I have checked the circuitry and it doesn't have any problems in connection. However could it be the problem of the op-amps i am using?
http://www.datasheetcatalog.org/datasheet/texasinstruments/lm358.pdf

Just to get it straight, U1 is U2 right.. Not a separate op-amp right?

 

Look at the datasheet of an LM358 dual opamp to see which pins do what.

 

Look at the datasheet of an LM358 dual opamp to see which pins do what.

Yup i have done it and circuitry seems fine to me without any connection error.

 

In the last schematic Bill Marsden posted, U1 is a 7809 regulator. All of the U2's are in one LM324 opamp. Since your LM358 has only two operational amplifiers in the IC, you will use them both.

 

You have a wiring error, there isn't any doubt. There is no way you can have 13V on R4 and R5, anywhere. One side of R4 is 9V, the other is 6.3 Volts. This means R5 adjusts between 0V - 6.3 Volts.

Parts List:

U1 : 7809
U2 : LM324
Q1 : 2N2907
Q2,3 : 2N6427 Qty 1
CR1,3,4 : 1N4001
C1 : 100µF
C2 :0.1µF
R1 : 100Ω
R2,5 : 10KΩ Variable
R3 : 21KΩ
R4 : 4.3KΩ
R6 : 100KΩ
R7 : 10KΩ
R8 : 1KΩ

I'm not too sure about R6, it may need to be lowered a bit. But until we have the correct voltage on R2 and R3, and R4 and R5 we are spinning our wheels. I notice you didn't answer the question about CR1. You do have one, right? And what is the voltage on the output of U1?

Question for the other guys, since we are using an LM358, can we eliminate CR3 and CR4?

 

You have a wiring error, there isn't any doubt. There is no way you can have 13V on R4 and R5, anywhere. One side of R4 is

Parts List:

U1 : 7809
U2 : LM324
Q1 : 2N2907
Q2,3 : 2N6427 Qty 1
CR1,3,4 : 1N4001
C1 : 100µF
C2 :0.1µF
R1 : 100Ω
R2,5 : 10KΩ Variable
R3 : 21KΩ
R4 : 4.3KΩ
R6 : 100KΩ
R7 : 10KΩ
R8 : 1KΩ

I'm not too sure about R6, it may need to be lowered a bit. But until we have the correct voltage on R2 and R3, and R4 and R5 we are spinning out wheels. I notice you didn't answer the question about CR1. You do have one, right? And what is the voltage on U1?

yup for CR1,3 and 4 i am using a 1N4001 at the moment which is available for me. I will test it out tomorrow and see if it follows to the measurement again which you have stated or not maybe rebuild it.

But i am aware of the connection path from CR1 to R4 is not connected to the collector of Q1 but connected to R1. Other then that there is nothing really major to be taken note of.

For the voltage readings, i will have to get it done tomorrow straight away once i reach school. Maybe you can jot down the important points that i can measure so that a comparison can be done so that i know whether i am at the right path.

Other then
R2 wiper : 6.3 - 8.3 volts
R5 wiper : 0 - 6.2 volts, maybe i can do with a bit more important points to measure to compare..

 

The output of U1 is a key point for the rest of the circuit. It is the power supply for all the transistors. This is why I keep asking what the voltage is there. You have C1, C2, R8, R1, and the collectors of Q3, Q4 all connected to this point.

 

The output of U1 is a key point for the rest of the circuit. It is the power supply for all the transistors. This is why I keep asking what the voltage is there. You have C1, C2, R8, R1, and the collectors of Q3, Q4 all connected to this point.

Ok will update you on my measurement soon..

 

Yes, eliminate CR3 and CR4.

 

The output of U1 is a key point for the rest of the circuit. It is the power supply for all the transistors. This is why I keep asking what the voltage is there. You have C1, C2, R8, R1, and the collectors of Q3, Q4 all connected to this point.

Ok it works, wookie was correct with the elimination of the CR3 and CR4 which was stated optional but however i connected it. I am able to get a voltage of 5V and 20mA at the output now.

Anyway i was just like to know what are the pros and cons of this circuitry in stability if there is any?

 

CR3 and CR4 wouldn't have anything to do with the functioning, it was to allow less capable op amps to work. You can drop the power supply to around 12 V without them.

Do you have the full range of adjustment? I am a little concerned about R6 being too large, but if it works, good enough.

Something I held off mentioning, you can program the parameters with voltage only. This sounded like something you would like, the trick is to use the ranges I gave you on the + side of the op amps.

