Common collector load driver

Rich,

Max power will be about 9 watts at 30 volts. What is the number of the transistor you are using and the config. one side of the resistive load is connected to the collector and the other side to the 30 volt supply, the emitter is grounded or tie to the 30 volt power supply return. The control voltage is connected to the base thru a base resistor. Is this your config?

Den

Worst case, if you have to drop 300mA at up to 30V (30V supply, desired output of 0V (OK, your load is not likely to be drawing 300mA at 0V, but the principle doesn't change)), you have to dissipate 9W in your transistor - no way around that. A mosfet will have to dissipate the same 9W under the same conditions. If you can reduce your max supply voltage or load current, or if you can drive your load in a pulsed manner rather than an analog manner, you can reduce the power you need to dissipate.

BTW, I'm reading your load as a non-reactive load that draws 300mA regardless of voltage. If that's not correct (and I kind of doubt it is), then your maximum dissipated power amount will change based on the load V-I curve.

Mike

If you have a 2N4401 laying around you may want to try it.

I know they're good for 700ma but I'm just throwing that out.

Or beef up to a MJE3055 or something?

As long as you use a linear pass transistor you'll find your circuit very inefficient at output voltages lower than the supply voltage, and the power lost to the inefficiency will get burnt up in the pass transistor.

About the only way that you can get away from that situation is to use a switching amplifier, which is _much_ more involved to design, build and test.

Assuming a mild-mannered load, the power in your drive transistor will be (supply V - output V)(output current). For the extreme case where you're driving 300mA at 0V, with a supply voltage of 30V, that's 9 watts you've got to find a home for before it burns out your transistor.

Really, your only two choices involve either biting the bullet and using a pass transistor and heat sink that can handle the power dissipation, or biting the bullet and using a switching amplifier with all of its design complexity.

The power will always be Iout * (Input_voltage - Output_voltage) for a linear regulator. That's the same whether you use a MOSFET, a BJT or a rheostat operated by a servo.

If I take the worst-case interpretation of your requirements (that it might be required to supply 300mA into a 0-ohm load with a 30V supply) then you need to dissipate 9W, which will require a substantial heatsink.

You might consider using a switching regulator, but I can'tt think of a simple single chip solution off the top of my head. Check out Linear Technology's line, there may be something you can use if cost isn't a factor.

Best regards, Spehro Pefhany

Rich,

You can use a positive 28VDC reg chip such as a LM350TG 3 A, Adjustable Output 1.2 to 33 V, Voltage Regulator TO-220 case. Instead of connecting the gnd lead to gnd you would connect it to your controlling voltage. This will vary the output as you want. The bottom line is with a 300 ma load at 30 volts you will be generating 9 watts, thats the way it is. The output will be what ever you set it for even if the load may vary.

Good luck,

Dennis

Well, except it won't go below 1.25V (typical) without a negative input.

OTOH, it will be short-circuit proof, SOA-protected, and over-temperature protected (sort of).

Yup. Unless you have a switching regulator.

Best regards, Spehro Pefhany

Complexity? Switchers are pretty tame these days. Pick one of many available at your current rating, add an inductor, a few caps and a resistor divider. The only part that is not required with an LDO is the inductor! The manufacturer does all the work of design for you. All you have to do is follow their cookbook.

Switching supplies are that easy. You may find a switching amplifier is still a bit of a challenge.

If you know different, quote me a manufacturer and part number, I'd like to know about it.

that's because you're creating to much resistance in the transistor keep the voltage down on low levels when loaded.. This translates to heat.

When using lower input voltages, the transistor is at a lower resistant value and thus, less heat.

A Buck regulator will correct for this error.

Rich,

Can you give us a little more info as what the load is and what your driving it with. It seems to me the ans. are getting a little complex for all we know it might be just changing the brightness of a lamp.

Dennis