Transistors are CURRENT BASED devices, each with their OWN individual current gain figure. If the current gains (and tempcos and other things) are not exactly matched, having "the same matched voltage on the base" does NOT mean they will be sharing the load evenly. Their different current gains means each one will be drawing different amounts of load. Generally the highest one will heat up and then by its tempco gain even more current, heat up more, etc.... pop. You either need somewhat chonky 'ballast' resistors at the base of each one (same principle as balancing LEDs when they're fed a voltage, so they draw an even current), or you need a separate opamp for each transistor so each one has its feedback loop individually controlled in the opamp loop. Drawing them all in one loop will typically lead to you blowing a transistor (or one after another in a chain as each one fails) each time you up the power.
It is very important that each output power transistor MUST have its own emitter and base resistor ! Otherwise the one transistor that has the slightly tiniest highest Hfe (current gain) will take ALL the current by itself ! Think about it ! Thus most likely this single higher gain transistor will quickly be overloaded by the thermal junction run-away causing a lower B-E voltage. This implies you must use at least 4 times higher resistor values on R7 and R0, as they functionally are in parallel through the transistors. I suggest 10K ohm for R7 (on each transistor) and 0R47 for R0 (on each transistor). Simple to modify as you just have to solder in these resistors on each transistors base and emitter wires - effectively having them in series With existing R7 and R0 on the PCB respectively. The feedback to U2.1 is okay to still take from the PCB mounted R0 as this will still represent the average current through all transistors as the modification ensures equal current through all transistors. No modification to the PCB then ... The circuit - as is - is simply not suited for parallelled output transistors without this modification .... And may cause you endless headache to fault trace 😉
Interesting and I agree about balancing the transistors, but should that not happen if each has it's own base resistor but the emitters are connected together? Because each one then can still have it's own base-emitter voltage. As these are darlington transistors then they should not need much base current, so 10K base resistors sound reasonable. Regards R0, when I originally fitted 0R5 by mistake I could only get approx 0.5A load. When I fitted the 0R05 the load worked like the original one. So if another resistor was added to each emitter (0R47) wouldn't this mess with the load in such a way to limit the maximum current it can sink? I have some 0R22 resistors I could try as emitter resistors, and these are a common value in amplifiers etc. Regards the feedback, are you saying to put a resistor on each emitter, then connect the other ends of all these resistors together and connect that node to the 0R05 current sense resistor? I do have a couple more TIP122 so there is plenty scope to experiment here like you say. Thanks for the suggestions
The base current in the darlingtons are so small it is not the best place to try balancing out the transistors. You have to do it in the path where the current is high - and that is the emitter. Remember that the emitter resistors are effectively in parallel in the functional circuit and is only used to spread out the current by imposing a voltage drop in series with the B-E voltage. It may work perfectly fine with your available 0R22 resistors. If you convert to powerFETs you have to consider you now have a voltage amplifier setup.@@LearnElectronicsRepair
@@CXensation Thanks for that, I had not considered the base current amplification in the Darlington. What you say makes perfect sense to me. My only concern is putting resistors in the emitter circuit will decrease the current sink capabilities of the load. Seeing as they would have to be in series with the existing 0R05 otherwise how can the combined current be monitored. But then presumably the feedback circuit will just increase the drive to the Darlington transistors until it gets enough voltage across R050 to balance the circuit. Plenty of ideas to play with on a followup 😁
Besides what other commenters said about current sharing imbalances, you also have to take into account the SOA (safe operating area) of the transistors, they can only achieve the banner specs at ambient temperature and for a short time, as you increase the collector emitter voltage your maximum safe current is decreased significantly for DC(continuous) operation, the graph is usually near the end of the datasheet, before package information. For the TIP122 you can only pull 1.8A at 30v VCE which is below the max power(assuming you can dump that power while keeping the transistor cool, if not you have to reduce the power even more). For a low voltage load like this one the effect is not as pronounced, but you should take it into consideration.
