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I want know which is the correct way to approach problems where we have to identify CC and CV mode.

Let us consider a question -

For various load resistors, connected across \$Channel-1\$ of the DC power supply the following table is obtained-

\$R_L\$ Output Voltage (V) Load current (A) Mode of Operation
\$\infty\$ 5.00 0.00 a
100 5.00 0.05 b
200 5.00 0.25 c

Now we have to determine what modes are a, b, and c


I have done this problem in two ways-

First: Since V is constant and load current is changing with values of load resistance so, a,b, and c must all be in CV mode. But here I am not considering the fact that in the third reading load is drawing more current than the ratio \$V/R_L\$.

Second: Is based on this text that I have recently read, here they have given the graph-

enter image description here

I will plot the values of V and I and see which operation line it is following(CV or CC), but for the third reading, we find that \$R_L\$ is greater than \$R_c= V/R_L \$, it will lie on CV mode line. Checking for first and second reading as well, I get all a, b, and c in CV mode.


So, which method is more correct and general such that I can apply it on all such problems?
Please let me know if the second method is correct approach or not.

Also what do they mean by this?-

A power supply will operate in constant voltage (CV) mode provided the load does not require more current than the current limit setting.

Link for the image source for reference- Fundamentals of DC Power suppy

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  • \$\begingroup\$ I’m voting to close this question because homework needs an attempt at a solution \$\endgroup\$
    – Voltage Spike
    Commented Aug 18, 2021 at 21:17
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    \$\begingroup\$ @VoltageSpike♦ I have attempted the question, even the answer is correct and I am asking if the way I am thinking is correct or not. You need to re-evaluate your judgement. \$\endgroup\$
    – Mirae
    Commented Aug 18, 2021 at 21:48
  • \$\begingroup\$ I have mentioned in the very first line of the question, what I want to know, and question is just an example. \$\endgroup\$
    – Mirae
    Commented Aug 18, 2021 at 21:58
  • \$\begingroup\$ The voltage is 5 V in each case so it's constant, so ... \$\endgroup\$
    – Transistor
    Commented Aug 18, 2021 at 22:00
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    \$\begingroup\$ First task is to spot the mistake in the table. \$\endgroup\$
    – user16324
    Commented Aug 18, 2021 at 22:06

1 Answer 1

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There is an error in your table. 200 Ω on a 5 V supply will draw 25 mA (0.025 A), not 0.25 A.If it's drawing more than 0.025 A at 5 V then it's not 200 ohms. V = IR has to apply.

Say the value of current is 0.025A for same values of V and R, so do you think we need second method? Or is second method any useful?

If the voltage is constant then you haven't hit the current limit yet so no further analysis is required. You haven't given us the CC limit and we can't calculated it from the table so we can't calculate RC so you can't do the graphical analysis.

Also what do they mean by this?-

A power supply will operate in constant voltage (CV) mode provided the load does not require more current than the current limit setting.

They just mean exactly that.

enter image description here

Figure 1. A lab power supply. Image source: Instek.

If you have access to a lap supply such as shown above you can set a voltage limit and current limit on the device. You can then connect load resistors and see the effect.

For example: If you set U = 5 V and I = 100 mA then \$ R_C = \frac U I = \frac {5} {0.1} =50\ \Omega \$. For any load > 50 Ω the PSU will be able to run at constant voltage (5 V) and the load will draw < 100 mA. At any load < 50 Ω the load will draw 100 mA but the voltage will drop. At 50 Ω the load will draw 100 mA and the voltage will just hold steady at 5 V.

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  • \$\begingroup\$ Can you also mention one such example for constant current mode condition? I am a bit confused. \$\endgroup\$
    – Mirae
    Commented Aug 21, 2021 at 14:32
  • \$\begingroup\$ I did. "At any load < 50 Ω the load will draw 100 mA but the voltage will drop." \$\endgroup\$
    – Transistor
    Commented Aug 21, 2021 at 14:45
  • \$\begingroup\$ You did, I misread. It's clear now. \$\endgroup\$
    – Mirae
    Commented Aug 21, 2021 at 16:20

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