linear low voltage & current derived from low frequency signal from micro controller

Thread Starter

Peter@UNSW

Joined May 18, 2022
9
Hi there, I am a bit stuck with my project and would like to seek help or suggestions.

Aim: linear low voltage and current supply according to voltage signal form an arduino.

input: 0-5V, output: 0 to 12 up to 18 V (linear relation to input signal), 0 to 10mA (linear relation to input signal)
The input signal frequency is variable from 0 to approximately 30Hz.

Current approach is using a power pack (4xAA) to power a boost module, used as a single supply to a precision op amp.
That delivers the desired linear voltage output, but with constant current supply of about 10mA after a switching on period of about 0.6V
please see image attached.

Are there different devices available like FET or MOSFETs that are better to do the job?
The mosfet I tried has a delayed switching on behaviour that does not let any current through until a signal voltage of about 2V. output_V1_inrangeupto1V.png
 

Papabravo

Joined Feb 24, 2006
18,426
4 x AA = 4.8 VDC to 6 VDC, allowing for each cell to drop to 1.2V, and you want 12-18 Volts on the output from the boost module. The actual output of the boost module will be controlled by a 0-5 VDC control signal. Do I have that much correct so far?

In a Boost Converter the output voltage is related to the input voltage by the relationship:

\( {V_o}\;=\;\cfrac{V_{in}}{1-D}\text{, where D is the duty cycle} \)

This means for 18 volts on the output the duty cycle will be between 0.67 and 0.73, which means that it may be difficult to exert precise control over the output since small changes in duty cycle imply large changes in the output voltage. For 12 volts on the output we have a duty cycle range of 0.50 to 0.60, better but still rather limiting. I'll be surprised if you can get the precise control you are looking for.

I don't know if this is what you want, I kinda think you might have something else in mind.
 

Thread Starter

Peter@UNSW

Joined May 18, 2022
9
Hi Papabravo,
thanks for looking into the issue. I should have made it more clear and included a schematic of the circuit. The powerpack is supplying power to the boost module, that supplies the power to the op amp. (schematics attached)

The output of the op amp is controlled by an input signal. Then the output current is not linear to the output voltage of the op amp as shown in the diagram.
IMG_6517.jpg
Is there a device (instead of the op amp) that has a linear relation of output current and voltage?
 

Ian0

Joined Aug 7, 2020
5,147
Hi Papabravo,
thanks for looking into the issue. I should have made it more clear and included a schematic of the circuit. The powerpack is supplying power to the boost module, that supplies the power to the op amp. (schematics attached)

The output of the op amp is controlled by an input signal. Then the output current is not linear to the output voltage of the op amp as shown in the diagram.
View attachment 267748
Is there a device (instead of the op amp) that has a linear relation of output current and voltage?
Firstly, I don't know you expect to get a negative output form a positive supply.
Secondly, you can't control BOTH output current and voltage at the same time!
If you control the output voltage then the output current will depend only on the output voltage and the load.
If you control the output current then the output voltage will depend only on the output current and the load
 

Thread Starter

Peter@UNSW

Joined May 18, 2022
9
Hi Ian0,
I don't expect a negative output current from a positive supply. The output voltage is 0-15V, a range from 0 up to +15V. I understand, my expression can cause confusion.
And I measured the output of the op amp. While the voltage changed linearly the output current changed as pointed out in the diagram. That was with a load of 0 (zero).
The question is how to reduce the current to achieve a linear relation of voltage and current.
 

Ian0

Joined Aug 7, 2020
5,147
The question is how to reduce the current to achieve a linear relation of voltage and current.
It depends on the load. If the load is entirely resistive, then the relationship between voltage and current will be linear.
If the load is not resistive, then it won't.
 

Papabravo

Joined Feb 24, 2006
18,426
OK, that clarifies things a bit. You may still have some misconceptions about how things work.
  1. Regardless of the gain setting, the output of an opamp cannot go above the supply voltage. To even achieve that you have to specify a device with rail-to-rail outputs. In this context a "rail" refers to one of the two power supply voltages for the opamp. In your case it would be the output of the boost module and ground. You should make those connections explicit in your schematic/block diagram. Don't forget to include power supply bypass capacitors if you decide to build this device.
  2. In your schematic, the output current will be determined by the feedback resistors which is 15 Volts / 15 kΩ = 1 mA, since there is no other load on the output of the opamp.
  3. If you introduce another load on the output, that load will determine the additional amount of output current up to some limit. Most opamps can supply about 50 mA.
  4. For a fixed load, the output current from an opamp will be constant, you cannot set the current arbitrarily.
  5. With an AC voltage source that swings between 0V and 5V, the output voltage of your circuit will swing between approximately 0 and 15 Volts. Since it has a fixed load of 15 kΩ, the current in the "load" will vary linearly from 0 to 1 mA.
  6. Suppose we introduce and additional 300 Ω to ground load on the output. Now the total output load will be 300 Ω || 15 kΩ ≈ 294 Ω. Now the 15 V swing will produce currents from approximately 0 to approximately 51 mA. I say approximately because that may be beyond the capabilities of the chosen device.
 

