audioguru2
- Apr 6, 2004
- 12,026
- Joined
- Apr 6, 2004
- Messages
- 12,026
It is a current regulator, not a limiter. U3 reduces the output voltage so that the voltage across R7 (caused by output current) equals the voltage set by the current-setting pot.
Current regulator is a better term than current limiter. I stand corrected!
U3's output must go down to bring the difference between the inputs to 0. What I
cannot see is what is limiting U3's output to a mid level voltage. It seems to me that
any difference greater than 70uV should drive it to one rail or the other. The open loop
gain is too high.
Let me try a different question. If the power supply is in voltage regulation mode what voltage would you expect on the output of U3? What voltage would you expect when in current regulation mode?
Thanks,
Kevin
U3's output must go down to bring the difference between the inputs to 0. What I
cannot see is what is limiting U3's output to a mid level voltage. It seems to me that
any difference greater than 70uV should drive it to one rail or the other. The open loop
gain is too high.
Let me try a different question. If the power supply is in voltage regulation mode what voltage would you expect on the output of U3? What voltage would you expect when in current regulation mode?
Thanks,
Kevin
kcc
You are half right...
U3 "looks" like a comparator while it's not active. When U3 becomes active it behaves like a linear amplifier. At first glance, most would call it an integrator.
It does have a feedback loop... it's just a "big one" that includes D9, U2, R15, Q2, Q4 and R7.
The current limit pot set's a "reference" for U3. While the output current hasn't reached the the "setpoint", U3 indeed behaves like a comparator, sitting at the positive rail. As the output current increases, so does the voltage drop across R7. When the voltage at the negative input of U3, pin 2, reaches the "setpoint" the output comes out of the + rail and starts to go towards is minus rail. The rate at which it does this is controlled by the integrating cap C8 (C8 is also feedback). The gain of U3 in this circuit isn't as high as you might think when it's "active" and not sitting at the positive rail!! As the output falls, D9, a clamping diode, starts to pull down U2 pin 3, but the voltage at pin 3 is what controls the output voltage. So as the load tries to draw more current, U2 pin 3 is pulled lower and lower. This in turn starts to turn off Q4... via Q2. So, as the load tries to draw more current the output voltage will be reduced, just enough, such that both inputs to U3 are at held at the same voltage (feedback).
You are half right...
U3 "looks" like a comparator while it's not active. When U3 becomes active it behaves like a linear amplifier. At first glance, most would call it an integrator.
It does have a feedback loop... it's just a "big one" that includes D9, U2, R15, Q2, Q4 and R7.
The current limit pot set's a "reference" for U3. While the output current hasn't reached the the "setpoint", U3 indeed behaves like a comparator, sitting at the positive rail. As the output current increases, so does the voltage drop across R7. When the voltage at the negative input of U3, pin 2, reaches the "setpoint" the output comes out of the + rail and starts to go towards is minus rail. The rate at which it does this is controlled by the integrating cap C8 (C8 is also feedback). The gain of U3 in this circuit isn't as high as you might think when it's "active" and not sitting at the positive rail!! As the output falls, D9, a clamping diode, starts to pull down U2 pin 3, but the voltage at pin 3 is what controls the output voltage. So as the load tries to draw more current, U2 pin 3 is pulled lower and lower. This in turn starts to turn off Q4... via Q2. So, as the load tries to draw more current the output voltage will be reduced, just enough, such that both inputs to U3 are at held at the same voltage (feedback).
Indulis:
I've been pondering on this all weekend.
For the feedback loop to include D9, U2, R15, Q2, and Q4 wouldn't it have to include
the load as well?? I don't see any path back to U3(-) input that doesn't include the load?
Any guess as to what the gain would be?
I'm currently looking hard at the LED branch since R19, R20, Q3, R22, D12 AND R7 would make a feedback loop. There was a comment in one of the threads that all of those components were part of the current regulator (of course I can't find it now :-( )
Any thoughts on that path?
Thanks,
Kevin "Still thoroughly confused but trying hard" C.
I've been pondering on this all weekend.
For the feedback loop to include D9, U2, R15, Q2, and Q4 wouldn't it have to include
the load as well?? I don't see any path back to U3(-) input that doesn't include the load?
Any guess as to what the gain would be?
I'm currently looking hard at the LED branch since R19, R20, Q3, R22, D12 AND R7 would make a feedback loop. There was a comment in one of the threads that all of those components were part of the current regulator (of course I can't find it now :-( )
Any thoughts on that path?
Thanks,
Kevin "Still thoroughly confused but trying hard" C.
Let's see... how can I explain this...