I've reverse engineered the PCB myself from the layout in the designers TH-cam video. You haven't made any mistakes. Your schematic is correct. I have to ask where you bought the TIP122 transistors though. Were they from Ali Express? 🤔
The TIP122 from AliExpress... Probably 😉 Thanks for taking the to check our work (it was Detlef and Myself who did the reverse engineering). If you check the other comments i figured out what the op-amp around the TL431 is doing so the circuit makes sense to me now
The reason why you don’t have identical values on pin 5 and 6 is that the output of the OPAMP is saturated. According the datasheet the typical maximum output voltage at 5V supply is 3.5V. If you calculate the voltage divider, the OPAMP should deliver 3.86V which it simply can’t. Is this circuit meant to be a kind of pre-regulator to the TL431? Shouldn’t be necessary in the supply range from 5 to 24V.
Thanks for the info, I like this because with these sort of videos and the discussion we can all learn something new. I have to agree with you I honestly can't figure out what the op-amp is doing in this circuit other than it could be regulating the supply to TL431 in some way. It makes no sense to me to effectively 'amplify' the 2.5V refenrence voltage and feed that to the 10K pot. Anyway that is not what it is doing here if the reverse engineered schematic is correct. Why not just use a resistor divider to set whatever reference we want on TL431 and feed that directly into the pot? With 100K resistor in series the current flowing through the 100K+10K is very small, and the other opamp input draws basically no current anyway.
Some comments on the original design say they have done this and got 140W load, but they don't go into specifics how they did it. From what I was seeing, it did work with four in parallel until one failed. It has been suggested that transistors were not balanced because they should have individual emitter resistors as well as individual base resistors. The TIP122 datasheet says the rating is 100V, 5A, 65W. Even if only one transistor was passing all the current IMHO it should not have failed. I had 45W according to the USB analyzer so that's like 2.1A at 20V which does not exceed any of the ratings. Obviously I'll be trying out a lot of ideas from these comments on the next video so hopefully we can all learn more than we already thought we knew 🙂
@@LearnElectronicsRepair Richard, be careful about those seemingly simple figures fron the datasheet. I can hardly imagine a TO220 package dissipating 65W (submerged in liquid nitrogen, possibly). The datasheet might say 65W at t_junction 25 Celsius, for example (haven’t checked).
@@inse001 It did say something like that, but the degradation per degree C was fairly small above that. However it did go short. Will experiment further on next video and also use thermal camera
Regarding the difference between the voltage-difference between the pot-output and pin 3 one question: Did you always get the GND-level for your measurements from the same point? The GND conductor is not too wide for a current of about 4A and there may be a remarkable voltage drop on the way from the 0R05 resistor to the USB-connector particularly with regard to the fact that it is reduced at the point whrere the OP-Amp is connected to GND. This may cause very different readings if the GND-level is taken from different points e.g. once at the USB-Connector and once directly at the pot. With your design (obviously a double-sided PCB with a GND-plane on one side) you will not have this problem...
Something odd-looking (to me, so maybe wrong) in the 2.5V reference circuit. The TL431 typically has a minimum cathode current of 1mA, typically 10mA. Although I nearly went doolally trying to figure out what that circuit was supposed to do, it seems to me that only microamps can flow via the 2M resistor, though there may be other components, connections not shown.
There is something odd with that part of the circuit. It looks like the 2M resistor feeds the TL431, but then the 680R resistor on the output of the op amp also feeds TL431 and the voltage going to the 10K potentiometer via the 100K resistor can only be te output of TL431 which is 2.5V in this configuration. So we get around 0.25V on the potentiomeeter which is what I was seeing. But in that case what is the op-amp even doing? I did think we have an error in the reverse engineered schematic, and we may well do, but the 680R and 100K resistors are also connected like this on the original PCB
@@LearnElectronicsRepair Yes, very odd. Almost as if the op-amp is attempting to be the two resistors required to set the reference on the TL431, depending on the supply voltage. Depending on the reverse-engineering revisit it may be worth just using a fixed reference component and not using that op-amp part. Also, may be worth investigating the drive current into the multiple TIP122's, there are some notes online about using pre-drivers or IRF1404's instead. Apologies for the duplicate comment, that's YT auto-deleting and re-appearing comments again.
