That is the current at which the zener impedance is measured. It has nothing to do with the zener current capacity.
Take a 250mW 12V zener. How much current at 12V will dissipate 250mW? That's the max in-service current allowed for a 250mW zener.
What is the max current that the zener will see in a superreg, in series with the opamp? What is the max current the opamp can draw through the zener? Hint: the current into the opamp output needs to come from the current source supplying the series device base/opamp output.
Jan
Take a 250mW 12V zener. How much current at 12V will dissipate 250mW? That's the max in-service current allowed for a 250mW zener.
What is the max current that the zener will see in a superreg, in series with the opamp? What is the max current the opamp can draw through the zener? Hint: the current into the opamp output needs to come from the current source supplying the series device base/opamp output.
Jan
Hi Jan,
Sorry, I am not an electronics engineer, just a retired audiophile. All I want is to hear a good sound before I kick the bucket. So my worst nightmare is to buy wrong parts and see my money and work going up in smoke. This is why I appreciate your help so much!
My guess is that the maximum opamp current is the input-output voltage difference divided by R1 (249). The article states 5V; so the answer is 20mA.
I get parts from Mouser whose search tool lists “test current” and “zener current”. I believe the latter refers to the maximum current through the diode.
So I probably need a zener with the maximum current about twice of expected, in this case 40-50mA.
Please let me know if this is correct or if I got it wrong. Thanks again for all your help!
Alex
Sorry, I am not an electronics engineer, just a retired audiophile. All I want is to hear a good sound before I kick the bucket. So my worst nightmare is to buy wrong parts and see my money and work going up in smoke. This is why I appreciate your help so much!
My guess is that the maximum opamp current is the input-output voltage difference divided by R1 (249). The article states 5V; so the answer is 20mA.
I get parts from Mouser whose search tool lists “test current” and “zener current”. I believe the latter refers to the maximum current through the diode.
So I probably need a zener with the maximum current about twice of expected, in this case 40-50mA.
Please let me know if this is correct or if I got it wrong. Thanks again for all your help!
Alex
If you look at for instance the superreg pos circuit in the diyaudio store, you see a current source around Q2. How much current? Assume the LED to have 2.6V, then across R1 you will have about 2V. 2V / 249R is about 8mA.
This 8mA divides between the base of the pass transistor or the opamp output via the zener, and by regulating this division, the thing regulates.
This also means of course that the max current the opamp/zener can absorb is 8mA. Any zener can handle that.
Be careful with specs in a Mouser list; the defining number and what it means is in the data sheet for the part, and even there you sometimes have errors.
Jan
This 8mA divides between the base of the pass transistor or the opamp output via the zener, and by regulating this division, the thing regulates.
This also means of course that the max current the opamp/zener can absorb is 8mA. Any zener can handle that.
Be careful with specs in a Mouser list; the defining number and what it means is in the data sheet for the part, and even there you sometimes have errors.
Jan
I know we have been discouraged in earlier posts to ask specific questions, but:
What resister and diode values do I need for +/-18V?
R5,6 and 7 & R12,13,14. D5 and D10
After 7 straight hours of reading this forum, comparing comments and suggestions to the supplied schematic on the store, reading the article, using the "Customizing the output voltage.." article to create a spreadsheet that I hoped would help me calculate, trying to model in LTSpice, I've hit the wall. Thanks in advance!!
What resister and diode values do I need for +/-18V?
R5,6 and 7 & R12,13,14. D5 and D10
After 7 straight hours of reading this forum, comparing comments and suggestions to the supplied schematic on the store, reading the article, using the "Customizing the output voltage.." article to create a spreadsheet that I hoped would help me calculate, trying to model in LTSpice, I've hit the wall. Thanks in advance!!
You don't have to change the diodes, but only one resistor, either R5 or R6. You can use a 2k linear pot instead (set initially to 1k) to change the output voltage to 18V. Then measure the pot and you'll have your value.
Also note that for 18V you need at least 23-24V at the input, so you may want to use 35V caps instead of 25V.
I don't think I have discouraged asking questions - I think that is the purpose of this thread, to answer specific questions. No problem for me.
The previous post gives you the answer. Also, I understand you read my writeup in the diyaudio store on the customizing, I tried to make it clear how to set Vo.
If you could quote me a specific para that is unclear I could improve it. You can PM me if you prefer.
Jan
Specifically, is this unclear to anyone, can anyone suggest an improved wording:
The most important insight is that in normal operation, the two inputs to the opamp are pretty much equal in voltage. As the stock unit uses a 6.9V LM329 reference diode (D5), connected to the non-inverting input, immediately you know those opamp inputs need both to be at 6.9V.
