robertradabaugh
- Sep 1, 2012
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What value of resistors do I need and a sanity check
- Thread starter jjanes
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OK thanks for that. I don't think it'll tell us anything though. The simulator says there's 1.93V across the LED and zener, and there's 1.93 uA leakage current through the zener. I doubt that current is accurate. The simulator may just be assuming a 1 megohm leakage resistance at that voltage.
I think you probably won't be able to simulate that circuit accurately unless you can find accurate models for those specific components (the LED and the zener). The LED is a special type, and has a typical forward voltage of 1.6V at 1 mA. The LED you're using in your simulation has a forward voltage of 1.93V so it's probably a generic red LED.
So I think you'll have to build the circuit, using a genuine HP/Avago LED for LEDC, and actually measure the leakage current in DZ.
If you're buying components, you might want to buy a few higher voltage zeners (in the lowest power rating you can find - usually 0.4 watts) in the hope that a higher zener voltage will have a lot less leakage current at 1.6V and therefore would be usable. For example, get 2.7V, 3V, 3.3V, 3.6V, 3.9V to give yourself some options.
I think you probably won't be able to simulate that circuit accurately unless you can find accurate models for those specific components (the LED and the zener). The LED is a special type, and has a typical forward voltage of 1.6V at 1 mA. The LED you're using in your simulation has a forward voltage of 1.93V so it's probably a generic red LED.
So I think you'll have to build the circuit, using a genuine HP/Avago LED for LEDC, and actually measure the leakage current in DZ.
If you're buying components, you might want to buy a few higher voltage zeners (in the lowest power rating you can find - usually 0.4 watts) in the hope that a higher zener voltage will have a lot less leakage current at 1.6V and therefore would be usable. For example, get 2.7V, 3V, 3.3V, 3.6V, 3.9V to give yourself some options.
If you intend working with ESD sensitive deviced into the future then it would be worthwhile spending the money on a antistatic benchtop mat and wrist strap.
For the occassional one off use, even just buying a proper antistatic wrist strap and having it properly earthed will take care of a lot of the possible discharge problems
cheers
Dave
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None of the parts in that circuit are really sensitive to ESD, even though some of them may come in protective packaging. The LEDs may be slightly ESD-sensitive; you can protect them by sliding a small piece of conductive foam, or a few coils of a very fine spring, onto the leads, and only removing it when the device has been soldered into the circuit.
With LEDC, it's important to ensure it can't become disconnected. I'd suggest the following: cut the leads off around 1/4" from the body, solder a twisted pair of flexible (stranded) wires to the lead ends (tin the leads and wires before soldering them together), inspect the joint carefully (make sure the solder is shiny and completely covers the lead and the wire) and pull the wires gently, and protect the join with hot melt glue or similar. Keep the lead stumps shorted together until the wires are soldered to the board.
With LEDC, it's important to ensure it can't become disconnected. I'd suggest the following: cut the leads off around 1/4" from the body, solder a twisted pair of flexible (stranded) wires to the lead ends (tin the leads and wires before soldering them together), inspect the joint carefully (make sure the solder is shiny and completely covers the lead and the wire) and pull the wires gently, and protect the join with hot melt glue or similar. Keep the lead stumps shorted together until the wires are soldered to the board.
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Hi everyone. I am new to the forum. I own a medical clinic in Tennessee and have been reading the research on TDCS. I have been looking at different units and would like to have one built for research purposes.
I know nothing about electronics but have read all your posts. I don't have the knowledge/experience at this time to build this circuit you all have created, but I am willing to pay someone to build it.
If anyone is interested, please let me know.
Thanks,
I know nothing about electronics but have read all your posts. I don't have the knowledge/experience at this time to build this circuit you all have created, but I am willing to pay someone to build it.
If anyone is interested, please let me know.
