FAQ
TL;DR: In this BJT H‑bridge, V_BE ≈ 0.7 V, and—as one expert put it—"treat transistors as switches" to see which devices conduct or cut off for each input. In this specific schematic the BJTs don’t actually reach saturation, impacting efficiency. [Elektroda, Dorin Dragan, post #21663924]
Why it matters: This FAQ helps beginners diagnose H‑bridge behavior, reduce heat, and choose BJTs vs MOSFETs for motor drivers.
Quick Facts
- Typical small‑signal BJT V_BE(on): ~0.6–0.7 V; use this to decide ON/OFF states. [Elektroda, Dorin Dragan, post #21663924]
- PN2222A can show V_CE(sat) ≈ 0.1 V, illustrating low drop when hard on. [Elektroda, Steve Lawson, post #21663934]
- In the shown complementary BJT stage, none of the transistors truly saturate, so efficiency suffers. [Elektroda, Steve Lawson, post #21663936]
- For large motor currents, designers prefer MOSFETs over BJTs due to easier low‑loss conduction. [Elektroda, Steve Lawson, post #21663930]
- IGBTs are viable mainly at higher voltages; check datasheets for suitability. [Elektroda, Steve Lawson, post #21663930]
How does this BJT H‑bridge decide which transistors turn on or off?
Model each BJT as a switch. If the base‑emitter junction is forward‑biased by about 0.7 V with the correct polarity, it turns on. With input A high, Q5 pulls the bases of Q1/Q2 low, turning Q2 on and Q1 off. With A low, R1 pulls bases high, turning Q1 on and Q2 off. The other half mirrors this. [Elektroda, Dorin Dragan, post #21663924]
Do the transistors in this specific schematic actually reach saturation?
No. In this arrangement, the emitters cannot rise or fall beyond about 0.6–0.7 V from their bases. That constraint prevents true saturation, so devices sit just out of saturation. It simplifies drive but wastes power compared with a fully saturated switch stage. [Elektroda, Steve Lawson, post #21663936]
Why avoid the active region in switching?
When a transistor sits half‑on, it drops significant voltage at substantial current. That dissipates power as heat instead of delivering it to the load. Biasing for hard on (near saturation) or hard off minimizes loss and keeps devices cooler. [Elektroda, Steve Lawson, post #21663926]
How can I estimate heat in a saturated BJT leg?
Use the datasheet’s V_CE(sat) and your load current: P ≈ V_CE(sat) × I. Then compare the result with the device’s P_tot rating. If P exceeds safe limits, add a heatsink or pick a device with higher ratings. [Elektroda, Steve Lawson, post #21663930]
Why can V_BE be ~0.7 V while V_CE(sat) is only ~0.1 V?
The base‑emitter junction is forward‑biased near 0.7 V, while the collector‑base junction is reverse‑biased in normal operation. That allows V_CE to be very small at saturation, sometimes around 0.1 V for devices like PN2222A. [Elektroda, Steve Lawson, post #21663934]
What determines top versus bottom device conduction (PNP vs NPN)?
Top PNP devices conduct when their bases are driven lower relative to their emitters (toward ground). Bottom NPN devices conduct when their bases are driven higher relative to their emitters (toward V+ return). The arrow on the symbol hints at required polarity. [Elektroda, Peter Evenhuis, post #21663923]
How do I reduce overheating in a motor H‑bridge?
Start with calculations: use V_CE(sat) and current for BJTs or R_DS(on) and I² for MOSFETs. Compare with thermal limits (P_tot, θJA/θJC) and add heatsinking. For high currents, MOSFETs often run cooler when properly driven. [Elektroda, Steve Lawson, post #21663930]
Should I oversize BJTs to avoid heatsinks?
Only if power calculations plus thermal resistance show safe junction temperatures. Oversizing helps margin, but verify with P_tot and V_CE(sat) at your current. Always validate against the datasheet and real measurements. [Elektroda, Steve Lawson, post #21663930]
Are MOSFETs actually lower loss than BJTs here?
Yes for high current stages. MOSFETs behave resistively and can achieve milli‑ohm R_DS(on), so I²R loss stays low. That translates to less heat than a BJT dropping a fixed V_CE(sat) at the same current. [Elektroda, Steve Lawson, post #21663933]
Are IGBTs a better switch for my DC motor H‑bridge?
Use IGBTs mainly at higher voltages. For typical low‑voltage DC motors, MOSFETs are preferred due to low R_DS(on) and easier gate drive. Always align the device with your bus voltage and current. [Elektroda, Steve Lawson, post #21663930]
Why might this complementary BJT stage be chosen despite lower efficiency?
It cuts complexity and cost. Not driving devices into deep saturation can also improve switching speed, which is helpful for driving MOSFET gates or light loads. It’s a simplicity–efficiency trade‑off. [Elektroda, Steve Lawson, post #21663937]
How do pull‑ups and the Q5 transistor influence base drive?
With input A high, Q5 pulls the bases of Q1/Q2 low more strongly than the pull‑up. That turns the appropriate device on and the other off. With input A low, the pull‑up biases the opposite state. Repeat symmetrically for the other half‑bridge. [Elektroda, Dorin Dragan, post #21663924]
Quick 3‑step: How can I trace current for each input state?
- Treat each BJT as a switch with ~0.7 V base‑emitter threshold.
- Set A/B high or low, then note which bases go low or high via Q5/pull‑ups.
- Mark ON devices and follow the supply‑to‑ground path through the motor.
[Elektroda, Dorin Dragan, post #21663924]
Can a switching BJT still get hot when used “correctly”?
Yes. Even fully on, a BJT drops some voltage, so it dissipates power as heat. “There is still a voltage drop and thus a power loss,” especially at higher currents. Ensure thermal design and margins. [Elektroda, Steve Lawson, post #21663933]
