Consider the following constant-current battery charging circuit (simplified for the purposes of exposition):

CC NiMH charger schematic

My understanding is that the current through the diode and battery will be held at 0.24 A by the regulator, which will increase or decrease its output voltage as necessary to maintain that current. (It does this here because by design the regulator adjusts its output voltage to maintain a constant 1.25 V difference between the OUT and ADJ pins. Since the regulator is holding the voltage drop over the resistor constant, and the resistor itself is a constant value, by I = V/R the current must be constant.)

Thus, with or without diode D1 the battery will be charged at the same rate. The diode will drop the voltage that the battery sees by 0.7 V as compared to the regulator output, but the regulator should automatically increase its output voltage by 0.7 V as compared to a circuit without the diode to compensate for this, right? Or am I missing a subtlety here about how diodes work and how the current is "used"?

Will this change the charging behaviour of the NiMH cell at all? (Assume the cell is not being overcharged but is charged in the usual recommended way: removed from the charger after reaching full charge—12 hours at C/10 if the cell started fully discharged.)

I also understand that the (reversed as compared to discharging) voltage across an NiMH battery in the charger will typically peak around 1.6 V as it reaches full charge. At that point, with the diode there, that would mean that the regulator would be be generating an output voltage of:

      1.25 V  drop over resistor, constant OUT/ADJ difference
      0.70 V  drop over diode (nominal)
    + 1.60 V  drop over NiMH battery
      3.55 V  regulator output

Is this correct? (And it thus implies that, with the ~2 V regulator drop-out, a 9 V input should provide plenty of "headroom" for maintaining the regulator output at 0.24 A, right?)

  • \$\begingroup\$ I agree. I prefer a charge stop at 1.44 V for NiMH cells. \$\endgroup\$
    – Jens
    Jan 18 at 15:17
  • \$\begingroup\$ @Jens Note that this question is about constant-current charging (typically terminated with a timer). There are of course other valid (and probably better in many respects) ways of charging NiMH cells, but that's not what I'm trying to learn about here. \$\endgroup\$
    – cjs
    Jan 18 at 15:45

1 Answer 1


Your observations are absolutely correct; I will just summarize them.

The circuit operation is based on the negative feedback principle as follows.

First, a current source is made by a voltage source (voltage regulator) and resistor in series. It would work perfectly if shorted (if the battery and diode were replaced by a piece of wire) because the entire voltage would be applied across the resistor and the current would be determined by Ohm's law.

When we insert the 1.2 V battery cell in the circuit, the voltage after the resistor increases and since this voltage is subtracted from the input voltage, the current decreases.

To compensate it, the voltage source (regulator) increases its voltage with the same value. As a result, the resulting voltage across the resistor and accordingly, the current does not change.

When we insert the diode in series, the voltage after the resistor increases by another 0.7 V and the current decreases again... but the regulator increases its voltage with the same 0.7 V. As a result, the resulting voltage across the resistor and accordingly, the current does not change.

So the battery and diode act as a disturbance to this negative feedback circuit that compensates this disturbance by raising its output voltage. You can add more and more cells and diodes... and whatever you want... and the circuit will compensate for it... if only there is enough voltage difference between the input and output voltage.

The clever trick here is that the current-creating voltage source is dynamic and follows the load voltage with a constant offset. This is only one of many possible ways to make a constant current source.

EDIT: An answer to the @cjs's comments below

... there's no way in this scenario that the battery could see any difference between the circuit with or without the diode, right?

Exactly! You can include as many diodes as you want and the battery will not "feel" it because the voltage regulator (forced to act as a current regulator) will compensate for the losses (voltage drops) in (across) them. Of course, a sufficient input voltage is necessary.

Your question is conceptual; so let's consider the conceptual arrangement for producing a constant current. It consists of three elements in a loop - "current-creating" voltage source V, "current-determining" resistor R and a load. In the "ideal" case, the load is just a "piece of wire" (e.g., an "ideal" ammeter). Then the current is simply I = V/R... and to obtain a constant current, both voltage and resistance can be constant. It is the simplest possible current source... but which is "ideal". More precisely speaking, it is not perfect; it is imperfect but is made to work under ideal load conditions (zero load resistance and voltage) and because of that it behaves as if it was perfect.

... with a constant current supply, the NiMH cell determines what the voltage will be...

The problem in your case is that the load is a bit peculiar - once it is a battery (something that somehow keeps the voltage of its terminals constant); then we add another thing (diode) that has the same behavior. If, on the one hand, we combine the input voltage source and the resistor... and, on the other hand - the battery and the diode, we can say that two sources - current and voltage - are connected to each other. In this pair, the current source sets the current through them and the voltage source sets the voltage across them. So the battery sets the voltage across itself.

... and nothing else in the circuit can affect that

Exactly! The battery voltage can be slightly influenced by the current if it was varying... but the current is constant. Why is it constant?

The clever trick here is that the total "current-creating" voltage (across the resistor) is kept constant by changing the input voltage (the regulator's output voltage); so the current is constant.

This configuration is always the same; only, in some (your) cases, the resistor is "floating" since it is connected in series to the input voltage source and the load is grounded while, in other cases (e.g., a transistor current source with emitrer resistor), the resistor is grounded and the load is floating.

Also, in some cases (e.g. op-amp inverting circuits), the constant "current-creating" voltage is achieved by adding an additional (op-amp output) voltage to the input voltage.

Another very common way to maintain a constant current is to vary the resistance at a constant voltage... but that is another story...

  • \$\begingroup\$ So just to be clear: there's no way in this scenario that the battery could see any difference between the circuit with or without the diode, right? (I ask because people keep mentioning voltage to me, and I don't see where voltage comes into this at all: it seems to me that with a constant current supply, the NiMH cell determines what the voltage will be and nothing else in the circuit can affect that, so long as the current supply doesn't sag). \$\endgroup\$
    – cjs
    Jan 18 at 16:53
  • \$\begingroup\$ @cjs, Your questions are very interesting; wait for me an hour (just to have dinner) and I will answer you. \$\endgroup\$ Jan 18 at 16:58
  • \$\begingroup\$ @cjs, I have tried hastily to outline the philosophy of the problem. I would appreciate it if you don't hesitate to ask more (conceptual) questions. \$\endgroup\$ Jan 18 at 18:44
  • 1
    \$\begingroup\$ Thanks for your further explanations. I do feel I understand pretty well how and why a voltage regulator wired up like this is a constant current source; I've expanded the question slightly to clarify that. I was just running into situations where it appeared someone was telling me that this changed how the battery was charged, and I couldn't reason my way to seeing how that could happen. As it turns out, my reasoning was correct: it can't. So most likely when someone's telling me that, I need to be more clear about the circuit I'm using (for which I now have this nice example). Thanks again. \$\endgroup\$
    – cjs
    Jan 19 at 0:37

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