audioguru2
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Pin 2 of U# is connected to the output 0V which is connected to the rectifier's 0V through R7. Check your wiring.
View attachment 37987
View attachment 37987
Hi , i have built the power supply and for several try it has worked but ( it has never show me current at the ampmete rconnected in series with the positive wire) when i loaded suddenly, i just can read 0 v in voltmeter output.
I dont know if it has to do with Q1 ant the shutdown protection or just Q1 has blown off.
Any idea?
Thanxs
I dont know if it has to do with Q1 ant the shutdown protection or just Q1 has blown off.
Any idea?
Thanxs
audioguru2
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Hi Zama,
Welcome to our forum. ;D
What did you use for a load that killed it?
Did you make the original project or the modified and improved one?
Welcome to our forum. ;D
What did you use for a load that killed it?
Did you make the original project or the modified and improved one?
I made the circuit as it is on the web.
The thing is, that there was no reading of the ampmeter ( i have not reading of the ampmeter before ) when the power source came down, but the led wasn´t on when it happened.
The load was a stepper motor of 1.4 amps per phase and 5 v, connected by a tanslator circuit of TIP120 transistors (that the have a protection diode), that were controlled via a circuit with the parallel port.(all conections were ok)
I think that the problem has to be with Q1 as it is alreay on ,and i read a 45v betwen the colector and the emiter ,probably in the saturation region.
I dont know if Q1 has burn out or just the negative rail is still at the the level of the shut down protecction circuit.
Thanks for your support.
The thing is, that there was no reading of the ampmeter ( i have not reading of the ampmeter before ) when the power source came down, but the led wasn´t on when it happened.
The load was a stepper motor of 1.4 amps per phase and 5 v, connected by a tanslator circuit of TIP120 transistors (that the have a protection diode), that were controlled via a circuit with the parallel port.(all conections were ok)
I think that the problem has to be with Q1 as it is alreay on ,and i read a 45v betwen the colector and the emiter ,probably in the saturation region.
I dont know if Q1 has burn out or just the negative rail is still at the the level of the shut down protecction circuit.
Thanks for your support.
audioguru2
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Hi Zama,
Q1 is supposed to be turned off when the project is operating. It turns on only when the mains is turned off. Q1 must conduct the short-circuit current of U2 which could be 20mA if your project's U2 is strong.
If the project's output voltage is set to 30V when the mains is turned off, then Q1 must dissipate up to 30V x 20mA= 600mW which is much too much for a little BC558 transistor which causes its destruction.
That's why in the forum I recommend using a TIP31A power transistor for Q1. ;D
View attachment 37996
Q1 is supposed to be turned off when the project is operating. It turns on only when the mains is turned off. Q1 must conduct the short-circuit current of U2 which could be 20mA if your project's U2 is strong.
If the project's output voltage is set to 30V when the mains is turned off, then Q1 must dissipate up to 30V x 20mA= 600mW which is much too much for a little BC558 transistor which causes its destruction.
That's why in the forum I recommend using a TIP31A power transistor for Q1. ;D
View attachment 37996
Hi audioguru,
reading this forum, you're the man for all my questions ... I think
As entioned before in this topic, my current control circuit won't work properly. I will give some measurement results. On Pin 3 I can control the voltage from 0 to 1,7V (no load) witch is correct, because R18 forms together with P2 andR17 a resistor devider. At max, the next equation gives us: (10k/(56k+10K+33))*10,7=1,62V
When I put a load on the circuit (Resistor) I can control pin # from -0,2 to 1,5V... Thats is already strange for me...
The Base of Q3 has to become more negative than his emitter voltage so the trasistor can switch on, what lights the LED. To get the Base voltage far below 34V, pin 6 of U3 must get Low. This is only possible when Pin 2 of U3 becomes higher then pin3 of the op-amp. And here is the mystery.. pin 2 only gets lower when a current is flowing trough the shunt resistor R7! ???
What am I missing?? I hope someone can help me!! thnx in advance!