@@ralphj4012 I think I have it. I'm pretty sure the opamp is acting like a constant voltage source for TL431. The output of TL431 sets the output of the op amp (pin 7) to something a volt or so above the reference voltage and this powers TL431 via the 680R resistor. The reference of 2.5V is then feeding into the pot. Obviously the load is meant to work on any voltage from about 5V to 30V but I never knew TL431 needed a fairly constant input voltage to give 2.5V ref.
@@LearnElectronicsRepair That's making some sense, though I think you mentioned that you had differing voltages on the pot (0.2V, 0.6V) between the different versions so, personally, I wouldn't trust the op-amp bit, especially as 2.5V never gets to the pot, due to the divider. But, hey-ho, any more analyses and my head would explode. Good luck with this project.
@@ralphj4012 Now that bit where I was seeing 0.6V on the original version does not make any sense to me either! More work required there. Without your head exploding, the 2.5V reference from TL431 and the 2M resistor sets the voltage on the non inverting input of U2.2 (pin 5) at startup. This forces the output of U2.2 to something like 3.7V because of the ratio of feedback resistors R3/R2. The 3.7V from U2.2 pin 7 then powers the TL431 via the 680R resistor and this supply will always be 3.7V regardless of Vcc (supply from PSU under load test). R6 (100K) then feeds the 2.5V reference to the 10K potentiometer, putting about 0.225V on the pot. It only looks like R5 680R and R6 100K are acting like two resistors in series on the output of U2.2 (voltage divider) because of the way we drew the schematic. The pot is connected to TL431 cathode via the 100K resistor and TL431 cathode is our reference voltage 2.5V. If there is something wrong with the readings I took, it is on the original version where I saw 0.2V/0.6V.
The schematics and Gerber will be made available once we get it to work correctly, this is the prototype. I did show the schematic in this video. I had to reverse engineer it as there is no schematic available for the original version so it could have some error in it. I don't think so, but it still could...
Impossible as they are darlington types. No matter how close you match them up - the single one with itty bitty highest current gain will ALLWAYS eat up ALL the current by itself ! You can ONLY do it with powerFETs as they have the opposite thermal characteristic thus balancing themselves out.
@@CXensation Even with one transistor passing all the current ,I had something like 45W on the usb analyzer, so that is 2.1A at 20V potentially flowing through one transistor rated at 5A 100V 65W. So why would it fail? Of course I'm going to try your other suggestions on previous comment regards emitter resistors and we will all see how that works out . And we can play around with the thermal camera too. Changing the design to powerFET could also be an interesting challenge. Thanks for your input, let's keep learning. 🙂
Something odd-looking (to me, so maybe wrong) in the 2.5V reference circuit. The TL431 typically has a minimum cathode current of 1mA, typically 10mA. Although I nearly went doolally trying to figure out what that circuit was supposed to do, it seems to me that only microamps can flow via the 2M resistor, though there may be other components, connections not shown.
@@inse001 It should, but what I am wondering is what is the point of the opamp in the TL431 circuit in the first place? Why not just use a resistor divider to set the output of TL431 to whatever we want and then feed that directly into the 100K resistor/10K pot? Very little current is going to flow that way. I work it out as 22.7uA assuming the input impedance of the other op-amp fed from the 10K pot is infinite. Max rating for TL431 is 37V and max Vref is 36V. LM358 is max 32V supply. More than enough for our load which was intended to be for USB PD, but looks like it could easily load 24V PSU and probably 30V is probably the limit.
I think I worked out what the opamp is doing, it's giving the TL431 a relatively stable supply voltage about 1.2V above the reference voltage, via 680R resistor, regardless of the output voltage of the PSU we are loading. Read the comments below in the duplicated ralphj4012 post for more info
Transistors are CURRENT BASED devices, each with their OWN individual current gain figure. If the current gains (and tempcos and other things) are not exactly matched, having "the same matched voltage on the base" does NOT mean they will be sharing the load evenly. Their different current gains means each one will be drawing different amounts of load. Generally the highest one will heat up and then by its tempco gain even more current, heat up more, etc.... pop.
