NEW: Learning electronics? Ask your questions on the new Electronics Questions & Answers site hosted by CircuitLab.
Basic Electronics » Current vs Voltage
March 09, 2013 by oshjdf 
Case 16V 1A 3V 1A 1.5V 1A Device: 1.5V 1A Q: What happens when I use 6V 1A and 3V 1A? Case 23A 1.5V 2A 1.5V 1A 1.5V Device: 1.5V 1A Q: What happens when I use 3A 1.5V and 2A 1.5V? 

March 09, 2013 by sask55 
I dont think there is a good answer to your question as you have asked it. I don’t understand what the voltage and amperage values you have listed represent. Often the amperage value listed for an electrical component or power supply represent the maximum current that the component will handle or in the case of a power supply the current that the supply is capable of producing while supplying the specified voltage. Without a voltage difference between two points in the circuit there will be no current. It is the voltage difference and resistance between points in a circuit that determines the current flow thru that resistance. Perhaps a look at an explanation of voltage and current would clarify the question. There are plenty of such examples on the web. One example. voltage 
March 09, 2013 by oshjdf 
Hi sask, I don't know why the format of my post went wrong. Thanks for the link. It provides more understanding. Actually I try to understand voltagecurrent relationship in the perspective of power supply and component. Right now, this is my understanding after reading several pages from the site you linked me to. When a power supply (e.g. battery) stated to have 9V 1A, it means that it can provide up to 1A at any volt (based on IV graph in nerdkit guide). When a component (e.g. LED) stated to operate at 1.82.2v and 50mA, this is voltage that the component will function. It's called forward voltage or should we say effective voltage. So, if I want my LED to function correctly using 9V 1A battery, then I have to find a way to reduce the voltage supplied by the battery to the LED in the range of 1.82.2v; so that the current flow through the LED doesn't exceed the 50mA threshold. Please clarify whether my understanding is correct or not. Another things that concern me are voltage drop and zero voltage. Is the only way to find voltage drop is by using voltmeter? I don't understand zero voltage at all. 
March 10, 2013 by sask55 
Oshjdf First of all I am certainly no expert on this subject. It is important to keep in mind that a voltage is always a measurement of electrical potential difference between two points. Even when voltage is stated as if it is one point it is always referenced from a second point, the second point is often ground. For example 110Volt Ac power is referenced from earth ground. A nine volt battery is a reference from one terminal on the battery to the other terminal. There is always two point and the difference in electrical potential between them is the voltage. An analogy can be made between voltage and pressure. A liquid will flow thru a open pipe from a high pressure point to a lower pressure point. Current will flow thru a circuit from a high voltage point to a lower voltage point. Zero volts just means that the two points have the same electrical potential, since there is no difference in the electric potential there will be no current flow if the two points where connected. When a voltage is placed across a resistor current will flow from the higher voltage point through the resistor to the lower voltage point. The greater the resistance the smaller the amount of current that will flow for any given voltage. Thinking about the liquid analogy resistance would be like a small pipe or a pipe filled with sand the smaller the pipe the higher the resistance to the flow and the smaller the flow rate would be for any given pressure. Voltage drop is the change in the voltage as the current moves through a resistor. Voltage drop can be measured with a volt meter or calculated using ohms law if the values of the resistor and the current are known. It can also be calculated by determining the ratios of resistances thru a circuit. If a number of resistors are placed in series across a voltage source each of the resistors will have a voltage drop through it. The total of all the individual voltage drops though each resistor will add up to the voltage of the supply source. Semi conductor such as LED’s, other diodes, transistors, IC’s ext do not handle electric charges the same as resistive loads do. When considering the current though a simple resistor type load Ohms law I=V/R can be used to determine the relationship between the voltage applied though the resistance and the current that will result. Semi conductors generally do not follow this law and must be considered differently. Simple common power supplies such as a battery supply a relatively constant voltage not a constant current. All power supplies will have a practical limit of how much current it is capable of supplying. A nine volt battery will be producing a voltage near 9 volts between the terminals even, when it is not connected to anything and no current is being produced. If we ignore the fact that all batteries and other power sources have internal resistance that affects the output voltage and as long as we do not attempt to draw more current from the battery then it is capable of supplying the voltage will remain relatively constant at near nine volts. Using ohms law I=V/R , or 1A= 9V/R or R=9/1 so resistance = 9 ohms. We can calculate that any resistor smaller then 9 ohms will draw more then 1 amp on a 9 volt supply. This is the limit of the supply and is much larger then the LED can handle You will require a larger resistance value and the battery will not be required to supply the 1 amp max that it is rated at. Here is an explanation of how to calculate the value of the restore for your LED. The LED will glow with less then 50mA of current though it but it will no be a bright as it is rated to be at its maximum current. In other word you can use a somewhat larger resistor and get a dimmer LED but too much current will likely shorten the life of the LED a great deal I hope this is somewhat helpful Darryl 
