Great explanation, with your graph it makes it very easy to understand why it's better to have the load resistor. The load resistor not only reduces the power going through the mosfet, because it's doing that it will also extend the adjustable current at the higher end that you are able to control and keep constant, so it helps stabilizing the circuit at the higher end up to the maximum level. Thanks for sharing this info online, to me it was very helpful visualizing what happens in the circuit. If I can't visualize what happens in a circuit I'm not able to make it in a practical and usable circuit so thanks again for all your effort. Ricardo Penders
Interesting to see the FET running cooler at 1.8 amps than 130mA. Of course this is to be expected if you think about it, as PWM for example runs the transistor either full-off (high volt drop but no current, stays cool) or full-on (high current but low volt drop, stays cool). This vid shows that in practice, plus all the bits in between.
I just had to say thank you so much for helping me understand how the circuit fully works because I now understand that basically without elaborating to much that the aprox. 1-2ohm load is swapped in essence from the Mosfet to the power resistor during loads. I now understand what i need to do to improve my Arduino Li-Ion (18650) discharger and how to drop temperatures on my Mosfet a bit (Currently using a small fan to cool). It is in the working prefboard semi completed state for my expectations on this project/tool. It is obvious that in your circuit your not using the resistor for current measurement just to keep the balance when required for the FET. I am going to get a better Power resistor tomorrow because my current one is not exactly 1ohm its 0.8 so it is not accurate enough and it is maybe a 1W proper power resistor, ill need to get a good 1% THD 1 Ohm at at least 5W because i am essentially using it as a current shunt to in return logicaly control the PWM output as the voltage drops. Ideally i should try to get a proper DALE/Vishay 1 Ohm shunt maybe? What are your thoughts? Again thanks so much for being such a good teacher and doing so much work to help teach us!
A second option to provide your top-end 2 amp limit, which I've done at present: add a resistor between your multiturn pot and the supply, effectively pushing your maximum voltage down on the pot. This is also good if you find you're going all the way up to your limit in just a turn or two of the pot. If adding a 1 ohm resistor it may be worth making it switchable, as at very low test voltages it might get in the way (you'll never drive a 1.5 volt battery supply above 1.5 amps, for example).
Thanks...there's a good chance I might be missing one thing ;-) Let me test what you are saying and I'll feedback in the final video. Thanks for your input.
Looking for an answer about dummy load and I stumble upon your post. Unfortunately, I didn't understand any of the technical terms you were using, obviously because I am not a n adept electronic enthusiast. What I noticed is and get interested in is the background, wow you have a huge collections of digital multitester. Can I have one, couldn't find any of that here in the Philippines with affordable price. LOL.. I think you are very informative, keep up the good work..
Mine doesn't use a 1 ohm resistor, but does have a digital ammeter which'll have a shunt resistor, and that's between the FET and ground where the 1 ohm would be anyway.
You should do the exact same test but with a 10 Ohm resistor instead, I think that a 10 Ohm resistor in this case may be helpful bringing the power dissipation in the mosfet down even more, another thing you can do is to use a different mosfet called a linear mosfet which is specially purposed for this kind of applications. Using linear mosfets you can make your load far more capable of handling huge amounts of current continuously.
The load resistor simply is a current shunt. A DC load is basically a constant current sink. The opamp 'measures' the voltage across the shunt and controls the MOSFET it matches the reference voltage. So if you drive the opamp with 1V and your shunt is 0.5 ohms, the current is set to 2A (1V/0.5R=2A). The shunt then dissipates 2W. The FET dissipiates (Vload-Vshunt)*Ishunt. 5V load at 2A is 10W into the DC load. 10-2=8W@FET.. A 12V load still means the shunt heats by 2W. The FET does (12-1)*2=22W
hee martin in the four scenario's i am missing some things. 1: 6V / 130mA = 46.15 ohm but the fet is only 43 ohm 2: 6V / 130mA = 46.15 ohm fet = 41.5 + the resistor = 42.5 ohm 3: 6V / 1373mA = 4.37 ohm fet = 0.45 + 1 = 1.45 ohm 4: 6V / 1373mA = 4.37 ohm fet = 1.238 ohm some ware in the circuit there is a ~3ohm resistor. i think its in the wires and contacts. this explains why the circuit is working with out the resistor. the wires act like a resistor and give feedback to the opamp. manfred
what load resistors value should i used for my cars rear brake/turn signal leds. in my mustang 2001, it has 8 total bulds, 2 reverse and 6 are L,R 3 on each side for the brake/turn/parking lights(3157). each buld on each side share the same wire for all three functions.