Thanks,
Thanks Kris
thanks a lot Kris for your circuit your submitted
i used LM334Z chip to build the circuit and it worked with me. but i faced problem which is i can't control the output current by changing the resistor value, and i overcome this problem by adding a potentiometer in series with the human body and that solved the problem.
thanks alot for your new circuit using Transistor
i like the circuit and i have a request
is it possible please to draw the cirucit witout all the LEDs, switches, zinnar diod and all variable resistors except one variable resistor to be used for changing the output current value
which circuit do you think has better performance? the one with LM334Z or the one you submitted with Transistor?
many thanks for you
thanks a lot Kris for your circuit your submitted
i used LM334Z chip to build the circuit and it worked with me. but i faced problem which is i can't control the output current by changing the resistor value, and i overcome this problem by adding a potentiometer in series with the human body and that solved the problem.
thanks alot for your new circuit using Transistor
i like the circuit and i have a request
is it possible please to draw the cirucit witout all the LEDs, switches, zinnar diod and all variable resistors except one variable resistor to be used for changing the output current value
which circuit do you think has better performance? the one with LM334Z or the one you submitted with Transistor?
many thanks for you
Last edited:
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Hi Zaid,
My design is based on the design in post #45 on this thread. I think it is better than the LM334Z-based design, but it is very important (for safety) that the red LED cannot be broken or become disconnected from the circuit.
You can use the circuit in post #45 with the following caveats and changes:
I recommend monitoring the output current with a milliammeter, as shown in my diagram. Without this, you can't be sure that the desired current is flowing. If you can't get enough current to flow, first try improving the contact of the electrodes, using salt water. If that doesn't work, you can increase the battery voltage from 9V to 12-18V.
The resistor marked 43k should be 10k (for a 12V battery). If you change the battery voltage, recalculate the resistor by subtracting 2 from the battery voltage and multiplying by 1000. For example, for an 18V battery, the resistor should be 16k.
You can delete the green LED. Replace it with the current meter.
The red LED should be type HLMP-D150 as shown in my design.
I suggest using the transistors listed in my design instead of the 2N2907.
For safety, you MUST use a fuse in the output path. I have recommended the Littelfuse 0279.005V with a rating of 5 mA.
My design is based on the design in post #45 on this thread. I think it is better than the LM334Z-based design, but it is very important (for safety) that the red LED cannot be broken or become disconnected from the circuit.
You can use the circuit in post #45 with the following caveats and changes:
I recommend monitoring the output current with a milliammeter, as shown in my diagram. Without this, you can't be sure that the desired current is flowing. If you can't get enough current to flow, first try improving the contact of the electrodes, using salt water. If that doesn't work, you can increase the battery voltage from 9V to 12-18V.
The resistor marked 43k should be 10k (for a 12V battery). If you change the battery voltage, recalculate the resistor by subtracting 2 from the battery voltage and multiplying by 1000. For example, for an 18V battery, the resistor should be 16k.
You can delete the green LED. Replace it with the current meter.
The red LED should be type HLMP-D150 as shown in my design.
I suggest using the transistors listed in my design instead of the 2N2907.
For safety, you MUST use a fuse in the output path. I have recommended the Littelfuse 0279.005V with a rating of 5 mA.
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current limiter
hi Kris
thanks for your reply
i would like to know if there is something called current limiter (not regulator) by using transistor and feedback from the output?
what do you think of the attached circuit? How can i make the current limited to 5mA? or the other one?
many thanks in advanced
hi Kris
thanks for your reply
i would like to know if there is something called current limiter (not regulator) by using transistor and feedback from the output?
what do you think of the attached circuit? How can i make the current limited to 5mA? or the other one?
many thanks in advanced
Attachments
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Hi Zaid.
A current regulator is the same thing as a current limiter. Current in a circuit, and its interaction with voltage and resistance, is a little difficult to explain.
Your second attachment is actually the same as the design in post #45, but it uses an NPN transistor, so it looks upside-down, and it uses two diodes in series instead of the LED.
Topologically they are the same in all other respects. The "load" (the TCDS probes) connects between the collector and the side of the power supply opposite the emitter end. The difference is not important since the circuit is powered from an isolated battery.
ANY CIRCUIT THAT YOU INTEND TO CONNECT TO A HUMAN BODY MUST BE ELECTRICALLY SELF-CONTAINED AND POWERED FROM A BATTERY. THE BATTERIES MUST BE CHARGED EXTERNALLY; NO CHARGER MAY BE CONNECTED TO THE UNIT WHILE IT IS IN USE.
Your first attachment is a similar design to the second attachment, but with the addition of Q2, which increases the circuit's incremental resistance - another way of saying that it makes the circuit more accurate.
You can use the design in the first attachment with a few changes.
1. The circuit now uses NPN transistors such as 2N3904 or BC547B.
2. The "Load" in the first diagram represents the TDCS patient. Because the circuit uses NPN transistors, it is now connected on the positive side of the circuit.