March 12, 2013 by oshjdf 
Thanks for the explanation. It's easy to understand the theory just by reading it. Let me give you an example. Let's say the IV relationship of the component is linear. 2.5V ~ 0.025A, 4.0V ~ 0.040A, 5.5V ~ 0.055A I have 9V battery and I design the circuit to have power supply 4V. The resistor required: R = V / I = (9  4) / 0.04 = 125 ohm 1st question is, when the component is in the mode of using only 0.025A (2.5V), what happens to the current no being used (0.015A = 0.040A  0.025A)? Does it turned to heat? 2nd question is, what happened when the component requires 0.055A (5.5V) while the current provided is only 0.040A (4.0V)? 
March 12, 2013 by sask55 
I believe that for the purposes of this discussion with static DC power there are only two general types of components.
Component #1 “2.5V ~ 0.025A, 4.0V ~ 0.040A, 5.5V ~ 0.055A” I don’t know for sure what you mean by these numbers. It appears that you are quoting the voltage drop across the component for various current levels. If that is the case keep in mind that the voltage and amperage values are not independent they are linked, you cannot change one without changing the other. The component you are describing in your example is a resistor or resistive load with a value of 100 ohms. What I think you are saying is this component has a voltage drop of 2.5 Volts when 0.025A of current are going through it, a 4Volt voltage drop when 0.040A are going through it and a 5.5V voltage drop when 0.055Aamps are going through it, that is the very definition of a 100 ohm resistive load. All resistors have that linear IV relationship, if you like to phrase it that way. Ohms law can be used to calculate how much the voltage will change from one side of the resistor to the other if the current level is known. Or conversely we can calculate the current level that is flowing through the resistor if the drop in the voltage across the resistor is known. “R = V / I = (9  4) / 0.04 = 125 ohm” This type of calculation is used for semiconductor type components where the 4Volt drop across the component would be a constant and not change as the current through the component is changed. Because the component that you described does not have a set constant voltage drop, as a LED would, your calculation will only hold true for a .04 amp current. When other current levels are passed though the component the voltage drop will not be 4 volts The .04 A current will result in a voltage drop across the component #1 of 4.0V. The remainder of the circuit ( the other resistor) will have a voltage drop of 5volts because the total voltage of the battery is 9 volts. You have a 9 volt supply. You want to have a .04Amp current thru your circuit. R=9V/.04A or total resistance is 9/.04 = 225 ohms In order to get a total resistance of 225 ohms you will place the 125 ohm resistor in series with the 100 ohm component #1. or 225100= 125 Your answer is correct because you know the voltage drop of component #1 at .04A is 4 volts. It would be incorrect for any other current level where the voltage drop across component #1 would not be 4 volts. “1st question is, when the component is in the mode of using only 0.025A (2.5V), what happens to the current no being used (0.015A = 0.040A  0.025A)? Does it turned to heat?” There is no mode involved with component #1. As a current runs though the component the voltage level is reduced (used up if you like) by the components effective resistance. The more current passed though it the greater the voltage drop through the component. The electrical energy may be going into heat, sound, mechanical movement, light, radio waves,ect It depends on the function of component. The only way to get a 2.5 volt drop across your component #1 is to have a .025A current running through it. So the question is how much total resistance do you require to get a current flow of .025A from a 9 Volt source? You want to have a .025Amp current thru your circuit. R=9V/.025A or total resistance is 9/.025 = 360 ohms In order to get a total resistance of 360 ohms you will place the 260 ohm resistor in series with the 100 ohm component #1. or 360100= 260 The same principle applies for question #2. The only way to get a 5.5Volt drop across the component is to pass .055Amps though it. The question becomes how much total resistance will produce 0.055amps from a 9 volt source. Semi conductor type components like diodes, including LED’s, do NOT have a linear IV relationship at all. Diodes have a more or less fixed forward voltage drop regardless of the current flow. That is to say a LED may have a 1.5Volt forward voltage drop. That is the voltage level on one side of the LED will be 1.5 volts lower then the other end regardless of the current flowing through it. 