The output voltage is divided down by R14 and R13 for the inverting input. So if you want 10V output, R14 and R13 need to be selected so that this 10V is divided down to 6.9V at the inverting input. So there must be 6.9V across R14 and (10-6.9=3.1) across R13. This give us a direct and simple equation: R13/R14=3.1/6.9.
Assuming you select R14 first, then R13=(3.1/6.9)*R14. Like when you select R14=4.9k, then R13 needs to be (3.1/6.9)*4.9=2.2k. (You may need to play a bit with the R14 value so R13 comes out an existing value).
Jan
The most important insight is that in normal operation, the two inputs to the opamp are pretty much equal in voltage. As the stock unit uses a 6.9V LM329 reference diode (D5), connected to the non-inverting input, immediately you know those opamp inputs need both to be at 6.9V.
The output voltage is divided down by R14 and R13 for the inverting input. So if you want 10V output, R14 and R13 need to be selected so that this 10V is divided down to 6.9V at the inverting input. So there must be 6.9V across R14 and (10-6.9=3.1) across R13. This give us a direct and simple equation: R13/R14=3.1/6.9.
Assuming you select R14 first, then R13=(3.1/6.9)*R14. Like when you select R14=4.9k, then R13 needs to be (3.1/6.9)*4.9=2.2k. (You may need to play a bit with the R14 value so R13 comes out an existing value).
Jan
I feel it would help some, or many, people if, as well as that explanation of the means of calculation the values, a table was listed showing the values of the parts required for a range of output voltages. Op-amps, voltages refs, resistors, and anything else.
That ideally would be added to post one of this thread by any admin or moderator with the vBulletin permissions to edit that post.
That ideally would be added to post one of this thread by any admin or moderator with the vBulletin permissions to edit that post.
I think a 10 year old would be able to to the calculations but we are lazy. Personally I have a created a spreadsheet for this but the difference is that I have trimming resistors in parallel and in series so it requires more work. The spreadsheet becomes very handy when I use the LM329 which have a low tolerance and I want a precise voltage.
Something like this ?
(Sqrt(power(10;log(((6.9+8)/6.9)*1000)))^2)-1000=1159R
Ya, give me the table with some common values... I never was very good with rocket surgery equations..
Alright, I started this mess, allow me to weigh in again. DIY is awesome, but I'm new to this. My recent dive into to circuit design/concepts was last winter when I built the Pass Korg B1 for my father. I know what the components on their own do and how they react to inputs. Understanding larger circuits beyond a simple opamp circuit is coming, slowly. Life is busy so when I come back to start another project, like this one, I need to refresh myself on some of the concepts. From "Customizing the output voltage" article I believe the following equations I've worked out are correct:
R14 = R13/((Vout-D5)/D5) and thus R13 = ((Vout-D5)/D5)xR14
So, to get 18V with a 6.9V reference diode, one possible solution is R13 = 1.6k, R14 = 1k
So, @peranders is correct, grade 5 stuff.
To confirm, and expand my understanding, I opened LTspice for the first time in a few months to build the circuit, using an example posted on this forum and that didn't give the expected results. Far from it. Then, comes the task of sourcing components. Resisters, of course, are easy. When the originally spec'd transistors and diodes can't be found, I didn't have the experience needed to select proper alternatives or the confidence to know where the rest of the circuit might need to be adapted to work with the alternate. As with most of you, no doubt, my goal is to learn, get the correct components and build carefully, then enjoy the music.
Someday, I hope to be one of those that can offer assistance to others on this forum. I'm not there yet.
R14 = R13/((Vout-D5)/D5) and thus R13 = ((Vout-D5)/D5)xR14
So, to get 18V with a 6.9V reference diode, one possible solution is R13 = 1.6k, R14 = 1k
So, @peranders is correct, grade 5 stuff.
To confirm, and expand my understanding, I opened LTspice for the first time in a few months to build the circuit, using an example posted on this forum and that didn't give the expected results. Far from it. Then, comes the task of sourcing components. Resisters, of course, are easy. When the originally spec'd transistors and diodes can't be found, I didn't have the experience needed to select proper alternatives or the confidence to know where the rest of the circuit might need to be adapted to work with the alternate. As with most of you, no doubt, my goal is to learn, get the correct components and build carefully, then enjoy the music.
Someday, I hope to be one of those that can offer assistance to others on this forum. I'm not there yet.
I can understand the difficulty to source parts. For instance, we here in Europe use the BC546 etc. series for almost anything that needs a small signal transistor, but they are unknown elsewhere on the world.
Specifying 'or equivalent' only is useful if you know other parts and relevant specs.
So here we can definitely do better.
Jan
Specifying 'or equivalent' only is useful if you know other parts and relevant specs.
So here we can definitely do better.
Jan