You either need somewhat chonky 'ballast' resistors at the base of each one (same principle as balancing LEDs when they're fed a voltage, so they draw an even current), or you need a separate opamp for each transistor so each one has its feedback loop individually controlled in the opamp loop. Drawing them all in one loop will typically lead to you blowing a transistor (or one after another in a chain as each one fails) each time you up the power.
It is very important that each output power transistor MUST have its own emitter and base resistor !
Otherwise the one transistor that has the slightly tiniest highest Hfe (current gain) will take ALL the current by itself !
Think about it !
Thus most likely this single higher gain transistor will quickly be overloaded by the thermal junction run-away causing a lower B-E voltage.
This implies you must use at least 4 times higher resistor values on R7 and R0, as they functionally are in parallel through the transistors.
I suggest 10K ohm for R7 (on each transistor) and 0R47 for R0 (on each transistor).
Simple to modify as you just have to solder in these resistors on each transistors base and emitter wires - effectively having them in series With existing R7 and R0 on the PCB respectively.
The feedback to U2.1 is okay to still take from the PCB mounted R0 as this will still represent the average current through all transistors as the modification ensures equal current through all transistors.
No modification to the PCB then ...
The circuit - as is - is simply not suited for parallelled output transistors without this modification ....
And may cause you endless headache to fault trace 😉
Interesting and I agree about balancing the transistors, but should that not happen if each has it's own base resistor but the emitters are connected together? Because each one then can still have it's own base-emitter voltage. As these are darlington transistors then they should not need much base current, so 10K base resistors sound reasonable. Regards R0, when I originally fitted 0R5 by mistake I could only get approx 0.5A load. When I fitted the 0R05 the load worked like the original one.
So if another resistor was added to each emitter (0R47) wouldn't this mess with the load in such a way to limit the maximum current it can sink? I have some 0R22 resistors I could try as emitter resistors, and these are a common value in amplifiers etc.
Regards the feedback, are you saying to put a resistor on each emitter, then connect the other ends of all these resistors together and connect that node to the 0R05 current sense resistor?
I do have a couple more TIP122 so there is plenty scope to experiment here like you say. Thanks for the suggestions
The base current in the darlingtons are so small it is not the best place to try balancing out the transistors.
You have to do it in the path where the current is high - and that is the emitter.
Remember that the emitter resistors are effectively in parallel in the functional circuit and is only used to spread out the current by imposing a voltage drop in series with the B-E voltage.
It may work perfectly fine with your available 0R22 resistors.
If you convert to powerFETs you have to consider you now have a voltage amplifier setup.@@LearnElectronicsRepair
@@CXensation Thanks for that, I had not considered the base current amplification in the Darlington. What you say makes perfect sense to me. My only concern is putting resistors in the emitter circuit will decrease the current sink capabilities of the load. Seeing as they would have to be in series with the existing 0R05 otherwise how can the combined current be monitored. But then presumably the feedback circuit will just increase the drive to the Darlington transistors until it gets enough voltage across R050 to balance the circuit. Plenty of ideas to play with on a followup 😁
Besides what other commenters said about current sharing imbalances, you also have to take into account the SOA (safe operating area) of the transistors, they can only achieve the banner specs at ambient temperature and for a short time, as you increase the collector emitter voltage your maximum safe current is decreased significantly for DC(continuous) operation, the graph is usually near the end of the datasheet, before package information.
For the TIP122 you can only pull 1.8A at 30v VCE which is below the max power(assuming you can dump that power while keeping the transistor cool, if not you have to reduce the power even more). For a low voltage load like this one the effect is not as pronounced, but you should take it into consideration.
I've reverse engineered the PCB myself from the layout in the designers TH-cam video. You haven't made any mistakes. Your schematic is correct. I have to ask where you bought the TIP122 transistors though. Were they from Ali Express? 🤔
The TIP122 from AliExpress... Probably 😉 Thanks for taking the to check our work (it was Detlef and Myself who did the reverse engineering). If you check the other comments i figured out what the op-amp around the TL431 is doing so the circuit makes sense to me now
The reason why you don’t have identical values on pin 5 and 6 is that the output of the OPAMP is saturated.
According the datasheet the typical maximum output voltage at 5V supply is 3.5V.
If you calculate the voltage divider, the OPAMP should deliver 3.86V which it simply can’t.