3. R1 needs to be replaced by a potentiometer with a 150 ohm resistor in series, to set the current. The current in the "load" is related to the resistance by the following formulas:
3A. "Load" current (in milliamps) is roughly 650 divided by the sum of the potentiometer's resistance, at its current position, plus the 150 ohm resistor, in ohms.
3B. To achieve a certain current, the sum of those resistances must be roughly 650 divided by the desired current in milliamps.
4. With the potentiometer at its lowest resistance (0 ohms), the current will be about 4.3 mA. This is the maximum for this circuit; it is limited by the 150 ohm resistor.
5. If you use a 1000 ohm potentiometer, at its maximum resistance end, the current will be about 0.6 mA (calculated as 650 / 1150, as per point 3B above).
6. R2 in your first attachment corresponds to the 43k resistor in post #45, which should be changed to about 10k (assuming a 12V power supply).
7. Note: Q1 in the first attachment corresponds to the main transistor in the simpler circuit; Q2 can be the same type of transistor. A 2N3904 or a BC547B are suitable at these low currents.
8. There is no place for an LED (to indicate whether the circuit is in regulation or not) in the two-transistor circuit.
9. As I said in post #88, I recommend monitoring the output current with a milliammeter connected in series with the "load".
10. For safety, you MUST use a fuse in series with one of the TDCS probes. I recommended the Littelfuse 0279.005V with a rating of 5 mA.
A current regulator is the same thing as a current limiter. Current in a circuit, and its interaction with voltage and resistance, is a little difficult to explain.
Your second attachment is actually the same as the design in post #45, but it uses an NPN transistor, so it looks upside-down, and it uses two diodes in series instead of the LED.
Topologically they are the same in all other respects. The "load" (the TCDS probes) connects between the collector and the side of the power supply opposite the emitter end. The difference is not important since the circuit is powered from an isolated battery.
ANY CIRCUIT THAT YOU INTEND TO CONNECT TO A HUMAN BODY MUST BE ELECTRICALLY SELF-CONTAINED AND POWERED FROM A BATTERY. THE BATTERIES MUST BE CHARGED EXTERNALLY; NO CHARGER MAY BE CONNECTED TO THE UNIT WHILE IT IS IN USE.
Your first attachment is a similar design to the second attachment, but with the addition of Q2, which increases the circuit's incremental resistance - another way of saying that it makes the circuit more accurate.
You can use the design in the first attachment with a few changes.
1. The circuit now uses NPN transistors such as 2N3904 or BC547B.
2. The "Load" in the first diagram represents the TDCS patient. Because the circuit uses NPN transistors, it is now connected on the positive side of the circuit.
3. R1 needs to be replaced by a potentiometer with a 150 ohm resistor in series, to set the current. The current in the "load" is related to the resistance by the following formulas:
3A. "Load" current (in milliamps) is roughly 650 divided by the sum of the potentiometer's resistance, at its current position, plus the 150 ohm resistor, in ohms.
3B. To achieve a certain current, the sum of those resistances must be roughly 650 divided by the desired current in milliamps.
4. With the potentiometer at its lowest resistance (0 ohms), the current will be about 4.3 mA. This is the maximum for this circuit; it is limited by the 150 ohm resistor.
5. If you use a 1000 ohm potentiometer, at its maximum resistance end, the current will be about 0.6 mA (calculated as 650 / 1150, as per point 3B above).
6. R2 in your first attachment corresponds to the 43k resistor in post #45, which should be changed to about 10k (assuming a 12V power supply).
7. Note: Q1 in the first attachment corresponds to the main transistor in the simpler circuit; Q2 can be the same type of transistor. A 2N3904 or a BC547B are suitable at these low currents.
8. There is no place for an LED (to indicate whether the circuit is in regulation or not) in the two-transistor circuit.
9. As I said in post #88, I recommend monitoring the output current with a milliammeter connected in series with the "load".
10. For safety, you MUST use a fuse in series with one of the TDCS probes. I recommended the Littelfuse 0279.005V with a rating of 5 mA.
hi Kris
many thanks for your suggestions Kris
I have simulated the circuit in multisim and it worked perfectly
but i have 2 problems
1- the load i use is 6k ohm and it doesn't give me the right output current values unless i use 600 ohm as a load (I have measured the resistance between the two electrodes placed on the head (sponge with salted water) and it gave me around 6k ohm (see attachment please). Is that the value of the load you have considered when you designed the circuit?
