Current Source Question - Page 1
Forum section: electronics / beginners

author: fsd66
author_rank: Newbie
timestamp: August 18, 2012, 06:03:48 am
content: Hi, I am new to electronics, and I was wondering what would happen if I put a reverse voltage as a load to a constant current source. Will the current source increase it's voltage by the amount of the reverse voltage?Thanks

author: OndraSter
author_rank: Contributor
timestamp: August 18, 2012, 06:10:01 am
content: It will go lower - as in further away from zero.Ohm's law applies everywhere:U = I*RI is constant (let's say -5A)R is:a) 5Ohm => U = -5 * 5 = -25Vb) 20Ohm => U = -5 * 20 = -100V

author: IanB
author_rank: Super Contributor
timestamp: August 18, 2012, 06:34:43 am
content: Quote from: fsd66 on August 18, 2012, 06:03:48 amYour question lacks a little clarity.In the theoretical world there are ideal current sources and ideal voltage sources.A current source has infinite impedance and will simply reflect whatever voltage is placed across it.A voltage source has zero impedance and will simply pass through any current in the circuit.So if you connect a voltage source to a current source, the voltage will be the voltage imposed by the voltage source--no higher, no lower. At the same time the current in the circuit will be the current imposed by the current source. This current will simply flow through the voltage source unimpeded.In real life there these idealized circuit elements don't exist and so the situation will vary slightly from the ideal case. But it will still be close.

author: Psi
author_rank: Super Contributor
timestamp: August 18, 2012, 07:05:02 am
content: I think he means which way will it go with a negative voltage on the input, Positive or Negative?eg.. negative 12V supply connected to current source set at 1A,Will it simply draw 1A at -12V or try to pull the load positive.Do any desktop current sources even have a signed input control for the current? Will they fight the load to produce current in the direction that's set?Or will it detect the load direction and draw 1A from whatever voltage it gets(i don't have a desktop current source so i dunno how they work)

author: Rerouter
author_rank: Super Contributor
timestamp: August 18, 2012, 07:49:31 am
content: well it greatly depends what type of current source you have, as some types can easily become over-biased and increase in the amount they source,best to simulate your circuit and make sure no matter what it cannot be under or over biased, so far playing around with the conventional methods looks to be an interesting dillemma,this is the only fail safe implemtentation i can come up with so farhttp://tinyurl.com/bvw426j

author: IanB
author_rank: Super Contributor
timestamp: August 18, 2012, 02:36:42 pm
content: Quote from: Psi on August 18, 2012, 07:05:02 amA current source will source (supply) the expected current regardless of the voltage present on its output terminals (positive, negative, zero, doesn't matter).Therefore concepts like "fighting the load" or "detecting the load direction" are not applicable. The current source will simply provide the current it is programmed to provide no matter what voltage happens to be present.Obviously real implementations of current sources have limits on the range of terminal voltages they can tolerate, but within those limits the voltage will be irrelevant.

author: IanB
author_rank: Super Contributor
timestamp: August 18, 2012, 02:47:24 pm
content: To make this clearer, try the following thought experiment.You could construct a 1 A current source by placing a 1000 ohm resistor in series with a 1000 V DC supply.To conduct the thought experiment, place a 6 V battery across the output of the current source and use Ohm's law to calculate the current in the circuit. First place the battery one way round (+/-), then place it the other way round (-/+). Calculate the current either way.Then for grins, place a 5 ohm resistor on the output, then place a 5 ohm resistor in series with the 6 V battery and place that combination on the output both ways round.

author: IanB
author_rank: Super Contributor
timestamp: August 18, 2012, 06:47:43 pm
content: A further observation on this.In my previous example I used a 1000 ohm resistor. This gives the current source an impedance of 1000 ohms, which is quite high but still only an approximation of the ideal infinite impedance.We could make a better approximation if we used a higher voltage like 1 megavolt in series with a 1 megohm resistance. Now the source impedance is 1 megohm which is getting closer to the ideal.It occurred to me that a real example of a current source is available to play with. It is a Van de Graaff generator. Get yourself one of these and do some experiments.Firstly discharge the dome and then measure the resistance between the dome and the ground terminal. You should hopefully find no current path and any typical resistance meter will show open circuit. So we have the first test of a current source satisfied. Infinite source impedance.Next connect a wire from the dome, pass it through a microammeter, and then to ground. Turn on the generator and see what current flows through the meter. This is the nominal output of the current source.Now put some conductive objects in the current path, such as resistors, small batteries, your hand, etc. (Make sure the generator is stopped and discharged when you do this.) In each case, turn the generator back on and look at the current on the microammeter. You should find that no matter what conductive item you put in the current path the current stays the same.If you are curious, also measure the voltage difference across each conductive item when the current is flowing. Then you can see exactly what happens with batteries.

author: Psi
author_rank: Super Contributor
timestamp: August 18, 2012, 11:02:17 pm
content: I think the key thing to understand here is the differences and similarities between a current sink/source and an electronic load.

author: fsd66
author_rank: Newbie
timestamp: August 18, 2012, 11:38:03 pm
content: Wow, thanks for all of the replies! Maybe I should build a constant current source and just see what happens.