DC-DC input resistance for LTSpice simulation

Thread Starter

andrew74

Joined Jul 25, 2022
248
Hi!
I’m simulating a three-phase rectifier with diodes and an R-78HE5.0-0.3 DC-DC converter in LTSpice.
The three sine waves come from a three-phase motor which I assume runs at 15 Hz, so with 8 pole pairs that’s approximately 111 rpm (this is just an example)

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Can I simulate the DC-DC converter by replacing it with an equivalent resistor?

If I know its output voltage (5V) and current (300mA), assuming an efficiency of 85% and a rectified input voltage of 48V, can I calculate a resistor to ‘replace’ it in the simulation?

Po = Vo * Io = 5 * 0.3 = 1.5W
Pi = Po / n = 1.5 / 0.85 = 1.765W
Ri = Vi^2 / Pi = 48 / 1.765 = 1305 ohm

I am aware that real DC-DC converters behave like a negative differential resistor (if the input voltage rises, the input current falls to keep the power constant), while a fixed resistor behaves in the opposite way... but I am assuming fixed values to maintain a constant equivalent resistance under such conditions

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If my calculation is correct, I end up with an output that hasn’t been rectified properly:
View attachment 370225

Apart from increasing the capacitance, what can I do to reduce ripple?
(I’ve included L1 (which has no effect in this circuit) because I’d like to test and understand which CLC filter to use at the input to the DC-DC converter, given that it’s recommended in the event of EMI)

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I’m afraid I might be doing something wrong when setting up the three input generators
 

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Last edited:

crutschow

Joined Mar 14, 2008
38,638
A fixed load resistor should be okay for your simulation purposes.

You reduce ripple (if needed) by increasing the filter capacitance and/or adding a larger inductor (but with the inductor at the input to the capacitors).
There's no other magic to reduce ripple.
 

Thread Starter

andrew74

Joined Jul 25, 2022
248
A fixed load resistor should be okay for your simulation purposes.

You reduce ripple (if needed) by increasing the filter capacitance and/or adding a larger inductor (but with the inductor at the input to the capacitors).
There's no other magic to reduce ripple.
Thank you for your reply.

Have I modelled the three phases of the motor correctly? Because I set it to 48V, but in fact the voltage between phase and neutral fluctuates between -48V and +48V, producing a DC output of around 90V .. which would destroy my DC-DC converter
I’m getting a bit confused by engine theory…

If I remember correctly, in a three-phase system, the peak voltage between two phases is sqrt(3) times the phase voltage … therefore, the three-phase diode bridge brings the DC output to the peak value of the concatenated voltage minus the voltage drop across the two conducting diodes
 
Last edited:

Thread Starter

andrew74

Joined Jul 25, 2022
248
Why are you adding a 48Vdc offset to the sources?
That's why you are getting about 80V in the sim.
The offset should be 0.
Power AC sources have zero offset.
You’re right, offset = 0.

If the voltage formula between phase and neutral is Vdc/2*sin(...) … and my Vdc is 48V … I should set the amplitude to 24V on my three generators, right? … and not 48V

1786209559660.png
 

Papabravo

Joined Feb 24, 2006
22,101
A suggestion. Do not use the default diode for this application. Choose an actual part suitable for this application as if you were going to build an actual prototype.
 

eetech00

Joined Jun 8, 2013
4,719
Thank you for your reply.

Have I modelled the three phases of the motor correctly? Because I set it to 48V, but in fact the voltage between phase and neutral fluctuates between -48V and +48V, producing a DC output of around 90V .. which would destroy my DC-DC converter
I’m getting a bit confused by engine theory…

If I remember correctly, in a three-phase system, the peak voltage between two phases is sqrt(3) times the phase voltage … therefore, the three-phase diode bridge brings the DC output to the peak value of the concatenated voltage minus the voltage drop across the two conducting diodes
1. Set each sine voltage source to about 36.9v*sqrt(2).
36.9 equals the AC peak of about 48vdc +3v for diode drops.
2. Place a ground at the junction of the voltage sources (Ltspice doesn't care)

1786221371645.png
 
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