2- i use 18V batteries, do you think that make difference?
i would like to ask about the 5mA fuse
is there any possibility to have a small circuit that does the same job of that fuse working as a breaker which cuts off the circuit as soon as the current reaches 5mA, because the price in that site is little bit high for some people comparing to the price of IC or any other components.
many thanks for your suggestions Kris
I have simulated the circuit in multisim and it worked perfectly
but i have 2 problems
1- the load i use is 6k ohm and it doesn't give me the right output current values unless i use 600 ohm as a load (I have measured the resistance between the two electrodes placed on the head (sponge with salted water) and it gave me around 6k ohm (see attachment please). Is that the value of the load you have considered when you designed the circuit?
2- i use 18V batteries, do you think that make difference?
i would like to ask about the 5mA fuse
is there any possibility to have a small circuit that does the same job of that fuse working as a breaker which cuts off the circuit as soon as the current reaches 5mA, because the price in that site is little bit high for some people comparing to the price of IC or any other components.
Attachments
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You need to understand voltage, current, resistance, and current regulators. Read some Wikipedia articles and look for tutorials on those keywords.
The TCDS circuit tries to make current flow in a load. In this application, this load is the subject's cranium, and salt water to ensure good connections (minimum resistance). I will call this a "wired cranium"
The electrical behaviour of biological material is far too complex to fully describe as a resistance, but it can be loosely approximated as a resistance. You measured 6 kilohms. That sounds about right.
Voltage, current, and resistance are related through an extremely important - probably the most fundamental - law used in electronics: Ohm's Law. It states:
I = V / R. (Rearranges to . . R = V / I . . . and V = I x R.)
I is current (which flows THROUGH wires and components) measured in amps;
V is voltage (which is measured BETWEEN two points in a circuit), in volts, and
R is resistance (which is a characteristic of resistors and some similar components), measured in ohms.
Notice that current is always measured in amps in that formula. This is inconvenient because we're dealing with milliamps. But if we measure resistance in kilohms instead of ohms, we can measure current in milliamps - the opposite rescalings of resistance and current cancel each other out and Ohm's Law still works.
So let's plug some numbers in. Let's consider the wired cranium to be a 6k resistor and say we want to feed a DC current of 1 mA through it. We can calculate the voltage using the rearrangement of Ohm's Law: V = I x R. Using milliamps and kilohms, we get:
V = I x R
= 1 * 6
= 6V.
This means that if we apply 6V DC between those electrodes, 1 mA of current will flow through the subject's cranium. If we apply 9V DC, 1.5 mA of current will flow. If we want to cause more than 1.5 mA to flow, we need more voltage. A single 9V battery cannot cause more than 1.5 mA to flow in a wired cranium with a resistance of 6k. The current is limited by the resistance of the wired cranium.
Actually, adding the current regulator circuit drops some voltage, so using a 9V power source, you might only get 1.4 mA to flow through the wired cranium.
Let's say you have set your current regulator for 4 mA test current. But if your circuit has a 9V battery, and the wired cranium resistance is 6k, it's impossible to cause 4 mA of current to flow. The actual current flow will never be more than about 1.4 mA.
In that situation, the current regulator can be described as "saturated", or "out of regulation". The current regulator tries to pass 4 mA but the load resistance is so high that it can't provide enough voltage (because it is limited by its battery voltage) to cause 4 mA to flow.
In this state, with the original design with the LED, the LED would go out, indicating that the current regulator is "out of regulation", and the actual cranial current is not as high as the value selected on the control potentiometer in the current regulator.
There are two ways to get more current to flow: increase the battery voltage, or decrease the wired cranium resistance.
I would try the second idea first. See what you can do with larger contact areas, and more salt water. I have no experience with attaching electrodes to people, so please look elsewhere for advice on that subject.
Increasing the battery voltage might cause subjects to feel an electric shock sensation when the electrodes are first applied, unless the circuit is switched off until the probes are firmly attached to the subject. In any case, the idea makes me uncomfortable.
The values I suggested for the current-setting components in the two-transistor circuit - about 150 ohms for the fixed resistor and 1k for the variable resistor - will give a current range from about 0.6 mA to about 4.3 mA.
If you want to test wtih 4 mA and your circuit uses a 9V battery, you will need a wired cranium resistance of 2 kilohms or less. I calculated this from another rearrangement of Ohm's Law: R = V / I, where R is resistance in kilohms, V is voltage (I'll assume about 8V because of the 1V dropped by the current regulator) and I is current in millamps.
