# DC/DC Converter - Duty Cycle vs. Current

Discussion in 'Electronics Resources' started by boiledbeans, Feb 5, 2015.

1. ### boiledbeans Thread Starter New Member

Sep 5, 2014
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Hi all

I have a question about using a DC/DC Converter to regulate current during battery charging/discharging. How is this done mathematically?

Let's say I have a rechargeable battery connected to a DC bus via a DC/DC converter. An example is given here on page 14, using Fig 14
In the paragraph below Fig 14, it says "The magnitude and polarity of the battery current can be controlled via adjustment of the duty cycle D".

So how is this battery current control achieved in terms of mathematics and formulas?

My understanding of the problem is as follows,
For example, at a particular instant, the battery registers 10V. So I charge it from a 50V bus, which means buck mode, then the duty ratio formula for an ideal buck converter would be

V_batt / V_bus = D = 10/50 = 0.2

However, current is also regulated by this formula

I_bus / I_batt = D = 0.2

So how can I specify that I want to charge the battery at say, 1A (I_batt = 1A) using the duty ratio D since the duty ratio is already locked at 0.2?

2. ### crutschow Expert

Mar 14, 2008
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You can either regulate current or voltage but not both simultaneously.
For battery charging, typically the voltage is set to the value you want at the end of charge and a current-limit portion of the circuit limits the current to the maximum charge current you want during the charge (in practice this occurs by lowering the output voltage until the desired current is achieved).
Both voltage and current are regulated by negative feedback from the output voltage and current (typically by measuring the voltage across a small shunt resistor in series with the load) to control the PWM duty-cycle.

So the duty-cycle will be adjusted to regulate the voltage by the current-limit circuit during charging time and then by the voltage setting at the end of charge. The duty-cycle is thus not "fixed".

Make sense?

3. ### boiledbeans Thread Starter New Member

Sep 5, 2014
8
0
So let's say I'm just focusing on the constant current battery charging part (ignoring the constant voltage charging part).

Does it mean that the equation for an ideal buck converter
V_batt / V_bus = I_bus / I_batt = D = 10/50 = 0.2
is invalid in this case?

If so, then what is the equation relating current to duty cycle in this case? For example, if I want to charge the battery at 1A, what duty cycle should I select theoretically, assuming open-loop control?

4. ### crutschow Expert

Mar 14, 2008
13,501
3,375
The duty-cycle determines the voltage, not current. The current then follows the voltage.
Thus the current is determined by the source voltage, as determined by the converter duty cycle, minus the battery voltage divided by the battery resistance.

So to determine the correct converter open-loop duty-cycle you would need to know the input voltage, the battery voltage, and the battery resistance. That's why such converters are operated closed-loop.

The ratio of input current to output current is still related to the duty-cycle of course.

5. ### boiledbeans Thread Starter New Member

Sep 5, 2014
8
0
So how can I apply this formula correctly to this situation?

V_batt / V_bus = I_bus / I_batt = D

where
battery voltage V_batt = 10V
desired battery charging current I_batt = 1A
DC bus voltage V_bus = 50V

Because it seems if V_batt and V_bus is fixed, then the duty cycle is also fixed and it cannot be varied to charge at the desired current.

Assuming we know all the parameters such as battery resistance (say R=0.01 ohms) for a theoretical study to calculate the duty cycle.

Is V_bus not a fixed value, but variable as you said "current is determined by the source voltage, as determined by the converter duty cycle"?

6. ### ScottWang Moderator

Aug 23, 2012
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The formula is to describe that how high the voltage you can get, it depends on the duty cycle of the percentage, it means that if the duty cycle is more then the voltage you can get is higher, otherwise the voltage will be lower .

If the voltage is going down then the current also will be going down, since the resistor is fixed, and the I = V/R.

7. ### boiledbeans Thread Starter New Member

Sep 5, 2014
8
0
But in my situation as described above, the voltage input and voltage output are fixed.

For example, voltage input is fixed by the DC bus at 50V. Voltage output is fixed by the battery voltage (at that moment) at 10V. Using the formula V_batt / V_bus = D, then duty cycle also becomes fixed at 0.2. Now, I can no longer use duty cycle to vary current?

8. ### ScottWang Moderator

Aug 23, 2012
4,930
777
Although using the pwm can reducing the voltage, but if you reducing from 50V to 10V then I don't think it is a good idea, and the current only 1/5 A comparing with the original, and if the pwm circuit damaged then the voltage may too high and could kill the battery.

So if you trying to using the pwm circuit to do the charging then you also need a current limiting circuit and check the current enough or not when the current through the pwm circuit.

The current limiting circuit will be as a mosfet and adding a resistor to detecting the current, if the current over the setting then the mosfet will be turn off.

Last edited: Feb 6, 2015
9. ### crutschow Expert

Mar 14, 2008
13,501
3,375
You seem to fixated on the output voltage being fixed. The output voltage is not fixed.
The battery voltage is fixed but if you have the output voltage exactly equal to the battery voltage no current will flow.
So for a 10V battery, 1A of charging current and .01Ω battery resistance for example, you need to have a PWM duty-cycle that gives an output voltage equal to (1A*.01Ω) + 10V = 10.01V.

Last edited: Feb 6, 2015
10. ### boiledbeans Thread Starter New Member

Sep 5, 2014
8
0
Ah, I see, so the converter voltage output is not fixed.

It seems the ability to control current flow depends on the fact that the battery is non-ideal? If the battery was ideal with no internal resistance, then it wouldn't work because the converter output must be exactly equal to the battery voltage of 10V?

11. ### crutschow Expert

Mar 14, 2008
13,501
3,375
Yes, you need some resistance in the circuit, either the battery or the source.
Otherwise the slightest difference in voltage will give infinite current, along with infinite power, and the universe, as we know it, will go up in a puff of smoke.

12. ### boiledbeans Thread Starter New Member

Sep 5, 2014
8
0
Ah ok. Thanks to both of you for your help!