I am sorry but the resistors sole function is to provide feedback for the opamp even at 3A constant current the the power in the resistor is 9W and the power in this resistor is constant what ever the test voltage is so in your case of 10V the rest of the voltage is across the FET so that is 7V @ 3 A = 21W. Now if your test voltage is 100V it will be 97V @ 3A = 291W in the FET.
Could the reason the power and heat decrease as you increase current be because of connecting the feedback(ed) input to ground when you remove the 1ohm resistor ? The opamp adjusts it's output to change the input it's connected to to make that input voltage equal the other inputs voltage. It could be raising ground above 0 volts, trying to do this, thus decreasing the overall voltage across the load so decreasing power according to P = V * I with V decreasing. Maybe the 1 ohm resistor is important for isolating the opamp input/output from ground.
The total power dissipated by the FET and resistor goes up with the current, at low current the FET dissipates most of the heat, at high current the resistor dissipates most of the heat. Without an actual resistor you are using the voltage drop across your wires as current feedback(not as stable) for the op-amp. But your wire cannot dissipate much power so the FET is dissipating the majority of the power all of the time. As such as the current goes up the resistorless FET's power dissipation goes up, as the total power dissipation only goes down when the current goes down the knee of the resistorless FET's graph cannot happen, that was an error on the video.
hi, jd here very nice. i have a (?) my truck alarm have a smoll siren i want to install 30watt big siren. what should i do for more watt load, ? the load resister will work?
If I got it right, you connected the negative input of the opamp to ground. The resistor is used as a shunt resistor. It is the way you have to measure the current sinked by measuring the voltage drop of the shunt resistor and using Ohm's Law. What you've made by not using shunt resistor is that you shorted out the negative input of the opamp to ground. You are able to control the current the same way because probably the cables have a few hundred miliohms or who knows, some ohms. The resistor is not there to limit the current nor to split the power dissipation between the resistor and the FET. His reason to be there is to measure the actual current consumption. So, the behaviour of the circuit without the resistor is a little odd, but surely bad designed since the current will probably vary with the input voltage, there is no way to actualy keep the current constant and it is going to vary as the FET temperature changes and the opamp will probably be saturated. The reason why your circuit works deserves further investigation, because it shouldn't work.
Also, there is no reason to have that voltage follower opamp stage... The opamp has high impedance inputs, the voltage follower is lowering the signal impedance, but since the second opamp has high impedance inputs, it will not affect the output voltage of the potentiometer. You should further investigate what is happening when you remove the voltage follower and how are you able to control the current without the shunt resistor, because that is not expected at all.
And again, you should further investigate those voltage drops you are getting in the FET. That 0.2V voltage drop with the shunt resistor is acceptable because you got a resistor between the source and the ground that probably has the 5.8V remaining (I guess your input voltage was 6V). Without the shunt resistor, you are connecting the drain to 6V and the source to GND, so effectively the voltage across the FET is 6V, not 1.7V unless your cables have a significant resistance. That would also explain the reason why it works without the shunt resistor.
I am answering as I watch the video, sorry about so much comments. But the answer to your car battery question is in the shunt resistor. The shunt is going to limit the current even if the FET is fully conducting, but it will not happen, since the opamp is going to make all it can to ensure that the voltage across the resistor is equal to the potentiometer voltage. It means that no more than the current set will pass through the circuit, even if you connect that battery.
And my final comment. I've still not viewed the rest of the videos, but you are not testing the main goal of a dummy load. You should vary the input voltage and see if the current is constant, that is the main goal of the dummy load.
Thanks for the feedback and input. I'll certainly look into the points you make. Cheers, Martin.
Thanks for your post and information. I'll cover what you have posted in the next video. Cheers, Martin.
Great explanation, with your graph it makes it very easy to understand why it's better to have the load resistor.
The load resistor not only reduces the power going through the mosfet, because it's doing that it will also extend the adjustable current at the higher end that you are able to control and keep constant, so it helps stabilizing the circuit at the higher end up to the maximum level.
Thanks for sharing this info online, to me it was very helpful visualizing what happens in the circuit.
If I can't visualize what happens in a circuit I'm not able to make it in a practical and usable circuit so thanks again for all your effort.
Ricardo Penders
Yes, good points let me pick this up in the final video. Cheers, Martin.
