Collector bias for a BJT?

[rmjr]

Hello,

I am new to this forum and I would like to ask a simple question about BJT collector biasing configuration. In the following transistor configuration I require the input to be a ramp starting at 2.0VDC to 3.3VDC. At the output (Collector Vc) I require a downwards ramp starting at 1.2 down to 0.4VDC. Please note, I am new to transistors and even though this can be done with op-amps, I would still prefer to use transistors since this is what I am learning right now.

Therefore, in the attachment below, excluding R1 and R3 for a moment, I have a transistor which draws approximately 995ua at collector (Measured). I also measured approximately 7.9ua at the base. Therefore:

hFE = Ic/Ie = 995ua/7.9ua = 125

I calculated Rb, like this:

Rb = (Vc - Vbe)/Ib
Rb = (1.2 - 0.642) / 7.9ua = 70.632 K ohms > Used a 75K!

I then add in R1 and R3 to tne mix and here is pretty much where I start to get confused!

I know the circuit works with R1 at aproximately 130K and R3 at 75K. But I don't know how to calculate these resistor values.

Can some please help with the calculations/rationalizations for calculating R1 and R3!
Thanks all for your help!
r

[pebe]

You are talking about a 'ramp' between two voltages, but a ramp implies a voltage that changes over a period of time. To generate a ramp you need an inductor or capacitor, and a resistor in order to get a 'time constant'.

Did you mean you want to get 1.2V out for an input of 2.0V, and 0.4 out for 3.3V input? If so, I'm not at all sure that it can be calculated because Hfe will probably get smaller as the collector voltage falls to the low level of 0.4V, and the current through R2 will reverse.

[rmjr]

Hello pebe,

The 2V, comes from a dac, which has its time component all by itself. In other words, the dac ramps from 2V to 3.3 vdc at a specific frequency. Therefore, time is in the signal itself, all I want to know is how to calculate R1 and R2 in the circuit shown... I tried many ways but I don't seem to get the right values!

[quote]
Did you mean you want to get 1.2V out for an input of 2.0V, and 0.4 out for 3.3V input?
[/quote]
Yes:
IN:
2->3.3
OUT:
1.2->0.4

thanks

[pebe]

Hi rmjr,

Using your figure of 125 for Hfe, I came up with the following using Thevenin.

R1 – 122.5K
R2 – 114.7K
R3 – 75K

They won’t be to high accuracy because:
1. They don’t allow for Hfe spreads
2. They use a figure of 0.6V for Vbe, and don’t allow for changes due to base current.
3. Vbe is temperature dependant

I found then easiest way was to assume:
1. The point at 2.97V on the ramp (from your information) would give 0.6V output. That means no current flows through R2 and it can be ignored at this time.
2. Base current at that Ic is equivalent to a 55.5K resistor across R3
3. R1 can now be simply calculated to give 0.6V at the base with 2.97V input.
4. R2 feeds back to the base of the transistor. It ‘looks into’ an effective resistance of R1, R3, and the 55.5K resistors in parallel, so R2 can now be calculated from the relative swings on the input and output.

I hope that makes sense.