This is part of a series of articles "Let’s talk about series that can be found in my index." Since as a moderator I can insert my post to make them contiguous. If you comment before I am finished I will wait until It is convenient, to reply.


Wendy's Index.

What is constant current?

It uses a regulator similar to a voltage regulator, except as the name suggests, it will vary the voltage to create a fixed current in a load. As with voltage regulators it breaks down when power supply limits are reached. I will use the cc symbol to show a trace where the cc condition exist, it might also be used to show the circuit and its function.

The standard symbols for a constant current circuits have some variation (similar to the EU and USA symbols for resistors) so I will use this one.


F1,,,,,,,,,,,,,,,,,,,,,,

I have picked 20ma because that is my standard personal default value for 5mm LEDs. That will be my goal for all the circuits I will show hereafter.

Because I am planning on introducing another design that has been forming in my head I am going to create a new symbol. I will explain why later.


F2,,,,,,,,,,,,,,,,,,,,,,
I will add these symbols to my PD schematic drawing tool called PaintCAD which can be found here.


Conventional current Flow vs. Electron Current Flow


The arrow shows the direction of current. Most CC circuits only regulate current in one direction. There are exceptions, which is why I made the dual facing arrow symbol. In this case I went with the Conventional Current flow standard.

The built in Constant Current in Op Amps


Figure 3 shows a typical classic op amp circuit. Virtual ground is a key concept in these circuits because the + input is forced by the output of the op amp to match the - input. An interesting thing happens with this circuit as the current in R feedback is forced to match the current in R Input, to the limits of the op amps ability.


F3,,,,,,,,,,,,




Using Constant Current to Charge Capacitors

Something interesting happens when you charge a capacitor with a Constant Current Source. The charge becomes linear, instead of the classic log function of an RC circuit. It does something similar when used on inductors. Anyone who is interested in the math can review this thread.This means we can create sweep functions as was used in older TVs and oscilloscopes. It also, when used with Op Amps becomes a true calculus derivative function. This can be quite handy if you are making an analog computer. This was used before digital computers overtook analog computers for math calculations. Figure 4 shows an example of such a circuit.


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It still is very useful in electronics today. For example, turning a triangle wave into a sine wave is fairly straight forward. Many analog to digital conversion methods rely on translating a fixed analog voltage into a digital pulse whose duration is directly equivalent to an input voltage. This is called a dual slope to digital convertor, and is the heart of most digital voltmeters. It can do a similar conversion. In measuring the value of a capacitance or inductance by converting these values into a pulse whose duration is directly proportional to these values. I will show examples later in this article. Figure 5 shows a Square/triangle wave generator.


F5,,,,,,,,,,,,


Note the phase relationships shown in the drawing. This is accurate. The 7555 changes state when the output of the op amp integrator reaches 1/3 or 2/3 of Vcc.


Examples of Constant Current Sources


I will show some examples of constant current sinks and sources. Whether a constant current regulator is a sink or source is a matter of perspective. The voltage that all the other voltages are referenced to is called ground. My personal preference is the negative side of the power source, this bias is reflected in many of my schematics. A current source limits the current from the voltage source to the load. A current sink limits the current going to ground. Polarity of the power supply is fairly irrelevant.


BJT Constant Current Regulators


Figure 6 show the Classic Current Mirror, it uses a little used theory for biasing a Bipolar Junction Transistor (BJT), that of the BE responding to a fixed voltage. In this case the voltage set by a fixed current in Q1 cause a mirror current in Q2.


This concept will work, but I would advise against using it. It is frequently used by IC (integrated circuit) designers because two transistors on the same die are very close to being the same electrically and are very close to being the same temperature. Both being very critical to the concept. The variation in temperature can lead to thermal runaway. Getting two matched transistors can also be a problem. Typically transistors can have significant variations, even within the same part number family.

F6,,,,,,,,,,,,,,


Common Emitter Current Regulators


Though the circuit shown in Figure 7 may resembles a Common Collector design the fact that the load will be on the collector puts it firmly in the Common Emitter configuration. A stable fixed voltage on the base will create a very stable constant current regulator. It has the distinct advantage of being able to handle very large currents if the transistor is rated for the loads. I have used a variation of this circuit to create an electronic load to test power supplies.