My pleasure, thanks for the feedback.
I learnt many things from this tutorial. I'm going to watch one more time for better understanding.
Thanks.
Excellent analogy of the belt sander. Thanks for showing how you proved it out.
Interesting to see the FET running cooler at 1.8 amps than 130mA. Of course this is to be expected if you think about it, as PWM for example runs the transistor either full-off (high volt drop but no current, stays cool) or full-on (high current but low volt drop, stays cool). This vid shows that in practice, plus all the bits in between.
I just had to say thank you so much for helping me understand how the circuit fully works because I now understand that basically without elaborating to much that the aprox. 1-2ohm load is swapped in essence from the Mosfet to the power resistor during loads. I now understand what i need to do to improve my Arduino Li-Ion (18650) discharger and how to drop temperatures on my Mosfet a bit (Currently using a small fan to cool). It is in the working prefboard semi completed state for my expectations on this project/tool. It is obvious that in your circuit your not using the resistor for current measurement just to keep the balance when required for the FET.
I am going to get a better Power resistor tomorrow because my current one is not exactly 1ohm its 0.8 so it is not accurate enough and it is maybe a 1W proper power resistor, ill need to get a good 1% THD 1 Ohm at at least 5W because i am essentially using it as a current shunt to in return logicaly control the PWM output as the voltage drops. Ideally i should try to get a proper DALE/Vishay 1 Ohm shunt maybe? What are your thoughts? Again thanks so much for being such a good teacher and doing so much work to help teach us!
Hi Manfred,
Yes, your assumption is correct, the rest of the resistance is in the wires and contacts. I cover this in Part 4.
Cheers,
Martin.
very grateful for your explanation. Thank you very much
A second option to provide your top-end 2 amp limit, which I've done at present: add a resistor between your multiturn pot and the supply, effectively pushing your maximum voltage down on the pot. This is also good if you find you're going all the way up to your limit in just a turn or two of the pot.
If adding a 1 ohm resistor it may be worth making it switchable, as at very low test voltages it might get in the way (you'll never drive a 1.5 volt battery supply above 1.5 amps, for example).
Thanks...there's a good chance I might be missing one thing ;-)
Let me test what you are saying and I'll feedback in the final video.
Thanks for your input.
Looking for an answer about dummy load and I stumble upon your post. Unfortunately, I didn't understand any of the technical terms you were using, obviously because I am not a n adept electronic enthusiast. What I noticed is and get interested in is the background, wow you have a huge collections of digital multitester. Can I have one, couldn't find any of that here in the Philippines with affordable price. LOL.. I think you are very informative, keep up the good work..
Mine doesn't use a 1 ohm resistor, but does have a digital ammeter which'll have a shunt resistor, and that's between the FET and ground where the 1 ohm would be anyway.
You should do the exact same test but with a 10 Ohm resistor instead, I think that a 10 Ohm resistor in this case may be helpful bringing the power dissipation in the mosfet down even more, another thing you can do is to use a different mosfet called a linear mosfet which is specially purposed for this kind of applications.
Using linear mosfets you can make your load far more capable of handling huge amounts of current continuously.
The load resistor simply is a current shunt. A DC load is basically a constant current sink. The opamp 'measures' the voltage across the shunt and controls the MOSFET it matches the reference voltage. So if you drive the opamp with 1V and your shunt is 0.5 ohms, the current is set to 2A (1V/0.5R=2A). The shunt then dissipates 2W. The FET dissipiates (Vload-Vshunt)*Ishunt. 5V load at 2A is 10W into the DC load. 10-2=8W@FET.. A 12V load still means the shunt heats by 2W. The FET does (12-1)*2=22W
hee martin
in the four scenario's i am missing some things.
1: 6V / 130mA = 46.15 ohm but the fet is only 43 ohm
2: 6V / 130mA = 46.15 ohm fet = 41.5 + the resistor = 42.5 ohm
3: 6V / 1373mA = 4.37 ohm fet = 0.45 + 1 = 1.45 ohm
4: 6V / 1373mA = 4.37 ohm fet = 1.238 ohm
some ware in the circuit there is a ~3ohm resistor. i think its in the wires and contacts.
this explains why the circuit is working with out the resistor. the wires act like a resistor and give feedback to the opamp.
manfred
what load resistors value should i used for my cars rear brake/turn signal leds. in my mustang 2001, it has 8 total bulds, 2 reverse and 6 are
L,R 3 on each side for the brake/turn/parking lights(3157). each buld on each side share the same wire for all three functions.