The values you have picked are very low, the stability should be excellent.

 

CR3 and CR4 wouldn't have anything to do with the functioning, it was to allow less capable op amps to work. You can drop the power supply to around 12 V without them.

Do you have the full range of adjustment? I am a little concerned about R6 being too large, but if it works, good enough.

Something I held off mentioning, you can program the parameters with voltage only. This sounded like something you would like, the trick is to use the ranges I gave you on the + side of the op amps.

The values you have picked are very low, the stability should be excellent.

Well the ranges i am getting with a full knot turn on the potentiometer is 0v to 21.09mA for the current range and 0v to 5.32V for the voltage range. If i had not remove CR3 and CR4, i would have got a voltage of 14v across the output of the op-amps.

Anyway just for my understanding, The 2 op-amp as u mention are connected to as voltage followers or buffers, while R3,R2, CR1 , R5 and R4 are connected together to act as a voltage divder between the current regulator and the voltage regulator.

While the front part of the circuitry is a 9v regulator which provides a higher voltage for the voltages on the PN junctions of the transistor as 5V will be too little.

It is also possible to adjust the voltage ranges (say to 24V) by adjusting R6 and still leaving the current range of 20mA right?

 

While the front part of the circuitry is a 9v regulator which provides a higher voltage for the voltages on the PN junctions of the transistor as 5V will be too little.

U1 is also the voltage reference, which makes the rest of the circuit stable. If you go through the logic the voltage is pretty well used up, the 100Ω sense resistor R1 uses 2V at 20ma, and Q2/3 uses 1.2 volts for the base emitters.

Anyway just for my understanding, The 2 op-amp as u mention are connected to as voltage followers or buffers, while R3,R2, CR1 , R5 and R4 are connected together to act as a voltage divder between the current regulator and the voltage regulator.

That pretty well sums it up.

Well the ranges i am getting with a full knot turn on the potentiometer is 0v to 21.09mA for the current range and 0v to 5.32V for the voltage range. If i had not remove CR3 and CR4, i would have got a voltage of 14v across the output of the op-amps.

I suspect one of the diodes CR3 or CR4 was either open or backwards. They really don't make that much difference except to a marginal op amp. They would each drop 0.6 volts, and that was all.

It is also possible to adjust the voltage ranges (say to 24V) by adjusting R6 and still leaving the current range of 20mA right?

Yes, each control is basically independent.

For what it's worth, I think I like the LM358 as a alternate. May use it in some of my other designs.

 

Anyway thanks alot!!! =) ur designs were brilliant.. Anyway your idea came out from a power supply right, if i recalled correctly. Power supply course 102?

 

Anyway as mention of adjusting R6 to tune the voltage of the output to a higher voltage . Is there any other components that i must take note that must be changed to say a higher volt of 10V or 24V. changing the 7809 to a 7815 must be use in a way for a 10V Output right?

Just to make sure i understand the circuitry, i have a few questions about the components which are used.

1.
U2 forms a very good high impedance buffer, which allow us to use resistors we want as opposed to letting the transistors gain force us to use low values.
Do you mean that U2 actually produces a high impedances which is what we need. And this will have an effect to R6 value that was selected?
A voltage follower can function as an interface impedance. (Connecting high impedance to low impedance) Does R6 in this case not only help the voltage follower to be in a stable state but also prevent the transistor from gaining any more current through the base?

2.
Can the Current Regulator and the Voltage Regulator be swoop in the opposite direction. Meaning from the left, the 7809 will come first followed by the voltage regulator and then the current regulator?

3.
U1 plays an important role as the a voltage reference just like the 7809, however the voltage follower has a task to supply the current regulator with a base current and is also linked to R5 through the voltage divider to U2. I would just like to know in detail in a easier way what is the importances of having U1 beside just providing a voltage references and also the voltage divider.

 

Those are a lot deeper questions than I think you realize, encompasing several types of electronics. I'll do my best to answer them, but to understand the answers you'll have to do some serious reading.

The voltage divider is the basis for almost all controls, one way or another. It uses ohms law, E=IR. E=Voltage, I = Current, and R= Resistance. You can tweak this formula so if you have any two values the 3rd just drops out. With a voltage divider you can create any stable voltage using a fixed voltage you need.

This may seem simplistic, but without a complete understanding of this root principle you will be stuck when it comes to electronics, or figuring out anything else related to electronics. The AAC book online has many chapters devoted to teaching these principles and their applications.