View attachment 37998
reading this forum, you're the man for all my questions ... I think
As entioned before in this topic, my current control circuit won't work properly. I will give some measurement results. On Pin 3 I can control the voltage from 0 to 1,7V (no load) witch is correct, because R18 forms together with P2 andR17 a resistor devider. At max, the next equation gives us: (10k/(56k+10K+33))*10,7=1,62V
When I put a load on the circuit (Resistor) I can control pin # from -0,2 to 1,5V... Thats is already strange for me...
The Base of Q3 has to become more negative than his emitter voltage so the trasistor can switch on, what lights the LED. To get the Base voltage far below 34V, pin 6 of U3 must get Low. This is only possible when Pin 2 of U3 becomes higher then pin3 of the op-amp. And here is the mystery.. pin 2 only gets lower when a current is flowing trough the shunt resistor R7! ???
What am I missing?? I hope someone can help me!! thnx in advance!
View attachment 37998
audioguru2
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Hi Smit,
The current regulator works by using U3 as a comparator to compare the voltage across R7 caused by load current, with the setting of the current pot P2 and causing the output voltage to be reduced through D9 to maintain the set current.
Since the reference voltage is +11.2V, the output of U3 goes from being high when it is not regulating the current, to at most +10.6V during current regulation. Of course when the output of U3 is only +10.6V or less then it turns on the LED through Q3.
3A though R7 produces only 1.41V at pin 2 of U3 but pin 3 is 1.70V when the pot is at max. Why? Because pots have a wide tolerance and if your 10k pot is actually 8k then the project will produce very close to 3A.
Another member reduced the value of R18 and added a trimpot in series to adjust the max setting of P2 to be exactly 3.0A. He also added a trimpot in series with the voltage-adjusting pot to adjust the max setting of P1 to be exactly 30.0V.
The schematic of the current regulator is confusing to look at since the output 0V is not the same as the rectifier 0V with R7 is in between. It is not confusing if you look at all voltages with the rectifier 0V as a common reference.
Your 11.2V reference is low at 10.7V probably because you used a zener diode for D8 that is rated at a high current. The original circuit uses 4.7k for R4 which sets a zener diode current of only 1.2mA. I recommend using a BZX79C5V6 zener diode which is rated at 5mA, and changing R4 to 1k for a zener diode current of 5.6mA.
In my sketch the load current is 3A and the pot is max. Reduce the setting of the pot P2 and you will see how it regulates the current. ;D
View attachment 37999
The current regulator works by using U3 as a comparator to compare the voltage across R7 caused by load current, with the setting of the current pot P2 and causing the output voltage to be reduced through D9 to maintain the set current.
Since the reference voltage is +11.2V, the output of U3 goes from being high when it is not regulating the current, to at most +10.6V during current regulation. Of course when the output of U3 is only +10.6V or less then it turns on the LED through Q3.
3A though R7 produces only 1.41V at pin 2 of U3 but pin 3 is 1.70V when the pot is at max. Why? Because pots have a wide tolerance and if your 10k pot is actually 8k then the project will produce very close to 3A.
Another member reduced the value of R18 and added a trimpot in series to adjust the max setting of P2 to be exactly 3.0A. He also added a trimpot in series with the voltage-adjusting pot to adjust the max setting of P1 to be exactly 30.0V.
The schematic of the current regulator is confusing to look at since the output 0V is not the same as the rectifier 0V with R7 is in between. It is not confusing if you look at all voltages with the rectifier 0V as a common reference.
Your 11.2V reference is low at 10.7V probably because you used a zener diode for D8 that is rated at a high current. The original circuit uses 4.7k for R4 which sets a zener diode current of only 1.2mA. I recommend using a BZX79C5V6 zener diode which is rated at 5mA, and changing R4 to 1k for a zener diode current of 5.6mA.
In my sketch the load current is 3A and the pot is max. Reduce the setting of the pot P2 and you will see how it regulates the current. ;D
View attachment 37999
Last edited by a moderator:
U3 is "not really" used as a comparator, at least not in a "true sense". It is true, while the output current draw is below the current set point level, the output of U3 is sitting at the positive rail, but that's where the comparator similarity ends.
