is drawn WRONG with the transistor symbol upside down and the pin numbers are also WRONG. All datasheets agree that pin 2 is the collector but the schematic shows it as the emitter even though there is a faint little "C" there.I never looked at the logic of using an emitter-follower as shown on the schematic to turn on the fan when the temperature gets COLD.

author: Dave
author_rank: Super Contributor
timestamp: April 12, 2016, 12:22:14 am
content: Quote from: Audioguru on April 11, 2016, 02:47:04 pmI don't know which circuit you're looking at, the circuit that he pasted doesn't have an emitter follower anywhere in sight.Quote from: Audioguru on April 11, 2016, 02:47:04 pmNo doubt about that.

author: PSanf
author_rank: Contributor
timestamp: April 12, 2016, 01:33:33 am
content: OK, I'm redesigning things, and trying one op amp with hysteresis.I've redone what I have on my bread board, and now I'm trying to figure out how to get it to work the way I'd like.I'm seeing that I can vary the output of RV1 between GND and VCC. The output of RV2 will go from 0v to slightly less than 2.5v then back down to 0v as I dial the potentiometer. Not really sure what's up with that. By playing with my photometers I can get the op amp to rail in either director, but I'm not sure which op amp is controlling which part of my design. Is RV1 the temperature set point, my low threshold, or is RV2?

author: Audioguru
author_rank: Super Contributor
timestamp: April 12, 2016, 02:45:02 am
content: I spotted a few errors and fixed them:

author: PSanf
author_rank: Contributor
timestamp: April 12, 2016, 12:41:39 pm
content: Quote from: Audioguru on April 12, 2016, 02:45:02 amWhy 200k for the pot?

author: Dave
author_rank: Super Contributor
timestamp: April 12, 2016, 01:15:38 pm
content: You need a higher resistance to nudge the non-inverting output just a bit, not a third of the supply voltage. Also add a 47k resistor in series with RV2, so you don't short the input to the output, should you turn the dial to the end.The temperature setpoints will be set by both trimmers. RV1 sets the temperature somewhere between the two temperature switching points (more or less), RV2 sets the hysteresis. For example: If you set the trimmers so that the fan would turn on at 60°C and turn back off when the temperature dropped back to 50°C, and you disconnected RV2, you would find that the circuit now toggles somewhere around 55°C.The settings of the trimmers will be trial and error if you tried to set the exact temperatures. Perhaps you could temporarily replace the thermistor with a decade resistor and simulate the resistance at various temperatures.

author: PSanf
author_rank: Contributor
timestamp: April 12, 2016, 11:10:49 pm
content: Quote from: Dave on April 12, 2016, 01:15:38 pmOK. Don't have a 200K on hand. I have some 5Ks, 10Ks, and 50Ks. Can you point me to a resource that goes over the math being used? How about a "non-technical" article that goes over the idea? Also, can you check to make sure I'm understanding this? RV1 will be the main set point, and RV2 basically pushes that value in either direction to create the upper and lower thresholds (which is why I saw the bell curve voltage last time)?

author: Audioguru
author_rank: Super Contributor
timestamp: April 13, 2016, 12:25:52 am
content: I used simple Ohm's Law and a 50k resistor for RV2:

author: PSanf
author_rank: Contributor
timestamp: April 13, 2016, 02:05:03 am
content: Quote from: Audioguru on April 13, 2016, 12:25:52 amI've put that together on my breadboard, and it works great. Does what it's advertised to do, and probably what I'm going to use.For the sake of the conversation, though, it doesn't do what I'm wanting. I was hoping to create a circuit that would stay on for a set period of time (in second) after the temperature drops back below the "low" threshold. This will keep it running until the temperature falls back below the low threshold, but that's also when the fan will stop. How would I go about making sure the fan keeps running for a period after that?

author: Audioguru
author_rank: Super Contributor
timestamp: April 13, 2016, 02:57:24 am
content: You would need to add a timer somewhere in the circuit.

author: ludzinc
author_rank: Unknown
timestamp: April 13, 2016, 04:25:35 am
content: Quote from: PSanf on April 13, 2016, 02:05:03 amI'm thinking you are failing to grasp the fact that the hysteresis will give you what you want.For arguments sake, lets assume you want your fan to turn off when the temperature is below 30 degrees. With a plain comparator you'd need a time delay as the fan would switch on / off rapidly as the temperature varied from 29.9 to 30.1 degrees etc.With Hysteresis, you could set the on temp to be 30 degrees, but the off temp to be 28 degrees, and the fan will switch on and off infrequently.

author: PSanf
author_rank: Contributor
timestamp: April 13, 2016, 12:35:13 pm
content: Quote from: ludzinc on April 13, 2016, 04:25:35 amHysteresis provides an effective control method, that in a lot of ways is better (and what I'll be using), but not what I was originally shooting for.My idea was that the fan would kick on at (say) 30 degrees, and when the temperature cooled to 29.9 the fan would keep running for a period to push it lower, say down to 25.After other options where raised, shortcomings pointed out, and the circuit analysed with a multimeter, it's clear that it was going to have some long term issues down the road. I didn't have issues with the fan kicking on and off frequently, but found that I wasn't able to accurately control the period of time the fan wold stay on.Somehow, my original design managed to function nearly as intended. Now I'm curious to find out what the correct circuit would have been.

author: Audioguru
author_rank: Super Contributor
timestamp: April 13, 2016, 01:09:00 pm
content: Simply reduce the resistance of RV2 for more hysteresis which will allow the fan to stay turned on longer. But the high temperature threshold will then increase so RV1 will need resetting.If you have too much hysteresis then it will switch only one time maybe when the power supply is turned on but will never switch back.

author: PSanf
author_rank: Contributor
timestamp: April 13, 2016, 02:43:42 pm
content: Quote from: Audioguru on April 13, 2016, 01:09:00 pmBasically, going back to what I said in an earlier post. RV1 controls the temperature set point, and RV2 would "split" that set point into a higher and lower threshold while pushing them away from the set point.