design example

Apr 23, 2005 2 Replies

Hi all, i need your help for understanding this design example



Problem: A full bridge speed controller circuit is designed to control a 12v motor. The switching frequency must be above the audible limit (20kHz). The motor has a total resistance of 0.12 Ohms. Choose suitable MOSFETs for the bridge circuit, within a reasonable price limit, and suggest any heatsinking that may be required. The ambient temperature is assumed to be 25ºC. Solution: Lets have a look at the IRF3205 and see if it is suitable. First the drain current requirement. At stall, the motor will take 12v / 0.12 Ohms = 100 Amps. We will first make a guess at the junction temperature, at 125ºC We must find what the maximum drain current is at 125ºC first. The graph of figure 9 shows us that at 125ºC, the maximum drain current is about 65 Amps. Therefore 2 IRF3205s in parallel should be capable in this respect. How much power will the two parallel MOSFETs be dissipating? Lets start with the power dissipation whilst ON and the motor stalled, or just starting. That is the current squared times the on-resistance. What is RDS(on) at 125ºC? Figure 4 shows how it is derated from its front-page value of 0.008 Ohms, by a factor of about 1.6. Therefore, we assume RDS(on) will be 0.008 x 1.6 = 0.0128. Therefore PD = 50 x 50 x 0.0128 = 32 Watts. How much of the time will the motor be either stalled or starting? This is impossible to say, so we will have to guess. 20% of the time is quite a conservative figure - it is likely to be a lot less. Since the power causes heat, and the heat conduction is quite a slow process, the effect of power dissipation tends to get averaged out over quite long time periods, in the region of seconds. Therefore we can derate the power requirement with the quoted 20%, to arrive at an average power dissipation of 32W x 20% = 6.4W. Now we must add the power dissipated due to switching. This will occur during the rise and fall times, which are quoted in the Electrical Characteristics table as 100ns and 70ns respectively. Assuming the MOSFET driver can supply enough current to fulfill the requirements of these figures (gate drive source resistance of 2.5 Ohms = pulse output drive current of 12v / 2.5 Ohms = 4.8 Amps), then the ratio of switching time to steady-state time is 170ns * 20kHz = 3.4mW which is negligable. These on-off timings are a bit crude however, for more information about on-off times, see here. Now what are the switching requirements? The MOSFET driver ship we use will cope with most of these, but its worth checking. The turn-on voltage, Vgs(th), from the graphs of Figure 3 is just over 5 Volts. We have already seen that the driver should be able to source 4.8 Amps for a very short period of time. Now what about the heatsink. You may want to read the chapter on heatsinks before this section. We want to keep the temperature for the semiconductor junction below 125ºC, and we have been told that the ambient temperature is 25ºC. Therefore, with a MOSFET dissipating 6.4W on average, the total thermal resistance must be less than (125 - 25) / 6.4 = 15.6 ºC/W. The thermal resistance from junction to case makes up for 0.75 ºC/W of this, typical case to heatsink values (using thermal compound) are 0.2 ºC/W, which leaves 15.6 - 0.75 - 0.2 = 14.7 ºC/W for the heatsink itself. Heatsinks of this θjc value are quite small and cheap. Note that the same heatsink can be used for both MOSFETs to the left of or to the right of the load in the H- bridge, since these two MOSFETs are never both on at the same time, and so can never both be dissipating power at the same time. The cases of them must be electrically isolated however. See the heatsinks page for more information on the required electrical isolation.



These are my doubts:


1) "Assuming the MOSFET driver can supply enough current to fulfill the requirements of these figures (gate drive source resistance of 2.5 Ohms = pulse output drive current of 12v / 2.5 Ohms = 4.8 Amps)". What does it means ? Shouldn't the choise of the mosfet driver be based on Qg= I*T --> I=Qg/T --> I=146ns*20KHz = 2,92mA ?


2) Which are the familiar rules to be followed in choosing the proper mosfet driver ?



thanks in advance



mauri



I take it you mean micro-seconds there.

The gate has to be charged and discharged very quickly, by short pulses taking of the order of 100 nanoseconds or less, each. I think you are using the wrong value for time in this calculation, but I'm a bit confused by your use of 'ns' for units of charge. I take it this is meant to be 146 nano-Coulombs total gate charge?

These are really nano seconds, like described in data sheet.

Sorry 146nC. I've read this value from the data sheet --> Qg max

waiting for reply

thanks

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