It's configured as an integrator.......... as the voltage across R7 increases, and for the sake of simplicity, I'll say U3 starts to "turn-on" (go towards it's negative rail), the clamp diode D9 starts to pull down the output voltage set point votage at U2 pin 3. It will only pull it down as far as it needs to so that the current limit point is not exceeded (foldback). C8 controls the rate at which the level changes. U3 does not slam to the negative rail like a comparator would, although it could go there if it needed too. In fact, if U3 was being uesd as a true comparator, the power supply would go into a "hic-up mode" as U3 went rail-to-rail.
It's configured as an integrator.......... as the voltage across R7 increases, and for the sake of simplicity, I'll say U3 starts to "turn-on" (go towards it's negative rail), the clamp diode D9 starts to pull down the output voltage set point votage at U2 pin 3. It will only pull it down as far as it needs to so that the current limit point is not exceeded (foldback). C8 controls the rate at which the level changes. U3 does not slam to the negative rail like a comparator would, although it could go there if it needed too. In fact, if U3 was being uesd as a true comparator, the power supply would go into a "hic-up mode" as U3 went rail-to-rail.
audioguru2
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Hi Indulis,
Welcome to our forum. ;D
You are correct by stating that U3 is not being used as a true comparator. It is using its inputs to compare the voltage across R7 that is caused by load current, with the voltage from the current-setting pot. Then it uses its internal gain of a few hundred thousand to very accurately reduce the output voltage so that the load current is exactly what is needed.
U3 has negative feedback and is used as an error amplifier. Comparator ICs never have negative feedback because they would oscillate.
The time constant of R21 (10k) and C8 (330pF) is pretty quick at only 3.3us so I would say that it is used for loop compensation to avoid oscillation at a high frequency, instead of being used as an integrator. ;D
Welcome to our forum. ;D
You are correct by stating that U3 is not being used as a true comparator. It is using its inputs to compare the voltage across R7 that is caused by load current, with the voltage from the current-setting pot. Then it uses its internal gain of a few hundred thousand to very accurately reduce the output voltage so that the load current is exactly what is needed.
U3 has negative feedback and is used as an error amplifier. Comparator ICs never have negative feedback because they would oscillate.
The time constant of R21 (10k) and C8 (330pF) is pretty quick at only 3.3us so I would say that it is used for loop compensation to avoid oscillation at a high frequency, instead of being used as an integrator. ;D
Yeah, I'm sure C8 is used as a loop compensation cap (limit/control U3 output slew, but that's what feedback is all about), but that doesn't stop it from being a integrator, which IT IS!!!!
I saw a post on another fourm that covered feedback in comparators circuits and why it is used!!!
I saw a post on another fourm that covered feedback in comparators circuits and why it is used!!!
audioguru2
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OK.indulis said:
Comparator ICs don't have built-in frequency compensation so they can have gain at high frequencies. Therefore they never use negative feedback with a resistor or integrating capacitor in order to avoid oscillation. Usually comparator circuits use a small amount of positive feedback to avoid oscillation at the threshold voltage caused by stray capacitance or noise. ;D
Hi audioguru,
I changed R4 to a 1K value, and my Voltage value has climed from 10.7 tp 11.2V!! I did not used the BZX zener, but this one will be delivered soon I hope. I looked at you're schmatic and see that the voltage acros R7 becomes 1,4V when there is a current flow of 3A.
I tried it and what happens.... When I measure the voltage accros R7 when there is a current of 3Amp, the voltage is -1.4V!!!!!!! How can it become negative???
I changed R4 to a 1K value, and my Voltage value has climed from 10.7 tp 11.2V!! I did not used the BZX zener, but this one will be delivered soon I hope. I looked at you're schmatic and see that the voltage acros R7 becomes 1,4V when there is a current flow of 3A.
I tried it and what happens.... When I measure the voltage accros R7 when there is a current of 3Amp, the voltage is -1.4V!!!!!!! How can it become negative???
audioguru2
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Hi Smit,
Don't connect the negative lead of your meter to the output's 0V terminal, connect it to the rectifier's 0V at the negative wire of C1 like this:
View attachment 38009
Don't connect the negative lead of your meter to the output's 0V terminal, connect it to the rectifier's 0V at the negative wire of C1 like this:
View attachment 38009
Last edited by a moderator:
audioguru2
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I think I made an error on my previous calculation which I corrected.
