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dissipating heat on incandescent lamps instead of the MOSFET

qwertyqwq

Member
hi.
i was surfing on internet and found that scheme which calls "electronic load" .
NV_0716_Oliveira_Figure02.jpg
And the author wrote for this circuit as follows ;
"It’s important to note that — unlike in the traditional approach — most of the heat in this circuit is dissipated in the incandescent lamps instead of the MOSFET. Since the MOSFET is either turned OFF (close to infinite resistance) or turned ON (close to zero resistance), the power dissipated in the device is much lower than with the traditional circuit. The incandescent lamps do the heavy lifting here and dissipate most of the heat. Plus — unlike the MOSFETs — incandescent light bulbs do not need large heatsinks! "
I didnt understand why this would happens, cause of the same current will flow through the mosfet and same switching period will mosfet's base fall into. Lamb is just limiting the current on second circuit.
Is that circuit true ? If its true how does it possible? Can you explaine how heat is going to bulb instead of mosfet ? Thanks.

Edit: Can we think like;
for example we need 1 amp load current. On 12v supply voltage we need 12W power. Lets say our mosfet has 1ohm Rds(on). The power dissipation on mosfet will 1W. And the other (11W) dissipation will be on lamb. Is my syntesis true ?
 
Last edited:

Pommie

Well-Known Member
Most Helpful Member
The first circuit is stupid as it has no load and the MOSFET shorts the power supply (and unsurprisingly gets hot). In the second, the lamp is the load. It's basically nonsense.

Mike.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
Do the maths - W = V x I

In a linear circuit you might have 5V across the FET and 10A through it - so 50W of heat in the FET (10 x 5).

In PWM the FET (or transistor) is either fully ON or fully OFF - so assuming a 10V supply:

FET ON - current = 10A, voltage = 0V (or close to it), W = 0 x 10 (so zero watts).

FET OFF - current = 0, voltage = 10, W = 10 x 0 (so zero watts).

Add OFF and ON together 0 + 0 = 0, so zero watts total.

In practice the voltage drop across the FET will be slightly above zero, but still very low, and the FET will be briefly in linear mode as it switch between ON and OFF, and OFF and ON - which is why it's important to switch as fast as possible.

This is why PWM and switch-mode have taken over the electronics world.
 

dr pepper

Well-Known Member
Most Helpful Member
Not what you were asking but I'll mention it, I tried something like this to test smps's, but I found all kinds of weird happened, the lamp and fet togther present a complex load that can upset smps's, or a supply that has stability compensation, esp if its not well designed like a prototype.
Fine for discharging bateries and the like.
The stupid way can work if done correctly, but it'd be w hole lot better with a current sense resistor & associated circuit.
 

Externet

Active Member
Since the MOSFET is either turned OFF (close to infinite resistance) or turned ON (close to zero resistance),
The incandescent lamp does have a resistance at all times; that may be what confuses you. OFF or ON mosfet is like shorted or open conditions which translate nothing to ohm's law for dissipation calculations.
 

gophert

Well-Known Member
Most Helpful Member
The first circuit is stupid as it has no load and the MOSFET shorts the power supply (and unsurprisingly gets hot). In the second, the lamp is the load. It's basically nonsense.

Mike.
No, the first circuit makes a fine load for a dummy load. It is not nonsense. It is only nonsense as a dummy load if the user applies too much voltage to the gate. These MOSFETS are used in their linear range and an array of them can be set up to creat a bunch of heat (on purpose).
 

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