F7,,,,,,,,,,,,,

 

Show and Tell

This chapter will give an example how the various measurement techniques, mentioned in the previous chapter. First we will show an op amp integrator. Take the integrator shown in Figure 8. If the frequency of the square wave is too slow the op amp output will clip on the power supplies (Output 3). At the frequency where R Input equals the capacitive reactance you will get a perfect triangle wave (Output 2).As the frequency increases the peak to peak value will shrink until it would be very hard to measure(Output3). It is important to note the slope of the output while changing does not change in all 3 cases.

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Why is this important? It can be used as an important component in analog to digital conversion. Figure 9 shows how a voltage can be converted to digital information which can now be used by DVMs. This is the technique that most DVMs use. The pulse width out is a direct reflection of the voltage in. This technique is called dual slope D/A conversion, and is quite accurate.

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IC Current Regulators

Now we get to my favorite current regulators. The LM317 (positive voltage) and LM337 (negative voltage) three terminal voltage regulators also make excellent current regulators. The new LD (Low Dropout) regulators are even better for this application. These voltage regulators work by maintaining 1.25V across the output and adjustment pins. This creates voltage across R1 that will stay at 1.25V. This is a current through R1 that will be a constant current. Instead of going to ground, takes the other side of R1 and use it as a Constant current source or sink. Figure 10 shows how this arrangement would work. Unlike for voltage regulation, constant current regulators require no capacitors.

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Op Amp Current Regulators

This is not my favorite circuit. Basically done by modifying voltage regulator similar to the one shown in Figure 10. Figure 11 shows how this might be done. The extra complexity is not worth the effort in my opinion. It will allow the regulation of much higher currents than the previous examples. The extra complexity isn’t worth it. I = 2 VDC / 100Ω = 20ma.

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Hello Wendy,

The following page shows the symbol of a current source:
http://electronicsbyexamples.blogspot.com/2013/05/adjustable-constant-current-source-1a12v.html

These are used here in the eBook:
https://www.allaboutcircuits.com/textbook/reference/chpt-9/power-sources/

Bertus

 

So a project like this always gnaws at me if I don't finish it successfully. I recently came across this schematic in this website. It was shown to me by other people elsewhere on AAC and on this website:

https://circuitdigest.com/tutorial/current-regulators-construction-working-and-design-types

​

Now I have to figure out how to program specific currents. Any ideas? R1 seems to be a constant, while R2 programs the current. Now I have to figure out how to calculate R2 to get the desired current I want.

 

Now I have to figure out how to calculate R2 to get the desired current I want.

Current will be about 0.7V/R2.

 

So R2 =0.7/I? Excellent, that gives me something to test.

 

Hi W,
Does this help, as stated Vbe is not linear.
E

@Wendy
If you post the transistors you are using and the desired fixed current options, Vinput and the Load resistance, I will rerun/post the LTspice simulation.
E

 

@Wendy,

In your first post, Fig. 3 shows 20 mA going toward the inverting input of the opamp from the input signal and 20 mA going toward the inverting input of the opamp through the feedback resistor. By KCL, that requires 40 mA of current flowing into the inverting input of the opamp.

Also, you don't need a new symbol with a bidirectional arrow to indicate that the constant current is about to go in either direction. That is handled perfectly well by the value of the current, itself, being able to be either positive or negative. But by having a bidirectional arrow, it is now underfined which direction current is flowing through the source when it is positive.

 

The current in question can flow through the capacitor in either direction, and is a constant when it does. This creates a triangle wave.

As to the graph, I am more interested in the current over voltage of the regulator.

 

The current in question can flow through the capacitor in either direction, and is a constant when it does. This creates a triangle wave.

So?



During the portion of the cycle that the current is positive, which way is it flowing?

You can't tell. Your symbol is ambiguous.

Schematic symbols need to be unambigious.

Consider a isolated DC power supply that can output either a positive or a negative voltage. They don't put ± on both terminals. One is marked at positive (or has a red connector) and the other negative (or black). The voltage it outputs can be either positive or negative, and whatever it is, it is relative to the defined polarity established by the markings on the supply terminals.

 

Constant current diodes, also known as current regulating diodes, are commonly available over a range of 1mA to 20 mA.