BobTPH

Joined Jun 5, 2013
4,907
What does the graph you posted represent? You say you aren’t trying to produce negative voltages or currents, and yet, the graph shows just that. And, as stated by others l, your circuit cannot do that, so it clearly is not the output of your circuit. What is it?

Bob
 

LowQCab

Joined Nov 6, 2012
2,054
Hold on for a minute ..........
Please state the problem that You are trying to solve FIRST.

There may be a hundred different ways to achieve this end,
but without a full explanation of what you're trying to accomplish,
the conversation just goes all over the place with guesses and assumptions,
none of which is very useful to anyone, and just plain confusing.
.
.
.
 

Thread Starter

Peter@UNSW

Joined May 18, 2022
9
What does the graph you posted represent? You say you aren’t trying to produce negative voltages or currents, and yet, the graph shows just that. And, as stated by others l, your circuit cannot do that, so it clearly is not the output of your circuit. What is it?

Bob
Hi Bob,
true, my apologies. The circuit diagram presents just parts that need improvement. (likely the other parts as well, but they work ok for me at the moment).
The reason is the difficulty to achieve negative current supply for the op amp from a single power supply and also to get a negative signal from a micro controller.
The desired controlled input signal can be anything from -5 to + 5V. Using a D/A converter or a micro controller, we distributed the input signal over two pins, were one presents a positive voltage for negative values and the other pin positive voltage for positive values. The pins also regulate the H bridge reversing the current accordingly and then the result is as shown in the diagram.
I thought I make it easier, when I present the op amp circuit, but I just caused more confusion.
 

Thread Starter

Peter@UNSW

Joined May 18, 2022
9
[/QUOTE]
Hi LowQCab,
sorry for all the confusion. May I bring your attention to just the positive part of the diagram? I would like a linear relation between output voltage and current out of a battery driven device. Roughly in numbers 0-12V and 0-10mA. I am very open to try several things.
OK, that clarifies things a bit. You may still have some misconceptions about how things work.
  1. Regardless of the gain setting, the output of an opamp cannot go above the supply voltage. To even achieve that you have to specify a device with rail-to-rail outputs. In this context a "rail" refers to one of the two power supply voltages for the opamp. In your case it would be the output of the boost module and ground. You should make those connections explicit in your schematic/block diagram. Don't forget to include power supply bypass capacitors if you decide to build this device. - I understand that
  2. In your schematic, the output current will be determined by the feedback resistors which is 15 Volts / 15 kΩ = 1 mA, since there is no other load on the output of the opamp. Correct. I have a stupid question, If I want separate the signal input and opamp output circuit. Is that possible?
  3. If you introduce another load on the output, that load will determine the additional amount of output current up to some limit. Most opamps can supply about 50 mA. true, op amp max current is about 30mA, I our case it is reduced to 10mA via a resistor.
  4. For a fixed load, the output current from an opamp will be constant, you cannot set the current arbitrarily. is it possible to use the signal input to regulate output current as well?
  5. With an AC voltage source that swings between 0V and 5V, the output voltage of your circuit will swing between approximately 0 and 15 Volts. Since it has a fixed load of 15 kΩ, the current in the "load" will vary linearly from 0 to 1 mA. I experienced that input signals larger that 0.6 V the output current is limited to the max
  6. Suppose we introduce and additional 300 Ω to ground load on the output. Now the total output load will be 300 Ω || 15 kΩ ≈ 294 Ω. Now the 15 V swing will produce currents from approximately 0 to approximately 51 mA. I say approximately because that may be beyond the capabilities of the chosen device. the extra Load - between output and ground of boost module will be approximately something like 3k - 6k Ohm.
 

Thread Starter

Peter@UNSW

Joined May 18, 2022
9
Dear all,
thanks for looking into the issue and my apologies for all the confusion. I very much appreciate all your comments and suggestions! I'll try to get a complete circuit of mine and post it again to make the picture more complete.
 

BobTPH

Joined Jun 5, 2013
4,907
What you are asking for is impossible.