If the output of a device (in this case the opamp U3) can somehow influence the levels at it's input pin(s), it's feedback.
Because the voltage level at U3 pin 6 can infuence/control the voltage on U3 pin 2, all the elements in that string are part of te feedback. Since the load is what's causing the current limit circuit (sorry, current regulator) to become active, yes, the load should be an element of the loop.
I'm afraid that my days of writing transfer functions with that many variables is long gone (killed off those brain cells LONG AGO)!!! Aside from an academic exercise, why would you want to know the gain? If for some reason I "had to know", I'd run it through a simulator.
If your having trouble understanding how the circuit works, break it up into parts. For example, redraw the schematic without D12, R22,Q3, R19, R20, C8, U3, R17, R21, P2, R17 and D9. Look only at the "voltage loop", then add in the other suff. BTW... guess what, that circuit will work just fine with all those components removed, you just won't have "current limit".
If the output of a device (in this case the opamp U3) can somehow influence the levels at it's input pin(s), it's feedback.
Because the voltage level at U3 pin 6 can infuence/control the voltage on U3 pin 2, all the elements in that string are part of te feedback. Since the load is what's causing the current limit circuit (sorry, current regulator) to become active, yes, the load should be an element of the loop.
I'm afraid that my days of writing transfer functions with that many variables is long gone (killed off those brain cells LONG AGO)!!! Aside from an academic exercise, why would you want to know the gain? If for some reason I "had to know", I'd run it through a simulator.
How can any of those components "control/change" the voltage on the input pins to U3? They can't, so they are not part of U3's feedback. Those components only control an indicator. For the sake of argument, lets rationalize it was in the loop. For that to be true, the current flowing thru D12 (LED) would have to go back towards R7... it doesn't, it flow towards C1. In fact, you could rip out all those components and the circuit would work just fine!!
If your having trouble understanding how the circuit works, break it up into parts. For example, redraw the schematic without D12, R22,Q3, R19, R20, C8, U3, R17, R21, P2, R17 and D9. Look only at the "voltage loop", then add in the other suff. BTW... guess what, that circuit will work just fine with all those components removed, you just won't have "current limit".
I was asking the gain in hopes of getting to understand what is confusing me a litte better. To that extent it IS an academic exercise. Unfortunately I've never worked with a simulator so I don't even know how difficult that would be.indulis said:
Yes. I realized that mistake after I had posted. I see it is not part of the feedback as the current would be flowing the wrong way
I understand how the voltage regulation works. What I am questioning is what keeps U3 from acting as a comparator when current regulation kicks in. Your description about it being at the positive rail when R7 voltage drop is low makes sense but left me with the question about current regulation. I keep getting stuck on how far can U3's output come down from the voltage rail before the pass transistor is cut off. It has to be greater than 0.7 (otherwise Q3 would not turn on and the LED wouldn't lite) but shouldn't go all the way down to negative voltages. This is how I keep coming back to the gain in linear mode. Unfortunately, despite excellent descriptions by both you and audioguru, I'm still not quite cathching it.
For the moment I'm going off to review the Art of Electronics chapter on op-amps and pay particular attention to the integrator circuit portion (since C8 seems to be the configuration for the most obvious feedback). Hopefully this will trigger some understanding! Wish me luck!
When I can describe this to myself I will be happy. Until then, the pondering continues.
Thanks for the help!!!
Kevin
Kevin
You say you understand how the "voltage loop" functions... good!! Looking at just the voltage loop, consider the function of D9. It's connected to the U2 pin 3 node, and the voltage at this node is what controls the output voltage. Once the voltage on the the cathode of D9 falls (is pulled down) a "diode drop" below the "programed voltage" on U2 pin 3 voltage, the "clamping action" will start. At this point, the voltage on the cathode has taken over control of the output voltage. The lower the cathode voltage goes the lower the output voltage goes.
Consider what can change the current through the load. If you have a fixed load, and the output voltage is going down, the current flowing through the load will also go down. If you have a fixed output voltage, changing the load will cause a change in the output current... decrease the load resistance, current goes up and increasing the load restistance makes the current go down (Ohms Law).
Back to the circuit... say you set the output voltage to some value. That will equate to some voltage at U2 pin 3. Now, if you were to connect a variable power supply to the cathode of D9 and the votlage of that supply was set to be more than a diode drop above the nodal voltage at U2 pin 3 (or a voltage like a opamp at it's + rail) , it would have no affect because the diode is back biased (it's off). But if you were to start decreasing that supply connected to D9 at some piont the diode will turn on and "clamp" the U2 nodal value to that voltage and cause a decrease in the output voltage. I hope you can rationalize this action in your head.
