Why be concerned with pulser power dissipation?

Because by understanding power dissipation, you'll be able to trade frequency for voltage (and the reverse) while staying within the limits of the instrument.

Let's start with current. The amount of current needed to charge a capacitor to a given voltage over a given time is:

I = C • V • F

where I is the current in amperes, C is the capacitance of the load, the output cable, and the internal capacitance of the pulser's output circuit in farads, V is the output voltage in volts, and F is the pulse repetition frequency in hertz.

Now, let's move on to power. The pulser dissipates power in the process of delivering this current, and each pulser has a maximum power dissipation specification. Exceeding this specification may damage the instrument, so we need to know whether our setup is safe.

The amount of power (in watts) dissipated by the instrument is found by multiplying both sides of the above equation by voltage:

P = C • V2 • F

Let's try an example. Assume the pulser is a PVX-4110 operating at 10 kV. The load consists of a 100 pF load capacitor, an additional 1200 pF from a 60-foot coaxial cable at 20 pF per foot, and the pulser's internal capacitance of 200 pF, for a total of 1500 pF. What's the power dissipation at 100 Hz?

P = (1.5 x 10-9)(1 x 108)(1 x 102) = 15 W

Because the PVX-4110's maximum power dissipation is 100 W, there is no problem. The frequency is low so the dissipation is low. BUT, if we were to increase the frequency to 1000 Hz, the power dissipation becomes 150 W -- well above the maximum specification.

Here's another example. We're using a PVM-4210 differential pulser to generate 500 V pulses, and we want to output a 500 kHz burst ten percent of the time. The load is 50 pF, the pulser's internal capacitance is 125 pF, and three feet of coaxial cable adds 65 pF. The total is 240 pF.

P = C • V2 • F = (2.4 x 10-10)(2.5 x 105)(5 x 105) = 30 W

Since the burst is present 10% of the time, the average power dissipation per channel is 3 W. The two internal power supplies are each rated at 4 W, so we are within the limit. However, increasing the voltage to, say, 600 V exceeds the limit, which means the frequency would have to be reduced. (Voltage is squared, so changing the voltage makes a big difference in power.)

So, we can balance voltage and frequency, keeping the load capacitance fixed. We can also balance with capacitance, keeping in mind that increasing the load capacitance has the additional result of increasing rise time.
Updated: 20 Feb 2019 02:20 AM