March 12, 2013 by oshjdf 
The differentiation with examples between resistive load components and semiconductor components has made my confusion become more clear. Here is my understanding. Semiconductor components by themselves can be categorized into two parts. (a) Constant current. This type consumes current constantly. This type has single operation after turned on. As long as you provide the components with current in range (e.g. 2550mA; from datasheet), the components will work. The current you choose, however, will affect the performance. For example is LED. For 2550mA LED, using 45mA will be brighter than 27mA. The way to manage current in this type of semiconductor components is easier. Just use resistor. The formula: R = (Vin  Vcomponent) / I Vin = voltage of your power supply Vcomponent = voltage to produce the desired current (have to look at description or IV graph of the component for the forward voltage) I = desired current I think servo motor can also be categorized into this type. (b) Fluctuate current. This type does not consume current constantly. The current consumption fluctuates depend on the operation. This type has many modes of operation e.g. idle, sleep, transmitting, receiving. For example microcontroller and communication components (e.g. bluetooth, xbee, GPS, GSM). There is no way to manage current because the current drawn will depend on the mode of operation. The only way to handle this type is by using components that can regulate voltage to be in the range of the forward voltage of the component. But the maximum current that can be handled by the regulator must be more than maximum current may be drawn by the component. For example, component A, 2.55.5V, 40uA1.2A. In logic, 3.3V voltage regulator can be used. But, if looking back at maximum current may be drawn by component A, 3.3V voltage regulator cannot be used because the maximum current the 3.3V voltage regulator can handle is 0.8A. 5V voltage regulator cannot also be used because the maximum current it can handle is 1A. Diode can also be used to regulate voltage but also have to look up at maximum current it can handle. For example 1N4001 (0.72V voltage drop) cannot be used because the maximum current it can handle is 1A. But 1N5401 (1.2V voltage drop) can be used because the maximum current it can handle is 3A. So, assuming 9V battery is the power supply, 3 1N5401 will reduce the voltage into 5.4V which is in the forward voltage range of the component. Summary: Use resistor for semiconductor that use constant current to operate. Regulate voltage for semiconductor that has many modes of operation; not constant current to operate. Please comment. 
March 14, 2013 by sask55 
We are getting into a lot of different areas in this thread. I believe I may have inadvertently confused the primary question at hand by making references to other semiconductor components. I was thinking about voltage drops between certain leads in some discrete semiconductor components not electrical system or devices. For this discussion it would likely be best to just consider the voltage drop on LED,s and other diodes. I think we have to make it clear if we are talking about a simple electrical component or an electric devices or system that contains many components. My comments about paragraph (a) Current is not consumed at all in a circuit, current moves through a circuit. The same amount of current is leaving the higher voltage power terminal (+) as is returning to the lower voltage terminal (),all the time. The current can be divided into parallel portions of the circuit but must be recombined and returned to the power supply to close the current loop. “This type consumes current constantly” If you wish to use the word consume it would be more understandable to say this type consumes voltage constantly. It is the voltage that is changing as the current moves through the circuit. We are discussing what is happening when a number of components are placed in series and a voltage difference placed across the series (ie connected to a battery). Each component in series will be responsible from some portion of that total voltage change. The sum of all the individual voltage drops for each individual component in the series must add up to the voltage of the power supply, there is no other way to get from the voltage level of the + terminal to the – terminal through the circuit. For example we know from a LED data sheet it has a constant forward voltage drop of say 1.5 volts. That leaves the remainder of the power supply voltage to be taken up by the voltage drop in other components in the series. This is the basis of the calculation to determine the value of the current limiting resistor in the series. Ok so it seams to me that the question that you are asking is how do we know when to make use of what method to reduce a voltage level if the voltage level we have has to be reduced to a lower voltage to be used by some device system or component.