Next... R7 in this circuit, by ohms law, developes a voltage across it that is propotional to the current that is flowing in the output (a.k.a. current sense resistor). The more current that flows, the bigger the voltage drop. Now we're back to U3... the arm of P2 applies a reference voltage to U3 pin 3, so as long as the voltage on U3 pin 2 is below that value, U3 looks like a comparator at it's + rail. As the load current is increased, the voltage across R7 increases until the voltage on U3 pin 2 goes just a little higher than the voltage on pin 3 and the output comes off the rail and starts down. You ask how far will it come down... the output of U3 will go down to such a value that the output voltage is reduced enough to cause the voltage drop across R7 to make the voltage at U3 pin 2 be EXACTLY the same as the voltage on pin 3.
Does this help?
You say you understand how the "voltage loop" functions... good!! Looking at just the voltage loop, consider the function of D9. It's connected to the U2 pin 3 node, and the voltage at this node is what controls the output voltage. Once the voltage on the the cathode of D9 falls (is pulled down) a "diode drop" below the "programed voltage" on U2 pin 3 voltage, the "clamping action" will start. At this point, the voltage on the cathode has taken over control of the output voltage. The lower the cathode voltage goes the lower the output voltage goes.
Consider what can change the current through the load. If you have a fixed load, and the output voltage is going down, the current flowing through the load will also go down. If you have a fixed output voltage, changing the load will cause a change in the output current... decrease the load resistance, current goes up and increasing the load restistance makes the current go down (Ohms Law).
Back to the circuit... say you set the output voltage to some value. That will equate to some voltage at U2 pin 3. Now, if you were to connect a variable power supply to the cathode of D9 and the votlage of that supply was set to be more than a diode drop above the nodal voltage at U2 pin 3 (or a voltage like a opamp at it's + rail) , it would have no affect because the diode is back biased (it's off). But if you were to start decreasing that supply connected to D9 at some piont the diode will turn on and "clamp" the U2 nodal value to that voltage and cause a decrease in the output voltage. I hope you can rationalize this action in your head.
Next... R7 in this circuit, by ohms law, developes a voltage across it that is propotional to the current that is flowing in the output (a.k.a. current sense resistor). The more current that flows, the bigger the voltage drop. Now we're back to U3... the arm of P2 applies a reference voltage to U3 pin 3, so as long as the voltage on U3 pin 2 is below that value, U3 looks like a comparator at it's + rail. As the load current is increased, the voltage across R7 increases until the voltage on U3 pin 2 goes just a little higher than the voltage on pin 3 and the output comes off the rail and starts down. You ask how far will it come down... the output of U3 will go down to such a value that the output voltage is reduced enough to cause the voltage drop across R7 to make the voltage at U3 pin 2 be EXACTLY the same as the voltage on pin 3.
Does this help?
Hi all,
I have a few question regarding to the modified version PSU. I noticed that the out for U1 is 11.2V and it's operating in fixed configuration. Can a high precision OPA 27 or OPA37 being used instead? another question is regarding to the TIP31 NPN Transistor for Q1 and Q2, a compatible part is BD131 NPN. But can a BD132 PNP be used instead? BD132 is the complementary part of BD131. Thanks in advance.
I have a few question regarding to the modified version PSU. I noticed that the out for U1 is 11.2V and it's operating in fixed configuration. Can a high precision OPA 27 or OPA37 being used instead? another question is regarding to the TIP31 NPN Transistor for Q1 and Q2, a compatible part is BD131 NPN. But can a BD132 PNP be used instead? BD132 is the complementary part of BD131. Thanks in advance.
audioguru2
- Apr 6, 2004
- 12,026
- Joined
- Apr 6, 2004
- Messages
- 12,026
An OPA27 might work but a de-compensated OPA37 cannot be used. It will oscillate.adiankhoo said:
No, the circuit is designed for an NPN transistor.
Matzegrufti
- Feb 18, 2007
- 1
- Joined
- Feb 18, 2007
- Messages
- 1
only one short question: must R7 be 0,47 Ohm or can it also be 0,51 Ohm?
audioguru2
- Apr 6, 2004
- 12,026
- Joined
- Apr 6, 2004
- Messages
- 12,026
If R7 is 0.47 ohms then the max output current is supposed to be 3A but people say it is higher. 0.51 ohms will reduce the max current which might be 3A or a little higher. Try it.
Indulis
Sorry to drop out like that but the family and I nicked off for a short holiday to New Zealand.
But this is exactly what is my trouble (and why I started this by asking about the gain) because it seems to me that you'd hit the negative rail pretty quickly if the gain is not fairly low.