R = V / I
= 8V / 4 mA
= 2 kilohms
You will need to make much better contact with the subject to reduce the resistance from 6k to 2k.
My opinion on the 5 mA fuse is that it is absolutely essential. Remember, you are connecting this device to a human being. Any of a number of things could concievably go wrong with the device and cause excessive current to flow. Neglecting to install and replace safety components because of their cost is highly irresponsible.
The TCDS circuit tries to make current flow in a load. In this application, this load is the subject's cranium, and salt water to ensure good connections (minimum resistance). I will call this a "wired cranium"
The electrical behaviour of biological material is far too complex to fully describe as a resistance, but it can be loosely approximated as a resistance. You measured 6 kilohms. That sounds about right.
Voltage, current, and resistance are related through an extremely important - probably the most fundamental - law used in electronics: Ohm's Law. It states:
I = V / R. (Rearranges to . . R = V / I . . . and V = I x R.)
I is current (which flows THROUGH wires and components) measured in amps;
V is voltage (which is measured BETWEEN two points in a circuit), in volts, and
R is resistance (which is a characteristic of resistors and some similar components), measured in ohms.
Notice that current is always measured in amps in that formula. This is inconvenient because we're dealing with milliamps. But if we measure resistance in kilohms instead of ohms, we can measure current in milliamps - the opposite rescalings of resistance and current cancel each other out and Ohm's Law still works.
So let's plug some numbers in. Let's consider the wired cranium to be a 6k resistor and say we want to feed a DC current of 1 mA through it. We can calculate the voltage using the rearrangement of Ohm's Law: V = I x R. Using milliamps and kilohms, we get:
V = I x R
= 1 * 6
= 6V.
This means that if we apply 6V DC between those electrodes, 1 mA of current will flow through the subject's cranium. If we apply 9V DC, 1.5 mA of current will flow. If we want to cause more than 1.5 mA to flow, we need more voltage. A single 9V battery cannot cause more than 1.5 mA to flow in a wired cranium with a resistance of 6k. The current is limited by the resistance of the wired cranium.
Actually, adding the current regulator circuit drops some voltage, so using a 9V power source, you might only get 1.4 mA to flow through the wired cranium.
Let's say you have set your current regulator for 4 mA test current. But if your circuit has a 9V battery, and the wired cranium resistance is 6k, it's impossible to cause 4 mA of current to flow. The actual current flow will never be more than about 1.4 mA.
In that situation, the current regulator can be described as "saturated", or "out of regulation". The current regulator tries to pass 4 mA but the load resistance is so high that it can't provide enough voltage (because it is limited by its battery voltage) to cause 4 mA to flow.
In this state, with the original design with the LED, the LED would go out, indicating that the current regulator is "out of regulation", and the actual cranial current is not as high as the value selected on the control potentiometer in the current regulator.
There are two ways to get more current to flow: increase the battery voltage, or decrease the wired cranium resistance.
I would try the second idea first. See what you can do with larger contact areas, and more salt water. I have no experience with attaching electrodes to people, so please look elsewhere for advice on that subject.
Increasing the battery voltage might cause subjects to feel an electric shock sensation when the electrodes are first applied, unless the circuit is switched off until the probes are firmly attached to the subject. In any case, the idea makes me uncomfortable.
The values I suggested for the current-setting components in the two-transistor circuit - about 150 ohms for the fixed resistor and 1k for the variable resistor - will give a current range from about 0.6 mA to about 4.3 mA.
If you want to test wtih 4 mA and your circuit uses a 9V battery, you will need a wired cranium resistance of 2 kilohms or less. I calculated this from another rearrangement of Ohm's Law: R = V / I, where R is resistance in kilohms, V is voltage (I'll assume about 8V because of the 1V dropped by the current regulator) and I is current in millamps.
R = V / I
= 8V / 4 mA
= 2 kilohms
You will need to make much better contact with the subject to reduce the resistance from 6k to 2k.
My opinion on the 5 mA fuse is that it is absolutely essential. Remember, you are connecting this device to a human being. Any of a number of things could concievably go wrong with the device and cause excessive current to flow. Neglecting to install and replace safety components because of their cost is highly irresponsible.
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You're welcome
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I'm sorry, I don't want to try to advise any further on this subject. I want nothing to do with people who are using home-made circuits to pass current through live people.
