Subject:
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RE: h-bridges, current info
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Newsgroups:
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lugnet.robotics.handyboard
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Date:
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Mon, 15 Apr 1996 22:50:03 GMT
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Original-From:
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Chuck McManis <CMCMANIS@NETCOMspamcake.COM>
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Viewed:
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2177 times
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The answer is yes, but it isn't as useful.
BJTs (Bipolar Junction Transistors) are current based devices, if you look
at the 'base' input to the transistor as a pipe, for NPN devices there needs
to be current flowing "into" the base for them to turn on, for PNP transistors
you need current flowing "out" of the base for them to turn on. Given a typical
schematic like: NPN (PNP)
Collector (Emitter)
R /
Input ----\/\/\/\---- Base |
\ Emitter (Collector)
For an NPN transistor the Voltage at "Input" has to be higher than the
voltage at "Base" for a current to flow "into" the transistor. For a PNP
transistor the situation is reversed (Input at a lower voltage than Base
to get a current flowing "out" of the transistor)
In an H-Bridge configuration, one half of the bridge looks like this:
Motor Voltage In
+
| NPN (PNP)
| Collector (Emitter)
R /
Q1 Input ----\/\/\/\---- Base |
\ Emitter (Collector)
|
+ (A)
MOTOR
+ (B)
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| NPN (PNP)
| Collector (Emitter)
R /
Q2 Input ----\/\/\/\---- Base |
\ Emitter (Collector)
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Motor Voltage GROUND
Q1 is the "upper" transistor and Q2 is the "lower" transistor. The
voltages of interest are:
Vcb1 Voltage between collector and base of Q1
Vbe1 Voltage between the base and emitter of Q1
Vab Voltage across the motor itself
Vcb2 Voltage between collector and base of Q2
Vbe2 Voltage between the base and emitter of Q2
Now this equation must be satisfied:
Vcb1 + Vbe1 + Vab + Vcb2 + Vbe2 = Vmotor
(all the voltages sum to your motor voltage)
Since we're using the transistors effectively as switches, we only
care about their "saturation" voltage. This is the voltage drop from
their collector to their emitter when they are fully on. In our example
this is equal to Vcb1 + Vbe1 and Vcb2 + Vbe2. In the data
sheet this will be Vcesat. A nominal value is .7v but check your
data sheet to be sure! Let's say we're using transistors with this
nominal saturation voltage, and a motor supply of 9.6volts, then
Our equation something like .7 + Vab + .7 = 9.6, Vab = 8.2volts.
The design question becomes what to put on the 'input' of Q1 and Q2 to
turn them on. Since Q1's emitter (collector) is at the 'top' of the motor
the voltage difference between the input and Vmotor is about .7 volts.
So if Q1 was an NPN transistor and you connected "input" to Vmotor,
and the required turn on current was 70mA, the resistor would be
.7/.070 = 10 ohms. But Vbe is probably .5v. so its more like 6 ohms.
However if you connect the "input" to motor ground, the voltage
difference is nearly 9 volts so you can get a lot more current. Since
"motor ground" is below the transistor "ground" (on the emitter (collector))
the voltage is seen by the transistor as being 'negative' (or -9v).
Mosfets work the same way except that they want a 'voltage' on the
input (gate) rather than a current into the base. Thus to use an Nchannel
mosfet the gate voltage must be *higher* than the motor voltage. That
is what "high side" drivers do with charge pumps, create an artifically
high voltage to turn on N-channel mosfets.
You can use four NPN transistors if your transistor control voltage is
higher than your motor voltage.
--Chuck
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From: Jeff Keyzer[SMTP:jkeyzer@calweb.com]
Sent: Friday, April 12, 1996 8:46 AM
To: handyboard@media.mit.edu
Subject: h-bridges, current info
I'm trying to figure out a couple things. First, can four NPN
transistors be used to form an h-bridge? I was looking in a book, the Robot
Builder's Bonanza, which features a circuit which uses four TIP41
transistors. However, don't most h-bridges use two PNP and two NPN
transistors? What's the difference in the two designs? I was reading over
the article on h-bridges just posted, and found it a tad confusing, as the
diagram was incomplete...
Second, what's the maximum current the handy board can provide
through its motor power connector? Surely someone knows!
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