Diodes and transistors are related. They both have PN junctions. Again, this is way over simplified, but in general a PN junction using a silicon device (most of what is out there) drops .7V. They drop the same amount too. Since the voltage is constant, and they drop the same amount, you can use diodes to offset the Base Emitter of a transistor. This is why there is CR1 matching Q1. When calculating the voltage divider you subtract CR1 drop (a constant voltage) from the power supply, and use the remander to calculate the voltage divider R2 and R3. Even the wiper of R2 is further use of the voltage divider, giveing us our range. Same with R4 and R5.

Op Amps are another course in themselves. They are high input impedance and low output impedance. The voltage follower is taking this to the extreme.

Op ams are so high impedance they have to have a resistance to ground just for reference. Figure billions of ohms input, and milliohms out. That and not all op amps are created equal, which is why I kept playing around with CR3 and CR4. By the way, that resistor shown above can be a megohm or more.

Then there are transistors. We used two different configurations here, but they are related. I wrote a related article here that covers some of it. A transistor is fundimentally a constant current device, but it can be made into a voltage follower also.

Using this diagram as a guide...

Re seen through the base is the gain of the transistor ( β , or beta) times Re. So if you put a constant voltage on the base the emitter is a constant voltage. This is a voltage follower. If you vary Re the voltage trys to stay the same, but remember, the resistance at base also varies. This is why R6 was an issue.

The same diagram also shows the constant current aspect of transistor design. If the voltage across Re stays the same then the current on the collector (through Rc) doesn't vary, no matter what Rc value is.

All of this is in the AAC Book. Well worth the read.

I learned something doing this, we always do. To learn, teach. You can access stuff I find interesting through my blog.

 

Anyway as mention of adjusting R6 to tune the voltage of the output to a higher voltage . Is there any other components that i must take note that must be changed to say a higher volt of 10V or 24V. changing the 7809 to a 7815 must be use in a way for a 10V Output right?

Yes.

Just to make sure i understand the circuitry, i have a few questions about the components which are used.

1.
U2 forms a very good high impedance buffer, which allow us to use resistors we want as opposed to letting the transistors gain force us to use low values.
Do you mean that U2 actually produces a high impedances which is what we need. And this will have an effect to R6 value that was selected?
A voltage follower can function as an interface impedance. (Connecting high impedance to low impedance) Does R6 in this case not only help the voltage follower to be in a stable state but also prevent the transistor from gaining any more current through the base?

Like I said above, the Re is transfered through. If Re is 0Ω and you're current regulator is set for 0 amps, the the only current source is what comes through R6. We want to get this as low as possible, so we set R6 for as much as we can get by with.

Q2 and Q3 are a Darlington pair. Since the resistance of Re transfers though depending on the gain of the transistor, we raise the gain of the transistor as high as possible. If the gain of each transistor is 50, then the gain of the pair is 50 X 50, or 2500.

2.
Can the Current Regulator and the Voltage Regulator be swoop in the opposite direction. Meaning from the left, the 7809 will come first followed by the voltage regulator and then the current regulator?

No, the current regulator output absorbs a variable voltage. The amount of drop between the input (R1) and output (Collector of Q1) varies a lot depending on the setting of the current regulator.

3.
U1 plays an important role as the a voltage reference just like the 7809, however the voltage follower has a task to supply the current regulator with a base current and is also linked to R5 through the voltage divider to U2. I would just like to know in detail in a easier way what is the importances of having U1 beside just providing a voltage references and also the voltage divider.

You need a voltage reference, no getting around it. This is basic math, we need to translate as many variables into constants as possible, otherwise you could never get repeatable controls.

 

Hey thanks alot,

i will read up more to know in detail how the circuit works, but in the meantime can i also ask u for the calculation u did on the resistors and how u manged to get the 5V range and the 20mA. I roughly got the idea theoritically, but need to have a references to know if i am right.

 

Q2 and Q3 are voltage followers. Voltage in = Voltage Out minus Base Emitter drop (in this case it is 1.2 Volts). So if you want 5 volts out you need 6.2 volts in. This sets the voltage you need for R5, from there calculating R4 is easy.

The current regulator is a bit harder. The emitter resistor and the base voltage programs the current. With a 100Ω resistor 20 ma is 2Volts across the resistor. You just figure from there.

In both cases this goes back to understanding voltage dividers. Have you read the AAC book on resistors?

 

hmm alright will be doing it so.. Quite a good book to be found online.. However many things trigger me to know more, like the example of why the circuitry are connected in such ways and also the resistance values that play a part.