I am certain that there are entire books written on power supplies, what to consider, how to deal with various issues ext, so the amount of possible detail would very large. Keep in mind that many devices will have there own built in power supply circuits. Many laptops,phones and tablets can be charged by plugging directly into any of the common AC power grids North American 120volt European 220volt it does not mater the internal voltage regulators in the devices will convert the voltage to what is required. Also there is a very large selection of inexpensive easy to use integrated circuit chips available to reduce voltage levels. Virtually any voltage output with any maximum current rating can be purchased very reasonably. Relatively simple feed back circuits can be placed around these chips to produce good quality variable voltage output. These voltage regulators are designed to constantly monitor the output voltage they are producing and very quickly adjust to keep the voltage as constant as possible through a considerable range of current. When compared to using resistors in a voltage divider type scenario a voltage regulator is often the best (or the only real) option to reduce the voltage level for two reasons.
As you have pointed out many components will have a variable load characteristic. The current drawn by the Micro in the Nerdkit will change from second to second depending on what the micro is doing and what other components are connected to it. The power supply has to be capable of maintaining a steady output voltage as the resistance (more accurately the impedance, but I am not going to go into that now) is constantly changing. It is the resistance (again impedance) of many types of load that is fluctuating. Changes in impedance directly make changes to the current flow to the component. A servo motor for example, which is not a semiconductor load but a high inductance resistive type load, will have a wide range of possible ohm values depending on a number of factors. Therefore the current drawn through a servo motor varies a great deal even when the supply voltage is not changing. Again the only way that the current drawn though a circuit can change is to change ether the voltage or the resistance (impedance) I=V/R, since the voltage is fixed at the power source (battery) any change in current is a direct result of a change in resistance of the load. We could easily reduce a 9 volt battery voltage to 5 volts with a voltage divider type setup by placing a couple of resistors ( a 400 ohm and a 500 ohm would work) in series across the battery terminal. If we used a high impedance volt meter we could verify the voltage drop across the 500 ohm resistor is indeed 5 volts. Why can we use this 5Volt drop to power the micro? Because anything we add to this circuit changes the balance of the voltage divider. With the micro in parallel to the 500 ohm resistor the effective resistance of the entire voltage divider circuit is changed and the ratio of the resistance levels change. Resistor type voltage dividers are very useful but only in certain circumstances, primarily very low current ie high resistance components with a known resistance value. Even in the practice of using a current limiting resistor with a LED the setup can be very wasteful as far as power use is concerned. If we wish to use a LED with a 1.2Volt forward voltage drop on a nine volt battery we have to waste 86% of the power the battery is producing. All that wasted electrical energy is going into heat at the current limiting resistor. This can be calculated by the ratio of the voltage drop across the LED as compared to the voltage drop across the current limiting resistor. Of course there are times that this is not significant loss of power, we just do not care. But to be more energy efficient a voltage regulator chip to bring the voltage down to a value much closer to 1.2V would be helpful. Or possibly use LED’s in series would make use of more of the power. This is far from complete, many factors may be considerd when reducing voltage levels. Darryl 
Please log in to post a reply.
Did you know that you can build a circuit to convert the "dit" and "dah" of Morse code back into letters automatically? Learn more...