I am sorry but the resistors sole function is to provide feedback for the opamp even at 3A constant current the the power in the resistor is 9W and the power in this resistor is constant what ever the test voltage is so in your case of 10V the rest of the voltage is across the FET so that is 7V @ 3 A = 21W. Now if your test voltage is 100V it will be 97V @ 3A = 291W in the FET.
Did I hear you correctly. Did you say this circuit could be used as a adjustable current control on a home made Bench Top Power Supply?
Great work! I learned a lot. Thank you!
youtube will not let me give you the web site for the plasma speaker
very nice video great info tnx
Could the reason the power and heat decrease as you increase current be because of connecting the feedback(ed) input to ground when you remove the 1ohm resistor ?
The opamp adjusts it's output to change the input it's connected to to make that input voltage equal the other inputs voltage. It could be raising ground above 0 volts, trying to do this, thus decreasing the overall voltage across the load so decreasing power according to P = V * I with V decreasing.
Maybe the 1 ohm resistor is important for isolating the opamp input/output from ground.
The total power dissipated by the FET and resistor goes up with the current, at low current the FET dissipates most of the heat, at high current the resistor dissipates most of the heat.
Without an actual resistor you are using the voltage drop across your wires as current feedback(not as stable) for the op-amp.
But your wire cannot dissipate much power so the FET is dissipating the majority of the power all of the time.
As such as the current goes up the resistorless FET's power dissipation goes up, as the total power dissipation only goes down when the current goes down the knee of the resistorless FET's graph cannot happen, that was an error on the video.
hi, jd here very nice. i have a (?) my truck alarm have a smoll siren i want to install 30watt big siren. what should i do for more watt load, ? the load resister will work?
If I got it right, you connected the negative input of the opamp to ground. The resistor is used as a shunt resistor. It is the way you have to measure the current sinked by measuring the voltage drop of the shunt resistor and using Ohm's Law.
What you've made by not using shunt resistor is that you shorted out the negative input of the opamp to ground. You are able to control the current the same way because probably the cables have a few hundred miliohms or who knows, some ohms. The resistor is not there to limit the current nor to split the power dissipation between the resistor and the FET. His reason to be there is to measure the actual current consumption.
So, the behaviour of the circuit without the resistor is a little odd, but surely bad designed since the current will probably vary with the input voltage, there is no way to actualy keep the current constant and it is going to vary as the FET temperature changes and the opamp will probably be saturated. The reason why your circuit works deserves further investigation, because it shouldn't work.
Also, there is no reason to have that voltage follower opamp stage... The opamp has high impedance inputs, the voltage follower is lowering the signal impedance, but since the second opamp has high impedance inputs, it will not affect the output voltage of the potentiometer. You should further investigate what is happening when you remove the voltage follower and how are you able to control the current without the shunt resistor, because that is not expected at all.
And again, you should further investigate those voltage drops you are getting in the FET. That 0.2V voltage drop with the shunt resistor is acceptable because you got a resistor between the source and the ground that probably has the 5.8V remaining (I guess your input voltage was 6V).
Without the shunt resistor, you are connecting the drain to 6V and the source to GND, so effectively the voltage across the FET is 6V, not 1.7V unless your cables have a significant resistance. That would also explain the reason why it works without the shunt resistor.
I am answering as I watch the video, sorry about so much comments. But the answer to your car battery question is in the shunt resistor. The shunt is going to limit the current even if the FET is fully conducting, but it will not happen, since the opamp is going to make all it can to ensure that the voltage across the resistor is equal to the potentiometer voltage. It means that no more than the current set will pass through the circuit, even if you connect that battery.
And my final comment. I've still not viewed the rest of the videos, but you are not testing the main goal of a dummy load. You should vary the input voltage and see if the current is constant, that is the main goal of the dummy load.
Sir i make that circuit and use same ic lm324 but it dont work why
hello sir this is mahesh How to make the ( Ac 11.5v & 1.6amps ) circuit & components
Great. I learned much.
I didn't realise there was a wattage drop off at that mid point despite continued current ramp
100amp load how to make for testing 54v/120amp smps module use in telecoms.
Has anyone made a dc load without a resistor, say using a hall effect ic for the feedback ?
Strong accent, French ?
answer me sir
wow if youu don't understand the function. if that resistor I need to quit watching