Is this circuit meant to be a kind of pre-regulator to the TL431?
Shouldn’t be necessary in the supply range from 5 to 24V.
Thanks for the info, I like this because with these sort of videos and the discussion we can all learn something new. I have to agree with you I honestly can't figure out what the op-amp is doing in this circuit other than it could be regulating the supply to TL431 in some way. It makes no sense to me to effectively 'amplify' the 2.5V refenrence voltage and feed that to the 10K pot. Anyway that is not what it is doing here if the reverse engineered schematic is correct. Why not just use a resistor divider to set whatever reference we want on TL431 and feed that directly into the pot? With 100K resistor in series the current flowing through the 100K+10K is very small, and the other opamp input draws basically no current anyway.
@@LearnElectronicsRepair maybe it’s just making use of the other half of the LM358..?
@@inse001 Just for the fun of it 😅Nahh I figured it out, see the comments on the RalphJ4012 threads
Are you sure you can use 4 darlingtons in parallel without adjusting any other values ??...cheers.
Some comments on the original design say they have done this and got 140W load, but they don't go into specifics how they did it. From what I was seeing, it did work with four in parallel until one failed. It has been suggested that transistors were not balanced because they should have individual emitter resistors as well as individual base resistors. The TIP122 datasheet says the rating is 100V, 5A, 65W. Even if only one transistor was passing all the current IMHO it should not have failed. I had 45W according to the USB analyzer so that's like 2.1A at 20V which does not exceed any of the ratings. Obviously I'll be trying out a lot of ideas from these comments on the next video so hopefully we can all learn more than we already thought we knew 🙂
@@LearnElectronicsRepair Richard, be careful about those seemingly simple figures fron the datasheet.
I can hardly imagine a TO220 package dissipating 65W (submerged in liquid nitrogen, possibly).
The datasheet might say 65W at t_junction 25 Celsius, for example (haven’t checked).
@@inse001 It did say something like that, but the degradation per degree C was fairly small above that. However it did go short. Will experiment further on next video and also use thermal camera
Thanks !@@LearnElectronicsRepair
Regarding the difference between the voltage-difference between the pot-output and pin 3 one question: Did you always get the GND-level for your measurements from the same point? The GND conductor is not too wide for a current of about 4A and there may be a remarkable voltage drop on the way from the 0R05 resistor to the USB-connector particularly with regard to the fact that it is reduced at the point whrere the OP-Amp is connected to GND. This may cause very different readings if the GND-level is taken from different points e.g. once at the USB-Connector and once directly at the pot.
With your design (obviously a double-sided PCB with a GND-plane on one side) you will not have this problem...
Something odd-looking (to me, so maybe wrong) in the 2.5V reference circuit. The TL431 typically has a minimum cathode current of 1mA, typically 10mA. Although I nearly went doolally trying to figure out what that circuit was supposed to do, it seems to me that only microamps can flow via the 2M resistor, though there may be other components, connections not shown.
There is something odd with that part of the circuit. It looks like the 2M resistor feeds the TL431, but then the 680R resistor on the output of the op amp also feeds TL431 and the voltage going to the 10K potentiometer via the 100K resistor can only be te output of TL431 which is 2.5V in this configuration. So we get around 0.25V on the potentiomeeter which is what I was seeing. But in that case what is the op-amp even doing? I did think we have an error in the reverse engineered schematic, and we may well do, but the 680R and 100K resistors are also connected like this on the original PCB
@@LearnElectronicsRepair Yes, very odd. Almost as if the op-amp is attempting to be the two resistors required to set the reference on the TL431, depending on the supply voltage. Depending on the reverse-engineering revisit it may be worth just using a fixed reference component and not using that op-amp part. Also, may be worth investigating the drive current into the multiple TIP122's, there are some notes online about using pre-drivers or IRF1404's instead. Apologies for the duplicate comment, that's YT auto-deleting and re-appearing comments again.
@@ralphj4012 I think I have it. I'm pretty sure the opamp is acting like a constant voltage source for TL431. The output of TL431 sets the output of the op amp (pin 7) to something a volt or so above the reference voltage and this powers TL431 via the 680R resistor. The reference of 2.5V is then feeding into the pot. Obviously the load is meant to work on any voltage from about 5V to 30V but I never knew TL431 needed a fairly constant input voltage to give 2.5V ref.