These are often used to provide operating power to industrial machinery monitoring accelerometers with integral electronics. A 2 conductor cable (usually a shielded cable) provides ~4mA to a remote sensor over the same conductor carrying the vibration signal back from the sensor. The electronics in the sensor bias to current to around 12 VDC (the supply or compliance voltage is usually 18 - 24 volts) and the AC signal voltage rides on the 12 volt offset. Signal levels as high as ±7 volts peak-peak are possible. At the instrumentation end, a DC-blocking capacitor removes the DC offest from the signal.

A few details about constant current sources are helpful for understanding the operation.

● A constant current source will do everything in its power to drive the rated current to the load. It will varying to voltage as required to maintain that current. (I suppose this is the effect rather than the cause, but it's clearer to think about it this way.)

○ In an open circuit, the constant current source will increase the voltage to maximum (supply voltage) in a (futile) effort to keep the current flowwing.

○ In a short circuit, the constant current source will decrease to near zero to limit the current to the specified value.

○ For a resistive load load, Ohm's Law comes into play: V = I × R. As resistance increases, the voltage increases to maintain the current up to the supply voltage. As resistance decreases, the voltage falls.

○ For active devices like LEDs or the above mentioned accelerometers, the voltage is the bias voltage set by the device. For an LED, it will be Vf. For the accelerometer, it's set by the integral electronics.

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Given the above info, it's easy to understand why this system is used to power industrial accelerometers:

> If the DC voltage at the instumentation end is approximately 12VDC, the circuit to the accelerometer is intact.

> If the DC voltage at the instrumentation end is 0VDC, the cable or accelerometer is shorted.

> If the DC voltage at the instrumentation end equals the suppy voltage, the cable is disconnected or open or the accelerometer is defective.

 

As to which way the current is flowing, that is a context issue based on the rest of the circuit around it.

 

As to which way the current is flowing, that is a context issue based on the rest of the circuit around it.

Which is precisely what makes it a poor symbol!

Imagine that the symbol for a battery had no polarity indication and the person that came up with it said that it's a context issue based on the rest of the circuit around it.

 

So I built this circuit using this layout.

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I recorded the following results.
0.5V 0.0ma
1.0V 5ma
1.5V 13ma
2.0v 25ma
2.5V 34ma
3.0V 44ma
3.5V 57ma
4.0V 73ma
4.5A 85ma
5.0V 102ma

This is seriously not a very good current regulator. More to come.

 

So I built this circuit using this layout.

View attachment 371764

View attachment 371765​

I recorded the following results.
0.5V 0.0ma
1.0V 5ma
1.5V 13ma
2.0v 25ma
2.5V 34ma
3.0V 44ma
3.5V 57ma
4.0V 73ma
4.5A 85ma
5.0V 102ma

This is seriously not a very good current regulator. More to come.

Which circuit are your measurements for? The schematic or the wiring diagram? They don't match -- for instance, in the wiring diagram the base of Q2 is tied directly to the negative side of the circuit and the collector-emitter current of Q2 goes through both R1 and R2.

 

I don"t see it. Q1 and Q2 both look okay to me. I'm using a variable power supply with a current meter and get the reading off of it to get these numbers. Tomorrow I will be trying another circuit, and publishing the results. Oops, I found an error. I will publish it tomorrow.

 

Here's the LTspice sim of the post #14 circuit:
It's not a great constant-circuit circuit, but it's simple and gives adequate performance for many applications, such as for a current-limiter.
You test results with the real circuit should be similar.

It should be noted that constant-current circuits used for current limiting are often simpler with poorer performance than those used to generate a constant-current through a varying load resistance.
And any discussion of constant-current sources should include the Howland current-pump which is a single op amp circuit that can generate a bidirectional output.

 

R2 is either defective or high value resistor (colors can be sometimes hard to tell) or it does not make good connection with breadboard.
value should be determined using R2=Vbe/I so for 5mA current and 600mV Vbe, R2 should be 120Ohm.

btw. Q1 in post2 (OpAmp current regulator) need to be NPN or NMOS. PNP or PMOS can also be used but the circuit need to be adapted (use high side). i prefer to use MOSFETs here as source and drain current is the same so load can connect to drain of MOSFET. it is always nice when one side of the load can connect to one of power rails directly.

 

while current regulation of this simple circuit less than perfect, it is better when load is connected directly to collector (R3 rather than R4)

 

it is better when load is connected directly to collector (R3 rather than R4)

Yes, that eliminates the error due to the bias current through R1 (sim below for a 100Ω load).

But what's the purpose of R4?