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[...] but I don't see how the voltage output from the LM350 is being properly adjusted to 1.25 V above the battery voltage
- there is a diode between the battery and the LM350 to prevent draining the battery, but this would also keep the LM350 from seeing the battery voltage, right?
So the LM350 is adjusting the output voltage to 1.25 V above ground, not above the positive terminal of the battery?
The datasheet for the LM350 has the attached diagram and this equation:Vout = 1.25 V * (1 + R2/R1) + Iadj * R2
It also says that Iadj is small so that term can usually be neglected. So when the PWM signal is high, the transistor grounds the Adj pin on the LM350. This is like saying R2 = 0, so Vout = 1.25 V.
When the PWM signal is low, the transistor does not allow any current through, and Vout should jump to its maximum for the LM350 (or to the supply voltage, which in my case is ~15 VDC).
In either case, the voltage is not simply 1.25 V above the battery voltage.
Any ideas for things to try? Please let me know if you need more information, want to see the code - whatever. I'm trying to understand this circuit better so I can figure out what to expect and what needs to be corrected.
-When PWM is high, Adjust pin is at close to 0 voltage. Vout is at 1.25 V (confirmed this with measurement), so as far as I can tell, there is still close to 1.25/0.62 = 2 A through the power resistor. Is all of this going to ground through the transistor?
- When PWM is low, Adjust pin should reach close to the supply voltage (adjust voltage will be close to Vout, which will in turn make Vout higher, etc etc until it is saturated).
I measure this voltage at close to 6V without a battery present (~7V with a battery present), even though my supply is close to 15V.
I put a power resistor between the diode and the positive terminal of the battery, and measured the voltage drop across it thinking that this would be an indication of the current flowing into the battery. Is this valid?
The voltage drop across the 10 Ohm resistor was only ~0.02 V. This works out to 0.002 A, which is about 1000 times less than I was expecting.
Can someone explain the constant current source? In the datasheet for the LM350 there is a schematic (see a couple of posts above) showing a resistor between the Adj and Out pins on the LM350, which provides a constant current. In my schematic, there is an additional 1000KOhm resistor between where I need the constant current source and the Adj pin. Does this explain the factor of 1000 that my current was off? I don't know how to analyze the circuit with the LM350 in the loop - the sum of the currents at the junctions between the power resistor and the diode gets confusing...
What exactly is happening when the PWM toggles the transistor? I provided some thoughts in a previous post, but now I'm not sure they were correct. If the PWM signal is high, is this "almost grounding" the Adj pin of the LM350? Does any of the current provided by the constant current source get routed through the transistor to ground?
And then when the PWM signal is low, it functions just as if the transistor isn't there? Just the constant current source as shown in the datasheet?
Or maybe this is the wrong way to think about it, since the PWM is really approximating an analog signal, it's more like applying some analog signal to the transistor?