Greetings,

I'm getting more famiiar with how to choose my resistors when setting up an H-bridge to switch a larger current. However, from what I can tell, I need to know the current of the driven circuit in order to do so.

Looking at one quadrant of my bridge:

I know I'll be drawing ~40mA, and can size R1 and R2 appropriately so that both transistors are just barely saturated when the +6v signal is on and both are completely off when the signal is off.

What's giving me trouble is that the resistance of my load will vary between 67ohms and 670 ohms. ...and as the resistance changes, the stauration threshhold changes. How should I approach building an H-bridge in this scenario?

Thank you,
-masked

 

You have to plan for the 67 Ohm load.
Rb = (Vin-Vbe) / (Ic/10)

Ic(Q2) = 27v/67 Ohms = 0.403A
R2 = (27-0.7)/.0403 = 26.3/0.0403 = 652.6 Ohms.
A table of standard resistor values is available here: http://www.logwell.com/tech/components/resistor_values.html
Bookmark that page. Refer to the E24 (green) columns.
620 Ohms and 680 Ohms are the closest values. You can actually use either one, but 680 is the best match for R2.

R1 should be roughly 5 * R2, or around 3.4k. You could use 3.3k or 3.6k.

Q1 needs a base resistor.
We also know from the above calculation that Ic=40.3mA, since that's the base current for Q2. The current through R1 will be inconsequential and can be ignored.

Rb = (6v-0.7v)/(40.3mA/10) = 5.3v/0.00403 = 1315 Ohms. 1.3k is just about perfect for Q1's base resistor.

 

So the idea is to plan for the highest current in our calcs, and then get the proper ratio of the resistors. After that, we know the same ratio will work for a lower current.

After this sinks in, I'm sure I'll have more questions but just to confirm:
The divide by ten is because of the transistor's beta rating, right?
i.e. to control 40mA of current across Q2ce, we only need 4mA at Q2b.
So this is just Ohms law, but we know the current at Q2 base is 4mA, and do the same thing for Q1 to get .4ma

Also, throw away R1 for a minute, it's still not intuitive to me why we are using the current at Q2c to perform resistance calculations for Q2b.
Is the operation of a PNP transistor such that the .7 voltage "leak" across cb travels through with the same amperage coming in to c?

If so, it seems that this effects efficiency. Those amps would flow across Q1 and "out" the Q1 emitter. So I should disconnect my Q1 emitter from ground, and instead connect it to the collector pin of the transistor in the lower half of the h-bridge. Does that sound better?

Thanks SgtWookie!