PNP bias/voltage divider

I have problems vieving this, The view that looks the best appears to show the emitter of the transistor connected to the + supply. If that's wrong, you have other problems and ignore the first two paragraphs.

You have some basic problems here. First, having a base voltage of -.7V (wrong number) does not mean an absolute -.7V, but rather -.7 with respect to the emitter. So, if the emitter was at +6.7V, the base would have to be at +6.0. Your resistor values are interchanged.

You don't bias a bipolar trtansistor that way. At 100 ma, the base-emitter voltage is going to be some number between about .60 and .68. Haven't looked at your transistor, but you want it to work with all transistors having betas between about 50 and 300. You could probably get it to work somewhat by returning R2 to the collector of the transistor instead of ground. and assuming the collector voltage is 1/2 of VCC. You would have to make R2 a variable resistor, and hope the temperature does not change a whole lot.

If this is a switching regulator, why is there *any* bias on the transistor?

Tam

To begin with, the base current is an exponentail function of the b-e voltage, and in addition it's quite temperature dependent. That means, for a particular base current, it's not a good idea to try to achieve a particular b-e voltage. The nominal b-e voltage just lets you know the base voltage above ground (that is, about 0.7V lower than the supply), so you can find the value of R2 to give you the required base current. Then R1 just insures that the base current goes to essentially zero when you are NOT pulling R2's bottom end down to ground potential, assuming you want to turn it off. R2 must supply the required transistor base current AND whatever current is in R2 when it has about

0.7V across it.

Second, the transistor DC current gain (HFE) is a strong function of the particular sample, and also of temperature. If you really want the collector to source 100mA to some fairly close tolerance, there are better ways to do it. If you just want the collector to pull up to the supply voltage (less a small drop) and the maximum current will be

100mA, then you're OK. But it's usual to "overbias" the base, and go for a ratio of base to collector current of 20:1 or even 10:1.

You didn't say anything about turning the transistor on and off, but if you want to do that, there may be some other considerations: speed, and how much voltage swing the thing that drives the on/off signal can supply.

Cheers, Tom

Resistor values interchanged?? 6 volts across his 9k R2 value does give about 667uA, and the current in R1 at 90k and 0.7V would be about

7uA, leaving 660uA for the base. There is some question then about how fast it would turn off, if that's the desire, if you open the connection to R2, and even some question about just what the purpose of the transistor is, but looks to me like the calcs for the values are OK with the given assumptions.

The assumptions, on the other hand, may be off-base. Should the transistor saturate, or should it act like a 100mA current source? Will it be turning on and off? Occasionally, or rapidly? Over what temperature range must it operate?

Cheers, Tom

You are right. I guess it is a coincidence, but it struck out at me that if you reversed the R values, you would get the correct base voltage if you ignored the loadg. Still, current biasing is not the way to do it either. I've been there, and had the factory scream bloody murder when the nominal beta=40 transistors started arriving with betas of 385.

Tam

There is some question then about how

First of all, thanks for all the comments, it really helps out and I appreciate everyone's time.

The regulator is modeled after an idea that I read about a few years ago. That circuit could only source 20mA or so and required a higher supply voltage, somewhere around 12V I believe. Using the same concept, I've created a regulator that can be driven from a 9V battery (6.5V to

8.xV), is fairly efficient (95% or so simulated) that provides 5.49V with

Transistors are not precise devices. You've a cat in hell's chance of pinning down that 100ma. It's usual to swamp out the Vbe voltage variations by adding some emitter resistance. Changes in Vbe due to varying temperature and currents will now be small compared to the larger voltage across the emitter resistor. A la ...

6.5 to 8.9 ---o---------o | | .-. .-. | | | | | | | |4.7 '-' '-' |220 | | |6V | 1ma |< o-------| BC327 |5.3V |\\ | | 5ma| | | o .-. | | 100ma load | | o '-' | | 1k | 0V---o---------o (created by AACircuit v1.28 beta 10/06/04

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First of all, the transistor will have a reduced gain as it saturates. The lower you need the saturation drop to be, the more base current you have to provide. So lets say you provide a base current that is 1/50th of the collector current, to drive a normally higher gain (say, beta 150) device well into saturation, to keep the on state conduction losses low. This would be 100 mA/50= 3mA.

Your R3 has to supply that current ans also the current that bleeds through R2 while the transistor is on. The only purpose for R2 is to drain the base stored charge away, to quickly turn the transistor off when the current through R3 is cut off. I usually try to keep the on state R2 current at least 1/10th of the on state base current. That would put it around .3 mA. If the on state base emitter voltage is about .7 volts, that means that R2 would have a resistance of about .7/.3 mA = 2.3k. Lets say we use 2.2k, a standard value. So R3 has to provide about 3.3 mA at the lowest supply voltage minus the Q3 base emitter drop and the Q1 collector to emitter drop. Lets say that is 6.5 - 0.7

-0.3 = 5.5 volts.

So R3 is about 5.5/3.3 mA =1.67 k. Lets say you use a 1.6k

5% standard value.

Unfortunately, at the fresh battery voltage, R3 supplies the base of Q1 with more current than is needed, and this hurts the efficiency of the supply a little. Replacing Q1 and R3 with a switchable current regulator would reduce this effect.

In addition to what John Popelish wrote in reply to this, I'd add that it could help you in getting it to switch a bit more crisply to put a large value of resistance between Q3's collector and Q1's base. I'd suggest perhaps 100k ohms, or a bit more. Also, R5 can be eliminated; it does practically nothing. There's already ample path from the base of Q2 to ground through R4 and R6.

Cheers, Tom

So, the 9K does not go to ground as you showed, but is a pulse input. I have two suggestions:

Look at replacing the PNP with a P channel FET. This is going to save a lot of base current since you are concerned about power. You need a FET that is fully ON at 6V. You don't have to limit the gate voltage to limit the drain current; that is limited by the inductor and the switching frequency. What you are proposing to build is called a buck regulator. You can buy chips to do that from Digikey and Mouser, made by outfits like National, Maxim, and TI.

Or, forget about the switching regulator and look at LOW DROPOUT series regulators. The efficiency would be very close to V(out)/V(in). Probably less efficient than the switcher with a new battery, and more efficient with an old battery. The 7805 puts out 5V, and requires a minimum input of 7V;

7.5 on a bad day (This is not low dropout)..

I don't know what you mean by 9V battery, but if it is what I think, did you ever calculate how long it would last with a 60 - 85 ma load?

Tam