@@LearnElectronicsRepair That's making some sense, though I think you mentioned that you had differing voltages on the pot (0.2V, 0.6V) between the different versions so, personally, I wouldn't trust the op-amp bit, especially as 2.5V never gets to the pot, due to the divider. But, hey-ho, any more analyses and my head would explode. Good luck with this project.
@@ralphj4012 Now that bit where I was seeing 0.6V on the original version does not make any sense to me either! More work required there.
Without your head exploding, the 2.5V reference from TL431 and the 2M resistor sets the voltage on the non inverting input of U2.2 (pin 5) at startup. This forces the output of U2.2 to something like 3.7V because of the ratio of feedback resistors R3/R2.
The 3.7V from U2.2 pin 7 then powers the TL431 via the 680R resistor and this supply will always be 3.7V regardless of Vcc (supply from PSU under load test).
R6 (100K) then feeds the 2.5V reference to the 10K potentiometer, putting about 0.225V on the pot. It only looks like R5 680R and R6 100K are acting like two resistors in series on the output of U2.2 (voltage divider) because of the way we drew the schematic.
The pot is connected to TL431 cathode via the 100K resistor and TL431 cathode is our reference voltage 2.5V. If there is something wrong with the readings I took, it is on the original version where I saw 0.2V/0.6V.
Is there schematics and PCB layout for this project ? Good job, learned a lot from the video.
The schematics and Gerber will be made available once we get it to work correctly, this is the prototype. I did show the schematic in this video. I had to reverse engineer it as there is no schematic available for the original version so it could have some error in it. I don't think so, but it still could...
Heya nice diy a lot of "playing" and leaarning why some things are as they are
Need a hair stylist for them fingers man. 😂 do some braids or something. Lol.
Just you wait until the next full moon!!!
@@LearnElectronicsRepair 😳
timestamp 1:37 Richard you can / could / should addan amp meter and a volt meter to it .. that shouldnt be hard to do ....
The completed version will have an optional daughter board that displays volts amps and watts. Detlef is designing that part
maybe the transistors were too much out of balance try with close matched pairs
Impossible as they are darlington types.
No matter how close you match them up - the single one with itty bitty highest current gain will ALLWAYS eat up ALL the current by itself !
You can ONLY do it with powerFETs as they have the opposite thermal characteristic thus balancing themselves out.
@@CXensation Even with one transistor passing all the current ,I had something like 45W on the usb analyzer, so that is 2.1A at 20V potentially flowing through one transistor rated at 5A 100V 65W. So why would it fail? Of course I'm going to try your other suggestions on previous comment regards emitter resistors and we will all see how that works out . And we can play around with the thermal camera too. Changing the design to powerFET could also be an interesting challenge. Thanks for your input, let's keep learning. 🙂
Something odd-looking (to me, so maybe wrong) in the 2.5V reference circuit. The TL431 typically has a minimum cathode current of 1mA, typically 10mA. Although I nearly went doolally trying to figure out what that circuit was supposed to do, it seems to me that only microamps can flow via the 2M resistor, though there may be other components, connections not shown.
The TL431 is also connected the output of the OPAMP which should allow a cathode current of 2mA.
@@inse001 It should, but what I am wondering is what is the point of the opamp in the TL431 circuit in the first place? Why not just use a resistor divider to set the output of TL431 to whatever we want and then feed that directly into the 100K resistor/10K pot? Very little current is going to flow that way. I work it out as 22.7uA assuming the input impedance of the other op-amp fed from the 10K pot is infinite. Max rating for TL431 is 37V and max Vref is 36V. LM358 is max 32V supply. More than enough for our load which was intended to be for USB PD, but looks like it could easily load 24V PSU and probably 30V is probably the limit.
I think I worked out what the opamp is doing, it's giving the TL431 a relatively stable supply voltage about 1.2V above the reference voltage, via 680R resistor, regardless of the output voltage of the PSU we are loading. Read the comments below in the duplicated ralphj4012 post for more info