Hero999
Banned
How much cheaper?ikalogic said:I can easily find TIP transistors, and they are much cheaper.
Where I live cheap MOSFETs are the same price as BJTs.
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How much cheaper?ikalogic said:I can easily find TIP transistors, and they are much cheaper.
Are you sure about the logic? It looks to me like you designed it for NPN pullups (Q5 and Q7). Keep in mind that NPN pullups don't invert, but PNPs do.ikalogic said:Thanks everybody,
now here is the final version, after all your comments:
**broken link removed**
Leave the logic part.. i know its fine, please just concentrate on the red shaded part and veryfy its free of errors.
I calculated all the resistances so that the base current in the 4 TIP transistors be around 38 to 43 mA.
thx a lot
i_build_stuff said:Here are the logic functions I came up with, for the base drives of Q1 - Q4:
b_Q1 <= P1 and not P2 and PWM;
b_Q2 <= not ( P2 and PWM );
b_Q3 <= P2 and not P1 and PWM;
b_Q4 <= not ( P1 and PWM );
I'm not sure if this is exactly the same as the original, but the inputs for different actions would be:
P1P2 -- action
00 -- freewheeling
01 -- forward
10 -- reverse
11 -- braking
(disclaimer: I'm half asleep right now, so you might want to run it through a simulator first)
Ron H said:Regarding your bridge-related questions - we need to know the load current before they can really be answered.
Ron H said:It's still going to smoke. More on the resistor values later.
Edit: I forgot to take the 220 out of the Q4 collector circuit. This doesn't change the fact that it won't work.
Styx said:why is Q8 being driven differently?
Ron H said:Your emitter follower outputs (Q1, Q3) will only go to about 3V. They need to go to 12V, which means the base will need to go to about 12.7V.
Your resistor values are lower than they need to be. You only need base drive of ~Ic/250,or about 12mA.
You need dead time during switching, as I mentioned in a previous post.
Yes, because the transistors do not turn of instantaneously. I ran a sim that suggested you should turn one half of a leg off about a microsecond before you turn the other half on.ikalogic said:Yes, i understand this point.. i don't know how you manage to detect errors that fast in a circuit.. i guess its a matter of experience.. anyway,
now here is the only solution i see to that specific problem, but sadly this would mean to lose the Active-turn off capability..
For the deat time, its noted. but when you say dead time, you mean a time where all transistors are OFF before running the bridge in the oposit current direction? to make sure no short circuit happens right?
Where did you find this? The OnSemi datasheet has all saturation curves with IC/IB = 250.And for the resistor.. the datasheet says i need 30 mA.. sometimes more.. anyway, i'll relie on testing to adjust the exact values..
But most important, did this solve the problem for (Q1, Q3)?
thx for your patience ron h
Noted. thxRon H said:Yes, because the transistors do not turn of instantaneously. I ran a sim that suggested you should turn one half of a leg off about a microsecond before you turn the other half on.
Also noted, i'll make the base current about 20 mA max...Ron H said:Where did you find this? The OnSemi datasheet has all saturation curves with IC/IB = 250.
Ron H said:No, the p-p output swing of an emitter follower (Q1, Q3) will always be less than the input swing.
ikalogic said:just one last hint for today:
http://www.mayothi.com/transistors.html
please tell me this guy is wrong about PNP transistors, when he says that connecting the base of a PNP to ground means no current is flowing !!? shouldn't there be current flowing out from the base?? right?