It looks like the current-setting pot has scaling resistors such that the max current the project will try to give is much too high at 4.1A. It will be even higher if the 10k pot is actually higher than 10k. If the pot is 12k then the max current the project will try to produce is 4.74A!
The project will melt or smoke if the output voltage is low. The transformer will be overloaded.
I think a trimpot should be added in series with R18 to reduce the max current to 3A. A 4.7k or 5k trimpot wired as a rheostat would work well. ;D
View attachment 38010
It looks like the current-setting pot has scaling resistors such that the max current the project will try to give is much too high at 4.1A. It will be even higher if the 10k pot is actually higher than 10k. If the pot is 12k then the max current the project will try to produce is 4.74A!
The project will melt or smoke if the output voltage is low. The transformer will be overloaded.
I think a trimpot should be added in series with R18 to reduce the max current to 3A. A 4.7k or 5k trimpot wired as a rheostat would work well. ;D
View attachment 38010
Hi!!
I replaced some components with the better components mentioned by audioguru. The BZX zener and the both TIP31A's also some resistors. But my LED won't light!! The output from U3 (pin6) is 34V when pin2 is in voltage value above pin3 Pin6 remains 34V. but when I turn back P2, pin2 of U3 gets below pin3 and the ouput, pin6, is decreasing til say, 26V. Is this in theory enough to light the LED?
I replaced some components with the better components mentioned by audioguru. The BZX zener and the both TIP31A's also some resistors. But my LED won't light!! The output from U3 (pin6) is 34V when pin2 is in voltage value above pin3 Pin6 remains 34V. but when I turn back P2, pin2 of U3 gets below pin3 and the ouput, pin6, is decreasing til say, 26V. Is this in theory enough to light the LED?
audioguru2
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Hi Smit,
Your project's U3 and Q3 aren't working properly.
Check the pins layout of Q3 and the values of R19 and R20.
Q3 should turn on when the output of U3 drops 5V and more below the positive supply voltage.
Test Q3 by connecting a 10k resistor from its base to ground and it should turn on.
Maybe U3 is busted from having too high a supply voltage. Since its output is 34V then its positive supply must be about 35.5V, plus its negative supply of -5.6V equals a total supply for it of 41.1V. If it is a TL081 then its spec'd absolute max supply voltage is only 36V.
U2 has the same high supply voltage problem.
That's why I recommend using ICs that have a higher supply voltage rating in this project.
What is your project's positive supply voltage without a load? If it is low enough when added to the negative supply voltage then common opamps rated for a 44V supply voltage could be used. ;D
Your project's U3 and Q3 aren't working properly.
Check the pins layout of Q3 and the values of R19 and R20.
Q3 should turn on when the output of U3 drops 5V and more below the positive supply voltage.
Test Q3 by connecting a 10k resistor from its base to ground and it should turn on.
Maybe U3 is busted from having too high a supply voltage. Since its output is 34V then its positive supply must be about 35.5V, plus its negative supply of -5.6V equals a total supply for it of 41.1V. If it is a TL081 then its spec'd absolute max supply voltage is only 36V.
U2 has the same high supply voltage problem.
That's why I recommend using ICs that have a higher supply voltage rating in this project.
What is your project's positive supply voltage without a load? If it is low enough when added to the negative supply voltage then common opamps rated for a 44V supply voltage could be used. ;D
with a 10k resistor on the base of Q3 the LED lights! I've tested my opamp (U3) little comparator test, and this one still works. (AD711 Analog Devices) But my Supply voltage goes down... the voltage connected to the collector of the 2N3055...
audioguru2
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If your AD711 opamp has more thn its absolute max supply voltage of 36V, it might stop working.
How many volts is your positive and negative suppies without a load on the project?
Why does your supply voltage drop and how many volts does it drop? It is normal for it to drop a couple of voltas when the project has a 3A load.
How many volts is your positive and negative suppies without a load on the project?
Why does your supply voltage drop and how many volts does it drop? It is normal for it to drop a couple of voltas when the project has a 3A load.