The current-set pot appears to be a simple voltage divider between 11.2V and 0V which allows the non-inverting input to vary from 5.6mV to 1.7V (Assuming perfect resistors, etc. etc). This is close to the range that R7 should drop (2mA to 3A should drop 940uV to 1.41). At the low end (2mA setting) the op-amp cannot have a gain of more than about 6 (5.6mV / 940uV) or U3 rides the rails.
However when you look at the U3 from the output end, this gain doesn't make sense. In order for current regulation to be active, the output of U3 is limited to Vmax - 4V (Anything greater and Q3 is in cutoff due to biasing of R19 and R20) or in this case 38V - 4V). If U3 output is 28V then the gain of U3 is more like 5000. Of course if you stuck with a gain of 6 then the output of U3 would be about 33mV and the circuit would still work.
It is these different possibilities that still have me wondering what the actual gain is and how it can be analysed.
Thanks,
Kevin
Sorry to drop out like that but the family and I nicked off for a short holiday to New Zealand.
The entire statement makes perfect sense. The basic workings of an op-amp reduce the difference between the inputs to zero thru the feedback network. For an op-amp the output must be at a rail voltage if the difference between the inputs cannot be taken to zero. (Isn't that the second of the five ideal op-amp rules?)indulis said:
But this is exactly what is my trouble (and why I started this by asking about the gain) because it seems to me that you'd hit the negative rail pretty quickly if the gain is not fairly low.
The current-set pot appears to be a simple voltage divider between 11.2V and 0V which allows the non-inverting input to vary from 5.6mV to 1.7V (Assuming perfect resistors, etc. etc). This is close to the range that R7 should drop (2mA to 3A should drop 940uV to 1.41). At the low end (2mA setting) the op-amp cannot have a gain of more than about 6 (5.6mV / 940uV) or U3 rides the rails.
However when you look at the U3 from the output end, this gain doesn't make sense. In order for current regulation to be active, the output of U3 is limited to Vmax - 4V (Anything greater and Q3 is in cutoff due to biasing of R19 and R20) or in this case 38V - 4V). If U3 output is 28V then the gain of U3 is more like 5000. Of course if you stuck with a gain of 6 then the output of U3 would be about 33mV and the circuit would still work.
It is these different possibilities that still have me wondering what the actual gain is and how it can be analysed.
Thanks,
Kevin
audioguru2
- Apr 6, 2004
- 12,026
- Joined
- Apr 6, 2004
- Messages
- 12,026
Hi Kevin,
U3 has an open-loop voltage gain of about 200,000. It has plenty of negative feedback from the current in R7 for it to be linear and reduce the voltage to the input of the output section to be less than 11.2V. The output section has a voltage gain of nearly 3.1.
U3 doesn't switch, it is a linear amplifier. If the current is too high for the setting of the pot then it reduces the output voltage so the current is what is needed. If the current is less than the setting of the pot then the output of U3 goes high so it doesn't do anything.
U3 has an open-loop voltage gain of about 200,000. It has plenty of negative feedback from the current in R7 for it to be linear and reduce the voltage to the input of the output section to be less than 11.2V. The output section has a voltage gain of nearly 3.1.
U3 doesn't switch, it is a linear amplifier. If the current is too high for the setting of the pot then it reduces the output voltage so the current is what is needed. If the current is less than the setting of the pot then the output of U3 goes high so it doesn't do anything.
Hi Kevin
Just got back from a 10 day holiday myself.
Forget Q3, R19 and 20. That part of the circuit only controls the D12 indicator (LED). It has nothing to do with current regulation or the gain. U3 is configured (when active) like a gain limited integrator. For simplicity just think of it as some resistance (impedance) that is in parallel with C8. C8 controls the rate at which the output of U3 changes, make C8 bigger and the dv/dt at the output of U3 will be slower, make the cap smaller and the output will swing faster. All of the other stuff in the loop just reduces the gain and for DC purposes can be thought of as a resistor (the loop also has AC gain). The transfer function for the loop is quite complex and I have no desire to "hurt my brain" that much trying to derive it. For this circuit it
View attachment 40383
View attachment 40384
Just got back from a 10 day holiday myself.
Forget Q3, R19 and 20. That part of the circuit only controls the D12 indicator (LED). It has nothing to do with current regulation or the gain. U3 is configured (when active) like a gain limited integrator. For simplicity just think of it as some resistance (impedance) that is in parallel with C8. C8 controls the rate at which the output of U3 changes, make C8 bigger and the dv/dt at the output of U3 will be slower, make the cap smaller and the output will swing faster. All of the other stuff in the loop just reduces the gain and for DC purposes can be thought of as a resistor (the loop also has AC gain). The transfer function for the loop is quite complex and I have no desire to "hurt my brain" that much trying to derive it. For this circuit it
View attachment 40383
View attachment 40384
Hi to everyone
I've decided to build the modified version of 0-30V power supply. I've got 30V 160VA transformer and the op-amps OPA445AP.
