For the following situation:

http://i.imgur.com/f7DM34A.png

I'm driving a solenoid and need to know what my power supply requirements are.

I measured with clamp ammeter and saw ~2A.

Can you tell me how to compute this value, and how to use it to determine my PSU requirements?

 

You just use Ohm's Law. The current equals the supply voltage divided by the solenoid resistance, V / R.

 

But it's a modulated PWM signal. V/R in this case is 27A and it's obviously not 27A average.

 

But it's a modulated PWM signal. V/R in this case is 27A and it's obviously not 27A average.

Any supply must be able to handle the peak loads of the system.

 

Some clamp ammeter may not response to the real current of pwm.

Normally the I = V/R = 26.5A.
Calculating the 50%/50% PWM, I = (V/R)/(100%/50%)
So calculating the I_Pwm = (V/R)/(100%/Hi duty%)

exp: pwm Hi duty 100%
I_Pwm = (V/R)/(100%/Hi duty%)
I_Pwm = 26.5A/(100%/100%)
I_Pwm = 26.5A/1
I_Pwm = 26.5A

exp: pwm Hi duty 80%
I_Pwm = (V/R)/(100%/Hi duty%)
I_Pwm = 26.5A/(100%/80%)
I_Pwm = 26.5A/1.25
I_Pwm = 21.2A

exp: pwm Hi duty 50%
I_Pwm = 26.5A/(100%/50%)
I_Pwm = 26.5A/2
I_Pwm = 13.25A

exp: pwm Hi duty 20%
I_Pwm = 26.5A/(100%/20%)
I_Pwm = 26.5A/5
I_Pwm = 5.3A

exp: pwm Hi duty 5%
I_Pwm = (V/R)/(100%/Hi duty%)
I_Pwm = 26.5A/(100%/5%)
I_Pwm = 26.5A/20
I_Pwm = 1.325A

 

Sorry, but is it not that simple. Since the load is an inductor, it will depend on the period of the PWM.

Bob

 

ScottWang's calculations are basically correct if the PWM frequency is high enough that the inductor ripple current never goes to zero. Anything above 10kHz should be fine for an inductance of 155μH.

Note that a free wheeling diode needs to be added across the solenoid (cathode to plus) for the circuit to work properly and avoid large voltage spikes across the transistor.

 

There is a diode, I didn't include it because it's not a part of the current calculation, what is shown is stripped down

 

There is a diode, I didn't include it because it's not a part of the current calculation, what is shown is stripped down

But it is part of the current calculation for a PWM signal. If the inductive current has nowhere to go when the transistor shuts off then the current must stop, which generates a large inductive spike at the transistor collector.

 

Cool fact. I have a protection diode from drain to Vcc that is not shown in the simplified schematic.

 

OH jeeze! I just realized that my drawing is very deceptive! During the period of time labeled T2, there is NO SIGNAL, it doesn't continue as the three dots would have you thinking.

That is, it's not just a simple continuous 50% duty cycle signal

It's a square wave at 33.33Hz, with a duty cycle of 2.67%, modulating another square wave at 31.25kHz with a duty cycle of about 50% (though this varies, so a solution in general terms is preferable).

 

OH jeeze! I just realized that my drawing is very deceptive! During the period of time labeled T2, there is NO SIGNAL, it doesn't continue as the three dots would have you thinking.

That is, it's not just a simple continuous 50% duty cycle signal

It's a square wave at 33.33Hz, with a duty cycle of 2.67%, modulating another square wave at 31.25kHz with a duty cycle of about 50% (though this varies, so a solution in general terms is preferable).

As a rough method ......

\(\text{I_{\small{AV}}=0.0267[LF_{duty}]x\frac{0.5[HF_{duty}]x53[Volts]}{2[Ohms]}=0.354 Amp}\)

 

That is what i calculated before too.Interesting. Why do you suppose clamp ammeter would read 1.6a

 

What clamp ammeter are you using?
Do you own a DVM? If so what DC voltage do you measure across the solenoid?