If the load is a resistor,there will always be a linear relationship between voltage and current.

If the load is anything that does not act like a resistor, you cannot force that linear relationship. You can supply a voltage, but the characteristics of the load determines the current.

An LED is common example of a non linear load. You cannot make it linear in voltage and current. You can set either one of them, but the other you cannot control.

Bob
 

LowQCab

Joined Nov 6, 2012
2,054
"" I would like a linear relation between output voltage and current out of a battery driven device ""
This statement makes no sense.
There is always a Linear relationship between Output-Voltage and Current,
it depends on the Load characteristics,
it can be no other way,
that's what Ohms-Law is for calculating.

As I asked earlier .........
Please state the problem that You are trying to solve FIRST.

You have a non-ideal situation, what is it ?
( specific details please )
( Pictures, and/or, Schematics, are much more useful than words ).
.
.
.
 

MisterBill2

Joined Jan 23, 2018
11,888
Again, L Q C is asking for more information, because without an adequate explanation of what the desired function actually is, the very best that you will get is random guesses.
So far the explanations make no sense to me in that, as already stated clearly, current into a load is dependent on both the applied voltage and the load resistance. That is fundamental physics. In diodes the resistance depends on the applied voltage because diodes are non-linear devices.
 

Thread Starter

Peter@UNSW

Joined May 18, 2022
9
Hi all, many thanks for all your tips. Please find my up-to-date circuit design. Sorry for the delay, I had to learn Eagle first, until I got it done.
Aim is to change the current flow direction and intensity, regulated by a input signal (1&2) from a micro controller.
Input signal is thereby limited to +0 up to +5V. The signals are of very short nature and of very low frequency, sub 5Hz. There load resistance between Out1 and Out2 is approximately 3kOhm up to 6kOhm. It can vary in that range a bit, but should be constant.
Since I do not know exactly the resistance of the load, I was tempted to believe that there is a way to regulate the max current going through the load.

Also, if you see obvious short comings in my schematic, could you please point them out to me?
Many things in advance.
LowCurrentSupply_circuit.png
 

MisterBill2

Joined Jan 23, 2018
11,888
The circuit will be easier to follow and understand if R6 is moved to be alongside R5, vertically oriented instead of being up among the diodes. There is room for it there, and as R5 and R6 constitute a voltage divider in the feedback loop it will make circuit evaluation ab it easier.
Are the signals IN1 and IN2 variable voltages or digital commands? Those two 1n4004 diodes certainly make ihe input response very non-linear, and the two LEDs make the input resistance even more non-linear in respect to input voltage.

So the circuit has a not-constant input resistance feeding an op-amp with a DC gain of about 2 feeding a current regulator, powering an H-bridge of fairly low current transistors that is set up to have a nominally linear operation mode. Just how the output will respond to the input is not either fully digital nor will the relationship be linear, although it will be "proportional."

An explanation with a description of the required output versus input voltage would answer a lot of questions. The circuit shown is analog but not really linear, it appears.
 

Thread Starter

Peter@UNSW

Joined May 18, 2022
9

Thread Starter

Peter@UNSW

Joined May 18, 2022
9
The circuit will be easier to follow and understand if R6 is moved to be alongside R5, vertically oriented instead of being up among the diodes. There is room for it there, and as R5 and R6 constitute a voltage divider in the feedback loop it will make circuit evaluation ab it easier.
Are the signals IN1 and IN2 variable voltages or digital commands? Those two 1n4004 diodes certainly make ihe input response very non-linear, and the two LEDs make the input resistance even more non-linear in respect to input voltage.

So the circuit has a not-constant input resistance feeding an op-amp with a DC gain of about 2 feeding a current regulator, powering an H-bridge of fairly low current transistors that is set up to have a nominally linear operation mode. Just how the output will respond to the input is not either fully digital nor will the relationship be linear, although it will be "proportional."

An explanation with a description of the required output versus input voltage would answer a lot of questions. The circuit shown is analog but not really linear, it appears.
Thanks for the tips on designing the schematic. I updated.
INT1 and INT2 are analog outputs from an Arduino-like microcontroller. With 0-3.3V at a 10-bit resolution. For the signal in, INT1 and INT2 should never be active at the same time. But to make sure that they do not interact with each other and cause something weird I put the two 1n4004 there. That no current will flow to both sides of the H-bridge. Does that make sense?
is there a better way to separate both input signals?
The LEDs are just there for visual control and could be removed. I was not aware that the input current to an op-amp has an influence on the output, as long as the voltage is maintained constant.
Great review! thanks for that again!
 
Top