I have BZX85C5V6 zener diodes at home. Could I use this diode instead of BZX79C5V6?
Also I'm not sure what type of heatsink to use for the output transistors. I usually shop on www.rswww.com but I don't know what thermal resistance C/W parameter to choose. I would be very glad if someone can suggest some type.
Thank you in advance.
I've decided to build the modified version of 0-30V power supply. I've got 30V 160VA transformer and the op-amps OPA445AP.
I have BZX85C5V6 zener diodes at home. Could I use this diode instead of BZX79C5V6?
Also I'm not sure what type of heatsink to use for the output transistors. I usually shop on www.rswww.com but I don't know what thermal resistance C/W parameter to choose. I would be very glad if someone can suggest some type.
Thank you in advance.
audioguru2
- Apr 6, 2004
- 12,026
- Joined
- Apr 6, 2004
- Messages
- 12,026
Hi Valorous,
Welcome to our forum. ;D
I recommended using the BZX79C zener diode because it is spec'd at a current of 5mA and when the value of R4 is reduced to 1k as I recommended then it operates very well at 5.6mA.
The original value of 4.7k for R4 had the zener diode operating at a very low current of only 1.2mA where most don't regulate properly.
Your BZX85 zener diodes are spec'd at a very high current of 45mA and their datasheet shows that they barely regulate at less than about 20mA. The opamp cannot drive it at 20mA or more.
The two 2N3055 output transistors must dissipate a max power of 60W each. Their max internal temperature is 200 degrees C and the ambient might be 30 degrees C so their temperature rise can be no more than 170 degrees C. They have a thermal resistance of 1.5 degrees C/W and thermal grease will add 0.5 degrees C. The max thermal resistance of the heatsink can be 170 - (2 x 60)/60= 0.83 degrees C/W for each transistor. Use two heatsinks, one for each transistor and insulate the heatsinks not the transistors. You can use smaller heatsinks if a fan is used.
View attachment 40385
Welcome to our forum. ;D
I recommended using the BZX79C zener diode because it is spec'd at a current of 5mA and when the value of R4 is reduced to 1k as I recommended then it operates very well at 5.6mA.
The original value of 4.7k for R4 had the zener diode operating at a very low current of only 1.2mA where most don't regulate properly.
Your BZX85 zener diodes are spec'd at a very high current of 45mA and their datasheet shows that they barely regulate at less than about 20mA. The opamp cannot drive it at 20mA or more.
The two 2N3055 output transistors must dissipate a max power of 60W each. Their max internal temperature is 200 degrees C and the ambient might be 30 degrees C so their temperature rise can be no more than 170 degrees C. They have a thermal resistance of 1.5 degrees C/W and thermal grease will add 0.5 degrees C. The max thermal resistance of the heatsink can be 170 - (2 x 60)/60= 0.83 degrees C/W for each transistor. Use two heatsinks, one for each transistor and insulate the heatsinks not the transistors. You can use smaller heatsinks if a fan is used.
View attachment 40385
Thank you audioguru for your answers. I've decided to use a 12V fan. Would a heatsinks with thermal resistance of roughly 1 C/W be enough?
I'm going to use the unregulated voltage from the circuit to drive the fan, but I'm not sure how to bring the voltage down to 12V.
I'm going to use the unregulated voltage from the circuit to drive the fan, but I'm not sure how to bring the voltage down to 12V.
audioguru2
- Apr 6, 2004
- 12,026
- Joined
- Apr 6, 2004
- Messages
- 12,026
Two heatsinks of 1 degree C/W would be fine with a fan.
You will have 28V to throw away if you use a 12V fan on the 40V unregulated voltage. More heat.
You will have 28V to throw away if you use a 12V fan on the 40V unregulated voltage. More heat.
Hello, I'm new here, so I greet everyone!
I found this project,and accidentally found this thread, I began reading but there are 85+ pages
, so plz, can someone tell me is anything wrong with this circuit?!
In first few pages there was something about transformer and diodes,and in last some modifications?! ???
I found this project,and accidentally found this thread, I began reading but there are 85+ pages
, so plz, can someone tell me is anything wrong with this circuit?!In first few pages there was something about transformer and diodes,and in last some modifications?! ???
