Is there a practical example by using a circuit where a source better modelled as a current source if it has a high output impedance? I have been hesitating to ask here but couldnt figure out myself. Im trying to grasp the logic behind it.
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\$\begingroup\$ You mean like a photodiode transimpedance amplifier? \$\endgroup\$– DKNguyenCommented Sep 25, 2019 at 20:14
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\$\begingroup\$ I think the question is misunderstood. I need a very simple source model with Thevenin eq high source resistance. By using that I want to understand why high source impedance sources are modeled as current source. \$\endgroup\$– user1999Commented Sep 25, 2019 at 20:16
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5 Answers
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Suppose you have an idealized voltage source with a resistance in series with the output (we'll call this the output resistance).
Now if you place a wide variety of load resistances across this source+output resistance, then the output resistance will form a divider with the load resistance. What happens if...
...the output resistance is very high compared to all the load resistances?
The total resistance in the circuit doesn't change very much no matter what load resistance you use since it is negligible compared to the output resistance. That means the current doesn't change very much with varying loads so it is more convenient to model as a current source.
On the other hand, the voltage across the load resistance changes a moderate amount. Double the load resistance and you can double the load voltage since the current stays fairly constant.
...if the output resistance is very low compared to all the load resistances?
The total resistance in the circuit now almost entirely depends on the load resistance since the output resistance is now negligible. Unlike before, now it is the voltage across the load that remains fairly constant. So in this case it is more convenient to model it as a voltage source.
And unlike before, the current now has moderate changes with load. Double the load resistance and you get about half the current.
So you can see here that we even though we were actually using an idealized voltage source, it was sometimes more convenient to model it as a current source. In the same way, it is sometimes more convenient to model a current source as a voltage source. In the current source's case the output resistance is in parallel rather than in series.
So let's start all over again with an ideal current source with a parallel output resistance and parallel load resistance.
Output Resistance >> Load Resistance The load resistance dominates since the much smaller resistance always dominates for parallel resistors. Very little current bleeds through the output resistance. Most of the current flows through the load resistance. Therefore, the current through the load resistance is kept relatively constant by the source while the voltage across it will vary moderately as the load varies. Therefore, tt is more convenient to model it as a current source.
Output Resistance << Load Resistance The output resistance dominates. Changes in the load resistance will have little effect on the overall current output of the source. Most of the current flows through the output resistance so the output resistance determines most of the output voltage. In this case, it is more convenient to model it as a voltage source.
So you can see here it didn't really matter if it was actually a voltage source or a current source. What really determined the more convenient model was the relative magnitude of the load resistance and source output resistance.
In discrete transistor amplifiers, that's why they use high value resistors in series with the a voltage source. It's meant to approximate a current source since the load presented by the transistor is supposed to be much less than that presented by the high value resistor (R1).
Taken from: https://www.electronics-notes.com/articles/analogue_circuits/transistor/long-tailed-pair-circuit.php
In integrated circuits they just build an actual current source out of more transistors.
Taken from: https://www.researchgate.net/post/Can_we_present_the_long-tailed_pair_as_a_bridge_circuit
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\$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. Any conclusions reached should be edited back into the question and/or any answer(s). \$\endgroup\$ Commented Sep 26, 2019 at 16:12
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Suppose you have the following voltage divider.
Here \$Z_1\$ is your source impedance and \$Z_2\$ is your load. The current is then
$$I = \frac{V_\mathrm{in}}{Z_1 + Z_2} \approx \frac{V_\mathrm{in}}{Z_1},$$
since \$Z_1 >> Z_2\$. Hence the current \$I\$ is approximately a constant and is independent of the load \$Z_2\$. Thus the circuit can be approximated as a current source.
P.S. Have a look at transforming voltage sources into current sources. Especially Norton’s theorem.
answered Sep 25, 2019 at 22:44
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\$\begingroup\$ Thanks this is also very illustrative with the circuit example. \$\endgroup\$– user1999Commented Sep 26, 2019 at 0:01
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EDIT to respond to the OP's clarification:
It's all about how much the current changes with load. As DKNguyen stated in the comments, if there is a high impedance in series with a source it can swamp the resistance of any loads (in the right range) that are connected, meaning the current is effectively constant with load. That's the definition of a current source. But of course the load impedance has to be small compared to the source impedance.
Previous answer in case anyone searches for this thread and wants some current source examples:
There are lots of examples. A current mirror is one:
In the picture the collector of Q4 can be modelled as a current source. There are also more precise current source circuits made with op-amps. There are many topologies, here's just one:
A current mode switching power supply has an inner current loop that is often modelled as a current source. A single JFET can be configured as a current source, though part to part and temperature variation and compliance are not necessarily very good:
Those are just a few examples, but with such a general question it's hard to say more.
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\$\begingroup\$ Please see my comment under my question. Im asking not for real examples Im asking a simple model explaining why high output impedance sources are modeled as current source. What is the idea behind modeling them as current source but not voltage sources \$\endgroup\$– user1999Commented Sep 25, 2019 at 20:19
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When we want to make a cheap & simple (quick and dirty) current-source, we choose a large voltage and R value.
The convention for design purposes is due to specs for tolerances and ratio of source/load.
If the source/load impedance ratio << 1 =voltage source
If the source/load ratio >> 1 = current source.
When close to ideal or “regulated” we define the % tolerance and conditions.
Exception
When using cct analysis, we can call it anything we want using {Norton, Thevenin} equiv. ccts.
a practical model uses the actual Rs or ESR with V in series or CC with leakage in shunt Rp or is derived from the load regulation error.
Remember that %V load regulation of any voltage source is basically %load/Rs or %load/ESR.
- This applies to regulators, batteries and capacitors for switched loads e.g. SMPS.
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One possible reason is that if you consider the almost inevitable capacitive loading on the source, the voltage source + resistor will require you to know the voltage and the resistance to estimate the behavior, whereas if you use a current source with a high-value shunt resistor the resistor can be ignored in many cases.
answered Sep 25, 2019 at 22:38
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\$\begingroup\$ In other words the Cap ESR becomes that load impedance to a transient or step change \$\endgroup\$– D.A.S.Commented Sep 25, 2019 at 22:50